├── .gitignore ├── LICENSE ├── README.md ├── introduction_to_the_theory_of_distributions_2nd ├── def.tex ├── main.pdf └── main.tex ├── introductory_functional_analysis_with_applications ├── ch2_normed_spaces_banach_spaces.tex ├── ch3_inner_product_spaces_hilbert_spaces.tex ├── ch4_fundamental_theorems_for_normed_and_banach_spaces.tex ├── def.tex ├── main.pdf └── main.tex ├── linear_algebra_done_right_3rd ├── ch3_linear_map.tex ├── ch5_eigenvalues_eigenvectors_and_invariant_subspaces.tex ├── ch7_operators_on_inner_product_spaces.tex ├── ch8_opertators_on_complex_vector_spaces.tex ├── def.tex ├── image │ └── 3-d-8.png ├── main.pdf └── main.tex └── real_analysis_3rd ├── ch11_measure_and_integration.tex ├── ch12_measure_and_outer_measure.tex ├── ch3_lebesgue_measure.tex ├── ch4_the_lebesgue_integral.tex ├── ch5_differentiation_and_integration.tex ├── ch6_the_classical_banach_spaces.tex ├── ch7_metric_spaces.tex ├── ch8_topological_spaces.tex ├── ch9_compact_and_locally_compact_spaces.tex ├── def.tex ├── main.pdf └── main.tex /.gitignore: -------------------------------------------------------------------------------- 1 | *.aux 2 | *.fdb_latexmk 3 | *.fls 4 | *.log 5 | *.out 6 | *.synctex.gz 7 | *.thm 8 | *.toc 9 | .DS_Store 10 | -------------------------------------------------------------------------------- /LICENSE: -------------------------------------------------------------------------------- 1 | MIT License 2 | 3 | Copyright (c) 2018 Yunwei Ren 4 | 5 | Permission is hereby granted, free of charge, to any person obtaining a copy 6 | of this software and associated documentation files (the "Software"), to deal 7 | in the Software without restriction, including without limitation the rights 8 | to use, copy, modify, merge, publish, distribute, sublicense, and/or sell 9 | copies of the Software, and to permit persons to whom the Software is 10 | furnished to do so, subject to the following conditions: 11 | 12 | The above copyright notice and this permission notice shall be included in all 13 | copies or substantial portions of the Software. 14 | 15 | THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR 16 | IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, 17 | FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE 18 | AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER 19 | LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, 20 | OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE 21 | SOFTWARE. 22 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # solutions 2 | My personal solutions to some textbook problems 3 | 4 | ## F. G. Friedlander, M. Joshi, *Introduction to the Theory of Distributions*, 2nd ed 5 | * Chapter 1. Test Functions and Distributions (2, 5, 6, 7, 9) 6 | 7 | ## H. L. Royden, *Real Analysis*, 3rd ed 8 | 9 | **Ongoing** 10 | * Chapter 3. Lebesgue Measure 11 | * 3.1 Introduction (1, 2, 3) 12 | * 3.2 Outer Measure (5, 6, 7, 8) 13 | * 3.3 Measurable Sets and Lebesgue Measure (10, 11, 12, 13) 14 | * 3.5 Measurable Functions (19, 21, 22.d, 23, 24) 15 | * 3.6 Littlewood's Three Principles (30, 31) 16 | * Chapter 4. The Lebesgue Integral 17 | * 4.2 The Lebesgue Integral of a Bounded Function (2) 18 | * 4.3 The Integral of a Nonnegative Function (3, 5, 6, 7, 8, 9) 19 | * 4.4 The General Lebesgue Integral (12, 13, 15.a, 16, 18) 20 | * Chapter 5. Differentiation and Integration 21 | * 5.1 Differentiation of Monotone Functions (3, 4, 5.a) 22 | * 5.2 Functions of Bounded Variation (7, 8, 9, 10.a, 11) 23 | * 5.4 Absolute Continuity (12, 13, 14, 17.a, 18, 20, 21) 24 | * 5.5 Convex Functions (23, 24, 25.a, 27, 28) 25 | * Chapter 6. The Classical Banach Spaces 26 | * 6.1 The L^p Spaces (1, 2, 3, 4) 27 | * 6.2 The Minkowski and Hölder Inequalities (8) 28 | * 6.3 Convergence and Completeness (9, 10, 11, 13, 16, 17, 18) 29 | * 6.4 Approximation in L^p (19) 30 | * Chapter 7. Metric Spaces 31 | * 7.7 Compact Metric Spaces (27, 28, 29) 32 | * 7.8 Baire Category (31.a, 34, 35, 39) 33 | * 7.10 The Ascoli-Arzela Theorem (47, 49, 50) 34 | * Chapter 8. Topological Spaces 35 | * 8.1 Fundamental Notions (3, 7, 10.(a, b)) 36 | * 8.2 Bases and Countability (11, 13, 16) 37 | * 8.3 The Separation Axioms and Continuous Real-Valued Functions 38 | (18, 20, 23, 24, 26) 39 | * 8.4 Connectedness (32, 33, 35) 40 | * Chapter 9. Compact and Locally Compact Spaces 41 | * 9.1 Compact Spaces (2, 3, 6) 42 | * 9.2 Countably Compactness and the Bolzano-Weierstrass Property (9, 10) 43 | * 9.3 Products of Compact Spaces (13, 15) 44 | * 9.4 Locally Compact Spaces (18, 19(a), 24, 26, 29(a, b)) 45 | * Chapter 11. Measure and Integration 46 | * 11.1 Measure Spaces (1, 3, 7, 9.(a, b, c, d)) 47 | * 11.2 Measurable Functions (10, 11, 13, 16) 48 | * 11.3 Integration (19, 20, 21, 22) 49 | * 11.4 General Convergence Theorems (24) 50 | * 11.5 Signed Measures (27, 28, 31, 32) 51 | * 11.6 The Radon-Nikodym Theorem (33, 34.a, 35.d, 36, 40.a) 52 | * 11.7 The L^p Spaces (41, 42, 43, 44, 45) 53 | * Chapter 12. Measure and Outer Measure 54 | * 11.1 Outer Measure and Measurability (1, 2) 55 | * 11.2 The Extension Theorem (4, 7) 56 | 57 | 58 | ## Sheldon Axler, *Linear Algebra Done Right*, 3rd ed 59 | 60 | **Archived** 61 | * Chapter 3. Linear Maps 62 | * 3.A The Vector Space of Linear Maps (Odd) 63 | * 3.B Null Spaces and Ranges (Even - {28, 30}) 64 | * 3.D Invertibility and Isomorphic Vector Spaces (Odd - {17, 19}) 65 | * 3.E Products and Quotients of Vector Spaces (Even - {14, 18, 20}) 66 | * 3.F Duality ({1, 3, 9, 11, 15, 19, 25, 29, 31, 37}) 67 | * Chapter 5. Eigenvalues, Eigenvectors, and Invariant Subspaces 68 | * 5.B Eigenvectors and Upper-Triangular Matrices (Odd - {19}) 69 | * Chapter 7. Operators on Inner Product Spaces 70 | * 7.A Self-Adjoint and Normal Operators ({2, 4, 8, 10, 12, 16}) 71 | * 7.B The Spectral Theorem ({4, 6, 8, 12, 14}) 72 | * 7.C Positive Operators and Isometries (Even - {10, 12}) 73 | * 7.D Polar Decomposition and Singular Value Decomposition (ongoing) 74 | * Chapter 8. Operators on Complex Vector Spaces 75 | * 8.A Generalized Eigenvectors and Nilpotent Operators (3, 5, 7, 9, 13, 15) 76 | * 8.B Decomposition of an Operator (10) 77 | 78 | 79 | ## Erwin Kreyszig, *Introductory Functional Analysis with Applications* 80 | 81 | **Archived** 82 | * Chapter 2. Normed Spaces. Banach Spaces 83 | * 2.3 Further Properties of Normed Spaces (4, 7, 8, 9, 10, 14) 84 | * 2.4 Finite Dimensional Normed Spaces (3, 4, 5) 85 | * 2.5 Compactness and Finite Dimension (5, 6, 7, 8, 9, 10) 86 | * 2.7 Bounded and Continuous Linear Operators (2, 3, 4, 7, 12) 87 | * 2.8 Linear Functionals (8, 9, 10, 11, 12, 13, 14, 15) 88 | * 2.9 Operators on Finite Dimensional Spaces (8, 12, 13) 89 | * 2.10 Normed Spaces of Operators. Dual Space (8, 9, 10, 11, 13, 14) 90 | * Chapter 3. Inner Product Spaces. Hilbert Spaces 91 | * 3.1 Inner Product Spaces. Hilbert Spaces (2, 3, 7, 8) 92 | * 3.2 Further Properties of Inner Product Spaces (7, 8, 9, 10) 93 | * 3.3 Orthogonal Complements and Direct Sums (7, 8, 9) 94 | * 3.4 Orthonormal Sets and Sequences (3, 7, 8) 95 | * 3.5 Series Related to Orthonormal Sequences (1, 3, 4, 5, 7, 8, 9, 10) 96 | * 3.6 Total Orthonormal Sets and Sequences (4, 6, 7, 9, 10) 97 | * 3.8 Functionals on Hilbert Spaces (3, 4, 5, 12, 14) 98 | * 3.9 Hilbert-Adjoint Operator (1, 2, 3, 4, 5, 6, 7, 8, 9) 99 | * 3.10 Self-Adjoint, Unitary and Normal Operators (4, 6, 9, 10, 13, 15) 100 | * Chapter 4. Fundamental Theorems for Normed and Banach Spaces 101 | * 4.2 Hahn-Banach Theorem (4, 5, 6, 8, 9, 10) 102 | * 4.3 Hahn-Banach Theorem for Complex Vector Spaces and Normed Spaces (1, 2, 7, 8, 13, 14, 15) 103 | * 4.5 Adjoint Operator (9, 10) 104 | * 4.6 Reflexive Spaces (2, 3, 4, 5, 6, 7, 8, 9, 10) 105 | * 4.7 Uniform Boundedness Theorem (1, 5, 6, 7, 10, 11, 13, 14) 106 | * 4.8 Strong and Weak Convergence (1, 2, 4, 5, 6, 7, 8, 9, 10) 107 | * 4.9 Convergence of Sequences of Operators (4, 6, 7, 9, 10) 108 | * 4.12 Open Mapping Theorem (2, 4, 5, 6, 7, 9) 109 | * 4.13 Closed Linear Operator. Closed Graph Theorem (5, 6, 7, 8) 110 | -------------------------------------------------------------------------------- /introduction_to_the_theory_of_distributions_2nd/def.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt, a4paper]{article} 2 | 3 | \usepackage[margin=1in]{geometry} 4 | \usepackage{ 5 | color, 6 | clrscode, 7 | amssymb, 8 | amsmath, 9 | listings, 10 | xcolor, 11 | supertabular, 12 | multirow, 13 | mathtools, 14 | mathrsfs, 15 | amsthm, 16 | systeme, 17 | amsfonts 18 | } 19 | \definecolor{bgGray}{RGB}{36, 36, 36} 20 | \usepackage[ 21 | colorlinks, 22 | linkcolor=bgGray, 23 | anchorcolor=blue, 24 | citecolor=green 25 | ]{hyperref} 26 | 27 | \newcommand{\vep}{\varepsilon} 28 | 29 | \newenvironment{solution} 30 | {\begin{proof}[Solution]} 31 | {\end{proof}} 32 | \newcommand{\rd}{\,\mathrm{d}} 33 | \newcommand{\inv}{^{-1}} 34 | \newcommand{\hp}{^\prime} 35 | \newcommand{\mcal}{\mathcal} 36 | \newcommand{\ubar}[1]{\text{\b{$#1$}}} 37 | \newcommand{\mrm}{\mathrm} 38 | \newcommand{\mscr}{\mathscr} 39 | \newcommand{\R}{\mathbb{R}} 40 | 41 | 42 | \DeclareMathOperator*\lowlim{\underline{lim}} 43 | \DeclareMathOperator*\uplim{\overline{lim}} 44 | \DeclareMathOperator*\esssup{ess\,sup} 45 | \DeclareMathOperator\sgn{sgn} 46 | \DeclareMathOperator{\cl}{cl} 47 | \def\upint{\mathchoice 48 | {\mkern13mu\overline{\vphantom{\intop}\mkern7mu}\mkern-20mu}% 49 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 50 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 51 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 52 | \int} 53 | \def\lowint{\mkern3mu\underline{\vphantom{\intop}\mkern7mu}\mkern-10mu\int} 54 | 55 | -------------------------------------------------------------------------------- /introduction_to_the_theory_of_distributions_2nd/main.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Engineev/solutions/4e33274fe1ed9e46fd0e6671c57cb589704939bd/introduction_to_the_theory_of_distributions_2nd/main.pdf -------------------------------------------------------------------------------- /introduction_to_the_theory_of_distributions_2nd/main.tex: -------------------------------------------------------------------------------- 1 | \input{def.tex} 2 | 3 | \DeclareMathOperator{\supp}{supp} 4 | 5 | \title{Solutions to \\ \textit{Introduction to the Theory of Distirbutions}} 6 | \author{Yunwei Ren} 7 | \date{} 8 | 9 | \begin{document} 10 | \maketitle 11 | \tableofcontents 12 | 13 | \vspace{1cm} 14 | 15 | \section{Test Functions and Distributions} 16 | \paragraph{1.2} 17 | \begin{proof} 18 | It suffices to show that $f \equiv 0$ on an open set $O$ iff the restriction 19 | of the distribution $\langle f, \cdot\rangle$ onto $O$ is the zero 20 | distribution. Suppose that $\langle f, \cdot\rangle|_O \equiv 0$ since the 21 | other direction is obvious. Assume, to obtain a contradiction, that 22 | $f(x) > 0$ for some $x \in O$. Since $f$ is continuous, there is an open 23 | neighborhood $U \subset O$ s.t. $f > \vep$ on $U$ for some $\vep > 0$. Choose 24 | a small closed ball $B \subset U$ centered at $x$ and let $\psi$ be the 25 | cutoff function with $\supp\psi \subset U$, $0 \le \psi \le 1$ and 26 | $\psi \equiv 1$ on $B$. Then 27 | \[ 28 | 0 = \langle f, \psi\rangle = \int_U f\psi > \vep\mu(B) > 0, 29 | \] 30 | where $\mu(B)$ is the measure of $B$. Contradiction. Thus, $f \le 0$. 31 | Similarly, we can show $f \ge 0$. Therefore, $f \equiv 0$ on $O$. 32 | 33 | The result is not true for $f \in L_1^{\mrm{loc}}(\mathbb{R}^n)$ in general 34 | since add a function which is zero a.e. to $f$ does not change the 35 | distribution but will change the support of $f$. 36 | \end{proof} 37 | 38 | \paragraph{1.5} 39 | \begin{proof} 40 | For every compact $K \subset (0, \infty)$, there is an integer $N$ s.t. 41 | $1/k \notin K$ for all $k > N$. Hence, for every $\phi \in C^\infty_c(0, 42 | \infty)$ with $\supp \phi \subset K$, 43 | \[ 44 | |\langle u, \phi\rangle| 45 | = \left|\sum_{k=0}^N \partial^k \phi(1/k)\right| 46 | \le \sum_{k=0}^N \sup|\partial^k \phi|. 47 | \] 48 | Thus, $u$ is a distribution on $(0, \infty)$. 49 | 50 | Assume, to obtain a contradiction, that $u = v|_{(0, \infty)}$ for some 51 | $v \in \mscr{D}\hp(\R)$. Let $f \in C^\infty_c(\R)$ be a cutoff function 52 | with $f \equiv 1$ on $[-1, 1]$. Then, the distribution $fu$ (cf. Sec. 2.5) 53 | is of infinite order since its restriction to $1/m$ is $\delta^{(m)}$ for 54 | every positive integer $m$. However, since $fu$ is compactly supported, it 55 | must have a finite order (cf. Sec. 3.1). Contradiction. 56 | \end{proof} 57 | 58 | \paragraph{1.6} 59 | \begin{proof} 60 | It follows immediately from the Riesz-Markov theorem. 61 | \end{proof} 62 | 63 | \paragraph{1.7} 64 | I am not sure whether the second part can be proved since if we put 65 | $f_\vep \equiv 0$ for some $\vep \in (0, 1)$, the asymptotic behavior will 66 | not change. 67 | \begin{proof} 68 | It suffices to show $\int f_\vep\phi \to \phi(0)$ as $\vep\to 0$. Let 69 | $B_\vep = \{|x| \le \vep\}$. We have 70 | \begin{align*} 71 | \left|\int f_\vep \phi - \phi(0)\right| 72 | &= \left|\int_{B_\vep} f_\vep (\phi - \phi(0)) \right| \\ 73 | &\le \sup_{x \in B_\vep}|\phi(x) - \phi(0)| \int|f_\vep| \\ 74 | &\le \mu\sup_{x \in B_\vep}|\phi(x) - \phi(0)| . 75 | \end{align*} 76 | Since $\phi\in C^\infty(\R^n)$, $\sup_{x\in B_\vep}|\phi(x) - \phi(0)| 77 | \to 0$ as $\vep \to 0$. Thus, $f_\vep \to \delta$ in $\mscr{D}\hp(\R^n)$. 78 | \end{proof} 79 | 80 | \paragraph{1.9} 81 | \begin{proof} 82 | Let $u_n(x) := \sum_{k=-n}^n c_ke^{ikx}$. For every $\phi \in 83 | C^\infty_c(\R)$, by repeatedly using integration by parts, we have 84 | \begin{align*} 85 | \langle u_n, \phi\rangle 86 | &= \int\sum_{k=-n}^n c_k e^{ikx}\phi(x)\rd x \\ 87 | &= \sum_{k=-n}^n c_k \int e^{ikx}\phi(x)\rd x \\ 88 | &= \sum_{k=-n}^n c_k \left(\frac{-1}{ik}\right)^{m+2} 89 | \int e^{ikx}\partial^{m+2}\phi(x)\rd x. \\ 90 | \end{align*} 91 | Note that $c_k \left(\frac{-1}{ik}\right)^{m+2} \le O(1/k^2)$ and the 92 | $\int e^{ikx}\partial^{m+2}\phi(x)\rd x$ is bounded. Thus, $\lim 93 | \langle u_n, \phi\rangle$ converges for every $\phi$, whence $u$ converges 94 | in $\mscr{D}\hp(\R)$. 95 | \end{proof} 96 | 97 | \end{document} 98 | 99 | 100 | 101 | 102 | 103 | 104 | 105 | 106 | 107 | -------------------------------------------------------------------------------- /introductory_functional_analysis_with_applications/ch2_normed_spaces_banach_spaces.tex: -------------------------------------------------------------------------------- 1 | \section{Normed Spaces. Banach Spaces} 2 | \setcounter{subsection}{2} 3 | \subsection{Further Properties of Normed Spaces} 4 | \paragraph{4.} 5 | cf. Prob. 13, Sec 1.2 6 | \begin{proof} 7 | The continuity of addition and multiplication follows respectively from the 8 | inequalities 9 | \[ 10 | \|(x_1+y_1)-(x_2+y_2)\| \le \|x_1-x_2\|+\|y_1-y_2\| 11 | \] 12 | and 13 | \[ 14 | \|\alpha_1x_1-\alpha_2x_2\| = 15 | \|\alpha_1x_1-\alpha_1x_2+\alpha_1x_2-\alpha_2x_2\| 16 | \le |\alpha_1|\|x_1-x_2\| + |\alpha_1-\alpha_2|\|x_2\|. 17 | \] 18 | \end{proof} 19 | 20 | \paragraph{7.} 21 | \begin{proof} 22 | Let $Y$ and $y_n$ be defined as in the hint. Then $\|y_n\|=1/n^2$, 23 | constituting a convergent number series. However, 24 | \[ 25 | \sum_{n=1}^N y_n = (1,1/4, \dots, 1/N^2, 0,\dots), 26 | \] 27 | which is divergent as $N\to\infty$. 28 | \end{proof} 29 | 30 | \paragraph{8.} 31 | \begin{proof} 32 | Let $(x_n)$ be a Cauchy sequence in $X$. Hence, for every $n>0$, there 33 | exists some $K_n>0$ such that for all $p,q>K_n$, $\|x_p-x_q\|<1/n^2$. 34 | Without loss of generality, we may assume that $(K_n)$ is increasing. Since 35 | the series $\|x_{K_{n+1}}-x_{K_n}\|$ is bounded by $1/n^2$, it converges. 36 | By the hypothesis, the series $(x_{K_{n+1}}-x_{K_n})$ also converges. Hence, 37 | \[ 38 | x_{K_n} = x_{K_1} + \sum_{i=1}^{n-1}(x_{K_{i+1}}-x_{K_i}) \to x 39 | \quad\text{as } n\to\infty. 40 | \]\par 41 | Now we show that $(x_n)$ converges to $x$. For every $\vep>0$, since $(x_n)$ 42 | is a Cauchy sequence, there exists some $N_1$ such that for all $p,q>N_1$, 43 | $\|x_p-x_q\|<\vep$. Meanwhile, since $x_{K_n}\to x$, once $K_n$ is large 44 | enough, $\|x-x_{K_n}\|<\vep$. Let $K_n>N_1$. Then for every $n>K_n$ 45 | \[ 46 | \|x_n-x\| \le \|x_n-x_{K_n}\| + \|x_{K_n}-x\| \le 2\vep. 47 | \] 48 | Thus, $X$ is complete. 49 | \end{proof} 50 | 51 | \paragraph{9.} 52 | \begin{proof} 53 | Let $(x_n)$ be an absolutely convergent series in Banach space $X$. Let 54 | $s_n=\sum_{i=1}^n x_n$. Now we show that $s_n$ is a Cauchy sequence and 55 | therefore convergent. Since $\sum_{i=1}^\infty\|x_i\|<\infty$, for every 56 | $\vep>0$, there exists some $N>0$ such that for all $n>N$, $\sum_{i=n} 57 | ^\infty\|x_i\|<\vep$. Hence, for every $N0$, by the definition of Schauder basis, there 77 | exists $\beta_1,\dots,\beta_n\in\mathbb{K}$ such that $\|x-(\beta_1e_1+ 78 | \cdots+\beta_ne_n)\|<\vep$. Let $M=\max_{i}\|e_i\|$. If $M=0$, then there is 79 | nothing to prove. Otherwise, since $\mathbb{W}$ is dense in $\mathbb{K}$, 80 | for $i=1,\dots,n$, there exists $\alpha_i\in\mathbb{W}$ with $|\alpha_i- 81 | \beta_i|<\vep/2^iM$. Hence, 82 | \begin{align*} 83 | \left\|x-\sum_{i=1}^n\alpha_ie_i\right\| 84 | \le& \left\|x-\sum_{i=1}^n\beta_ie_i\right\| + 85 | \left\|\sum_{i=1}^n(\beta_i-\alpha_i)e_i\right\| \\ 86 | \le& \vep + \sum_{i=1}^n|\alpha_i-\beta_i|\|e_i\| \\ 87 | \le& 2\vep. 88 | \end{align*} 89 | Thus, $S$ is dense in $X$ and therefore $X$ is separable. 90 | \end{proof} 91 | 92 | \paragraph{14.} 93 | \begin{proof} 94 | Clear that $\|\cdot\|_0$ is nonnegative. And $\|\alpha\hat{x}\|_0= \inf_{x 95 | \in\hat{x}}\|\alpha x\| = |\alpha|\|\hat{x}\|_0$. Meanwhile, $\|\hat{x}+ 96 | \hat{y}\|_0=\inf_{z\in\hat{x}+\hat{y}}\|z\|\le\inf_{z\in\hat{x}}\|z\|+ 97 | \inf_{z\in\hat{y}}\|z\|=\|\hat{x}\|_0+\|\hat{y}\|_0$. Finally, we show that 98 | $\|\hat{x}\|_0=0$ implies $\hat{x}=Y$ and invoke Prob. 4, Sec 2.2 to 99 | complete the proof. Since $\|\hat{x}\|_0=0$, there exists $(x_n)\subset 100 | \hat{x}$ which converges to $0$. Since $Y$ is closed, $Y$ is complete and so 101 | is its cosets. Therefore, $0\in\hat{x}$, enforcing $\hat{x}$ to be $Y$. 102 | \end{proof} 103 | % end 104 | 105 | \subsection{Finite Dimensional Normed Spaces} 106 | \paragraph{3.} 107 | \begin{proof} 108 | The reflexive property clearly holds. If there are positive $a$ and $b$ such 109 | that $a\|x\|_0\le\|x\|_1\le b\|x\|_0$ for all $x\in X$, then $\|x\|_1/b\le 110 | \|x\|_0\le\|x\|/a$. Hence the relation is symmetric. Next we further suppose 111 | there exists positive $c$ and $d$ such that that $c\|x\|_1\le\|x\|_2\le 112 | d\|x\|_1$. Then $ac\|x\|_0\le \|x\|_2\le bd\|x\|_0$, giving the transitive 113 | property. Thus, the axioms of an equivalence relation hold. 114 | \end{proof} 115 | 116 | \paragraph{4.} 117 | \begin{proof} 118 | Suppose the norms $\|\cdot\|$ and $\|\cdot\|_0$ are equivalent. Let $E 119 | \subset X$ be any open set with respect to $\|\cdot\|$, i.e., for every $x_0 120 | \in E$, there exists some $\delta>0$ such that $A=\{x\in X:\, \|x-x_0\|< 121 | \delta\}\subset E$. Since $\|\cdot\|\sim\|\cdot\|_0$, there exists some 122 | positive $c$ such that $\|x-x_0\|\le c\|x-x_0\|_0$. Hence, $B=\{x\in X:\, 123 | \|x-x_0\|<\delta/c \}\subset A\subset E$. Namely, $E$ is also open with 124 | respect to $\|\cdot\|_0$. Interchanging the roles of $\|\cdot\|$ and 125 | $\|\cdot\|_0$ completes the proof. 126 | \end{proof} 127 | 128 | \paragraph{5.} 129 | \begin{proof} 130 | Suppose the norms $\|\cdot\|$ and $\|\cdot\|_0$ are equivalent. Then for 131 | every $x\in X$, there exists some $c>0$ such that $\|x\|_0\le c\|x\|$. Let 132 | $(x_n)$ be a Cauchy sequence with respect to $\|\cdot\|$, i.e., for every 133 | $\vep>0$, there exists some $N>0$ such that for all $n,m>N$, $\|x_n-x_m\|< 134 | \vep/c$. Hence, $\|x_n-x_m\|_00$ such 163 | that for every $k$, 164 | \[ 165 | \|v-y_{k,1}b_1-\cdots-y_{k,n}b_n\| \ge c(1+|y_{k,1}|+\cdots+|y_{k,_n}|). 166 | \] 167 | Hence, the sequence $(y_{k,1},\dots,y_{k,n})$ of $n$-tuples is bounded and 168 | therefore has a convergent subsequence. Consequently, $(y_k)$ also has a 169 | convergent subsequence. Suppose that it converges to $z\in Z$. Note that 170 | $\|v-z\|=a$ and as $Y$ is closed, $z\in Y$. Thus, $a$ can be attained in 171 | $Y$. 172 | \end{proof} 173 | 174 | \paragraph{8.} 175 | \begin{proof} 176 | Since the unit ball $B$ with respect to $\|\cdot\|_2$ in $\mathbb{R}^n$ and 177 | $\mathbb{C}^n$ is compact and $\|\cdot\|$ is continuous, by 2.5-7, $x\mapsto 178 | \|x\|$ can attains its minimum, denoted by $a$, on $B$. Due to the positive 179 | definite property of a norm, $a$ is positive. Hence, $00$, there exists $x_1\in X$ with $f(x_1)=1$ such that 353 | $\tilde{d}+\vep\ge\|x_1\|$. Hence, 354 | \[ 355 | \|f\|(\tilde{d}+\vep)\ge\|f\|\|x_1\| \ge |f(x_1)| = 1. 356 | \] 357 | Since the choice of $\vep>0$ is arbitrary, $\|f\|\tilde{d}\ge 1$. Meanwhile, 358 | there exists $x_2\in X$ with $\|x_2\|=1$ such that $|f(x_2)|\ge \|f\|-\vep$. 359 | Put $x_3 = x_2/f(x_2)$. Then $f(x_3)=1$. Hence, 360 | \[ 361 | (\|f\|-\vep)\tilde{d} \le |f(x_2)|\|x_3\| = \|x_2\| = 1, 362 | \] 363 | which implies $\|f\|\tilde{d}\le 1$. Thus, $\|f\|\tilde{d}=1$. 364 | \end{proof} 365 | 366 | \paragraph{15.} 367 | \begin{proof} 368 | For every $x$ with $\|x\|\le 1$, $f(x)\le \|f\|\|x\|\le c$. Hence, $x\in 369 | X_{c_1}$. Meanwhile, for every $\vep>0$, by the definition of the supremum, 370 | there exists a $x$ with $\|x\|=1$ such that $|f(x)|>\|f\|-\vep$. By the 371 | linearity of $f$, we may remove the $|\cdot|$ on the right side. Hence, 372 | $f(x)\notin X_{c_1}$ where $c=\|f\|-\vep$. 373 | \end{proof} 374 | % end 375 | 376 | \subsection{Operators on Finite Dimensional Spaces} 377 | \paragraph{8.} 378 | \begin{proof} 379 | Let $\{b_2,\dots,b_n\}$ be a basis of $Z$ and $\{b_1,\dots,b_n\}$ a basis of 380 | $X$. Define $f\in X^*$ to be $f(b_i)=\delta_{1i}$. Clear that $\mathcal{N} 381 | (f)=Z$. By Prob. 11, Sec 2.8, $f$ is uniquely determined up to a scalar 382 | multiple. 383 | \end{proof} 384 | 385 | \paragraph{12.} 386 | \begin{proof} 387 | Let $\varphi:X\to\mathbb{K}^p$ be defined by $x\mapsto [f_1(x),\dots, 388 | f_p(x)]^T$. It can be verified that $\varphi$ is a linear operator. Since 389 | $\dim X=n>p$, $\varphi$ can not be injective. Hence, there exists $0\ne x\in 390 | X$ such that $\varphi(x)=0$. 391 | \end{proof} 392 | 393 | \paragraph{13.} 394 | \begin{proof} 395 | Let $\{b_1,\dots,b_m\}$ be a basis of $Z$ and $\{b_1,\dots,b_n\}$ a basis of 396 | $X$. Define $\tilde{f}\in X^*$ to be identical with $f$ on $b_1,\dots,b_m$ 397 | and $0$ on $b_{m+1},\dots,b_n$. Clear that $\tilde{f}|_Z=f$. 398 | \end{proof} 399 | % end 400 | 401 | \subsection{Normed Spaces of Operators. Dual Space} 402 | \paragraph{8.} 403 | \begin{proof} 404 | First we construct a linear bijection $T$ between $c_0\hp$ and $l^1$. A 405 | Schauder basis for $c_0$ is $(e_k)$, where $e_k=(\delta_{kj})$. Then for 406 | every $f\in c_0\hp$, define $Tf=(\gamma_k)=(f(e_k))$. Clear that $T$ is 407 | linear. Now we show that $Tf=(\gamma_k)\in l^1$, that is, $\sum_{k=1}^n| 408 | \gamma_k|$ is bounded and therefore convergent. Define $x_n=(\xi_k^{(n)})$ 409 | with 410 | \[ 411 | \xi_k^{(n)}=\begin{cases} 412 | \sgn\gamma_k, & k\le n, \\ 413 | 0, & k>n. 414 | \end{cases} 415 | \] 416 | Clear that $x_n\in c_0$. By the linearity and boundedness of $f$, 417 | \begin{equation} 418 | \label{eq:2.10.8} 419 | f(x_n)=\sum_{k=1}^\infty \xi_k^{(n)}\gamma_k = \sum_{k=1}^n|\gamma_k|. 420 | \end{equation} 421 | Since $f$ is bounded, $|f(x_n)| \le \|f\|\|x_n\| \le \|f\|$. Hence, $\sum\| 422 | \gamma_k\|$ is bounded. Thus, $Tf\in l^1$.\par 423 | Meanwhile, for every $y=(\beta_k)\in l^1$, define $Sy=g$ to be the 424 | functional $g(x)=\sum_{k=1}^\infty\xi_k\beta_k$ for $x=(\xi_k)$. On $c_0$, 425 | the summation does converge and clear that $g$ is linear and bounded. Hence, 426 | $g\in c_0\hp$. It can be verify that $ST=TS=I$ and $T$ is linear. Thus, $c_0 427 | \hp$ and $l^1$ is isomorphic.\par 428 | Now we show that $T$ constructed preserve the norm to complete the proof. 429 | For $x\in c_0$ with $\|x\|=1$, 430 | \[ 431 | |f(x)| = \left|\sum_{k=1}^\infty \xi_k\gamma_k\right|\le 432 | \sum_{k=1}^\infty |\gamma_k| = \|Tf\|. 433 | \] 434 | Hence, $\|f\|\le\|Tf\|$. And \eqref{eq:2.10.8} implies $\sum_{k=1}^n||\gamma 435 | |\le \|f\|$. Letting $n\to\infty$ yields $\|Tf\|\le \|f\|$. Thus, $\|Tf\|= 436 | \|f\|$. 437 | \end{proof} 438 | 439 | \paragraph{9.} 440 | \begin{proof} 441 | Let $(b_k)$ be a Hamel basis of $X$ and suppose that $f,g\in X^*$ coincide 442 | on every $b_k$. Then for every $x=\sum_{k=1}^\infty\xi_k b_k\in X$, 443 | \[ 444 | f(x)-g(x) = \sum_{k=1}^n\xi_k(f(b_k)-g(b_k)) = 0. 445 | \] 446 | Thus, $f=g$. Namely, $f$ is uniquely determined. 447 | \end{proof} 448 | 449 | \paragraph{10.} 450 | \begin{proof} 451 | Let $(b_k)$ be a Hamel basis of $X$ and without loss of generality we may 452 | assume $\|b_k\|=1$. Justified by Prob. 9, we can define $T\in X^*$ with 453 | $Tb_k=k$, which is clearly unbounded. 454 | \end{proof} 455 | 456 | \paragraph{11.} 457 | \begin{proof} 458 | It follows immediately from Prob. 10. 459 | \end{proof} 460 | 461 | \paragraph{13.} 462 | \begin{proof} 463 | For any $f,g\in M^a$ and scalar $a,b$, $(af+bg)(x)=af(x)+bg(x)=0$ for every 464 | $x\in M$. Hence, $M^a$ is a vector space. For $(f_n)\subset M^a\subset 465 | X\hp$, suppose that $f_n\to f\in M^*$. Since $M\hp$ is complete, it is 466 | closed and therefore $f\in M\hp$. For every $0\ne x\in M$, since $f_n\to f$, 467 | \[ 468 | \frac{|f_n(x)-f(x)|}{\|x\|} \to 0,\,\text{as $n\to\infty$}. 469 | \] 470 | Hence, $f(x)=0$. Thus, $M^a$ is closed.\par 471 | $X^a = \{0\}$ and $\{0\}^a = X\hp$. 472 | \end{proof} 473 | 474 | \paragraph{14.} 475 | \begin{proof} 476 | Let $\{b_1,\dots,b_m\}$ be a basis of $M$ and $\{b_1,\dots,b_n\}$ a basis of 477 | $X$. And let $\{\beta_1,\dots,\beta_n\}$ be the dual basis. Clear that $b_1, 478 | \dots,b_m\notin M^a$ whereas $b_{m+1},\dots,b_n$ does. Together with Prob. 479 | 13, this implies $M^a=\spn(b_{m+1},\dots,b_n)$. Thus, $\dim M^a = n-m$. 480 | \end{proof} 481 | % end 482 | -------------------------------------------------------------------------------- /introductory_functional_analysis_with_applications/ch3_inner_product_spaces_hilbert_spaces.tex: -------------------------------------------------------------------------------- 1 | %---------%---------%---------%---------%---------%---------%---------%--------- 2 | \section{Inner Product Spaces. Hilbert Spaces} 3 | \subsection{Inner Product Spaces. Hilbert Spaces} 4 | \paragraph{2.} 5 | \begin{proof} 6 | \[ 7 | \|x+y\|^2 = \langle x+y, x+y\rangle = \|x\|^2+\|y\|^2+2\langle x, y\rangle 8 | =\|x\|^2+\|y\|^2, 9 | \] 10 | where the last equality comes from the hypothesis of orthogonality. Now we 11 | show that for mutually orthogonal $x_1,\dots,x_m$ 12 | \[ 13 | \left\|\sum_{i=1}^m x_i\right\|^2 = \sum_{i=1}^m\|x_i\|^2, 14 | \] 15 | by induction on $m$. The case where $m=2$ has already been showed and we 16 | assume that the equation holds for $m-1$. Since $x_m$ is orthogonal with 17 | each $i=1,\dots,m-1$, $x_m$ is orthogonal to $x_1+\cdots+x_{m-1}$. Hence, 18 | \[ 19 | \left\|\sum_{i=1}^m x_i\right\|^2 = 20 | \left\|\sum_{i=1}^{m-1}x_i\right\|^2 + \|x_m\|^2 = 21 | \sum_{i=1}^m\|x_i\|^2, 22 | \] 23 | completing the proof. 24 | \end{proof} 25 | 26 | \paragraph{3.} 27 | \begin{proof} 28 | The equation implies $\langle x, y\rangle+\langle y, x\rangle=0$. The 29 | symmetric property of real inner products implies $\langle x, y\rangle=0$. 30 | Let $X=\mathbb{C}$ and $x=1, y=i$. It is easy to verify that $\|x+y\|^2= 31 | \|x\|^2+\|y\|^2=2$ but $x$ and $y$ are not orthogonal. 32 | \end{proof} 33 | 34 | \paragraph{7.} 35 | \begin{proof} 36 | It suffices to show that the zero vector is the only vector orthogonal to 37 | all vectors. Suppose that $\langle x_0, x\rangle=0$ for all $x\in X$, then 38 | $\|x_0\|^2=\langle x_0, x_0\rangle=0$. By the definiteness of the inner 39 | product, $x_0=0$. 40 | \end{proof} 41 | 42 | \paragraph{8.} 43 | We show that any norm satisfying the parallelogram equality can be derived 44 | form an inner product. 45 | \begin{proof} 46 | The proof of (IP3) is trivial and (IP4) follows immediately from the 47 | positive-definiteness of the norm. Hence we only show the linearity in the 48 | first factor here. For every $u, v, y\in X$, from the parallelogram equality 49 | we can derive, after some computation, that 50 | \begin{align*} 51 | 4\langle u+v, y\rangle &= \|u+v+y\|^2-\|u+v-y\|^2 \\ 52 | &= \|u+y\|^2-\|u-y\|^2+\|v+y\|^2-\|v-y\|^2 \\ 53 | &= 4\langle u, y\rangle + 4\langle v, y\rangle. 54 | \end{align*} 55 | Namely, (IP1) holds. By induction we can show that $\langle nu, y\rangle= 56 | n\langle u, y\rangle$ for $n=1,2,\dots$. And since $\langle -u, y\rangle = 57 | \langle 0-u, y\rangle = \langle 0, y\rangle - \langle u, y\rangle = \langle 58 | u, y\rangle$, 59 | \[ 60 | \langle nu, y\rangle = n\langle u, y\rangle, 61 | \quad\text{for $n\in\mathbb{Z}$}. 62 | \] 63 | Furthermore, for any positive integer $m$, 64 | \[ 65 | m\left\langle\frac{n}{m}u, y\right\rangle = 66 | mn\left\langle\frac{1}{m}u, y\right\rangle = 67 | n\langle u, y\rangle. 68 | \] 69 | Dividing the both sides by $m$ yields 70 | \[ 71 | \langle qu, y\rangle = q\langle u, y\rangle, 72 | \quad\text{for $q\in\mathbb{Q}$}. 73 | \] 74 | For every $\alpha\in\mathbb{R}$, let $(q_n)\subset\mathbb{Q}$ converges to 75 | $\alpha$. Now we show that $f(t)=\langle tu, y\rangle$ is continuous at $t= 76 | 0$ and by the additivity we may conclude that $f$ is continuous on 77 | $\mathbb{R}$. Since 78 | \begin{align*} 79 | 4|f(t)| 80 | &= |\|tu+y\|^2 - \|tu-y\|^2| \\ 81 | &= (\|tu+y\|+\|tu-y\|)|\|tu+y\|-\|tu-y\|| \\ 82 | &\le 4t\|u\|(t\|u\|+\|y\|) \to 0 83 | \end{align*} 84 | as $t\to 0$, $f(t)$ is continuous. For every $\alpha\in\mathbb{R}$, let 85 | $(q_n)\subset\mathbb{Q}$ be a convergent sequence with limit $\alpha$. Then 86 | \[ 87 | \langle\alpha u, y\rangle= 88 | \lim\langle q_nu, y\rangle= 89 | \lim q_n\langle u, y\rangle= 90 | \alpha\langle u, y\rangle. 91 | \] 92 | Hence, $\langle\cdot,\cdot\rangle$ is linear in the first factor. Thus, it 93 | is an inner product. Meanwhile, it is easy to verify that the norm it 94 | introduces is exactly the original norm. 95 | \end{proof} 96 | % end 97 | 98 | \subsection{Further Properties of Inner Product Spaces} 99 | \paragraph{7.} 100 | \begin{proof} 101 | First we note that 102 | \[ 103 | f(\alpha)=\|x+\alpha y\|^2-\|x-\alpha y\|^2 104 | = 2\bar{\alpha}\langle x,y\rangle + 2\alpha\langle y,x\rangle. 105 | \] 106 | Clear that $x\perp y$ implies $f(\alpha)=0$ for all scalar $\alpha$. For the 107 | converse, we suppose $f(\alpha)=0$ and put $\alpha=\langle x, y\rangle$. 108 | Then $0=f(\alpha)=2|\langle x, y\rangle|$. Thus, $x\perp y$. 109 | \end{proof} 110 | 111 | \paragraph{8.} 112 | \begin{proof} 113 | Clear that $x\perp y$ implies $\|x+\alpha y\|\ge \|x\|$. Therefore we only 114 | show the converse here. Without loss of generality, we assume $\|y\|=1$. 115 | Then $\|x+\alpha y\|\ge \|x\|$ for all scalar $\alpha$ implies 116 | \[ 117 | |\alpha|^2 + 118 | \bar{\alpha}\langle x, y\rangle + \alpha\overline{\langle x, y\rangle} 119 | \ge 0. 120 | \] 121 | Put $\alpha=-\langle x, y\rangle$ and we get 122 | \[ 123 | 0 \le |\langle x, y\rangle|^2 - 2|\langle x, y\rangle|^2 124 | =-|\langle x, y\rangle|^2, 125 | \] 126 | which implies $\langle x, y\rangle=0$. Namely, $x\perp y$. 127 | \end{proof} 128 | 129 | \paragraph{9.} 130 | \begin{proof} 131 | For every $\vep>0$, put $\delta=\vep/\sqrt{b-a}$. Then for every $x_1,x_2\in 132 | V$ with $\|x_1-x_2\|_\infty<\delta$, 133 | \[ 134 | \|x_1-x_2\|_2^2 = \int_a^b|x_1(t)-x_2(t)|^2\rd t 135 | \le (b-a)\delta^2 = \vep^2. 136 | \] 137 | Hence, $x\mapsto x$ is continuous. 138 | \end{proof} 139 | 140 | \paragraph{10.} 141 | \begin{proof} 142 | For every $u, w\in X$, 143 | \begin{align*} 144 | \langle Tu, w\rangle = 145 | & \frac{1}{4}(\langle T(u+w), u+w \rangle - \langle T(u-w), u-w \rangle)\\ 146 | &+ \frac{i}{4} 147 | (\langle T(u+iw), u+iw\rangle - \langle T(u-iw), u-iw\rangle).\\ 148 | \end{align*} 149 | Note that each component of the right hand side is of form $\langle Tx, x 150 | \rangle$ and hence equals to $0$. Putting $w=Tu$ yields $\|Tu\|^2=0$ for all 151 | $u\in X$. Thus, $T=0$. 152 | \end{proof} 153 | % end 154 | 155 | \subsection{Orthogonal Complements and Direct Sums} 156 | \paragraph{7.} 157 | \begin{proof} 158 | $\,$\par 159 | (a) $x\in A^{\perp\perp}$ iff for all $y\in A^\perp$, $\langle x, y\rangle= 160 | 0$. By the definition of $A^\perp$, the identity holds if $x\in A$. Hence, 161 | $A\subset A^{\perp\perp}$.\par 162 | (b) For all $x\in B^\perp$ and $y\in A\subset B$, $\langle x, y\rangle=0$ by 163 | definition. Hence, $x\in A^\perp$. Namely, $B^\perp\subset A^\perp$.\par 164 | (c) We show that $A^\perp$ is closed (no matter whether $A$ is or not) and 165 | invoke Lemma 3.3-6 to complete the proof. Suppose that $(x_n)\subset 166 | A^\perp$ converges to $x$. For all $y\in A$, $\langle x_n, y\rangle=0$. By 167 | the continuity of the inner product, $\langle x, y\rangle=0$ and therefore 168 | $x\in A^\perp$. Hence, $A^\perp$ is closed. Thus, $A^\perp=A^{\perp\perp 169 | \perp}$. 170 | \end{proof} 171 | 172 | \paragraph{8.} 173 | \begin{proof} 174 | We have show this in Prob. 7. 175 | \end{proof} 176 | 177 | \paragraph{9.} 178 | \begin{proof} 179 | It has been shown in Lemma 3.3-6 that the closedness of $Y$ implies $Y=Y^{ 180 | \perp\perp}$. Hence we only show the converse here. For every convergent 181 | $(x_n)\subset Y$, $(x_n)\subset Y^{\perp\perp}$. Since $Y^{\perp\perp}$ is 182 | closed by Prob. 8, the limit $x$ of $(x_n)$ belongs to $Y^{\perp\perp}$ and 183 | hence belongs to $Y$. Thus, $Y$ is closed. 184 | \end{proof} 185 | 186 | \paragraph{10.} TODO 187 | % end 188 | 189 | \subsection{Orthonormal Sets and Sequences} 190 | \paragraph{3.} 191 | \begin{proof} 192 | The situation where $x$ and $y$ are linearly dependent is obvious and hence 193 | we assume they are linearly independent here. By the homogeneity of the 194 | Schwarz inequality, we may assume without loss of generality that $\|x\|= 195 | \|y\|=1$. Put $z=(y-x\langle y,x\rangle)/\|y-x\langle y,x\rangle\|$. Then 196 | $\{x,z\}$ is orthonormal and therefore by (12*) 197 | \[ 198 | |\langle y,x\rangle|^2+|\langle y,z\rangle|^2 \le \|y\|^2 = 1. 199 | \] 200 | Since $|\langle y,z\rangle|^2$ is nonnegative, this implies $|\langle x,y 201 | \rangle|^2\le 1$, the Schwarz inequality. 202 | \end{proof} 203 | 204 | \paragraph{7.} 205 | \begin{proof} 206 | For each positive integer $n$, by the Schwarz inequality and (12*), 207 | \[ 208 | \sum_{k=1}^n|\langle x,e_k\rangle\langle y,e_k\rangle|\le 209 | \sqrt{\sum_{i=1}^n|\langle x,e_k\rangle|^2} 210 | \sqrt{\sum_{i=1}^n|\langle y,e_k\rangle|^2} \le 211 | \|x\|\|y\|. 212 | \] 213 | Since all terms in the summation is nonnegative, this implies $\sum_{k=1} 214 | ^\infty|\langle x,e_k\rangle\langle y,e_k\rangle|\le\|x\|\|y\|$. 215 | \end{proof} 216 | 217 | \paragraph{8.} 218 | \begin{proof} 219 | It follows immediately from Bessel inequality. 220 | \end{proof} 221 | % end 222 | 223 | \subsection{Series Related to Orthonormal Sequences} 224 | \paragraph{1.} 225 | \begin{proof} 226 | By Theorem 3.5-2, $\alpha_k=\langle x,e_k\rangle$. Meanwhile by the 227 | definition of the norm, 228 | \[ 229 | \|x\|^2 = 230 | \left\langle \sum_{k=1}^\infty\langle x,e_k\rangle e_k, x\right\rangle = 231 | \sum_{k=1}^\infty \langle x,e_k\rangle\langle e_k, x\rangle = 232 | \sum_{k=1}^\infty|\alpha_k|^2, 233 | \] 234 | where the second equality follows from the continuity of the inner product. 235 | \end{proof} 236 | 237 | \paragraph{3.} 238 | \begin{solution} 239 | Put $x\equiv 1$ on $[-\pi,\pi]$ and $e_k=\sin kt$. Since $x$ is even but 240 | $e_k$ is odd for every $k$, the series does not converges to $x$. 241 | \end{solution} 242 | 243 | \paragraph{4.} 244 | \begin{proof} 245 | By the triangle inequality, $\|x_m+\cdots+x_n\| \le \|x_m\|+\cdots+\|x_n\|$ 246 | for every $n\ge m>0$. Hence the convergence of $\sum \|x_k\|$ implies that 247 | $s_n$ is a Cauchy sequence. 248 | \end{proof} 249 | 250 | \paragraph{5.} 251 | \begin{proof} 252 | By Prob. 4, $\sum_{k=1}^n x_k$ is a Cauchy sequence. And since $H$ is 253 | complete, $\sum_{k=1}^\infty x_k$ converges. 254 | \end{proof} 255 | 256 | \paragraph{7.} 257 | \begin{proof} 258 | The existence of $y$ follows from Theorem 3.5-2(c). And for each $k$, 259 | \[ 260 | \langle x-y,e_k\rangle = 261 | \langle x,e_k\rangle - 262 | \sum_{j=1}^\infty\langle x,e_j\rangle\langle e_k,e_j\rangle = 263 | \langle x,e_k\rangle - \langle x,e_k\rangle = 0, 264 | \] 265 | where the second equality comes from the fact that $(e_k)$ is orthonormal. 266 | \end{proof} 267 | 268 | \paragraph{8.} 269 | TODO: Show the validation of the change of the order of summation. Or maybe 270 | we can show the equality directly. 271 | \begin{proof} 272 | We suppose that $x\in\bar{M}$ here since the proof of the other direction is 273 | obvious. Then there exists $(p_n)\subset M$ such that $x=\sum_{n=1}^\infty 274 | p_n$. For each $n$, suppose $p_n=\sum_{k=1}^\infty \langle p_n,e_k\rangle 275 | e_k$. This is valid because $p_n\in M$ and therefore is a finite linear 276 | combination of $(e_k)$. In fact, there are only finitely many nonzero term 277 | in the summation. Then 278 | \[ 279 | x=\sum_{n=1}^\infty p_n = 280 | \sum_{n=1}^\infty\sum_{k=1}^\infty\langle p_n,e_k\rangle e_k= 281 | \sum_{k=1}^\infty\left(\sum_{n=1}^\infty\langle p_n,e_k\rangle\right)e_k. 282 | \] 283 | \end{proof} 284 | 285 | \paragraph{9.} 286 | \begin{proof} 287 | First we suppose $\bar{M}_1=\bar{M}_2$. Then by Prob. 8, each $e_n$ and 288 | $\tilde{e}_n$ can be represented by (a) and (b) respectively.\par 289 | For the converse, (a) implies, again by Prob. 8, $e_n\in\bar{M}_2$ and 290 | therefore $M_1\subset\bar{M}_2$. Since $\bar{M}_2$ is closed, $\bar{M}_1 291 | \subset\bar{M}_2$. \textit{Mutatis mutandis}, this also shows $\bar{M}_2 292 | \subset\bar{M}_1$. Thus, $\bar{M}_1=\bar{M}_2$. 293 | \end{proof} 294 | 295 | \paragraph{10.} 296 | \begin{proof} 297 | Note that for every $m>0$, there are only finite $e_\kappa$ such that $ 298 | \langle x,e_\kappa\rangle\ge 1/m$. Otherwise we may choose a countable 299 | subset of them, which will violate the result in Prob. 8, Sec 3.4. Hence, 300 | the collection of all nonzero Fourier coefficient 301 | \[ 302 | \bigcup_{m=1}^\infty\{ e_\kappa:\, \langle x,e_\kappa\rangle\ge 1/m \} 303 | \] 304 | is at most countable. 305 | \end{proof} 306 | % end 307 | 308 | \subsection{Total Orthonormal Sets and Sequences} 309 | \paragraph{4.} 310 | \begin{proof} 311 | Suppose that $x$ and $y$ satisfy (3). We only show the relation for real 312 | cases here. The complex cases can be proved in a similar way. Using (9), 313 | Sec 3.1 and (3), 314 | \[ 315 | \langle x,y\rangle = \frac{1}{4}(\|x+y\|^2-\|x-y\|^2) = 316 | \frac{1}{4}\sum_k 317 | \left(|\langle x+y,e_k\rangle|^2-|\langle x-y,e_k\rangle|^2 \right). 318 | \] 319 | Meanwhile, 320 | \[ 321 | |\langle x\pm y,e_k\rangle|^2= 322 | \langle x\pm y,e_k\rangle\overline{\langle x\pm y,e_k\rangle}= 323 | |\langle x,e_k\rangle|^2+|\langle y,e_k\rangle|^2\pm 324 | 2\langle x,e_k\rangle\overline{\langle y,e_k\rangle}. 325 | \] 326 | Hence, $\langle x,y\rangle=\sum_k\langle x,e_k\rangle\overline{\langle 327 | y,e_k\rangle}$. 328 | \end{proof} 329 | 330 | \paragraph{6.} 331 | \begin{proof} 332 | Suppose $M=(e_k)$. We collect the $e_k$ which does not belong to $\spn(e_1, 333 | \dots,e_{k=1})$ and denote the new sequence by $(\tilde{e}_k)$. Clear that 334 | $\spn(e_k)=\spn(\tilde{e}_k)$ and $(\tilde{e}_k)$ is linearly independent. 335 | Let $(f_k)$ be the sequence generated from $(\tilde{e}_k)$ by the 336 | Gram-Schmidt process. Then clear that $(f_k)$ is orthonormal. And since for 337 | every $n$, $\spn(\tilde{e}_1,\dots,\tilde{e}_n)=\spn(f_1,\dots,f_n)$, 338 | $M\subset\spn(\tilde{e}_k)=\spn(f_k)$. Finally, since $M$ is dense in $H$, 339 | $\overline{\spn(f_k)}=H$. Thus, $(f_k)$ is a total orthonormal sequence of 340 | $H$. 341 | \end{proof} 342 | 343 | \paragraph{7.} 344 | \begin{proof} 345 | It follows from the definition of the separable Hilbert space and Prob. 6 346 | immediately. 347 | \end{proof} 348 | 349 | \paragraph{9.} 350 | \begin{proof} 351 | $\langle v,x\rangle=\langle w,x\rangle$ implies $\langle v-w,x\rangle=0$ for 352 | all $x\in M$, that is, $x\perp M$. Since $M$ is total, by Theorem 3.6-2, $v 353 | -w=0$. 354 | \end{proof} 355 | 356 | \paragraph{10.} 357 | \begin{proof} 358 | It follows immediately from Theorem 3.6-2(b). 359 | \end{proof} 360 | % end 361 | 362 | \setcounter{subsection}{7} 363 | \subsection{Functionals on Hilbert Spaces} 364 | \paragraph{3.} 365 | \begin{proof} 366 | The linearity follows from the sesquilinearity of the inner product and the 367 | boundedness from the Schwarz inequality. Furthermore, the Schwarz inequality 368 | also implies $\|f\|\le\|z\|$. Meanwhile, $\|f\|\ge\|f(z/\|z\|)\|=\|z\|$. 369 | Thus, $\|f\|=\|z\|$. 370 | \end{proof} 371 | 372 | \paragraph{4.} 373 | \begin{proof} 374 | Clear that the mapping $z\mapsto f$ is an isomorphism since it is 375 | surjective. And by Theorem 2.10-4, $X\hp$ is a Hilbert space. Hence, $X$ is 376 | also a Hilbert space. 377 | \end{proof} 378 | 379 | \paragraph{5.} 380 | \begin{proof} 381 | Since $l^2$ is complete. By Theorem 3.8-1, we may define $I:(l^2)\hp\to l^2$ 382 | to be $f\mapsto z$. Clear that $I$ is linear and injective. Meanwhile, it 383 | preserves the norm. Furthermore, by Prob. 3, it is surjective. Hence, $I$ 384 | is an isomorphism. Thus, $l^2$ is isomorphic to its dual. 385 | \end{proof} 386 | 387 | \paragraph{12.} 388 | \begin{proof} 389 | For every $x\in X$ and $y\in Y$, 390 | \begin{align*} 391 | |h(x+\Delta x, y+\Delta y)-h(x, y)| 392 | &=|h(\Delta x, y)+h(x, \Delta y)+h(\Delta x, \Delta y)| \\ 393 | &\le|h(\Delta x, y)|+|h(x, \Delta y)|+|h(\Delta x, \Delta y)|. 394 | \end{align*} 395 | Since $h$ is bounded, 396 | \[ 397 | |h(x+\Delta x, y+\Delta y)-h(x, y)|\le 398 | \|h\|(\|\Delta x|\|\|y\|+\|\Delta y|\|\|x\|+\|\Delta x|\|\|\Delta y\|). 399 | \] 400 | Thus, $h$ is continuous. 401 | \end{proof} 402 | 403 | \paragraph{14.} 404 | \begin{proof} 405 | If $h(x,x)=0$, then for any $t\in \mathbb{R}$, 406 | \[ 407 | 0\le h(th(y,x)x+y,th(y,x)x+y)=2t|h(x,y)|^2+h(y,y). 408 | \] 409 | Hence, $h(x,y)=0$, otherwise we may choose some $t<0$ such that is right 410 | hand side is negative. Thus, the inequality holds if $h(x,x)=0$.\par 411 | Now suppose $h(x,x)\ne 0$. Put 412 | \begin{equation} 413 | \label{eq:3.8.14} 414 | z=y-x\frac{h(y, x)}{h(x, x)} 415 | \end{equation} 416 | It is easy to verify that $h(z,x)=0$. Multiplying $z$ on the both sides of 417 | \eqref{eq:3.8.14} yields 418 | \[ 419 | 0\le h(z,z)=h\left(z,y-x\frac{h(y, x)}{h(x, x)}\right)=h(z,y)= 420 | h(y,y)-\frac{h(x,y)h(y,x)}{h(x,x)}. 421 | \] 422 | Thus, $|h(x,y)|^2\le h(x,x)h(y,y)$. 423 | \end{proof} 424 | % end 425 | 426 | \subsection{Hilbert-Adjoint Operator} 427 | \paragraph{1.} 428 | \begin{proof} 429 | By Theorem 3.9-4, $0^*=(0+0)^*=0^*+0^*$. Hence, $0^*=0$. For every $x,y\in 430 | X$, 431 | \[ 432 | \langle(I^*-I)x, y\rangle = \langle I^*x, y\rangle-\langle Ix, y\rangle= 433 | \langle x, Iy\rangle-\langle Ix, y\rangle=0. 434 | \] 435 | Hence, by Lemma 3.9-3, $I=I^*$. 436 | \end{proof} 437 | 438 | \paragraph{2.} 439 | \begin{proof} 440 | By Theorem 3.9-4, $T^*(T\inv)^*=(T\inv T)^*=I^*=I$. Hence, $(T^*)\inv= 441 | (T\inv)^*$. 442 | \end{proof} 443 | 444 | \paragraph{3.} 445 | \begin{proof} 446 | Since $\|T_n^*-T^*\|=\|(T_n-T)^*\|=\|T_n-T\|$, $T_n^*\to T^*$ as long as 447 | $T_n\to T$. 448 | \end{proof} 449 | 450 | \paragraph{4.} 451 | \begin{proof} 452 | It suffices to show that for all $x_2\in T^*(M_2^\perp)$ and $x_1\in M_1$, 453 | $\langle x_1,x_2\rangle=0$. $x_2\in T^*(M_2^\perp)$ implies the existence 454 | of some $y_2\in M_2^\perp$ with $T^*y_2=x_2$. Then 455 | \[ 456 | \langle x_1,x_2\rangle = \langle x_1,T^*y_2\rangle= 457 | \langle Tx_1,y_2\rangle = 0, 458 | \] 459 | where the last equality comes from the fact that $T(M_1)\subset M_2$ and 460 | $y_2\in M_2^\perp$. Thus, $M_1^\perp\supset T^*(M_2^\perp)$. 461 | \end{proof} 462 | 463 | \paragraph{5.} 464 | \begin{proof} 465 | By Prob. 4, $T^*(M_2^\perp)\subset M_1^\perp$ implies $M_2^{\perp\perp} 466 | \supset T(M_1^{\perp\perp})$. Since $M_1$ and $M_2$ are closed, by Prob. 9, 467 | Sec 3.3, $M_i^{\perp\perp}=M_i$ for $i=1,2$. Thus, $T(M_1)\subset M_2$. 468 | The converse part has already been proved in Prob. 4. 469 | \end{proof} 470 | 471 | \paragraph{6.} 472 | \begin{proof} 473 | $\,$\par 474 | (a) Since $T(M_1)=\{0\}\subset H_2$, by Prob. 4, $T^*(H_2)\subset 475 | M_1^\perp$.\par 476 | (b) For every $y\in[T(H_1)]^\perp$, $\langle y,Tx\rangle=0$ for all $x\in 477 | H_1$. Hence, $\langle T^*y, x\rangle=0$. By Lemma 3.8-2, $T^*y=0$ and 478 | therefore $y\in\mathcal{N}(T^*)$. Thus, $[T(H_1)]^\perp\subset\mathcal{N} 479 | (T^*)$.\par 480 | (c) Since $T^{**}=T$, it follows from (b) that $[T^*(H_2)]^\perp\subset 481 | M_1$. And since $M_1$ is closed, $M_1^{\perp\perp}=M_1$. Therefore, (a) 482 | implies $[T^*(H_2)]^\perp\supset M_1$. Thus, $M_1=[T^*(H_2)]^\perp$. 483 | \end{proof} 484 | 485 | \paragraph{7.} 486 | \begin{proof} 487 | It follows immediately from Lemma 3.9-3. 488 | \end{proof} 489 | 490 | \paragraph{8.} 491 | \begin{proof} 492 | For every $x\in H$ with $\|x\|=1$, 493 | \begin{align*} 494 | \|(I+T^*T)x\| 495 | &=\|x+T^*Tx\|=\langle x+T^*Tx,x+T^*Tx\rangle\\ 496 | &=\|x\|^2+\|T^*Tx\|^2+\langle x,T^*Tx\rangle+\langle T^*Tx,x\rangle\\ 497 | &=\|x\|^2+\|T^*Tx\|^2+\|Tx\|^2 \\ 498 | &\ge 1. 499 | \end{align*} 500 | Then, by Prob 7, Sec 2.7, $I+T^*T$ is invertible. 501 | \end{proof} 502 | 503 | \paragraph{9.} 504 | \begin{proof} 505 | If $T$ can be represent by that form, then $\mathcal{R}(T)$ can be spanned 506 | by $w_1,\dots,w_n$. Hence, it is finite dimensional.\par 507 | Now we suppose that $T$ has a finite dimensional range. Let $\{w_1,\dots, 508 | w_n\}$ be a orthonormal basis of $\mathcal{R}(T)$. Then for every $x\in H$, 509 | \[ 510 | Tx=\sum_{j=1}^n\varphi_j(x)w_j. 511 | \] 512 | Now we show that for each $j$, $\varphi_j$ is a bounded linear functional 513 | and invoke Riesz's Theorem to complete the proof. It is easy to verify the 514 | linearity of $\varphi_j$. For every $x$ with norm $1$, since $T$ is bounded 515 | and $(w_j)$ is orthonormal, 516 | \[ 517 | \|T\|\ge\left\|\sum_{j=1}^n\varphi_j(x)w_j\right\|\ge |\varphi_j(x)| 518 | \] 519 | for each $j=1,\dots,n$. Hence, every $\varphi_j$ is a bounded linear 520 | functional and therefore can be represented by $\varphi_j(x)=\langle x,v_j 521 | \rangle$. 522 | \end{proof} 523 | % end 524 | 525 | \paragraph{Self-Adjoint, Unitary and Normal Operators} 526 | \paragraph{4.} 527 | We only show the uniqueness here. 528 | \begin{proof} 529 | $T_1+iT_2=S_1+iS_2$ implies $T_1-iT_2=S_1-iS_2$. Sum these two equations and 530 | we get $T_1=S_1$. Meanwhile, it also implies $i(T_1+iT_2)=i(S_1+iS_2)$. 531 | Summing these two gives $T_2=S_2$. 532 | \end{proof} 533 | 534 | \paragraph{6.} 535 | \begin{proof} 536 | $\,$\par 537 | (a) We argue by contradiction. Let $k$ be the smallest positive integer such 538 | that $T^{2k}=0$. Then for every $x\in H$ 539 | \[ 540 | 0=\langle T^{2k}x,x\rangle = \langle T^kx,(T^k)^*x\rangle= 541 | \langle T^kx,T^kx\rangle=\|T^kx\|^2. 542 | \] 543 | Hence $T^k=0$, which contradicts with the smallest assumption of $k$. Hence, 544 | $T^n\ne 0$ for all even positive integer $n$.\par 545 | (b) If $T^n=0$ for some positive, not necessarily even, integer $n$, then so 546 | is $T^{2n}=0$. Hence, by (a), $T^n\ne 0$ for all $n\in\mathbb{N}$. 547 | \end{proof} 548 | 549 | \paragraph{9.} 550 | \begin{proof} 551 | Since $T$ is isometric, it preserves the topology. Hence $T(H)$ is closed 552 | as $H$ is closed. 553 | \end{proof} 554 | 555 | \paragraph{10.} 556 | \begin{proof} 557 | It suffices to show that $T$ is surjective. Let $\{e_1,\dots,e_n\}$ be an 558 | orthonormal basis of $X$. Then $\{Te_1,\dots,Te_n\}$ is also an orthonormal 559 | basis since $T$ is isometric. Hence, $T$ is surjective and therefore is 560 | unitary. 561 | \end{proof} 562 | 563 | \paragraph{13.} 564 | \begin{proof} 565 | It can be verified that $T_n^*T_n\to T^*T$ and $T_nT_n^*\to TT^*$. Since 566 | $T_n$ are normal, $T_n^*T_n=T_nT_n^*$. Hence, 567 | \[ 568 | \|T^*T-TT^*\|\le \|T^*T-T_n^*T_n\|+\|T_nT_n^*-TT^*\|\to 0 569 | \] 570 | as $n\to\infty$. Thus, $T$ is normal. 571 | \end{proof} 572 | 573 | \paragraph{15.} 574 | \begin{proof} 575 | If $T$ is normal, clear that $\|T^*x\|=\|Tx\|$. Now we suppose $\|Tx\|=\|T^* 576 | x\|$ for all $x\in H$. Then $\langle TT^*x,x\rangle=\langle T^*Tx,x\rangle$. 577 | Since $X$ is complex, by Lemma 3.9-3, $TT^*=T^*T$. Namely, $T$ is normal. 578 | \par 579 | By (a), for every $x\in H$, $\|T^2x\|=\|T^*Tx\|$. Hence, 580 | \[ 581 | \|T^2\|=\sup_{\|x\|=1}\|T^2x\|=\sup_{\|x\|=1}\|T^*Tx\|=\|T^*T\|=\|T\|^2, 582 | \] 583 | where the last equality comes from Theorem 3.9-4(e). 584 | \end{proof} 585 | % end 586 | 587 | 588 | 589 | 590 | 591 | 592 | 593 | 594 | 595 | 596 | 597 | 598 | 599 | 600 | 601 | -------------------------------------------------------------------------------- /introductory_functional_analysis_with_applications/ch4_fundamental_theorems_for_normed_and_banach_spaces.tex: -------------------------------------------------------------------------------- 1 | %---------%---------%---------%---------%---------%---------%---------%--------- 2 | \section{Fundamental Theorems for Normed and Banach Spaces} 3 | \setcounter{subsection}{1} 4 | \subsection{Hahn-Banach Theorem} 5 | \paragraph{4.} 6 | \begin{proof} 7 | By the positive homogeneity, $p(2\times 0)=2p(0)$. Hence, $p(0)=0$. 8 | Consequently, $0=p(x+(-x))\le p(x)+p(-x)$. Thus, $-p(x)\le p(-x)$. 9 | \end{proof} 10 | 11 | \paragraph{5.} 12 | \begin{proof} 13 | For every $x,y\in M$ and $\lambda\in[0,1]$, 14 | \[ 15 | p(\lambda x+(1-\lambda)y)\le \lambda p(x)+(1-\lambda)p(y) 16 | \le \lambda\gamma+(1-\lambda)\gamma=\gamma. 17 | \] 18 | Hence, $\lambda x+(1-\lambda)y\in M$. Thus, $M$ is convex. 19 | \end{proof} 20 | 21 | \paragraph{6.} 22 | \begin{proof} 23 | For every $x,t\in X$, 24 | \[ 25 | p(x-t)\le p(x)+p(-t) \quad\Rightarrow\quad 26 | p(x-t)-p(x)\le p(-t), 27 | \] 28 | and 29 | \[ 30 | p(x)=p(x-t+t)\le p(x-t)+p(t) \quad\Rightarrow\quad 31 | -p(t)\le p(x-t)-p(x). 32 | \] 33 | Since $p(0)=0$ and $p$ is continuous at $0$, $p(t)\to 0$ and $p(-t)\to 0$ as 34 | $t\to 0$. Hence, $p(x-t)-p(x)\to 0$ as $t\to 0$, that is, $p$ is continuous 35 | on $X$. 36 | \end{proof} 37 | 38 | \paragraph{8.} 39 | \begin{proof} 40 | First, $p(0)\ge p(0+0)-p(0)=0$. For nonzero $x$, we argue by contradiction. 41 | Assume that there exists some $x$ with $0<\|x\|\le r$ such that $p(x)<0$. 42 | Then $np(x)<0$ for $n=1,2,\dots$. For $n$ sufficiently large, $n\|x\|>r$ and 43 | therefore $p(nx)\ge 0$. However, by the subadditivity, $p(nx)\le np(x)<0$. 44 | Contradiction. Thus, $p(x)\ge 0$ on $X$. 45 | \end{proof} 46 | 47 | \paragraph{9.} 48 | \begin{proof} 49 | For all $x_1=\alpha_1x_0, x_2=\alpha_2x_0\in Z$ and scalars $a_1$ and $a_2$, 50 | \begin{align*} 51 | f(a_1x_1+a_2x_2)&=f((a_1\alpha_1+a_2\alpha_2)x_0)= 52 | (a_1\alpha_1+a_2\alpha_2)p(x_0) \\ 53 | &=a_1\alpha_1p(x_0)+a_2\alpha_2p(x_0)=a_1f(x_1)+a_2f(x_2). 54 | \end{align*} 55 | Thus, $f$ is linear. Now we show that for $\alpha\in\mathbb{R}$, $\alpha p 56 | (x_0)\le p(\alpha x_0)$ to complete the proof. If $\alpha\ge 0$, then it 57 | follows from the positive homogeneity. For negative $\alpha$, $\alpha p 58 | (x_0)=-p(-\alpha x_0)$ and by Prob. 4, $-p(-\alpha x_0)\le p(\alpha x_0)$. 59 | Thus, $f(x)\le p(x)$ for all $x\in Z$. 60 | \end{proof} 61 | 62 | \paragraph{10.} 63 | \begin{proof} 64 | Let $Z$ and $f$ have the same meaning as in Prob. 9. By Hahn-Banach theorem, 65 | there exists a linear extension $\tilde{f}$ of $f$ to $X$ with $\tilde{f}(x) 66 | \le p(x)$ for all $x\in X$. Replacing $x$ with $-x$ gives $\tilde{f}(-x)\le 67 | p(x)$. Finally, the linearity of $\tilde{f}$ yields $-p(-x)\le\tilde{f}(x)$. 68 | \end{proof} 69 | % end 70 | 71 | \subsection{Hahn-Banach Theorem for Normed Spaces} 72 | \paragraph{1.} 73 | \begin{proof} 74 | By (2), $p(2\times 0)=2p(0)$. Hence, $p(0)=0$. And for every $x\in X$, by 75 | (1), 76 | \[ 77 | 0=p(0)\le p(x)+p(-x)=2p(x), 78 | \] 79 | that is, $p(x)\ge 0$. 80 | \end{proof} 81 | 82 | \paragraph{2.} 83 | \begin{proof} 84 | By (1), $p(x)=p(x-y+y)\le p(x-y)+p(y)$. Therefore, $p(x)-p(y)\le p(x-y)$. 85 | Interchange the roles of $x$ and $y$ and we obtain $p(y)-p(x)\le p(y-x)= 86 | p(x-y)$, where the equality comes from (2). Thus, $|p(x)-p(y)|\le p(x-y)$. 87 | \end{proof} 88 | 89 | \paragraph{7.} 90 | \begin{proof} 91 | Define $\tilde{f}$ to be $x\mapsto\langle x, x_0/\|x_0\|\rangle$. Clear that 92 | it is a bounded linear functional on $X$ and $\tilde{f}(x_0)=\|x_0\|$. And 93 | by Riesz's Theorem, $\|\tilde{f}\|=\|x_0/\|x_0\|\|=1$. 94 | \end{proof} 95 | 96 | \paragraph{8.} 97 | \begin{proof} 98 | It follows immediately from Theorem 4.3-3. 99 | \end{proof} 100 | 101 | \paragraph{13.} 102 | \begin{proof} 103 | Just put $\hat{f}=\tilde{f}/\|x_0\|$. 104 | \end{proof} 105 | 106 | \paragraph{14.} 107 | \begin{proof} 108 | By Prob 13, there exists a $\hat{f}\in X\hp$ such that $\|\hat{f}\|=1/r$ and 109 | $\hat{f}(x_0)=1$. Let hyperplane $H_0=\{x\in X:\, \hat{f}(x)=1\}$ and half 110 | space $S_0=\{x\in X:\, \hat{f}(x)\le 1\}$. Then clear that $x_0\in H_0$ and 111 | for all $x\in S(0;r)$, $f(x) \le \|f\|\|x\| = r/r =1$. Hence, $x\in S_0$. 112 | \end{proof} 113 | 114 | \paragraph{15.} 115 | \begin{proof} 116 | If $\|x\|=c+2\vep>c$, then by Corollary 4.3-4, there exists some $0\ne f\in 117 | X\hp$ such that $|f(x)|/\|f\|\ge c+\vep$. Consequently, the functional $g=f/ 118 | \|f\|$, which is of norm $1$, is such that $|g(x_0)|>c$. Contradiction. 119 | \end{proof} 120 | % end 121 | \setcounter{subsection}{4} 122 | \subsection{Adjoint Operator} 123 | \paragraph{9.} 124 | \begin{proof} 125 | Note that every bounded linear functional is continuous by Theorem 2.7-9. 126 | Hence, $M^a=(\mathcal{R}(T))^a$. Thus, $g\in M^a$ iff $g\in(\mathcal{R} 127 | (T))^a$ iff $g(Tx)=(T^\times g)(x)=0$ for all $x\in X$ iff $T^\times g=0$ 128 | iff $g\in\mathcal{N}(T^\times)$. Namely, $M^a=\mathcal{N}(T^\times)$. 129 | \end{proof} 130 | 131 | \paragraph{10.} 132 | \begin{proof} 133 | For every $y=Tx\in\mathcal{R}(T)$, $g(Tx)=(T^\times g)(x)=0$ for all $g\in 134 | \mathcal{N}(T^\times)$. Hence, $y\in ^a\mathcal{N}(T^\times)$. 135 | \end{proof} 136 | % end 137 | \subsection{Reflexive Spaces} 138 | \paragraph{2.} 139 | \begin{proof} 140 | Since $Y$ is a closed subspace of a Hilbert space, it is complete. By Lemma 141 | 3.3-2, there is some $y\in Y$ such that $\|x_0-y\|=\delta$ and $z=x_0-y$ is 142 | orthogonal to $Y$. Define $\tilde{f}$ by $x\mapsto\langle x,z\rangle/ 143 | \delta$. Then clear that $\tilde{f}\in X\hp$ and $\tilde{f}(y)=0$ for all $y 144 | \in Y$. Meanwhile, by Riesz's Theorem, $\|\tilde{f}\|=\|z\|/\delta=1$. 145 | Finally, 146 | \[ 147 | \tilde{f}(x_0)=\frac{\langle x_0,x_0-y\rangle}{\delta}= 148 | \frac{\langle x_0-y+y,x_0-y\rangle}{\delta}=\delta. 149 | \] 150 | The proof is then completed. 151 | \end{proof} 152 | 153 | \paragraph{3.} 154 | \begin{proof} 155 | We denote the canonical mapping from $X$ to $X^{\prime\prime}$ by $C$ and 156 | the one from $X\hp$ to $X^{\prime\prime\prime}$ by $D$. Our goal is to find 157 | a $f\in X\hp$ for every given $h\in X^{\prime\prime\prime}$ such that $D(f)= 158 | h$, that is, for every $g\in X^{\prime\prime}$, $D(f)(g)=h(g)$. Since $X$ is 159 | reflexive, there is some $x\in X$ such that $g=Cx$. Put $f=hC$, which is 160 | clearly an element of $X\hp$. Since 161 | \[ 162 | h(g) = h(Cx)=(hC)(x)=f(x)\quad\text{and}\quad 163 | D(f)(g)=g(f)=(Cx)(f)=f(x), 164 | \] 165 | $h=D(hC)$. Thus, $X\hp$ is reflexive. 166 | \end{proof} 167 | 168 | \paragraph{4.} 169 | \begin{proof} 170 | By Prob. 3, the reflexivity of $X$ implies the reflexivity of $X\hp$. Now we 171 | suppose $X\hp$ is reflexive. Hence, again by Prob. 3, $X^{\prime\prime}$ is 172 | reflexive and therefore, by Theorem 4.6-4, is complete. Since $X$ is 173 | isomorphic to $\mathcal{R}(C)\subset X^{\prime\prime}$ and $\mathcal{R}(C)$, 174 | a closed subspace of a reflexive Banach space, is reflexive, so is $X$. 175 | Thus, a Banach space $X$ is reflexive iff $X\hp$ is reflexive. 176 | \end{proof} 177 | 178 | \paragraph{5.} 179 | \begin{proof} 180 | It suffices to show that $\delta>0$ and then putting $h=\tilde{f}/\delta$ 181 | will complete the proof. If $\delta=0$, then by the definition of the 182 | infimum, there exists $(y_n)\subset Y$ which converges to $x_0$. Then $x_0 183 | \in Y$ since $Y$ is closed, which contradicts our choice of $x_0$. Thus, 184 | $\delta>0$. 185 | \end{proof} 186 | 187 | \paragraph{6.} 188 | \begin{proof} 189 | We may assume without loss of generality that $Y_2\setminus Y_1$ is 190 | nonempty. Choose arbitrary $x_0\in Y_2\setminus Y_1\subset X\setminus Y_1$. 191 | By Prob. 6, there exists some $h\in X\hp$ such that $h(x_0)=1$ and $h\in 192 | Y^a$. Thus, the annihilators of $Y_1$ and $Y_2$ are different. 193 | \end{proof} 194 | 195 | \paragraph{7.} 196 | \begin{proof} 197 | If $Y$ is proper, then by Prob. 7, the annihilators of $Y$ and $X$ do not 198 | coincide, which contradicts our hypothesis. Hence, $X=Y$. 199 | \end{proof} 200 | 201 | \paragraph{8.} 202 | \begin{proof} 203 | If $x\in A$, then for every $f\in X\hp$ whose restriction to $M$ is $0$, 204 | $f(x_0)=0$ since $f$, being bounded, is continuous. For the converse, note 205 | that $f|_M=0$ implies $f|_A=0$. If $x_0\notin A$, then Prob. 5 guarantees 206 | the existence of some $f\in X\hp$ which vanishes on $A$ and is nonzero at 207 | $x_0$. Contradiction. Thus, $x_0\in A$. 208 | \end{proof} 209 | 210 | \paragraph{9.} 211 | \begin{proof} 212 | If $M$ is total, then clear that every $f\in X\hp$ vanishing on $M$ is zero 213 | everywhere on $X$. And the converse part follows immediately from Prob. 8. 214 | \end{proof} 215 | 216 | \paragraph{10.} 217 | \begin{proof} 218 | Let $\{b_1,\dots,b_n\}$ be a linearly independent subset of $X$ and define 219 | \[ 220 | \beta_i:\spn\{b_1,\dots,b_n\}\to\mathbb{F} 221 | \quad\text{by}\quad 222 | b_j\mapsto\delta_{ij} 223 | \] 224 | for $i=1,\dots,n$. By Hahn-Banach Theorem, we can extend them to linear 225 | functionls $\tilde{\beta}_i$ on $X$. Suppose that $f=x_1\tilde{\beta}_1+ 226 | \cdots+x_n\tilde{\beta}_n=0$. Then $0=f(b_i)=x_i$ for all $i$. Thus, $\{ 227 | \tilde{\beta}_1,\dots,\tilde{\beta}_n\}$ is linearly independent. 228 | \end{proof} 229 | % end 230 | \subsection{Uniform Boundedness Theorem} 231 | \paragraph{1.} 232 | \begin{solution} 233 | Meager, since $\mathbb{Q}$ is the union of all singleton of rational 234 | numbers. 235 | \end{solution} 236 | 237 | \paragraph{5.} 238 | \begin{proof} 239 | First we suppose $M$ is rare and argue by contradiction. If $(\bar{M})^c$ is 240 | not dense in $X$, i.e., there exists some $x\in X$ and $r>0$ such that $B(x; 241 | r)\cap(\bar{M})^c=\varnothing$. Hence, $B(x;r)\subset\bar{M}$, which 242 | contradicts the definition of rare subsets. Thus, $(\bar{M})^c$ is dense in 243 | $X$.\par 244 | Now we suppose $(\bar{M})^c$ is dense in $X$. Then for all $x\in\bar{M}$ and 245 | $r>0$, there exists some $y_r\notin\bar{M}$ but $y_r\in B(x;r)$. Hence, $x$ 246 | is not an interior point. Thus, $M$ is rare. 247 | \end{proof} 248 | 249 | \paragraph{6.} 250 | \begin{proof} 251 | If both $M$ and $M^c$ are meager, then so is their union $X$, but Baire's 252 | theorem says that a complete metric space is nonmeager in itself. Hence, 253 | $M^c$ is nonmeager if $M$ is. 254 | \end{proof} 255 | 256 | \paragraph{7.} 257 | \begin{proof} 258 | We argue by contradiction. Assume that for all $x\in X$, $\sup_n\|T_nx\|< 259 | \infty$. Then by the uniform boundedness theorem, there exists some $c$ 260 | such that $\|T_n\|\le c$ for all $n$. Hence, $\sup_n\|T_n\|\le c$. 261 | Contradiction. 262 | \end{proof} 263 | 264 | \paragraph{10.} 265 | \begin{proof} 266 | We may assume without loss of generality that $\eta_1\ne 0$. Define $T_n: 267 | c_0\to\mathbb{C}$ by $(\xi_j)\mapsto \sum_{j=1}^n\xi_j\eta_j$. Clear that 268 | $T_n$ are linear functionals. And since 269 | \begin{equation} 270 | \label{eq:4.7.10} 271 | |T_nx|=\left|\sum_{j=1}^n\eta_j\xi_j\right|\le 272 | \max_{j=1,\dots,n}|\xi_j|\sum_{j=1}^n|\eta_j|\le 273 | \|x\|\sum_{j=1}^n|\eta_j|, 274 | \end{equation} 275 | $T_n$ are bounded and $\|T_n\|\le\sum_{j=1}^n|\eta_j|$. Meanwhile, define 276 | $y=(\gamma_j)$ by 277 | \[ 278 | \gamma_j=\begin{cases} 279 | \sgn\eta_j, & j\le n, \\ 280 | 0 , & j>n. 281 | \end{cases} 282 | \] 283 | Clear that $y\in c_0$ and $\|y\|=1$. Since $|T_ny|=\sum_{j=1}^n|\eta_j|$, 284 | together with \eqref{eq:4.7.10}, we conclude $\|T_n\|=\sum_{j=1}^n|\eta_j|$. 285 | \par 286 | By Prob. 2, Sec 2.3, $c_0$ is a Banach space. And for each $x=(\xi_j)\in 287 | c_0$, since $\sum\xi_j\eta_j$ converges, $\|T_nx\|$ is bounded for $n$ large 288 | enough and therefore bounded for all $n$. Hence, by the uniform boundedness 289 | theorem, $\sum_{j=1}^n|\eta_j|=\|T_n\|\le c$ for some fixed $c$. Thus, $\sum 290 | |\eta_j|<\infty$. 291 | \end{proof} 292 | 293 | \paragraph{11.} 294 | \begin{proof} 295 | By Prob. 4, Sec 1.4, the Cauchy sequence $(T_nx)$ is bounded. Thus, by the 296 | uniform boundedness theorem, $(\|T_n\|)$ is bounded. 297 | \end{proof} 298 | 299 | \paragraph{13.} 300 | \begin{proof} 301 | Let $C:X\to X^{\prime\prime}$ be the canonical embedding and $(\varphi_n)= 302 | (Cx_n)$. By Lemma 4.6-1, $\|x_n\|=\|\varphi_n\|$. Note that $X^{\prime 303 | \prime}$, the dual space of $X\hp$, is complete and $f(x_n)=\varphi_n(f)$. 304 | Thus, by the uniform boundedness theorem, $(\|x_n\|)=(\|\varphi_n\|)$ is 305 | bounded. 306 | \end{proof} 307 | 308 | \paragraph{14.} 309 | \begin{proof} 310 | $\,$\par 311 | (a)$\Rightarrow$(c): It follows immediately from $|g(T_nx)|\le\|g\|\|x\| 312 | \|T_n\|$.\par 313 | (c)$\Rightarrow$(b): For fixed $x\in X$, let $\varphi_n=C(T_nx)$, where $C: 314 | Y\to Y^{\prime\prime}$ is the canonical embedding. For every $g\in Y\hp$, by 315 | (c), $|\varphi_n(g)|=|g(T_nx)|\le c_g$. Since $Y\hp$ is complete, by the 316 | uniform boundedness theorem, $(\|\varphi_n\|)=(\|T_nx\|)$ is bounded.\par 317 | (b)$\Rightarrow$(a): It is just what the uniform boundedness theorem states. 318 | \end{proof} 319 | % end 320 | \subsection{Strong and Weak Convergence} 321 | \paragraph{1.} 322 | \begin{proof} 323 | The mapping $\varphi_t:C[a,b]\to\mathbb{F}$, $x\mapsto x(t)$ is a bounded 324 | linear functional on $C[a,b]$. Hence, by the definition of weak convergence, 325 | $x_n(t)\to x(t)$. 326 | \end{proof} 327 | 328 | \paragraph{2.} 329 | \begin{proof} 330 | For every $f\in Y\hp$, $fT\in X\hp$. Since $x_n\wto x_0$, $(fT)(x_n)\to (fT) 331 | (x_0)$, that is, $Tx_n\wto Tx_0$. 332 | \end{proof} 333 | 334 | \paragraph{4.} 335 | \begin{proof} 336 | If $x_0=0$, then it is trivial. Otherwise, by Theorem 4.3-3, there exists 337 | some $f\in X\hp$ such that $f(x_0)=\|x_0\|$ and $\|f\|=1$. Since $x_n\wto 338 | x$, $|f(x_n)|\to |f(x_0)|=\|x_0\|$. Meanwhile, $|f(x_n)|\le \|f\|\|x_n\|=\| 339 | x_n\|$. Thus, $\lowlim\|x_n\|\ge \|x_0\|$. 340 | \end{proof} 341 | 342 | \paragraph{5.} 343 | \begin{proof} 344 | If $\bar{Y}=X$, then there is nothing to be proved. Otherwise we argue by 345 | contradiction. Assume that $x_0\in X\setminus\bar{Y}\ne\varnothing$. Then by 346 | Lemma 4.6-7, there exists some $f\in X\hp$ such that $f(Y)=\{0\}$ and $f 347 | (x_0)=\delta>0$. However, since $x_n\in Y$ and $x_n\wto x_0$, $f(x_0)$ must 348 | be $0$. Contradiction. Thus, $x_0\in\bar{Y}$. 349 | \end{proof} 350 | 351 | \paragraph{6.} 352 | \begin{proof} 353 | It follows immediately from Prob. 5. 354 | \end{proof} 355 | 356 | \paragraph{7.} 357 | \begin{proof} 358 | It follows immediately from Prob. 5. 359 | \end{proof} 360 | 361 | \paragraph{8.} 362 | \begin{proof} 363 | For every $f\in X\hp$, by the definition, $|f(x_n)|0$, there is an $N$ such that for all 420 | $n>N$, $\|T_n-T\|<\vep$. Hence, for all $x$ with $\|x\|=1$, 421 | \[ 422 | \|T_nx-Tx\|\le \|T_n-T\|\|x\|<\vep. 423 | \]\par 424 | Now we suppose the converse. Since $\|T_n-T\|=\sup_{\|x\|=1}\|T_nx-Tx\|$, 425 | there is some $x$ of norm $1$ such that 426 | \[ 427 | \|T_n-T\|-\vep<\|T_nx-Tx\|<\vep. 428 | \] 429 | Thus, $\|T_n-T\|\le 2\vep$, which implies $T_n\to T$. 430 | \end{proof} 431 | 432 | \paragraph{7.} 433 | \begin{proof} 434 | For every $x\in X$, since $T_nx$ converges, $(\|T_nx\|)$ is bounded. As $X$ 435 | is complete, this implies $(\|T_n\|)$ is bounded by the uniform boundedness 436 | theorem. 437 | \end{proof} 438 | 439 | \paragraph{9.} 440 | \begin{proof} 441 | For every $x\in X$ with $\|x\|=1$, 442 | \[ 443 | \|Tx\|=\lim_{n\to\infty}\|T_nx\|\le \lowlim_{n\to\infty}\|T_n\| 444 | \] 445 | where the first equality comes from the continuity of the norm. Thus, $\|T\| 446 | \le \lowlim_{n\to\infty}\|T_n\|$. 447 | \end{proof} 448 | 449 | \paragraph{10.} 450 | \begin{proof} 451 | Since $X$ is separable, we can find a total sequence $(b_n)$ of $X$. Now we 452 | choose a subsequence of $(f_n)\subset X\hp$ as follows. First we choose a 453 | subsequence of $(f_k(x_n))$ for each fixed $n$ inductively. Since $M$ is 454 | bounded, $(f_k(x_1))$ is a bounded sequence in $\mathbb{R}$ and therefore 455 | has a convergent subsequence $f_{k_i}(x_1)$. For $x_2$ we choose the 456 | subsequence of $(f_{k_i}(x_2))$ which is convergent. We proceed in this way 457 | and choose a sequence of $(f_k(x_n))$ for each $n$. Now, consider the 458 | subsequence $(g_k)$ of $(f_n)$ whose $n$-th element is the $n$-th functional 459 | in the $n$-th choice. It can be verified that it is a subsequence. And by 460 | our construction, $(f_k(x_n))$ converges for every $x_n$. Hence, by 461 | Corollary 4.9-7, it is weak$^*$ convergent. 462 | \end{proof} 463 | % end 464 | \setcounter{subsection}{11} 465 | \subsection{Open Mapping Theorem} 466 | \paragraph{2.} 467 | \begin{solution} 468 | Consider $f:\mathbb{R}\to\mathbb{R}$, $x\mapsto e^x$. Clear that $f$ is 469 | open. However, it maps $\mathbb{R}$, a closed set, onto $(0,\infty)$, an 470 | open set. 471 | \end{solution} 472 | 473 | \paragraph{4.} 474 | \begin{proof} 475 | It follows from the fact that $\|x_k\|<1/2^k$. 476 | \end{proof} 477 | 478 | \paragraph{5.} 479 | \begin{proof} 480 | The linearity and boundedness is clear. Meanwhile, it is easy to verify that 481 | $T\inv((\xi_k))=(k\xi_k)$. Let $x_n=(\xi_k^{(n)})$ be defined by $\xi_k 482 | ^{(n)}=\delta_{nk}$. Then $\|T\inv x_n\|=n$. Hence, $T\inv$ is unbounded. 483 | \par 484 | This does not contradict with 4.12-2 since $X$ is not complete. For example, 485 | The sequence $(x_n)\subset X$ where $x_n=(\xi_k^{(n)})$ is defined by 486 | \[ 487 | \xi_k^{(n)}=\begin{cases} 488 | 1/k, & k0$ such that 510 | \[ 511 | \|x\|=\|T\inv y\|\le c\|y\|=c\|Tx\|. 512 | \] 513 | Thus, putting $a=1/c>0$, we obtain $a\|x\|\le\|Tx\|$ for all $x\in X$. 514 | \end{proof} 515 | 516 | \paragraph{9.} 517 | \begin{proof} 518 | Let $J:X_2\to X_1$ be the identity map. Clear that $J$ a bijective linear 519 | operator. Then $\|x\|_1\le c\|x\|_2$ implies 520 | \[ 521 | \frac{\|Jx\|_1}{\|x\|_2}\le c\quad\text{for all nonzero $x\in X_2$}. 522 | \] 523 | Namely, $J$ is bounded. Since $X_1$ and $X_2$ are complete, by the bounded 524 | inverse theorem, the inverse $I$ of $J$ is also a bounded linear operator. 525 | Thus, for all $x\in X_2$ 526 | \[ 527 | \|x\|_2=\|Ix\|_2\le \|I\|\|x\|_1. 528 | \] 529 | \end{proof} 530 | % end 531 | \subsection{Closed Linear Operators. Closed Graph Theorem} 532 | \paragraph{5.} 533 | \begin{proof} 534 | The closedness of $T$ implies that $\mathcal{G}(T)=\{(x,Tx):\,x\in 535 | \mathcal{D}(T)\}$ is closed in $X\times Y$. Then $\mathcal{G}(T\inv)=\{(Tx, 536 | x):\,x\in\mathcal{D}(T)\}$ is closed in $Y\times X$. Thus, $T\inv$ is 537 | closed. 538 | \end{proof} 539 | 540 | \paragraph{6.} 541 | \begin{proof} 542 | Suppose $Tx_n\to y$ and $T\tilde{x}_n\to\tilde{y}$. Since $x_n\to x\in 543 | \mathcal{D}(T)$, by Theorem 4.13-3, $Tx=y$. Similarly, $Tx=\tilde{y}$. Thus, 544 | $y=\tilde{y}$. 545 | \end{proof} 546 | 547 | \paragraph{7.} 548 | \begin{proof} 549 | Since $X$ is a closed subset of itself, by Lemma 4.13-5, $T$ is closed and 550 | therefore, by Prob. 5, so is $T\inv$. Since $Y=\mathcal{R}(T)$ is closed in 551 | $Y$ itself, by the closed graph theorem, $T\inv$ is bounded and therefore 552 | continuous. 553 | \end{proof} 554 | 555 | \paragraph{8.} 556 | \begin{proof} 557 | $\,$\par 558 | (a) Let $(y_n)=(Tx_n)\subset A$ converge to $y\in Y$. Since $C$ is compact, 559 | $(x_n)$ has a convergent sequence, say, $(x_{n_k})$. Suppose $x_{n_k}\to x 560 | \in C$ as $k\to\infty$. By Theorem 4.13-3, $y=\lim Tx_{n_k}=Tx$. Therefore, 561 | $y\in A$. Thus, $A$ is closed.\par 562 | (b) Let $(x_n)\subset B$ converge to $x\in X$. Since $K$ is compact, the 563 | sequence $(y_n)=(Tx_n)\subset K$ has a convergent sequence $(y_{n_k})$. 564 | Suppose $y_{n_k}\to y\in K$. Since $T$ is closed, $y=Tx$. Hence, $x\in B$. 565 | Thus, $B$ is closed. 566 | \end{proof} 567 | % end 568 | 569 | 570 | 571 | 572 | 573 | 574 | 575 | 576 | 577 | 578 | 579 | 580 | 581 | 582 | 583 | 584 | -------------------------------------------------------------------------------- /introductory_functional_analysis_with_applications/def.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt, a4paper]{article} 2 | 3 | \usepackage[margin=1in]{geometry} 4 | \usepackage{ 5 | color, 6 | clrscode, 7 | amssymb, 8 | amsmath, 9 | listings, 10 | fontspec, 11 | xcolor, 12 | supertabular, 13 | multirow, 14 | mathtools, 15 | mathrsfs, 16 | amsthm, 17 | systeme, 18 | amsfonts 19 | } 20 | \definecolor{bgGray}{RGB}{36, 36, 36} 21 | \usepackage[ 22 | colorlinks, 23 | linkcolor=bgGray, 24 | anchorcolor=blue, 25 | citecolor=green 26 | ]{hyperref} 27 | \newfontfamily\courier{Courier} 28 | 29 | \newcommand{\vep}{\varepsilon} 30 | 31 | \newenvironment{solution} 32 | {\begin{proof}[Solution]} 33 | {\end{proof}} 34 | \newcommand{\rd}{\mathrm{d}} 35 | \newcommand{\inv}{^{-1}} 36 | \newcommand{\hp}{^\prime} 37 | \newcommand{\wto}{\xrightarrow{w}} 38 | \newcommand{\wsto}{\xrightarrow{w^*}} 39 | 40 | \DeclareMathOperator\sgn{sgn} 41 | \DeclareMathOperator\spn{span} 42 | \DeclareMathOperator\codim{codim} 43 | 44 | \DeclareMathOperator*\lowlim{\underline{lim}} 45 | \DeclareMathOperator*\uplim{\overline{lim}} 46 | \DeclareMathOperator*\esssup{ess\,sup} 47 | \def\upint{\mathchoice 48 | {\mkern13mu\overline{\vphantom{\intop}\mkern7mu}\mkern-20mu}% 49 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 50 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 51 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 52 | \int} 53 | \def\lowint{\mkern3mu\underline{\vphantom{\intop}\mkern7mu}\mkern-10mu\int} 54 | 55 | -------------------------------------------------------------------------------- /introductory_functional_analysis_with_applications/main.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Engineev/solutions/4e33274fe1ed9e46fd0e6671c57cb589704939bd/introductory_functional_analysis_with_applications/main.pdf -------------------------------------------------------------------------------- /introductory_functional_analysis_with_applications/main.tex: -------------------------------------------------------------------------------- 1 | \input{def.tex} 2 | 3 | \title{Solutions to\\ 4 | \textit{Introductory Functional Analysis with Applications}} 5 | \author{Yunwei Ren} 6 | \date{} 7 | 8 | \begin{document} 9 | \maketitle 10 | \tableofcontents 11 | \newpage 12 | 13 | \setcounter{section}{1} 14 | \input{ch2_normed_spaces_banach_spaces.tex} 15 | \newpage 16 | \input{ch3_inner_product_spaces_hilbert_spaces.tex} 17 | \newpage 18 | \input{ch4_fundamental_theorems_for_normed_and_banach_spaces} 19 | 20 | \end{document} 21 | -------------------------------------------------------------------------------- /linear_algebra_done_right_3rd/ch3_linear_map.tex: -------------------------------------------------------------------------------- 1 | \section{Linear Map} 2 | \subsection{The Vector Space of Linear Maps} 3 | \paragraph{1.} 4 | \begin{proof} 5 | If $T$ is linear, then $T(0,0,0)=0$ and therefore $b=0$. Meanwhile, 6 | $T(2,2,2)=2T(1,1,1)$ implies $12+8c=12+2c$. Hence, $c=0$. The proof of the 7 | converse part is trivial. 8 | \end{proof} 9 | 10 | \paragraph{3.} 11 | \begin{proof} 12 | Let $e_i$ be the $i$-th vector in the standard base of $\mathbb{F}^n$ and 13 | suppose that $Te_i = \sum_{j=1}^n A_{1,j}e_j$. Then for $x=(x_1,\dots,x_n)^T 14 | \in\mathbb{F}^n$, 15 | \[ 16 | Tx = T\left(\sum_{i=1}^n x_ie_i\right) = \sum_{i=1}^n x_iTe_i = 17 | \sum_{i=1}^n x_i\sum_{j=1}^nA_{j,i}e_j = 18 | \sum_{j=1}^n\left(\sum_{i=1}^nA_{j,i}x_i\right) e_j. 19 | \] 20 | \end{proof} 21 | 22 | \paragraph{5.} 23 | \begin{proof} 24 | Too lengthy to write it down... 25 | \end{proof} 26 | 27 | \paragraph{7.} 28 | \begin{proof} 29 | Let $\{x_0\}$ be a basis of $V$ and $\lambda$ be a scalar such that $Tx_0= 30 | \lambda x_0$. By the linearity of $T$, for every $x=kx_0$ in $V$, $Tx=kTx_0 31 | =k\lambda x_0=\lambda(kx_0)=\lambda x$. 32 | \end{proof} 33 | 34 | \paragraph{9.} 35 | \begin{solution} 36 | From the additivity condition we can derive that $\varphi(kz)=k\varphi(z)$ 37 | for any $k\in\mathbb{Q}$. Hence we can try some functions where $\varphi(iz) 38 | =i\varphi(z)$ fails. It turns out that $\varphi(z)=\Im(z)$ is one of the 39 | maps required. 40 | \end{solution} 41 | 42 | \paragraph{11.} 43 | \begin{proof} 44 | Let $\{\alpha_1,\dots,\alpha_p\}$ and $\{\alpha_1,\dots,\alpha_p,\beta_1, 45 | \dots,\beta_q\}$ be bases of $U$ and $V$ respectively. Then the linear map 46 | which maps $\alpha_i$ to $T\alpha_i$ and maps $\beta$ to $0$. Clear that 47 | it is the desired linear map. 48 | \end{proof} 49 | 50 | \paragraph{13.} 51 | \begin{proof} 52 | Suppose that $v_k$ is in the span of the other vectors and let $w_i=0$ for 53 | each $i\ne k$ and $w_k\ne 0$. No $T\in\mathcal{L}(V, W)$ can maps $v_i$ to 54 | $w_i$ since the linearity of $T$ would force $w_k$ to be $0$, leading to a 55 | contradiction. 56 | \end{proof} 57 | 58 | % end 59 | 60 | \subsection{Null Spaces and Ranges} 61 | \paragraph{2.} 62 | \begin{proof} 63 | Since $S$ maps every vector of $V$ into the null space of $T$, the map $TS$ 64 | is the zero map. Hence $(ST)^2 = S(TS)T = 0$. 65 | \end{proof} 66 | 67 | \paragraph{4.} 68 | \begin{proof} 69 | Suppose $S,T\in\mathcal{L}(\mathbb{R}^5,\mathbb{R}^4)$ maps and only maps 70 | $e_1,e_2,e_3$ and $e_3,e_4,e_5$ to the zero vector respectively. Then $e_1, 71 | e_2,e_4,e_5\notin\nul(S+T)$, implying that $\dim\nul(S+T)<2$. Hence $\{T\in 72 | \mathcal{L}(\mathbb{R}^5,\mathbb{R}^4\,:\,\dim\nul T>2)\}$ is not a subspace 73 | of $\mathcal{L}(\mathbb{R}^5,\mathbb{R}^4)$. 74 | \end{proof} 75 | 76 | \paragraph{6.} 77 | \begin{proof} 78 | It follows immediately from the rank-nullity theorem and the fact that $\dim 79 | \nul T$ and $\dim\range T$ are integers. 80 | \end{proof} 81 | 82 | \paragraph{8.} 83 | \begin{proof} 84 | Let $\{w_1,\dots,w_m\}$ be a basis of $W$ and $S,T\in\mathcal{L}(V,W)$ be 85 | two linear maps such that $\range S=\spn(w_1)$ and $\range T=\spn(w_2,\dots, 86 | w_n)$. Clear that $\range(S+T)=W$. Hence, the set described is not a 87 | subspace of $\mathcal{L}(V,W)$. 88 | \end{proof} 89 | 90 | \paragraph{10.} 91 | \begin{proof} 92 | For every $y\in\range T$ there exists some $x=\sum x_iv_i\in V$ such that 93 | \[ 94 | y=Ty = T\left(\sum_{i=1}^n x_iv_i\right) = \sum_{i=1}^n x_iTv_i. 95 | \] 96 | Hence, $\range T=\spn(Tv_1,\dots,Tv_n)$. 97 | \end{proof} 98 | 99 | \paragraph{12.} 100 | For readers who familiar with the orbit-stabilizer theorem or just the 101 | (group) homomorphism, the proof should be straightforward. 102 | \begin{proof} 103 | For every nonzero $y$ in $\range T$, there exists some $x\in V$ such that 104 | $Tx=y$. For each $y\ne 0$, we choose one such $x$, put them all together and 105 | put $0$ into them to get $U$. By the construction, clear that $T(U) = 106 | \range T$ and , $U\cap\nul T=\{0\}$. 107 | \end{proof} 108 | 109 | \paragraph{14.} 110 | \begin{proof} 111 | By the rank-nullity theorem, 112 | \[ 113 | \dim\nul T + \dim\range T = 8 \quad\Rightarrow\quad 114 | \dim\range T = 5 = \dim\mathbb{R}^5. 115 | \] 116 | Hence, $\range T = \mathbb{R}^5$ and therefore $T$ is surjective. 117 | \end{proof} 118 | 119 | \paragraph{16.} 120 | Actually, the cosets of the kernel partition the whole space. 121 | \begin{proof} 122 | Let $\{v_1,\dots,v_n\}$ be a basis of $\range T$ and $Tu_i=v_i$ for $i=1,2, 123 | \dots,n$. Denote $\spn(u_1,\dots,u_n)$ by $U$. We now prove that $V=U+\nul 124 | T$. For every $x\in V$, suppose that $Tx=y=\sum y_iv_i$ and $\tilde{x}=\sum 125 | y_iu_i$. Note that $\tilde{x}\in U$ and $T(x-\tilde{x})=Tx-T\tilde{x}=0$, 126 | i.e., $x-\tilde{x}\in\nul T$. Hence, $V=U+\nul T$. As both of $U$ and $\nul 127 | T$ are finite-dimensional, so is $V$. 128 | \end{proof} 129 | 130 | \paragraph{18.} 131 | \begin{proof} 132 | By the rank-nullity theorem, clear that $\dim V \ge \dim\range T =\dim W$ if 133 | there exists some surjective $T\in\mathcal{L}(V,W)$. \par 134 | Assume that $\dim V\ge\dim W$ and let $\{v_1,\dots,v_n\}$ and $\{w_1,\dots, 135 | w_m\}$ be bases of $V$ and $W$ respectively. Then the linear map which 136 | maps $v_i$ to $w_i$ for each $1\le i\le m$ is surjective. 137 | \end{proof} 138 | 139 | \paragraph{20.} 140 | \begin{proof} 141 | If $T$ is injective, then for every $y\in\range T$, there exists exactly 142 | one $x\in V$ such that $y=Tx$. Let $S$ be the map which maps $y$ to such 143 | $x$. It is linear since for every $y_1,y_2\in\range T$ and scalar $a,b$, 144 | supposing $Sy_i = x_i$, 145 | \[ 146 | T(ax_1+bx_2) = aTx_1 + bTx_2 = ay_1 + by_2. 147 | \] 148 | implying $S(ay_1+by_2) = ax_1+bx_2 = aSy_1 + bSy_2$. For every $x\in V$, 149 | $(ST)x = S(Tx) = x$. \par 150 | Suppose there exists some $S\in \mathcal{L}(W,V)$ such that $ST=I$. Then 151 | \[ 152 | Tx_1 = Tx_2 \quad\Rightarrow\quad 153 | STx_1 = STx_2 \quad\Rightarrow\quad 154 | x_1 = x_2. 155 | \] 156 | Hence, $T$ is injective. 157 | \end{proof} 158 | 159 | \paragraph{22.} 160 | \begin{proof} 161 | Let $\tilde{T}$ be the restriction of $T$ to $\nul ST$. It is still a linear 162 | map since $\nul ST$ is a subspace of $U$. Note that $x\in\nul ST$ iff $(ST)x 163 | =0$ iff $Tx\in\nul S$. Hence, $\range\tilde{T}\subset\nul S$. Thus, by the 164 | rank-nullity theorem, 165 | \[ 166 | \dim\range\tilde{T} \le \dim\nul S \quad\Rightarrow\quad 167 | \dim\nul ST - \dim\nul\tilde{T} \le \dim\nul S. 168 | \] 169 | Since $\nul\tilde{T} \le \nul T$, this implies $\dim\nul ST \le \dim\nul S + 170 | \dim\nul T$. 171 | \end{proof} 172 | 173 | \paragraph{24.} 174 | \begin{proof} 175 | If there exists $S\in\mathcal{L}(W,W)$ such that $T_2=ST_1$, then $\nul T_2 176 | =\nul ST_1$. Hence for every $x\in\nul T_1$, as $S(T_1x)=S0 = 0$, $x\in\nul 177 | T_2$. Therefore, $\nul T_1\subset\nul T_2$.\par 178 | Now we suppose $\nul T_1\subset\nul T_2$ and construct $S$. Note that all we 179 | concerns is its behavior on some basis of $\range T_1$. Let $\{w_1, \dots, 180 | w_n\}$ be a basis of $\range T_1$ and $T_1v_i = w_i$ for $i=1,\dots,n$. For 181 | each $x\in V$, let $U_x = \{x+y\,:\,y\in\nul T_2\}$ and $Sw_k=T_2x$ if $v_k 182 | \in U_x$. It can be verified that $S$ is well-defined and does satisfy the 183 | requirement as long as $\nul T_1\subset\nul T_2$. 184 | \end{proof} 185 | 186 | \paragraph{26.} 187 | \begin{proof} 188 | Let $\mathcal{P}_n(\mathbb{R})=\{p\in\mathcal{P}(\mathbb{R})\,\,:\,\deg p 189 | \le n\}$, which are some subspaces of $\mathcal{P}(\mathbb{R})$. We now 190 | prove that $D$ is a surjective linear map onto $\mathcal{P}_n(\mathbb{R})$ 191 | for every nonnegative integer $n$ by induction. \par 192 | Suppose $Dx = c_0 \ne 0$, then for any $0\ne c\in\mathcal{P}_0(\mathbb{R})$, 193 | $D(cx / c_0) = c$. Hence, $D$ is a surjective map onto $\mathcal{P}_0 194 | (\mathbb{R})$. Assume that $D$ is a surjective map onto $\mathcal{P}_{k-1} 195 | (\mathbb{R})$ and suppose $Dx^{k+1} = p=a_0 + a_1x + \cdots + a_kx^k$ where 196 | $a_k\ne 0$. For every nonzero $b_k$ and $q = b_0+b_1x+\cdots+b_kx^k \in 197 | \mathcal{P}_k(\mathbb{R})$, let $r$ be a polynomial with degree $\le k-1$ 198 | such that $q=b_k/a_k p + r$. By our induction hypothesis, there exists some 199 | polynomial $\tilde{r}$ such that $D\tilde{r}=r$. Then 200 | \[ 201 | D(b_k/a_k x^{k+1} + \tilde{r}) = \frac{b_k}{a_k}Dx^{k+1} + D\tilde{r} = 202 | \frac{b_k}{a_k}p+r = q. 203 | \] 204 | Hence, $D$ is also a surjective map onto $\mathcal{P}_k(\mathbb{R})$. Thus, 205 | $D$ is surjective. 206 | \end{proof} 207 | 208 | \paragraph{28.} TODO 209 | 210 | \paragraph{30.} TODO 211 | 212 | % end 213 | 214 | \setcounter{subsection}{3} 215 | \subsection{Invertibility and Isomorphic Vector Spaces} 216 | 217 | \paragraph{1.} 218 | \begin{proof} 219 | Clear that the linear map $T\inv S\inv$ is right and left inverse of $ST$ 220 | and therefore $ST$ is invertible. And by the uniqueness of the inverse, 221 | $(ST)\inv = T\inv S\inv$. 222 | \end{proof} 223 | 224 | \paragraph{3.} 225 | \begin{proof} 226 | First we suppose the existence of such an operator, then $T\inv$ is also the 227 | inverse of $S$. Hence $S$ is invertible and therefore injective. \par 228 | Now we suppose $S$ is injective. Let $\{u_1,\dots,u_m\}$ and $\{u_1,\dots, 229 | u_m, u_{m+1},\dots,u_n\}$ be bases of $U$ and $V$ respectively. $\{Su_1, 230 | \dots, Su_m\}$ is linearly independent as $S$ is injective and therefore we 231 | can expand it to a basis, $\{Su_1,\dots,Su_m,v_{m+1},\dots,v_n\}$, of $V$. 232 | Let $T\in\mathcal{L}(V)$ maps $u_i$ to $Su_i$ for $i=1,\dots,m$ and $u_j$ to 233 | $v_j$ for $j=m+1,\dots,n$. $T$ is obviously injective and therefore 234 | invertible as $V$ is finite-dimensional. 235 | \end{proof} 236 | 237 | \paragraph{5.} 238 | \begin{proof} 239 | Suppose that such an $S$ exists. Since $S$ is invertible, $\range S=V$. 240 | Hence, $\range T_2 = \range T_2S = \range T_1$. \par 241 | Now we suppose that $\range T_1=\range T_2$ and construct $S$ by defining 242 | its behavior on a basis of $V$. Let $\{v_1,\dots,v_m\}$ be a basis of $\nul 243 | T_1$. As $\range T_1=\range T_2$ implies $\dim\nul T_1=\dim\nul T_2$, we can 244 | set $Sv_i = u_i$ for $i=1,\dots,m$ where $\{u_1,\dots,u_m\}$ is a basis of 245 | $\nul T_2$. \par 246 | Let $\{v_1,\dots,v_m,v_{m+1},\dots,v_n\}$ be a basis of $V$. Clear that $\{ 247 | T_1v_{m+1},\dots,T_1v_n\}$ spans $\range T_1$. It is linearly independent 248 | since 249 | \begin{align*} 250 | & x_{m+1}T_1v_{m+1}+\cdots+x_nT_1v_n = 0 \\ 251 | \Rightarrow\quad& T_1(x_{m+1}v_{m+1}+\cdots+x_nv_n) = 0 \\ 252 | \Rightarrow\quad& x_{m+1}v_{m+1}+\cdots+x_nv_n \in \nul T_1 \\ 253 | \Rightarrow\quad& x_{m+1} = \cdots = x_n = 0. 254 | \end{align*} 255 | Hence, it is a basis of $\range T_1$. Since $\range T_1=\range T_2$, there 256 | exists $u_{m+1},\dots,u_n$ such that $T_2u_i = T_1v_i$ for $i=m+1,\dots,n$. 257 | It is easy to verify that $u_1,\dots,u_m,u_{m+1},\dots,u_n$ are linearly 258 | independent. Finally, for $i=m+1,\dots,n$, we also set $Sv_i = u_i$. Clear 259 | that $S$ is invertible and satisfies the requirement. 260 | \end{proof} 261 | 262 | \paragraph{7.} 263 | \begin{proof} 264 | $\,$\\ 265 | (a) For any $A,B\in E$ and scalar $a,b$, 266 | \[ 267 | (aA+bB)v = a(Av) + b(Bv) = 0. 268 | \] 269 | Hence, $E$ is a subspace of $\mathcal{L}(V,W)$. \\ 270 | (b) Since $v\ne 0$, putting $v_1=v$, there exists some vectors in $V$ such 271 | that $\{v_1,\dots,v_n\}$ is a basis of $V$. Let $U=\spn(v_2,\dots,v_n)$. It 272 | can be shown that $E$ is isomorphic to $\mathcal{U,W}$. Hence, $\dim E = 273 | (\dim V-1)\dim W$. 274 | \end{proof} 275 | 276 | \paragraph{9.} 277 | \begin{proof} 278 | If $S$ and $T$ are invertible, then clear that $T\inv S\inv$ is the inverse 279 | of $ST$. Meanwhile, if $S$ or $T$ is not invertible, therefore not 280 | surjective, then 281 | \[ 282 | \dim\range ST\le\min\{\dim\range S,\dim\range T\}<\dim V. 283 | \] 284 | Hence, $ST$ is not surjective and hence not invertible as $V$ is 285 | finite-dimensional. Thus, $ST$ is invertible iff $S$ and $T$ are invertible. 286 | \end{proof} 287 | 288 | \paragraph{11.} 289 | \begin{proof} 290 | Since $V$ is finite-dimensional and $S(TU)=(ST)U=I$, both $S$ and $U$ are 291 | invertible and the inverses of which are $TU$ and $ST$ respectively. Hence, 292 | \[ 293 | STU=I \quad\Rightarrow\quad T = S\inv U\inv, 294 | \] 295 | implying that $T$ is also invertible and $T\inv = US$. 296 | \end{proof} 297 | 298 | \paragraph{13.} 299 | \begin{proof} 300 | It follows almost immediately from Exercise 9 that all of $R$, $S$ and $T$ 301 | are invertible and therefore $S$ is injective. 302 | \end{proof} 303 | 304 | \paragraph{15.} 305 | \begin{proof} 306 | Let $\{e_1,\dots,e_n\}$ be the standard basis of $\mathbb{F}^{n,1}$ and 307 | suppose $Te_i = u_i$. It is easy to verify that $A=(u_1, \dots, u_n)$ is a 308 | $m$-by-$n$ matrix such that $Tx=Ax$ for every $x\in\mathbb{F}^{n,1}$. 309 | \end{proof} 310 | % end 311 | 312 | \subsection{Products and Quotients of Vector Spaces} 313 | \paragraph{2.} 314 | \begin{proof} 315 | We only prove the result for $m=2$. It is easy to prove it for arbitrary $m$ 316 | in a similar manner. Suppose that $V=V_1\times V_2$ is finite-dimensional. 317 | Then $V_1\times\{0\}$, a subspace of $V$, is finite-dimensional. Clear that 318 | $V_1$ is isomorphic to $V_1\times \{0\}$ and hence it is also of finite 319 | dimension. Similarly, $V_2$ is finite-dimensional. 320 | \end{proof} 321 | 322 | \paragraph{4.} 323 | \begin{proof} 324 | We construct the isomorphism $S:\mathcal{L}(V_1\times\cdots\times V_n,W)\to 325 | \mathcal{L}(V_1,w)\times\cdots\times\mathcal{L}(V_n,W)$ explicitly. For 326 | every $T\in\mathcal{L}(V_1\times\cdots\times V_n,W)$, suppose $T(v_1,\dots, 327 | v_n) = w$. Let $T_i(v_i)=w$ for $i=1,\dots,n$ and $ST = (T_1,\dots, T_n)$. 328 | Clear that $T_i\in\mathcal{L}(V_i, W)$ and $S$ is invertible. 329 | \end{proof} 330 | 331 | \paragraph{6.} 332 | \begin{proof} 333 | We may interpret the elements in $\mathcal{L}(\mathbb{F}^n,V)$ as mappings 334 | from the "coordinates" to "abstract vectors". With this in mind, we 335 | construct the isomorphism $S$. For every $(v_1,\dots,v_n)\in V^n$ and $(x_1, 336 | \dots,x_n)^T\in\mathbb{F}^n$, let 337 | \[ 338 | (S(v_1,\dots,v_n)) 339 | \begin{bmatrix} 340 | x_1 \\ \vdots \\ x_n 341 | \end{bmatrix} 342 | = \sum_{i=1}^n x_iv_i. 343 | \] 344 | It is easy to verify that $S$ does satisfy the requirement. 345 | \end{proof} 346 | 347 | \paragraph{8.} 348 | We may interpret the set of all possible $\lambda v+(1-\lambda w)$ as the 349 | "line" through $v$ and $w$. And the idea behind the proof is illustrated in 350 | the picture below. 351 | \begin{figure}[htbp] 352 | \centering 353 | \includegraphics[height=5cm]{./image/3-d-8.png} 354 | \end{figure} 355 | \begin{proof} 356 | If $A$ is an affine subset, i.e., there exists some subspace $U$ and $a\in 357 | V$ such that $A=a+U$, then for all $\lambda\in\mathbb{F}$ and $v,w\in A$, 358 | \[ 359 | \lambda v+(1-\lambda)w = \lambda(a+u_1) + (1-\lambda)(a+u_2) = 360 | a + (u_1+(1-\lambda)u_2) \in A 361 | \] 362 | where $u_1$ and $u_2$ are some elements in $U$. \par 363 | Now we suppose $\lambda v+(1-\lambda)w\in A$ holds, fix $a\in A$ and let $U= 364 | \{a_1-a \,:\, a_1\in A\}$. By the hypothesis, for every scalar $\lambda$ and 365 | $a_1\in A$, $a + \lambda(a_1-a)\in A$. Therefore, for every $u_1 = a_1-a \in 366 | U$, $\lambda u_1\in U$. Meanwhile, let $u_2=a_2-a\in U$, $(u_1 +u_2)/2\in U$ 367 | as 368 | \[ 369 | a+\frac{1}{2}(u_1+u_2) = a + \frac{1}{2}(a_1+a_2-2a) = 370 | \frac{1}{2}a_1+\frac{1}{2}a_2. 371 | \] 372 | Hence, by the previous result, $u_1+u_2\in U$. Thus, $U$ is a subspace and 373 | $A=a+U$ is an affine subset. 374 | \end{proof} 375 | 376 | \paragraph{10.} 377 | \begin{proof} 378 | Let $A$ be the intersection of every collection of affine subsets of $V$ and 379 | suppose $A$ is nonempty. Let $v,w\in A$ and $\lambda\in\mathbb{F}$. Then, by 380 | Exercise 8, for every affine subset $A_{\alpha}$ of $V$, $\lambda v+ (1- 381 | \lambda)w\in A_\alpha$. Hence it also belongs to $A$. Thus, $A$ is also an 382 | affine subset of $A$ (as long as nonempty). 383 | \end{proof} 384 | 385 | \paragraph{12.} 386 | \begin{proof} 387 | Let $\{a_1+U,\dots, a_m+U\}$ be a basis of $V/U$ and we first prove a small 388 | result: for every $v\in V$, there exists an unique list of $v_1,\dots,v_m\in 389 | \mathbb{F}$ such that $v - (v_1a_1 + \cdots v_ma_m)\in U$. Suppose that 390 | $v_1\hp,\dots, v_m\hp$ is such a list as well. Then 391 | \[ 392 | (v - (v_1a_1+\cdots+v_ma_m))-(v - (v_1\hp a_1+\cdots+v_m\hp a_m)) \in U. 393 | \] 394 | Therefore, 395 | \[ 396 | (v_1-v_1\hp)a_1+\cdots+(v_m-v_m\hp)a_m \in U = 0+U, 397 | \] 398 | Hence $v_i\hp = v_i$ for each $i=1,\dots,m$, completing the proof.\par 399 | Therefore, for every $v\in V$, denoting $v_1a_i+\cdots+v_ma_m$ as $a_v$, we 400 | may define $S$ to be map which maps $v$ to $(v-a_v,a_v+U)$. Now we show that 401 | $S$ is linear and bijective. For every $u,v\in V$ and scalar $a, b$, 402 | \begin{align*} 403 | aSu + bSv 404 | &= a(u-a_u,a_u+U) + b(v-a_v,a_v+U) \\ 405 | &= ((au+bv)-(aa_u+ba_v), (aa_u+ba_v) + U) \\ 406 | &= S(au+bv). 407 | \end{align*} 408 | $Su = 0$ iff $(u-a_u,a_u+U) = 0$ iff $u=a_u=0$ and therefore $S$ is 409 | injective. Clear that $S$ is surjective. Thus, $S$ is an isomorphism and 410 | $V$ is isomorphic to $U\times(V/U)$. 411 | \end{proof} 412 | 413 | \paragraph{16.} 414 | \begin{proof} 415 | Clear that every vector space with dimension $1$ over field $\mathbb{F}$ is 416 | isomorphic to $\mathbb{F}$. Hence, it suffices to prove there exists 417 | $\varphi\in\mathcal{L}(V,V/U)$ such that $\nul\varphi = U$ and the quotient 418 | map is just the map we want. 419 | \end{proof} 420 | % end 421 | 422 | \subsection{Duality} 423 | \paragraph{1.} 424 | \begin{proof} 425 | Suppose that $\varphi\in V\hp$ and is not the zero map. Then, $\varphi(v)=c 426 | \ne 0$ for some $v\in V$. By the linearity of $\varphi$, for every $0\ne a 427 | \in\mathbb{F}$, $\varphi(av/c)=a$ and $\varphi(0)=0$, completing the proof. 428 | \end{proof} 429 | 430 | \paragraph{3.} 431 | \begin{proof} 432 | It suffices to prove that there exists $\varphi\in V\hp$ which maps $v$ to 433 | a nonzero element of $\mathbb{F}$. We argue by contradiction. Assume that 434 | for all $\varphi\in V\hp$, $\varphi(v)=0$. Then $\{v\}^0 = V\hp$. Hence, 435 | $\dim \{v\} = \dim V-\dim \{v\}^0 = 0$, implying that $v=0$. Contradiction. 436 | \end{proof} 437 | 438 | \paragraph{9.} 439 | \begin{proof} 440 | For every $v=\sum x_iv_i \in V$ and $\psi\in V\hp$, 441 | \begin{align*} 442 | \psi(v) 443 | &= \psi(x_1v_1+\cdots+ x_nv_n) \\ 444 | &= \psi(v_1)x_1 + \cdots + \psi(v_n)x_n \\ 445 | &= \psi(v_1)\varphi_1(v) + \cdots + \psi(v_n)\varphi_n(v) \\ 446 | &= (\psi(v_1)\varphi_1 + \cdots + \psi(v_n)\varphi_n)(v), 447 | \end{align*} 448 | where the third equality comes from the definition of the dual space and the 449 | last one comes from the linearity of $\varphi_1,\dots,\varphi_n$. 450 | \end{proof} 451 | 452 | \paragraph{11.} 453 | \begin{proof} 454 | $\rank A = 1$ iff there exists some nonzero $\alpha\in\mathbb{F}^m$ such 455 | that $A = [d_1\alpha\,\dots\, d_n\alpha]$ iff $A = \alpha [d_1\, \dots \, 456 | d_n]$. 457 | \end{proof} 458 | 459 | \paragraph{15.} 460 | \begin{proof} 461 | $T\hp=0$ iff $\dim W\hp = \dim\nul T\hp$ iff $\dim W = \dim(\range T)^0$ iff 462 | $\range T = 0$ iff $T=0$. 463 | \end{proof} 464 | 465 | \paragraph{19.} 466 | \begin{proof} 467 | As $U\subset V$ and $V$ is finite-dimensional, $U=V$ iff $\dim U=\dim V$ iff 468 | $\dim U^0 = 0$ by 3.106 iff $U^0 = \{0\}$. 469 | \end{proof} 470 | 471 | \paragraph{25.} 472 | \begin{proof} 473 | Note that the RHS of the equality equals to 474 | \[ 475 | \tilde{U} = \bigcap_{\varphi\in U^0}\nul\varphi. 476 | \] 477 | For every $u\in U$, since $u\in\nul\varphi$ for every $\varphi\in U^0$ by 478 | definition. Hence, $U\subset\tilde{U}$. And let $\psi\in U^0$ be a linear 479 | functional such that $\nul\psi = U$. Then $\dim\tilde{U}\le\dim\nul\psi= 480 | \dim U$. Hence, $U=\tilde{U}$. 481 | \end{proof} 482 | 483 | \paragraph{29.} 484 | \begin{proof} 485 | By the hypothesis, for every $\psi\in W\hp$, $T\hp(\psi)=\psi\circ T 486 | =k\varphi$ for some scalar $k$. By 3.109, $\dim\range T =\dim\range T\hp=1$. 487 | Hence, there exists $\psi\in W\hp$ whose restriction to $\range T$ is an 488 | one-to-one map to $\mathbb{F}$. Thus, 489 | \[ 490 | \nul\varphi = \nul k\varphi = \nul(\psi\circ T) = \nul T. 491 | \] 492 | \end{proof} 493 | 494 | \paragraph{31.} 495 | In brief, we choose an arbitrary basis of $V$ and try to express the 496 | required basis with it by solving a system of linear equations. 497 | \begin{proof} 498 | Let $u_1,\dots,u_n$ be a basis of $V$ and $A=[\varphi_i(u_j)]$. Now we prove 499 | that $A$ is invertible. Suppose 500 | \[ 501 | x_1\begin{bmatrix} 502 | \varphi_1(u_1) \\ \vdots \\ \varphi_n(u_1) 503 | \end{bmatrix} 504 | + \cdots + 505 | x_n\begin{bmatrix} 506 | \varphi_1(u_n) \\ \vdots \\ \varphi_n(u_n) 507 | \end{bmatrix} 508 | = 0. 509 | \] 510 | and $u=x_1u_1+\cdots+x_nu_n$. Then, $\varphi_i(u) = 0$ for $i=1,\dots,n$. 511 | As $\varphi_1,\dots,\varphi_n$ is a basis of $V\hp$, this implies 512 | $(\spn(u))^0 = V\hp$. Hence, by 3.106, $\dim\spn(u)=0$ and therefore $u=0$. 513 | Thus, the columns of $A$ are linearly independent and therefore $A$ is 514 | invertible. \par 515 | Let 516 | \begin{equation} 517 | \label{eq:3.F.31.1} 518 | \begin{bmatrix} 519 | v_1 & \cdots & v_n 520 | \end{bmatrix} = 521 | \begin{bmatrix} 522 | u_1 & \cdots & u_n 523 | \end{bmatrix}A\inv 524 | \end{equation} 525 | and now we prove that $v_1,\dots,v_n$ is a basis of $V$ and the dual basis 526 | of it is exactly $\varphi_1,\dots,\varphi_n$. Since $u_1,\dots,u_n$ are 527 | linearly independent and $A\inv$ is nonsingular, so do $v_1,\dots,v_n$. 528 | Hence, $v_1,\dots,v_n$ is a basis of $V$. \eqref{eq:3.F.31.1} also implies 529 | \[ 530 | u_k = \varphi_1(u_k)v_1 + \cdots + \varphi_n(u_1)v_n. 531 | \] 532 | Applying $\varphi_i$ on the both sides for each $k=1,\dots,n$ yields 533 | \[ 534 | \begin{bmatrix} 535 | \varphi_i(u_1) \\ \vdots \\ \varphi_i(u_n) 536 | \end{bmatrix} = 537 | \begin{bmatrix} 538 | \varphi_1(u_1) & \cdots & \varphi_n(u_1) \\ 539 | \vdots & \ddots & \vdots \\ 540 | \varphi_1(u_n) & \cdots & \varphi_n(u_n) 541 | \end{bmatrix} 542 | \begin{bmatrix} 543 | \varphi_i(v_1) \\ \vdots \\ \varphi_i(v_n) 544 | \end{bmatrix}. 545 | \] 546 | Again, since $A$ is nonsingular, the system of linear equations has exactly 547 | one solution $\varphi_i(v_j) = 0$ for $i\ne j$ and $\varphi_i(v_i)=1$. 548 | Namely, $\varphi_1,\dots,\varphi_n$ is the dual basis of $v_1,\dots,v_n$. 549 | \end{proof} 550 | 551 | \paragraph{37.} 552 | \begin{proof} 553 | $\,$\\ 554 | (a) Since $\pi$ is surjective, $\pi\hp$ is injective by 3.108. \\ 555 | (b) $\range\pi\hp = (\nul\pi)^0 = U^0$. \\ 556 | (c) It follows immediately from (a) and (b). 557 | \end{proof} 558 | % end 559 | -------------------------------------------------------------------------------- /linear_algebra_done_right_3rd/ch5_eigenvalues_eigenvectors_and_invariant_subspaces.tex: -------------------------------------------------------------------------------- 1 | \section{Eigenvalues, Eigenvectors and Invariant Subspaces} 2 | \setcounter{subsection}{1} 3 | 4 | \subsection{Eigenvectors and Upper-Triangular Matrices} 5 | \begin{lemma} 6 | \label{lemma:eigenvalue,polynomial} 7 | If $\lambda$ is an eigenvalue of $T\in\mathcal{L}(V)$ and $p$ is a 8 | polynomial, then $p(\lambda)$ is an eigenvalue of $p(T)$. Note that unlike 9 | the statement in exercise 11, $\mathbb{F}$ does not required to be 10 | $\mathbb{C}$. 11 | \end{lemma} 12 | \begin{proof} 13 | Suppose that $Tv=\lambda v$ for some $0\ne v\in V$, then 14 | \[ 15 | p(T)v = \left(\sum_{k=0}^n a_kT^k \right)v = 16 | \sum_{k=0}^n a_kT^kv = \sum_{k=0}^n a_k\lambda^k v = p(\lambda)v. 17 | \] 18 | Hence, $p(\lambda)$ is an eigenvalue of $p(T)$. 19 | \end{proof} 20 | 21 | \paragraph{1.} 22 | \begin{proof} 23 | $\,$\\ 24 | (a) Since $T^n=0$ and 25 | \[ 26 | (I-T)(I+T+\cdots+T^{n-1}) = I+T+\cdots+T^{n-1} - T-\cdots-T^n 27 | = I - T^n = I, 28 | \] 29 | $I-T$ is invertible and $(I-T)\inv=I+T+\cdots+T^{n-1}$. \\ 30 | (b) The power series expansion of the function $(1-x)\inv$ at $x=0$ is 31 | $1 + x + \cdots + x^n + \cdots$. 32 | \end{proof} 33 | 34 | \paragraph{3.} 35 | \begin{proof} 36 | Since $1$ is the only eigenvalue of $T^2=I$ and $-1$ is not an eigenvalue of 37 | $T$, by \lemmaref{lemma:eigenvalue,polynomial}, $1$ is the only eigenvalue 38 | of $T$ and therefore $T=I$. 39 | \end{proof} 40 | 41 | \paragraph{5.} 42 | \begin{proof} 43 | Since $(STS\inv)^k = S(T(S\inv S)TS\inv\cdots ST)S\inv = ST^kS\inv$, 44 | \[ 45 | p(STS\inv) = \sum_{k=0}^n a_k(STS\inv)^k = \sum_{k=0}^n a_kST^kS\inv 46 | = Sp(T)S\inv. 47 | \] 48 | \end{proof} 49 | 50 | \paragraph{7.} 51 | \begin{proof} 52 | It follows immediately from \lemmaref{lemma:eigenvalue,polynomial}. 53 | \end{proof} 54 | 55 | \paragraph{9.} 56 | \begin{proof} 57 | Since $p(T)v=0=0v$, $0$ is an eigenvalue of $T$. Then by 58 | \lemmaref{lemma:eigenvalue,polynomial}, some of the zeros of $p$ are the 59 | eigenvalues of $T$. Assume that there exists some zero $x_0$ of $p$ that is 60 | not an eigenvalue of $p$. Then $q=p/(x-x_0)$ is a polynomial of degree less 61 | than $p$ and such that $q(T)v=0$. Contradiction. Hence, every zero of $p$ is 62 | an eigenvalue of $p$. 63 | \end{proof} 64 | 65 | \paragraph{11.} 66 | Note that the proof does not rely on 5.21. 67 | \begin{proof} 68 | Suppose that $\alpha$ is an eigenvalue of $p(T)$. If $p$ is a constant 69 | polynomial, then there is nothing to be proved. If $p$ is non-constant, then 70 | $p(x)-\alpha = c(x-\lambda_1)\cdots(x-\lambda_m)$ where $m\ge 1$. Since 71 | $\alpha$ is an eigenvalue of $p(T)$, 72 | \[ 73 | (p-\alpha)(T)=c(T-\lambda_1 I)\cdots(T-\lambda_m I) 74 | \] 75 | is singular. Hence, at least one of $T-\lambda_1 I,\dots,T-\lambda_m T$, 76 | denoted by $T-\lambda_k I$, is singular and therefore $\lambda_k$ is an 77 | eigenvalue of $T$ and $p(\lambda_k)=\alpha$. \\ 78 | The converse part is just \lemmaref{lemma:eigenvalue,polynomial}. 79 | \end{proof} 80 | 81 | \paragraph{13.} 82 | \begin{proof} 83 | Suppose that $U$ is a finite-dimensional $T$-invariant subspace of $W$. Then 84 | $T|_U$ is an operator on $U$, a complex vector space. Hence it has an 85 | eigenvalue as long as $U=\{0\}$. However, it does not and therefore $U = 86 | \{0\}$. 87 | \end{proof} 88 | 89 | \paragraph{15.} 90 | \begin{proof} 91 | $\begin{bsmallmatrix} 1 & 1 \\ 1 & 1 \end{bsmallmatrix}$. 92 | \end{proof} 93 | 94 | \paragraph{17.} 95 | \begin{proof} 96 | Let $\varphi$ be the map which takes $p\in\mathcal{P}_{n^2}(\mathbb{C})$ to 97 | $p(T)\in\mathcal{L}(V)$. It is linear since 98 | \[ 99 | \varphi(a_1p_1+a_2p_2) = (a_1p_1+a_2p_2)(T) = a_1p_1(T)+a_2p_2(T)= 100 | a_1\varphi(p_1)+a_2\varphi(p_2). 101 | \] 102 | Since $\dim\mathcal{P}_{n^2}(\mathbb{C})=n^2+1$ ans $\dim\mathcal{L}(V) = 103 | n^2$, $\varphi$ is not injective by 3.23. Namely, there exsits nonequal $p_1 104 | ,p_2\in\mathcal{P}_{n^2}(\mathbb{C})$ such that $\varphi(p_1)=\varphi(p_2)$. 105 | Hence, $\varphi(p_1-p_2)=(p_1-p_2)(T)=0$ where $p_1-p_2$ is a nonzero 106 | polynomial, having zeros in $\mathbb{C}$. Since $0$ is the eigenvalue of 107 | $(p_1-p_2)(T)$, one of its zeros is an eigenvalue of $T$ by exercise 11. 108 | \end{proof} 109 | 110 | % end 111 | 112 | \iffalse 113 | \subsection{Eigenspaces and Diagonal Matrices} 114 | \paragraph{1.} 115 | \begin{proof} 116 | Since $T$ is diagonalizable, $V$ has a basis $(v_1,\dots,v_n)$ consisting of 117 | eigenvectors of $T$. Suppose that $\lambda_1,\dots,\lambda_n$ are the 118 | eigenvalues corresponding to $v_1,\dots,v_n$ respectively and assume without 119 | loss of generality that $v_1,\dots,v_m\in\nul T$. For every $v\in V$, 120 | \[ 121 | v=x_1v_1+\cdots+x_nv_n 122 | =(x_1v_1+\cdots+x_mv_m) 123 | +T\left( \frac{x_{m+1}}{\lambda_{m+1}}v_{m+1}+\cdots 124 | +\frac{x_n}{\lambda_n}v_n \right). 125 | \] 126 | Hence, $V=\nul T + \range T$. \par 127 | Meanwhile, as $\range T\subset E(\lambda_{m+1},T)\oplus\cdots\oplus 128 | E(\lambda_n, T)$, $\dim\range T\le n-m$. And clear that $\dim\nul T=m$. 129 | Hence, $\dim V\ge \dim\nul T+\dim\range T$ and therefore $V=\nul T\oplus 130 | \range T$. 131 | \end{proof} 132 | 133 | \paragraph{3.} 134 | \begin{proof} 135 | Clear that (a) implies (b) and (c); (b), together with the rank-nullity 136 | theorem, implies (a). Meanwhile, if $\nul T\cap\range T=\{0\}$, then 137 | \begin{align*} 138 | \dim(\nul T+\range T) 139 | &= \dim\nul T+\dim\range T-\dim(\nul T-\cap\range T) \\ 140 | &= \dim\nul T+\dim\range T = \dim V. 141 | \end{align*} 142 | Namely, $V=\nul T\oplus\range T$. Hence, (a), (b) and (c) are equivalent. 143 | 144 | 145 | \end{proof} 146 | 147 | \paragraph{5.} 148 | \begin{proof} 149 | TODO 150 | \end{proof} 151 | 152 | \paragraph{7.} 153 | \begin{proof} 154 | TODO 155 | \end{proof} 156 | 157 | \paragraph{9.} 158 | \begin{proof} 159 | Since $T$ is invertible $v\in E(\lambda, T)$ iff $(T-\lambda I)v=0$ iff $Tv 160 | -\lambda v = 0$ iff $\lambda\inv v - T\inv v = 0$ iff $(T\inv-\lambda\inv)v 161 | =0$ iff $v\in E(\lambda\inv, T\inv)$. 162 | \end{proof} 163 | 164 | \paragraph{11.} 165 | \begin{proof} 166 | 167 | \end{proof} 168 | 169 | % end 170 | \fi 171 | -------------------------------------------------------------------------------- /linear_algebra_done_right_3rd/ch7_operators_on_inner_product_spaces.tex: -------------------------------------------------------------------------------- 1 | \section{Operators on Inner Product Spaces} 2 | Note that on the first page of this chapter, the author says that all the 3 | inner product spaces appeared in this chapter should be assumed to be 4 | finite-dimensional if not specified. 5 | 6 | \subsection{Self-Adjoint and Normal Operators} 7 | \paragraph{2.} 8 | \begin{proof} 9 | $\nul(T-\lambda I) = (\range(T-\lambda I)^*)^\perp = (\range(T^*- 10 | \bar{\lambda}I))^\perp$. Hence, $\dim\nul(T-\lambda I)>0$ iff $\dim\range( 11 | T^*-\bar{\lambda}I)<\dim V$ iff $\dim\nul(T^*-\bar{\lambda}I)>0$. Thus, 12 | $\lambda$ is an eigenvalue of $T$ iff $\bar{\lambda}$ is an eigenvalue of 13 | $T^*$. 14 | \end{proof} 15 | 16 | \paragraph{4.} 17 | \begin{proof} 18 | If follows immediately from 7.7. 19 | \end{proof} 20 | 21 | \paragraph{8.} 22 | \begin{proof} 23 | Let $S,T\in\mathcal{L}(V)$ be self-adjoint and $a,b\in\mathbb{R}$. Then 24 | by 7.5, 25 | \[ 26 | (aS+bT)^* = \bar{a}S^*+\bar{b}T^* = aS+bT. 27 | \] 28 | Hence, the set of self-adjoint operators on $V$ is a subspace of 29 | $\mathcal{L}(V)$. 30 | \end{proof} 31 | 32 | \paragraph{10.} 33 | \begin{proof} 34 | The $n$-by-$n$ matrices 35 | \[ 36 | A= 37 | \begin{bmatrix} 38 | 1 & & & \\ 39 | & \ddots & & \\ 40 | & & \ddots & \\ 41 | & & & 1 42 | \end{bmatrix} 43 | \quad\text{and}\quad 44 | B= 45 | \begin{bmatrix} 46 | 1 & 1 & & \\ 47 | & \ddots & \ddots & \\ 48 | & & \ddots & 1 \\ 49 | & & & 1 50 | \end{bmatrix} 51 | \] 52 | are both normal, but $C=B-A$, which is nilpotent, is not. 53 | \end{proof} 54 | 55 | \paragraph{12.} 56 | \begin{proof} 57 | Since $T$ is normal and $3$ and $4$ are two eigenvalues of $T$, there exists 58 | an orthonormal list $u_1,u_2$ such that $Tu_1=3u_1$ and $Tu_2=4u_2$. Let $v 59 | =u_1+u_2$. Then $\|v\|=\sqrt{2}$ and 60 | \[ 61 | \|Tv\| = \|3u_1+4u_2)\| = 5. 62 | \] 63 | \end{proof} 64 | 65 | \paragraph{16.} 66 | \begin{proof} 67 | 7.20 implies $\nul T=\nul T^*$ and therefore, by 7.7, $\range T=\range T^*$. 68 | \end{proof} 69 | % end 70 | 71 | \subsection{The Spectral Theorem} 72 | \paragraph{4.} 73 | \begin{proof} 74 | If $T$ is normal, then by the complex spectral theorem, clear that all pairs 75 | of eigenvectors corresponding to distinct eigenvalues of $T$ are orthogonal 76 | and $V=E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T)$.\par 77 | Now we show the converse. Clear that the union of the orthonormal bases of 78 | each $E(\lambda_i,T)$ is an orthonormal basis consisting of eigenvectors of 79 | $T$. 80 | \end{proof} 81 | 82 | \paragraph{6.} 83 | \begin{proof} 84 | Suppose $T\in\mathbb{L}(V)$ is self-adjoint and $Tu=\lambda u$ for some $u 85 | \ne 0$. Then by 7.21, $\lambda u = Tu=T^*u=\bar{\lambda}u$. Hence, $\lambda 86 | =\bar{\lambda}$ and therefore $\lambda$ is real.\par 87 | By the complex spectral theorem, $T$ has a matrix $M=\diag(\lambda_1,\dots, 88 | \lambda_n)$ with respect to some orthonormal basis. If $\lambda_i$, the 89 | eigenvalues of $T$, are all real. Then the conjugate transpose of $M$ is 90 | still $M$. Hence, $T=T^*$. Namely, $T$ is self-adjoint. 91 | \end{proof} 92 | 93 | \paragraph{8.} 94 | \begin{proof} 95 | Let 96 | \[ 97 | A = \begin{bmatrix} 98 | 0 & 1 & & \\ 99 | & \ddots & \ddots & \\ 100 | & & \ddots & 1 \\ 101 | & & & 0 102 | \end{bmatrix} 103 | \] 104 | be an $8$-by-$8$ matrix. $A^9=A^8=0$ but $A^2\ne A$. 105 | \end{proof} 106 | 107 | \paragraph{12.} 108 | \begin{proof} 109 | We argue by contradiction. Assume that for every eigenvalue $\lambda\hp$ of 110 | $T$, $|\lambda-\lambda\hp|>\vep$. Since $T$ is self-adjoint, there exists an 111 | orthonormal basis $e_1,\dots,e_n$ such that $Te_i=\lambda_ie_i$ for $i=1, 112 | \dots,n$. Then 113 | \[ 114 | Tv-\lambda v = \sum_{i=1}^n\langle v,e_i\rangle(\lambda_i-\lambda)e_i. 115 | \] 116 | Hence, 117 | \[ 118 | \|Tv-\lambda v\|^2 119 | = \sum_{i=1}^n |\langle v,e_i\rangle|^2|\lambda_i-\lambda|^2 120 | \ge \vep^2\sum_{i=1}^n |\langle v,e_i\rangle|^2 = \vep^2. 121 | \] 122 | Contradiction. Hence, $T$ has an eigenvalue $\lambda\hp$ such that $|\lambda 123 | -\lambda\hp|<\vep$. 124 | \end{proof} 125 | 126 | \paragraph{14.} 127 | \begin{proof} 128 | If there exists an inner product on $U$ which makes $T$ into a self-adjoint 129 | operator, then by the spectral theorem, $U$ has a basis consisting of 130 | eigenvectors of $T$. Now we suppose that such a basis, denoted by $u_1, 131 | \dots, u_n$, exists. Define $\langle e_i,e_j\rangle$ to be $0$ if $i\ne j$ 132 | and $1$ if $i=j$. It is easy to verify that it does define an inner product 133 | and $u_1,\dots,u_n$ is an orthonormal basis. 134 | \end{proof} 135 | % end 136 | 137 | \subsection{Positive Operators and Isometries} 138 | \paragraph{2.} 139 | \begin{proof} 140 | The hypothesis implies $T^2v=v$ and therefore $1$ is an eigenvalue of $T^2$ 141 | and $v$ is its associated eigenvector. Note that $T^2$ is still a positive 142 | operator and $T$ is its positive square root. Hence, $v$ is also an 143 | eigenvector of $T$ associated with eigenvalue $1$. Namely, $v=Tv=w$. 144 | \end{proof} 145 | 146 | \paragraph{4.} 147 | \begin{proof} 148 | $\langle T^*Tv,v\rangle=\langle Tv,Tv\rangle = \|Tv\|^2\ge 0$ for all $v\in 149 | V$. Hence, $T^*T$ is a positive operator on $V$. Similarly, $TT^*$ is a 150 | positive operator on $W$. 151 | \end{proof} 152 | 153 | \paragraph{6.} 154 | \begin{proof} 155 | If $T$ is positive, $T$ is self-adjoint and so does $T^k$. Meanwhile, $T$ 156 | has a diagonal matrix $\diag(\lambda_1,\dots,\lambda_n)$ with respect to 157 | some orthonormal basis and $\lambda_i\ge 0$ for each $i=1,\dots,n$. With 158 | respect to the same basis, $T^k$ has the matrix $\diag(\lambda_1^k,\dots, 159 | \lambda_n^k)$. Since $\lambda_i^k\ge 0$, $T^k$ is positive by 7.35. 160 | \end{proof} 161 | 162 | \paragraph{8.} 163 | \begin{proof} 164 | Suppose $T$ is positive and invertible. Then $\langle v,v\rangle_T=\langle 165 | Tv,v\rangle\ge 0$. By 7.35, $T=R^*R$ for some $R\in\mathcal{L}(V)$. Since 166 | $T$ is invertible, both $R^*$ and $R$ are invertible. Hence, $0=\langle v,v 167 | \rangle_T=\langle Tv,v\rangle=\|Rv\|^2$ iff $Rv=0$ iff $v=0$. Meanwhile, 168 | for every $u,v\in V$ and scalar $a,b$ 169 | \[ 170 | \langle au+bv,w\rangle_T = a\langle Tu,w\rangle + b\langle Tv,w\rangle = 171 | a\langle v,w\rangle_T + b\langle v,w\rangle_T. 172 | \] 173 | Finally, 174 | \[ 175 | \langle u,v\rangle_T = \langle Ru,Rv\rangle = 176 | \overline{\langle Rv,Ru\rangle} = \overline{\langle v,u\rangle_T}. 177 | \] 178 | Thus, $\langle\cdot,\cdot\rangle_T$ is an inner product on $V$.\par 179 | Now we suppose $\langle\cdot,\cdot\rangle_T$ is an inner product on $V$. 180 | Hence, by the positivity of inner product, $T$ is positive and by the 181 | previous discussion, $Tv=0$ iff $v=0$. Hence, $T$ is invertible. 182 | \end{proof} 183 | 184 | \paragraph{14.} 185 | \begin{proof} 186 | As $T$ is self-adjoint by Exercise 21 of Section 7.A, so does $-T$. 187 | Suppose that 188 | \[ 189 | f(x)=x_0 + x_1\cos x +\cdots+ x_n\cos nx + y_1\sin x +\cdots+ y_n\sin nx. 190 | \] 191 | Then 192 | \[ 193 | f^{\prime\prime}(x) 194 | = -\sum_{k=1}^nx_kk^2\cos kx - \sum_{k=1}^ny_kk^2\sin kx. 195 | \] 196 | Note that for all $i\ne j$, the integral of $(\cos ix\sin jx)$ is $0$. Thus, 197 | \begin{align*} 198 | \langle f,f^{\prime\prime}\rangle &= 199 | -\int_{-\pi}^\pi 200 | \left(x_0 + \sum_{k=1}^n(x_k\cos kx + y_k\sin kx) \right) 201 | \sum_{k=1}^n(k^2x_k\cos kx+ k^2y_k\sin kx)\rd x\\ 202 | &= -\int_{-\pi}^\pi\left( \sum_{k=1}^n x_k^2k^2\cos^2kx + 203 | \sum_{k=1}^ny_k^2k^2\sin^2kx \right)\rd x \\ 204 | &= -2\pi\sum_{k=1}^n k^2(x_k^2+y_k^2) \le 0. 205 | \end{align*} 206 | Therefore, $-T$ is a positive operator. 207 | \end{proof} 208 | % end 209 | 210 | \subsection{Polar Decomposition and Singular Value Decomposition} 211 | \paragraph{3.} 212 | \begin{proof} 213 | As $\sqrt{T^*T}$ is positive, $(\sqrt{T^*T})^*=\sqrt{T^*T}$. The polar 214 | decomposition asserts that there exists an isometry $S_1$ such that 215 | $T=S_1\sqrt{T^*T}$. Replacing $T$ with $T^*$ yields 216 | \[ 217 | T^*=S_2\sqrt{TT^*} \quad\Rightarrow\quad 218 | T=\sqrt{TT^*}S_2^*, 219 | \] 220 | where $S_2^*$ is also an isometry. 221 | \end{proof} 222 | 223 | \paragraph{8.} 224 | This result can be used to prove Exercise 1 in a simple way. 225 | \begin{proof} 226 | Note that $R^2$ and $T^*T$ are both positive (and thereby self-adjoint). 227 | Hence, $R^2-T^*T$ is also self-adjoint. For all $v\in V$, 228 | \begin{align*} 229 | \langle(R^2-T^*T)v,v\rangle 230 | = \langle R^2v,v\rangle - \langle T^*Tv,v\rangle 231 | = \|Rv\|^2 - \|Tv\|^2. 232 | \end{align*} 233 | Since $S$ is an isometry, $\|Rv\| = \|SRv\| = \|Tv\|$. Therefore, $\langle 234 | (R^2-T^*T)v,v\rangle=0$. Thus, $R^2=T^*T$ and so does their unique positive 235 | square roots. 236 | \end{proof} 237 | 238 | \paragraph{9.} 239 | \begin{proof} 240 | As $T=S\sqrt{T^*T}$ for some isometry $S$ and every isometry is invertible, 241 | $T$ is invertible iff $\sqrt{T^*T}$ is invertible. If $T$ is invertible and 242 | $T=S_1\sqrt{T^*T}=S_2\sqrt{T^*T}$, multiplying $(\sqrt{T^*T})\inv$ on the 243 | both sides yields $S_1=S_2$. \par 244 | Now we suppose that such an isometry $S$ is not unique and show that 245 | $\sqrt{T^*T}$ is not invertible to complete the proof. Suppose $T=S_1 246 | \sqrt{T^*T}=S_2\sqrt{T^*T}$ where $S_1\ne S_2$. Then, since $S_1-S_2\ne 0$ 247 | but $(S_1-S_2)\sqrt{T^*T}=0$, $\sqrt{T^*T}$ is not invertible and so is $T$. 248 | \end{proof} 249 | 250 | \paragraph{13.} 251 | \begin{proof} 252 | By the discussion in Exercise 9, $T$ is invertible iff $\sqrt{T^*T}$ is 253 | invertible iff $0$ is not an eigenvalue of $\sqrt{T^*T}$. 254 | \end{proof} 255 | % end 256 | -------------------------------------------------------------------------------- /linear_algebra_done_right_3rd/ch8_opertators_on_complex_vector_spaces.tex: -------------------------------------------------------------------------------- 1 | \section{Operators on Complex Vector Spaces} 2 | \subsection{Generalized Eigenvectors and Nilpotent Operators} 3 | \paragraph{3.} 4 | \begin{proof} 5 | Suppose $\dim V= N$. 6 | \begin{align*} 7 | v\in G(T\inv, \lambda\inv) 8 | &\Leftrightarrow (T\inv v-\lambda\inv I)^nv = 0 \\ 9 | &\Leftrightarrow 10 | \left(\sum_{k=0}^n\binom{n}{k}T^{-k}(-\lambda\inv)^{n-k}\right)v=0 \\ 11 | &\Leftrightarrow 12 | T^n(-\lambda)^n 13 | \left(\sum_{k=0}^n\binom{n}{k}T^{-k}(-\lambda)^{k-n}\right)v=0 \\ 14 | &\Leftrightarrow 15 | \left(\sum_{k=0}^n\binom{n}{k}T^{n-k}(-\lambda)^{k}\right)v=0 \\ 16 | &\Leftrightarrow (T-\lambda I)^nv = 0 \\ 17 | &\Leftrightarrow v\in G(T,\lambda). 18 | \end{align*} 19 | \end{proof} 20 | 21 | \paragraph{5.} 22 | Intuitively, $\nul T,\dots,\nul T^n$ is a sequence of subspaces where the 23 | preceding ones are contained by succeeding ones. And $T^{k}v$ lies in the 24 | additional part between two successive subspaces. 25 | \begin{proof} 26 | Note that $T^{m-1}v\ne 0$ but $T^mv=0$ implies for $k=1,\dots,m$ 27 | \[ 28 | T^{m-k}v\in\nul T^k,\quad T^{m-k}v\notin\nul T^{k-1}. 29 | \] 30 | Therefore, $T^{m-k}v\notin\spn(T^{m-1}v,\dots,T^{m-k+1}v)$; otherwise, 31 | $T^{m-k}v=x_1T^{m-1}v+\cdots+x_{m-k+1}T^{m-k+1}v$, implying that $T^{m-k}v 32 | \in\nul T^{k-1}$. Hence, $v, \dots, T^{m-1}v$ are linearly independent. 33 | \end{proof} 34 | 35 | \paragraph{7.} 36 | \begin{proof} 37 | It follows immediately from 8.19. 38 | \end{proof} 39 | 40 | \paragraph{9.} 41 | \begin{proof} 42 | Suppose that $v$ is nonzero and $(TS-\lambda I)v=0$. Then 43 | \[ 44 | (STS-\lambda S)v=0 \quad\Rightarrow\quad 45 | (ST-\lambda I)Sv=0. 46 | \] 47 | If $Sv=0$, then $0=(TS-\lambda I)v=T(Sv)-\lambda v$ implies $\lambda=0$. If 48 | $Sv\ne 0$, then $\lambda$ is an eigenvalue of $ST$ and therefore equals $0$ 49 | by Exercise 7. Hence, in every case, $\lambda=0$. Thus, $TS$ is also 50 | nilpotent. 51 | \end{proof} 52 | 53 | \paragraph{13.} 54 | \begin{proof} 55 | We are going to show that $N^{k-1}=0$ if $N^k=0$ for $k>1$ to conclude that 56 | $N=0$. Suppose that $N^k=0$ and let $M=N^{k-1}$. Then for every $v\in V$, 57 | \[ 58 | \|M^*Mv\|^2 = \langle M^*Mv, M^*Mv\rangle 59 | = \langle M^*MM^*Mv,v\rangle = \langle M^*M^*MMv,v\rangle. 60 | \] 61 | Since $M^2=N^{2k-2}=0$, this implies $M^*M=0$. Hence, with the polar 62 | decomposition, $M=0$. 63 | \end{proof} 64 | 65 | \paragraph{15.} 66 | \begin{proof} 67 | Suppose $\dim V=n$. Since $\nul N^{n-1}\ne\nul N^n$, 68 | \[ 69 | 0<\dim N < \cdots < \dim N^n. 70 | \] 71 | Hence, $\dim\nul N^j=j$ for $0\le j\le n$. Since $\dim\nul N^n=n$, $\nul N^n 72 | =V$ and therefore $N$ is nilpotent. 73 | \end{proof} 74 | % end 75 | 76 | \subsection{Decomposition of an Operator} 77 | \paragraph{10.} 78 | \begin{proof} 79 | By 8.29, there exists a basis of $V$ with respect to which $\mathcal{M}(T)$ 80 | has the form described in 8.29. Suppose $\mathcal{M}(D)$ be the diagonal 81 | matrix whose diagonal entries are the ones of $\mathcal{M}(T)$ and 82 | $\mathcal{M}(N)=\mathcal{M}(T)-\mathcal{M}(D)$. Clear that $N$ is nilpotent 83 | and 84 | \[ 85 | \mathcal{M}(N) = 86 | \begin{bmatrix} 87 | A_1-\lambda_1 I_1 & & 0 \\ 88 | & \ddots & \\ 89 | 0 & & A_m-\lambda_m I_m 90 | \end{bmatrix} 91 | \] 92 | where each $I_k$ is the identity matrix with corresponding size. Note that 93 | $\lambda_k I_k$ and $A_k-\lambda_k I_k$ are commuting. Hence, by Problem 9, 94 | $\mathcal{M}(N)$ and $\mathcal{M}(D)$ are commuting and so do $D$ and $N$. 95 | \end{proof} 96 | % end 97 | -------------------------------------------------------------------------------- /linear_algebra_done_right_3rd/def.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt, a4paper]{article} 2 | 3 | \usepackage[margin=1in]{geometry} 4 | \usepackage{ 5 | color, 6 | clrscode, 7 | amssymb, 8 | amsmath, 9 | listings, 10 | fontspec, 11 | xcolor, 12 | supertabular, 13 | multirow, 14 | mathtools, 15 | mathrsfs, 16 | amsthm, 17 | systeme 18 | } 19 | \definecolor{bgGray}{RGB}{36, 36, 36} 20 | \usepackage[ 21 | colorlinks, 22 | linkcolor=bgGray, 23 | anchorcolor=blue, 24 | citecolor=green 25 | ]{hyperref} 26 | \newfontfamily\courier{Courier} 27 | 28 | \DeclareMathOperator{\rank}{rank} 29 | \DeclareMathOperator{\adj}{adj} 30 | \DeclareMathOperator{\tr}{tr} 31 | \DeclareMathOperator{\diag}{diag} 32 | \DeclareMathOperator{\nul}{null} 33 | \DeclareMathOperator{\range}{range} 34 | \DeclareMathOperator{\spn}{span} 35 | % \DeclareMathOperator{\deg}{deg} 36 | 37 | \newenvironment{solution} 38 | {\begin{proof}[Solution]} 39 | {\end{proof}} 40 | \newcommand{\hp}{^\prime} 41 | \newcommand{\vep}{\varepsilon} 42 | \newcommand{\inv}{^{-1}} 43 | \newcommand{\rd}{\mathrm{d}} 44 | 45 | \renewcommand{\Im}{\text{Im}} 46 | \renewcommand{\Re}{\text{Re}} 47 | \renewcommand{\thesubsection}{\thesection.\Alph{subsection}} 48 | 49 | \newtheorem{lemma}{Lemma} 50 | \newcommand\lemmaref[1]{Lemma~\ref{#1}} 51 | 52 | -------------------------------------------------------------------------------- /linear_algebra_done_right_3rd/image/3-d-8.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Engineev/solutions/4e33274fe1ed9e46fd0e6671c57cb589704939bd/linear_algebra_done_right_3rd/image/3-d-8.png -------------------------------------------------------------------------------- /linear_algebra_done_right_3rd/main.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Engineev/solutions/4e33274fe1ed9e46fd0e6671c57cb589704939bd/linear_algebra_done_right_3rd/main.pdf -------------------------------------------------------------------------------- /linear_algebra_done_right_3rd/main.tex: -------------------------------------------------------------------------------- 1 | \input{def.tex} 2 | 3 | \title{Linear Algebra Done Right} 4 | \author{Yunwei Ren} 5 | \date{} 6 | 7 | \begin{document} 8 | \maketitle 9 | \tableofcontents 10 | 11 | \newpage 12 | \setcounter{section}{2} 13 | \input{ch3_linear_map.tex} 14 | \newpage 15 | \setcounter{section}{4} 16 | \input{ch5_eigenvalues_eigenvectors_and_invariant_subspaces.tex} 17 | \newpage 18 | \setcounter{section}{6} 19 | \input{ch7_operators_on_inner_product_spaces.tex} 20 | \newpage 21 | \input{ch8_opertators_on_complex_vector_spaces.tex} 22 | 23 | \end{document} 24 | -------------------------------------------------------------------------------- /real_analysis_3rd/ch12_measure_and_outer_measure.tex: -------------------------------------------------------------------------------- 1 | \section{Measure and Outer Measure} 2 | \subsection{Outer Measure and Measurability} 3 | \paragraph{1.} 4 | \begin{proof} 5 | Suppose that $E\subset X$ and there is a measurable $B$ with $\bar{\mu}B=0$ 6 | such that $E\subset B$. We show that $E$ is measurable, that is, for every 7 | $A\subset X$ with finite outer measure, $\mu^*(A)\ge\mu^*(A\cap E)+\mu^*( 8 | A\cap E^c)$. Since $A\cap E\subset E\subset B$ and $\mu^*$ is monotone, 9 | $\mu^*(A\cap E)\le\mu^*(B)=\bar{\mu}B=0$. Again by the monotonicity, $\mu^*( 10 | A)\ge\mu^*(A\cap E^c)$. Thus, $E$ is measurable, implying that $\bar{\mu}$ 11 | is complete. 12 | \end{proof} 13 | 14 | \paragraph{2.} 15 | \begin{proof} 16 | From the countable subadditivity we obtain that $\mu^*(A\cap E)\le\sum\mu^* 17 | (A\cap E_i)$. For the converse, first we consider just $E_1$ and $E_2$. 18 | Since $E_1$ is measurable, 19 | \[ 20 | \mu^*(A\cap E)=\mu^*(A\cap E\cap E_1)+\mu^*(A\cap E\cap E_1^c) 21 | \ge\mu^*(A\cap E_1)+\mu^*(A\cap E_2). 22 | \] 23 | By induction on $n$ we get $\mu^*(A\cap E)\ge\sum_{i=1}^n\mu^*(A\cap E_i)$. 24 | Let $n\to\infty$ and the proof is completed. 25 | \end{proof} 26 | % end 27 | \subsection{The Extension Theorem} 28 | \paragraph{4.} 29 | \begin{proof} 30 | $\,$\par 31 | (a) Since $\{D_j\}$ partitions $A$, $C_i\subset A$ for each $i$ and 32 | $\mcal{C}$ is closed under intersection, by condition (i), $\mu C_i=\sum_j 33 | \mu(C_i\cap D_j)$. Similarly for $\mu D_j$. Thus, 34 | \[ 35 | \sum_i\mu C_i=\sum_{i,j}\mu(C_i\cap D_j)= 36 | \sum_j\sum_i\mu(C_i\cap D_j)=\sum_j\mu D_j. 37 | \] 38 | This result implies that the definition of $\mu$ on $\mcal{A}$ is 39 | well-defined. \par 40 | (b) Since every $A\in\mcal{A}$ is a finite union of sets of $\mcal{C}$, it 41 | suffices to show that $\mu C\ge\sum\mu C_i$. Then, from this condition, the 42 | countably additivity follows. Since $\mu$ is nonnegative and monotone, 43 | $\mu C\ge\sum_{i=1}^n\mu C_i$ for all positive integer $n$. Let $n\to\infty$ 44 | and we get $\mu C\ge\sum\mu C_i$. 45 | \end{proof} 46 | 47 | \paragraph{7.} 48 | \begin{proof} 49 | To prove the "if" part, let $\vep_n=1/n$ and $A_n\in\mcal{A}_\delta$ be such 50 | that $\mu^*(E\setminus A_n)<\vep_n$. Put $A=\bigcup A_n$. Since the 51 | collection of $\mu^*$-measurable sets is a $\sigma$-algebra, $A$ is 52 | measurable. Meanwhile, since $\mu^*(E\setminus A)\le\mu^*(E\setminus 53 | A_n)<\vep_n$, $E\setminus A$ is of $\mu^*$-measure zero. Hence, it is 54 | measurable. Thus, $E=A\cup(E\setminus A)$ is also measurable.\par 55 | For the converse, suppose that $E$ is measurable and let $\vep>0$ be fixed. 56 | Note that $E^c$ is also measurable. Hence, by Prop. 6, there is a set $A\in 57 | \mcal{A}_\sigma$ with $E^c\subset A$ and 58 | \[ 59 | \vep>\mu^*(A\setminus E^c)=\mu^*(A\cap E). 60 | \] 61 | Then, $A^c\in\mcal{A}_\delta$, $A^c\subset E$ and $\mu^*(E\setminus A^c)= 62 | \mu^*(E\cap E)<\vep$. 63 | \end{proof} 64 | % end 65 | 66 | -------------------------------------------------------------------------------- /real_analysis_3rd/ch3_lebesgue_measure.tex: -------------------------------------------------------------------------------- 1 | \section{Lebesgue Measure} 2 | \subsection{Introduction} 3 | \paragraph{1.} 4 | \begin{proof} 5 | Since $\mathfrak{M}$ is an $\sigma$-algebra, $B\setminus A \in\mathfrak{M}$ 6 | as long as $A,B\in\mathfrak{M}$. Since $B\setminus A$ and $A$ are disjoint, 7 | $mB=mA+m(B\setminus A)\ge mA$ since $m$ is nonnegative. 8 | \end{proof} 9 | 10 | \paragraph{2.} 11 | \begin{proof} 12 | Let $A_0 = E_0$ and $E_k=A_k\setminus A_{k-1}$ for $k\ge 1$. Clear that 13 | $E_i$ and $E_j$ are disjoint for distinct $i$ and $j$, $\bigcup A_n=\bigcup 14 | E_n$ and $A_i\subset E_i$ for every $i$. Hence, 15 | \[ 16 | m\left(\bigcup E_n\right) = m\left(\bigcup A_n\right) 17 | = \sum mA_n \le \sum mE_n, 18 | \] 19 | where the last inequality comes from Exercise 1. 20 | \end{proof} 21 | 22 | \paragraph{3.} 23 | \begin{proof} 24 | Suppose that $mA<\infty$. Then $mA=m(A\cup\varnothing)=mA+m\varnothing$, 25 | implying that $m\varnothing=0$. 26 | \end{proof} 27 | % end 28 | 29 | \subsection{Outer Measure} 30 | \paragraph{5.} 31 | \begin{proof} 32 | We show that $\{I_n\}$ must cover the entire $[0,1]$ by contradiction. 33 | Assume that $x\notin I_k$ for $k=1,2,\dots,n$. Then, as $I_k$ are open and 34 | $n$ is finite, there exists some $\vep>0$ such that $(x-\vep,x+\vep)$ and 35 | $I_k$ are disjoint for every $k$. Since $\mathbb{Q}$ is dense in 36 | $\mathbb{R}$, there exists some rational number in $(x-\vep, x+\vep)$, 37 | contradicting with the hypothesis that $\{I_k\}$ covers all rational numbers 38 | between $0$ and $1$. 39 | \end{proof} 40 | 41 | \paragraph{6.} 42 | \begin{proof} 43 | By the definition of the outer measure, for every $\vep > 0$, there exists 44 | some collection $\{I_n\}$ of open intervals that covers $A$ and $\sum l(I_n) 45 | \le m^*A+\vep$. Let $O=\bigcup I_n$. $O$ is a countable union of open sets 46 | and therefore is also open. And by Proposition 2, $m^*O\le \sum l(I_n)$. 47 | Thus, $m^*O\le m^*A+\vep$. \par 48 | Let $\vep_n = 1/n$ and for each $n$, by the previous discussion, we can 49 | always get an open set $O_k$ such that $A\subset O_k$ and $m^*O\le m^*A+ 50 | \vep_m$. Let $G$ be the countable intersection of these open sets. Clear 51 | that $G$ is a $G_\delta$ set covering $A$ and $m^*A=m^*G$. 52 | \end{proof} 53 | 54 | \paragraph{7.} 55 | \begin{proof} 56 | If $m^*E=\infty$, it is trivial. Suppose that $m^*E\le\infty$. For any $x\in 57 | \mathbb{R}$, collection $\{I_n\}$ of open intervals covers $E+x$ iff $\{I_n 58 | -x\}$ covers $E$. Since the length of intervals is translation invariant, 59 | this implies $m^*(E+x)=m^*E$. 60 | \end{proof} 61 | 62 | \paragraph{8.} 63 | \begin{proof} 64 | Clear that $m^*A\le m^*(A\cup B)$. Meanwhile, $m^*(A\cup B) = m^*A + m^*B = 65 | m^*B$. Hence, $m^*(A\cup B)=m^*B$. 66 | \end{proof} 67 | % end 68 | 69 | \subsection{Measurable Sets and Lebesgue Measure} 70 | \paragraph{10.} 71 | \begin{proof} 72 | \begin{align*} 73 | mE_1+mE_2 74 | &= mE_1 + m(E_2\setminus E_1) + m(E_1\cap E_2) \\ 75 | &= m(E_1\cup(E_2\setminus E_1)) + m(E_1\cap E_2) \\ 76 | &= m(E_1\cup E_2) + m(E_1\cap E_2). 77 | \end{align*} 78 | \end{proof} 79 | 80 | \paragraph{11.} 81 | \begin{proof} 82 | $E_n = (n,\infty)$. 83 | \end{proof} 84 | 85 | \paragraph{12.} 86 | This is the countable version of Lemma 9. 87 | \begin{proof} 88 | It suffices to prove $m^*(A\cap \bigcup E_i) \ge \sum m^*(A\cap E_i)$. Since 89 | $\bigcup_{i=1}^\infty E_i\supset \bigcup_{i=1}^n E_i$ for every $n$, 90 | \begin{align*} 91 | m^*\left(A\cap \bigcup_{i=1}^\infty E_i\right) 92 | \ge m^*\left(A\cap \bigcup_{i=1}^n E_i\right) 93 | = \sum_{i=1}^n m^*(A\cap E_i), 94 | \end{align*} 95 | where the equality comes from Lemma 9. Since the left hand side is 96 | independent of $n$, we have 97 | \[ 98 | m^*\left(A\cap \bigcup_{i=1}^\infty E_i\right) \ge 99 | \sum_{i=1}^\infty m^*(A\cap E_i), 100 | \] 101 | completing the proof. 102 | \end{proof} 103 | 104 | \paragraph{13.} 105 | \begin{proof} 106 | First we suppose that $m^*E<\infty$. By Proposition 5, there exists some 107 | open set $O\supset E$ such that $m^*O\le m^*E+\vep$. If $E$ is measurable, 108 | then by the definition, 109 | \[ 110 | m^*(O\setminus E) = m^*O-m^*E \le \vep. 111 | \] 112 | Namely, (ii) holds. Meanwhile, $O\subset\mathbb{R}$ is a countable union of 113 | disjoint open intervals $\{I_n\}$. Since $mO=m^*O$ is bounded and $mO=\sum 114 | l(I_n)$, there exists some integer $N>0$ such that $mO-\sum_{n=1}^N l(I_n)< 115 | \vep$. Let $U=\bigcup_{n=1}^N I_n$. 116 | \begin{align*} 117 | m^*(U\bigtriangleup E) 118 | &= m^*((U\cup E) \setminus (U\cap E)) \\ 119 | & \le m^*(O\setminus (U\cap E)) \\ 120 | & = m^*((O\setminus U) \cup (O\setminus E)) \\ 121 | &\le m^*(O\setminus U) + m^*(O\setminus E) \\ 122 | &\le 2\vep. 123 | \end{align*} 124 | Hence, (ii) implies (vi). Now we show that (vi) implies (ii). If $m^*(U 125 | \bigtriangleup E)<\vep $, then there exists some countable collection $\{J_n 126 | \}$ of open interval such that 127 | \[ 128 | \sum l(J_n)\le m^*(U\bigtriangleup E)+\vep<2\vep. 129 | \] 130 | Let $J=\bigcup J_n$ and $O= U\cup J$. $m^*J < 2\vep$. And $O$ is open and 131 | covers $E$. Meanwhile, 132 | \[ 133 | m^*(O\setminus E) \le m^*(U\setminus E)+m^*(J\setminus E) < 3\vep. 134 | \] 135 | Hence, (ii) holds.\par 136 | Now, let $E$ be an arbitrary set and $E_n=E\cap(-n, n)$, which is a set with 137 | finite measure. Then by the previous discussion, there exists some open set 138 | $O_n\supset E_n$ with $m^*(O_n\setminus E_n)<\vep/2^n$. Let $O=\bigcup 139 | O_n$, an open set covering $E$ and 140 | \[ 141 | m^*(O\setminus E) \le \sum m^*(O_n\setminus E_n) < 2\vep. 142 | \] 143 | Hence, (i) implies (ii). Now we suppose (ii) holds and let $\vep_n=1/n$, 144 | then there exists a sequence of open sets $$ such that $m^*(O_n 145 | \setminus E)<1/n$. Let $G=\bigcap O_n\in G_\delta$. $m^*(G\setminus E)\le 146 | m^*(O_n\setminus E)\le 1/n$. Since the left hand side is independent of 147 | $n$, $m^*(G\setminus E)=0$. If (iv) holds, then by Lemma 6, $G\setminus E$ 148 | is measurable. Since $G\in G_\delta$ is also measurable, $E$ is measurable. 149 | Hence, (iv) implies (i).\par 150 | By the previous result, for any measurable $E$, there exists some closed set 151 | $F\subset E$ such that $\bar{F}$, which is open, contains $bar{E}$ and $m^*( 152 | \bar{F}\setminus \bar{E})<\vep$. Hence, $m^*(E\setminus F)<\vep$. We can 153 | proceed in a similar manner as we did in the last paragraph to prove that 154 | (iii) $\Rightarrow$ (v) $\Rightarrow$ (i), leading to the final conclusion. 155 | \end{proof} 156 | 157 | % end 158 | 159 | \setcounter{subsection}{4} 160 | \subsection{Measurable Functions} 161 | \paragraph{19.} 162 | \begin{proof} 163 | For every $\beta\in\mathbb{R}$, since $D$ is measurable, there exists a 164 | sequence of $\alpha_n\in D\cap(\beta-1/n,\beta)$. As 165 | \[ 166 | \{x:\, f(x)>r\} \quad\Leftrightarrow\quad 167 | \bigcup_{n=1}^\infty \{x:\, f(x)>r-1/n\} \quad\Leftrightarrow\quad 168 | \bigcup_{n=1}^\infty \{x:\, f(x)>\alpha_n\} 169 | \] 170 | and $\{x:\, f(x)>\alpha_n\}$ are measurable, so is $\{x:\, f(x)>r\}$. Hence, 171 | $f$ is measurable. 172 | \end{proof} 173 | 174 | \paragraph{21.} 175 | \begin{proof} 176 | $\,$\par 177 | (a) It follows immediately from $\{x:\, f(x)>\alpha\} = \{x\in D:\, f(x)> 178 | \alpha\}\cup \{x\in E:\, f(x)>\alpha\}$.\par 179 | (b) For $\alpha\ge 0$, the sets $\{x:\, f(x)>\alpha\}$ and $\{x:\, g(x)> 180 | \alpha\}$ are the same. And for $\alpha < 0$, 181 | \[ 182 | \{x:\, f(x)>\alpha\} = \{x:\, g(x)>\alpha\} \setminus \bar{D} 183 | \quad\text{and}\quad 184 | \{x:\, g(x)>\alpha\} = \{x:\, f(x)>\alpha\} \cup \bar{D}. 185 | \] 186 | Hence, $f$ is measurable iff $g$ is measurable. 187 | \end{proof} 188 | 189 | \paragraph{22.(d)} 190 | \begin{proof} 191 | Since $f$ and $g$ are finite almost everywhere, the set $A$ consisting of 192 | points where $f+g$ is of the form $\infty - \infty$ or $-\infty + \infty$ is 193 | of measure zero (and hence measurable). Therefore no matter how it is 194 | defined, $\{x\in A:\,f+g>\alpha\}$ is measurable for every $\alpha$. Namely, 195 | the restriction of $f+g$ to $A$ is measurable. Meanwhile, clear that the 196 | restriction to $D\setminus A$ is measurable where $D$ is the domain of $f$. 197 | Hence, by Exercise 21, $f$ is measurable. 198 | \end{proof} 199 | 200 | \paragraph{23.} 201 | \begin{proof} 202 | $\,$\par 203 | (a) Let $A_n=\{x:\,|f(x)|>n\}$, a sequence of measurable sets. As $A_{n+1} 204 | \subset A_n$, $mA_{n+1}\le mA_n$. Since $A=\bigcap A_n = \{x:\,|f(x)|=\infty 205 | \}$, $mA_1\le m[a,b]$ is finite and $mA=0$, by Proposition 14, there exists 206 | some $N$ such that for all $n\ge N$, $mA_n<\vep/3$. Set $M=N$ to complete 207 | the proof.\par 208 | (b) We consider the restriction of $f$ on to the set $E=[a,b]\setminus\{x:\, 209 | |f(x)|\ge M\}$, which is also a measurable real-valued function. To keep our 210 | notation simple, we denote the restriction by $f$ still. For every $\vep>0$, 211 | there exists some integer $N$ with $0<2M/N<\vep$. Let $E_n=\{x:\,x\in 212 | [-M+(n-1)\vep, -M+n\vep]\}$ ($n=1,2,\dots,N$) and define 213 | \[ 214 | \varphi(x) = \sum_{i=1}^N f(x_i)\chi_{E_i}, 215 | \] 216 | where $x_n\in E_n$ is arbitrary. Clear that $\varphi$ is a simple function 217 | and satisfy all the requirements.\par 218 | (c) Suppose that $\varphi(x)=\sum_{i=1}^n \alpha_i\chi_{E_i}$. For each $i 219 | =1,\dots,N$, $E_i$ is measurable and therefore by Proposition 15, there 220 | exists a finite union $U_i$ of open intervals such that $m(U_i\bigtriangleup 221 | E_i)<\vep$. Let 222 | \[ 223 | g(x) = \sum_{i=1}^N \alpha_i\chi_{U_i}. 224 | \] 225 | Clear that $g$ and $\varphi$ only may differ on a set with measure $N\vep$. 226 | (d) Suppose that $g(x)=\sum_{i=1}^N\alpha_i\chi_{U_i}$ is a step function. 227 | We may assume without loss of generality that $U_i$ are disjoint and 228 | $\bigcup U_i = [a,b]$. And suppose that $\{x_0=a < x_1 < \dots < x_N=b\}$ 229 | are the endpoints of the intervals. For each $i=1,\dots,N-1$, define 230 | \[ 231 | f(x) = (x-x_i+\vep)g(x_i-\vep) + (x_i+\vep - x)g(x_i+\vep),\quad 232 | x\in (x_i-\vep, x_i+\vep), 233 | \] 234 | and $f(x)=g(x)$ for the other points. (We assume that $\vep$ is small enough 235 | so that $f$ is well-defined.) Clear that $f$ is continuous and equals $g$ 236 | except on a set of measure less then $2N\vep$. 237 | \end{proof} 238 | 239 | \paragraph{24.} 240 | \begin{proof} 241 | For measurable $f$, we show that $\mathcal{A}=\{E:\, f\inv[E]\text{ is 242 | measurable}\}$ is a $\sigma$-algebra first. As the domain, denoted by $D$, 243 | of a measurable function is measurable, $\mathbb{R}\in\mathcal{A}$. If 244 | $E\in\mathcal{A}$, then since $f\inv[\bar{E}]= D\cap \overline{f\inv[E]}$, 245 | $f\inv[\bar{E}]$ is also measurable and therefore $\bar{E}\in\mathcal{A}$. 246 | Suppose that $$ is a sequence of sets of $\mathcal{A}$. Then, as 247 | \[ 248 | f\inv\left[\bigcup_{n=1}^\infty E_n\right] = 249 | \bigcup_{n=1}^\infty f\inv[E_n], 250 | \] 251 | $\bigcup E_n\in\mathcal{A}$. Hence, $\mathcal{A}$ is a $\sigma$-algebra.\par 252 | By the definition of a measurable function, every open interval belongs to 253 | $\mathcal{A}$. Since the collection of all Borel sets $\mathcal{B}$ is the 254 | $\sigma$-algebra generated by all open intervals, $\mathcal{B}\subset 255 | \mathcal{A}$. Namely, $f\inv[B]$ is measurable as long as $B\in\mathcal{B}$. 256 | \end{proof} 257 | 258 | 259 | % end 260 | 261 | \subsection{Littlewood's Three Principles} 262 | \paragraph{30.} 263 | \begin{proof} 264 | Let $\vep_n=1/n$ and $\delta_n =\eta/2^n$ ($n=0,1,\dots$). By Proposition 265 | 24, for each $n$, there exists some $A_n$ with measure less than $\delta_n$ 266 | such that for all $x\in E_n\setminus A_n$, $|f_m(x) - f(x)|<\vep_n$ for $m$ 267 | large enough. Let $A=\bigcup_{n=1}^\infty A_n$, the measure of which is less 268 | than $\sum \eta/2^n=\delta$. Meanwhile, for any $\vep>0$, by construction, 269 | for all $x\in E\setminus A$, $|f_m(x)-f(x)|<\vep$ for $m$ large enough. 270 | Namely, $f_n$ converges to $f$ uniformly on $E\setminus A$. 271 | \end{proof} 272 | 273 | \paragraph{31.} 274 | \begin{proof} 275 | Let $\vep_n=\delta/2^n$ ($n\ge 0$), then by Proposition 22, there exists 276 | continuous $g_n$ such that $E_n=\{x:\,|f(x)-g_n(x)|\ge\vep_n\}$ is of 277 | measure less than $\vep_n$. Let $E=\bigcup E_n$, the measure of which is 278 | less than $\delta$ and $g_n$ converges to $f$ on $[a,b]\setminus E$.\par 279 | By Egoroff's Theorem, there exists some $A\subset [a,b]\setminus E$ with $m 280 | A<\delta$ such that $g_n$ converges to $f$ uniformly on $[a,b]\setminus(E 281 | \cup A)]$. Since $E\cup A$ is measurable, by Proposition 15, there exists 282 | some open set $O\supset E\cup A$ such that $m(O\setminus(E\cup A))<\delta$. 283 | Let $F=[a,b]\setminus O$. We know that 284 | \begin{enumerate} 285 | \item $F$ is a closed set. 286 | \item $mF < 3\delta$. 287 | \item $g_n$ converges to $f$ uniformly on $F$. 288 | \end{enumerate} 289 | Hence, $f$ is continuous on $F$ And by Problem 2.40, there exists some 290 | continuous function on $\mathbb{R}$ such that $\varphi(x) = f(x)$ for $x\in 291 | F$.\par 292 | If $f$ is defined on $(-\infty,\infty)$, we can apply the previous result on 293 | each $[n,n+1]$ and "stick" the functions together as we did in Problem 23(c) 294 | to get the function required. 295 | \end{proof} 296 | % end 297 | -------------------------------------------------------------------------------- /real_analysis_3rd/ch4_the_lebesgue_integral.tex: -------------------------------------------------------------------------------- 1 | \section{The Lebesgue Integral} 2 | \setcounter{subsection}{1} 3 | 4 | \subsection{The Lebesgue Integral of a Bounded Function} 5 | \paragraph{2.} 6 | \begin{proof} 7 | $\,$\par 8 | (a) By Problem 2.51, $h$ is upper semicontinuous as $f$ is bounded and by 9 | Problem 2.50, ${x:\, h(x)<\lambda}$ is open and hence measurable for every 10 | $\lambda\in\mathbb{R}$. Thus, $h$ is measurable.\par 11 | Let $\varphi(x)\ge f(x)$ be a step function and $x_0$ any point other 12 | than the endpoints of the intervals occurring in $\varphi$. Then there 13 | exists some $\delta>0$ such that for all $x\in(x_0-\delta,x_0+\delta)$, 14 | $\varphi(x_0) = \varphi(x) \ge f(x)$. Hence, 15 | \[ 16 | h(x_0) = \inf_{\delta<0}\sup_{|x-x_0|<\delta}f(x) \le \varphi(x_0). 17 | \] 18 | Namely, $\varphi\ge h$ except at a finite number of points. Hence, $\int_a^b 19 | \varphi \ge \int_a^b h$ and therefore 20 | \[ 21 | R\upint_a^b f = \inf_{\varphi\ge f}\int_a^b\varphi(x)\rd x \ge \int_a^b h. 22 | \]\par 23 | We can also derive from the previous discussion that there is a sequence of 24 | $\langle\varphi_n\rangle$ of step functions satisfying $\varphi \downarrow h$. By 25 | Proposition 6, 26 | \[ 27 | \int_a^b h = \lim\int_a^b\varphi_n \ge R\upint_a^b f. 28 | \] 29 | Hence, $R\upint_a^b f = \int_a^b h$.\par 30 | (b) First suppose that $f$ is Riemann integrable and let $h$ and $g$ be the 31 | upper and lower envelope of $f$ respectively. By part (a), $f$ is Riemann 32 | integrable implies $\int_a^b(h-g) = 0$. Together with the fact that $h\ge 33 | g$, we conclude that $h=g$ a.e.. Therefore, by Problem 2.50, $f$ is 34 | continuous except on a set of measure zero.\par 35 | Note that the argument remains true if we reverse the order, verifying the 36 | converse part. Hence, the proposition holds. 37 | \end{proof} 38 | % end 39 | 40 | \subsection{The Integral of a Nonnegative Function} 41 | \paragraph{3.} 42 | \begin{proof} 43 | Suppose that $E_n=\{x:\, f(x)>1/n\}$. Then, 44 | \[ 45 | 0=\int f \ge \int_{E_n} f \ge \frac{mE_n}{n} 46 | \] 47 | implies $mE_n=0$. Hence, $m\{x:\, f(x)>0\}=m(\bigcup E_n)\le\sum mE_n=0$. 48 | Namely, $f=0$ a.e. 49 | \end{proof} 50 | 51 | \paragraph{5.} 52 | \begin{proof} 53 | For any fixed $x_0\in\mathbb{R}$, let $f_n(x) = f\cdot\chi_{(-\infty,x_0- 54 | 1/n]}$, which is a increasing sequence of nonnegative measurable function 55 | whose limit is $f\cdot\chi_{-\infty,x_0}$. Then by Theorem 10, 56 | \[ 57 | F(x_0)=\int_{-\infty}^{x_0} f = \int f\cdot\chi_{-\infty,x_0} 58 | = \lim\int f\cdot\chi_{(-\infty,x_0-1/n]} = \lim F(x_0-1/n). 59 | \] 60 | Meanwhile, since 61 | \[ 62 | |F(x_0)-F(x_0+1/n)| = \left|\int_{x_0}^{x_0+1/n}f(x)\rd x\right|= 63 | \left|\int_{-1/n}^0 g(x)\rd x \right|, 64 | \] 65 | where $g(x)=f(x_0-x)$, arguing on $g$ in a similar manner yields $F(x_0)= 66 | \lim F(x_0+1/n)$. Thus, $F$ is continuous. 67 | \end{proof} 68 | 69 | \paragraph{6.} 70 | \begin{proof} 71 | By Theorem 9, $\int f \le \lowlim\int f_n$. Meanwhile, $f_n\le f$ implies 72 | $\int f_n\le \int f$ and therefore $\uplim \int f_n \le \int f$. Hence, 73 | $\int f =\lim\int f_n$. 74 | \end{proof} 75 | 76 | \paragraph{7.} 77 | \begin{solution} 78 | $\,$\par 79 | (a) Let $f_n(x)=n\cdot\chi_{[0,1/n]}$. $f_n$ converges to $f=0$ except on 80 | $x=0$. For each $n$, $\int f_n = 1$ but $\int f=0$. Hence, the inequality 81 | could be strict.\par 82 | (b) Let $f_n(x)=\chi_{[n,\infty)}$. Then $\langle f_n\rangle$ is a 83 | decreasing sequence which converges to $f=0$, the integral of which is $0$. 84 | However, for every $n$, $\int f_n = \infty$. 85 | \end{solution} 86 | 87 | \paragraph{8.} 88 | \begin{proof} 89 | Let $g_n = \inf\{f_n,f_{n+1},\dots\}$. Clear that 90 | \begin{equation} 91 | \label{eq:4.3.8} 92 | \int g_n \le \int f_n. 93 | \end{equation} 94 | Meanwhile $\langle g_n\rangle$ is a increasing sequence converging to 95 | $\lowlim f_n$. Hence, by the Monotone Convergence Theorem and 96 | \eqref{eq:4.3.8} 97 | \[ 98 | \int \lowlim f_n = \int \lim_{n\to\infty}g_n = 99 | \lim_{n\to\infty}\int g_n \le \lowlim\int f_n. 100 | \] 101 | \end{proof} 102 | 103 | \paragraph{9.} 104 | \begin{proof} 105 | By Fatou's Lemma, 106 | \begin{equation} 107 | \label{eq:4.3.9} 108 | \int_E f \le \lowlim\int_E f_n. 109 | \end{equation} 110 | Similarly, $\int_{\bar{E}}f \le \lowlim\int_{\bar{E}} f_n$ and therefore 111 | \[ 112 | \int_E f_n = \int f_n-\int_{\bar{E}} f_n\quad\Rightarrow\quad 113 | \uplim\int_E f_n \le \int f- \int_{\bar{E}} f = \int_{\bar{E}} f. 114 | \] 115 | \eqref{eq:4.3.9} and the inequality above together implies $\int_E f_n\to 116 | \int f$. 117 | \end{proof} 118 | % end 119 | 120 | \subsection{The General Lebesgue Integral} 121 | \paragraph{12.} 122 | \begin{proof} 123 | Note that $\langle g+f_n\rangle$ is a sequence of nonnegative measurable 124 | functions. Hence by Problem 8, 125 | \[ 126 | \int_E\lowlim(g+f_n) \le \lowlim\int_E(g+f_n) \quad\Rightarrow\quad 127 | \int_E\lowlim f_n \le \lowlim\int_E f_n. 128 | \] 129 | The second inequality follows immediately from the definition of lower and 130 | upper limit. Replacing $g+f_n$ with $g-f_n$ and arguing in a similar manner 131 | gives the last inequality. 132 | \end{proof} 133 | 134 | \paragraph{13.} 135 | \begin{proof} 136 | $f_n\ge -h$ implies $f_n+h\ge 0$. Hence, $\int(f_n+h)$ always has a meaning. 137 | And since $g$ is integrable, $\int f_n = \int(f_n+h)-\int h$ also has a 138 | meaning. Similarly, $\int f$ has a meaning. Meanwhile, 139 | \[ 140 | \int f = \int(f+h) - \int h \le \lowlim\int(f_n+h) - \int h 141 | = \lowlim\int f_n. 142 | \] 143 | \end{proof} 144 | 145 | 146 | \paragraph{15.} 147 | \begin{proof} 148 | $\,$\par 149 | (a) By Problem 4, for every $\vep>0$, there exists some simple functions 150 | $\varphi_1\le f^+$ and $\varphi_2\le f^-$ such that 151 | \[ 152 | \int_E f^+ - \int_E\varphi_1 < \vep \quad\text{and}\quad 153 | \int_E f^- - \int_E\varphi_2 < \vep. 154 | \] 155 | Let $\varphi=\varphi_1-\varphi_2$, which is also a simple function. 156 | Meanwhile, 157 | \[ 158 | \int_E|f-\varphi| \le \int_E(f^+-\varphi_1) + \int_E(f^--\varphi_2) < 2\vep. 159 | \] 160 | \end{proof} 161 | 162 | \paragraph{16.} 163 | \begin{proof} 164 | For every integrable $f$, by Problem 15, there exists some step function $\psi 165 | =\sum_{k=1}^N c_k\chi_{E_k}$ such that $\int|f-\psi|<\vep$. Note that 166 | \begin{equation} 167 | \label{eq:4.4.16} 168 | \lim_{n\to\infty}\int_{-\infty}^\infty \psi(x)\cos nx\rd x = 169 | \lim_{n\to\infty}\sum_{k=1}^N c_k\int_{E_k}\cos nx\rd x = 0. 170 | \end{equation} 171 | Hence, 172 | \begin{align*} 173 | \left|\int_{-\infty}^\infty f(x)\cos nx\rd x \right| 174 | &= \left|\int_{-\infty}^\infty (f(x)-\psi(x)+\psi(x))\cos nx\rd x\right|\\ 175 | &\le \int_{-\infty}^\infty |f(x)-\psi(x)||\cos nx|\rd x + 176 | \left|\int_{-\infty}^\infty \psi(x)\cos nx\rd x\right| \\ 177 | &\le \vep + \left|\int_{-\infty}^\infty \psi(x)\cos nx\rd x\right| \\ 178 | &\to 0 \text{ as }n\to\infty. 179 | \end{align*} 180 | \end{proof} 181 | 182 | \paragraph{18.} 183 | \begin{proof} 184 | Let $\langle t_n\rangle$ be any sequence with $t_n\ne 0$ and tending to $0$. 185 | Then $\langle f(x,t_n)\rangle$ is sequence of functions satisfying the 186 | hypotheses of Lebesgue Convergence Theorem. Meanwhile, $f(x,t_n)\to f$ as 187 | $n\to\infty$. Hence, 188 | \[ 189 | \lim_{n\to\infty}\int f(x,t_n)\rd x = \int f(x)\rd x. 190 | \] 191 | Since the choice of $\langle t_n\rangle$ is arbitrary, by Problem 2.49f, 192 | \[ 193 | \lim_{t\to 0}\int f(x,t)\rd x = \int f(x)\rd x. 194 | \] 195 | If $f$ is continuous in $t$ for each $x$, then $\lim_{\Delta t\to 0}f(x,t+ 196 | \Delta t)=f(x,t)$ holds for every $t$. Therefore, replacing $t$ with 197 | $\Delta t$ in the previous result yields 198 | \[ 199 | \lim_{\Delta t\to 0}\int f(x,t+\Delta t)\rd x = \int f(x,t)\rd x. 200 | \] 201 | Namely, $h(t)$ is continuous. 202 | \end{proof} 203 | 204 | % end 205 | -------------------------------------------------------------------------------- /real_analysis_3rd/ch5_differentiation_and_integration.tex: -------------------------------------------------------------------------------- 1 | \section{Differentiation and Integration} 2 | \subsection{Differentiation of Monotone Functions} 3 | \paragraph{3.} 4 | "maximum" needs to be changed to "minimum" in both (a) and (b). 5 | \begin{proof} 6 | $\,$\par 7 | (a) We may assume without loss of generality that $c=0$. Since $f$ attains a 8 | local minimum at $x=0$, $f(h)\ge f(0)$ for every $h$ sufficiently small. 9 | Hence, for every small $h>0$, $(f(c+h)-f(c))/h>0$ and therefore $D_+f(c)\ge 10 | 0$. Meanwhile, by Problem 2.b, 11 | \[ 12 | -D_{-}f(0) = D^+f(0) \ge 0 \quad\Rightarrow\quad 13 | D_{-}f(0) \le 0. 14 | \] 15 | The other two inequalities follow immediately from the definitions of upper 16 | and lower limits.\par 17 | (b) If $f$ has a local minimum at $a$ or $b$, then we only have the right or 18 | left half of the inequalities. 19 | \end{proof} 20 | 21 | \paragraph{4.} 22 | \begin{proof} 23 | We first show this for $g$ with $D^+g\ge\vep>0$. For every $a\le x0$. 28 | Then $D^+g\ge\vep >0$. Hence $g$ is nondecreasing. Therefore, for every $a 29 | \le xc\}$ is bounded below and 56 | hence $A=\inf E$ is finite. For every $\vep>0$, there exists some $y>c$ such 57 | that $A\le f(c) 1/n\}$. Since $f$ is nondecreasing, $|f(x)- 61 | f(y)|\le f(b)-f(a)<\infty$ for every $x,y\in[a,b]$. Hence, $D_n$ is finite, 62 | otherwise we can choose a sequence $x_1<\dots(f(b)-f(a))/n$ 63 | such that $f(X_N)-f(x_1)> f(b)-f(a)$. Therefore, $\bigcup_{n=1}^\infty E_n$, 64 | the set of discontinuities, is countable.\par 65 | (b) Suppose $\{x_1,\dots,x_n,\dots\}=\mathbb{Q}\cap[0,1]$ and define $f(x)= 66 | \sum_{x_n0$, there exists some subdivision $a=x_0<\cdots0$, there exists a subdivision $a=x_0<\cdots0$, we have 161 | \[ 162 | g(x) = \int_0^\vep f\hp(t)\rd t+\int_\vep^x f\hp(t)\rd t + f(0) 163 | = \int_0^\vep f\hp(t)\rd t + f(x) - f(\vep) + f(0), 164 | \] 165 | where the second equality comes from the absolute continuity on $[\vep,1]$. 166 | By the continuity of $f$ at $x=0$, $f(\vep)\to f(0)$. Hence, letting $\vep 167 | \to 0$ yields $g(x)=f(x)$. Namely, $f$ is an indefinite integral. Thus, by 168 | Theorem 14, it is absolutely continuous. 169 | \end{solution} 170 | 171 | \paragraph{13.} 172 | \begin{proof} 173 | Since absolute continuity implies bounded variation, $\int_a^b|f\hp|\le 174 | T_a^b(f)$ by Problem 11. By the definiton of $T$, for every $\vep>0$, there 175 | exists some subdivision $a=x_0<\cdotsT_a^b(f)- 176 | \vep$. Meanwhile, for every $i=1,\dots,n$, 177 | \[ 178 | \int_{x_{i-1}}^{x_i}|f\hp| \ge 179 | \left|\int_{x_{i-1}}^{x_i}f\hp \right| = 180 | |f(x_i) - f(x_{i-1})|, 181 | \] 182 | where the second equality is guaranteed by the absolute continuity. Hence, 183 | $\int_a^b|f\hp|>T_a^b(f)-\vep$ for every $\vep>0$. Thus, $T_a^b(f)= 184 | \int_a^b|f\hp|$.\par 185 | By Lemma 4, $2P_a^b(f) = T_a^b(f) + f(b)-f(a)$. Hence, 186 | \[ 187 | P_a^b(f) = \frac{1}{2}\left(\int_a^b|f\hp| + f(b)-f(a)\right) 188 | =\frac{1}{2}\int_a^b (|f\hp|+f\hp) = \int_a^b[f\hp]^+. 189 | \] 190 | \end{proof} 191 | 192 | \paragraph{14.} 193 | \begin{proof} 194 | $\,$\par 195 | (a) Suppose that $f$ and $g$ are absolutely continuous. Then for every 196 | $\vep>0$, there exists some $\delta>0$ such that for all finite 197 | nonoverlapping $\langle(x_n,y_n)\rangle$ with $|x_n-y_n|<\vep$, 198 | \[ 199 | \sum |f(x_n)+g(x_n)-f(y_n)-g(y_n)| \le 200 | \sum |f(x_n)-f(y_n)|+|g(x_n)-g(y_n)| \le 2\vep. 201 | \] 202 | Hence, $f+g$ is also absolutely continuous. Since $-g$ is absolutely 203 | continuous as long as $g$ is, so is $f-g$. \par 204 | (b) Suppose that $f$ and $g$ are absolutely continuous. Then they are 205 | bounded, by $M$ for example. Hence for every $\vep>0$, there exists some 206 | $\delta>0$ such that for all finite nonoverlapping $\langle(x_n,y_n)\rangle$ 207 | with $|x_n-y_n|<\vep$, 208 | \begin{align*} 209 | &\sum |f(x_n)g(x_n)-f(y_n)g(y_n)| \\ 210 | =&\sum |f(x_n)g(x_n)-f(x_n)g(y_n) + f(x_n)g(y_n)-f(y_n)g(y_n)| \\ 211 | \le&\sum \{|f(x_n)||g(x_n)-g(y_n)| + |f(x_n)-f(y_n)||g(y_n)|\} \\ 212 | \le& M\vep. 213 | \end{align*} 214 | Thus, $fg$ is also absolutely continuous.\par 215 | (c) Since $f$ is continuous on $[a,b]$, $f$ can achieve its minimum in $[a, 216 | b]$. Hence, $|f(x)|\ge m>0$ as $f$ is never zero. Therefore for every $\vep> 217 | 0$, there exists some $\delta>0$ such that for all finite nonoverlapping 218 | $\langle(x_n,y_n)\rangle$ with $|x_n-y_n|<\vep$, 219 | \begin{align*} 220 | \sum\left|\frac{1}{f(x_n)}-\frac{1}{f(y_n)}\right| 221 | = \sum\left|\frac{f(x_n)-f(y_n)}{f(x_n)f(y_n)}\right| 222 | \le \frac{1}{m^2}\sum|f(x_n)-f(y_n)|\le \frac{\vep}{m^2}. 223 | \end{align*} 224 | \end{proof} 225 | 226 | \paragraph{17.} 227 | Part (a) is wrong. It can be fixed if we further require $g$ to be monotone 228 | increasing. 229 | \begin{proof} 230 | $\,$\par 231 | (a) For every $\vep>0$, let $\delta_1$ be the number in the definition of 232 | $F$ corresponding to $\vep$ and $\delta_2$ the number in the definition of 233 | $g$ corresponding to $\delta_1$. Then for every finite nonoverlapping 234 | $\langle (x_n, y_n)\rangle$ with $|x_n-y_n|<\delta_2$, $\sum |g(x_n)-g(y_n)| 235 | < \delta_1$. Since $g$ is monotone increasing, $(g(x_n), g(y_n))$ are 236 | nonoverlapping. Therefore, $\sum|F(g(x_n))-F(g(y_n))|<\vep$. Hence, $F\circ 237 | g$ is absolutely continuous. 238 | \end{proof} 239 | 240 | \paragraph{18.} 241 | \begin{proof} 242 | Without loss of generality, we assume that $g$ is nondecreasing. 243 | Since $mE=0$, for every $\vep>0$, by Proposition 3.15, there exists an open 244 | set $O\supset E$ with $mO<\vep$. Meanwhile, there exists a sequence of 245 | disjoint open intervals $\langle I_n=(a_n,b_n)\rangle$ such that 246 | $\bigcup_{n=1}^\infty I_n=O$ and $l(I_n)<\delta$ where $\delta$ is the 247 | number in the definition of absolute continuity. Then $g[E]\subset 248 | \bigcup_{n=1}^\infty g[I_n\cap[0,1]]$. Since $g$ is continuous, the image of 249 | an interval is still an interval and since $g$ is also nondecreasing, $g[I_n 250 | \cap[0,1]]=(g(a_n\hp), g(b_n\hp))$, where $a_n\hp=\max\{a_n,0\}$ and $b_n\hp 251 | =\min\{b_n,1\}$. Finally, 252 | \[ 253 | m(g[E]) \le \sum_{n=1}^\infty m(g[I_n]) = \sum_{n=1}^\infty|g(b_n\hp)- 254 | g(a_n\hp)| \le \vep, 255 | \] 256 | where the last inequality comes from the absolute continuity of $g$. Since 257 | the choice of $\vep$ is arbitrary, $m(g[E])=0$. 258 | \end{proof} 259 | 260 | \paragraph{20.} 261 | \begin{proof} 262 | $\,$\par 263 | (a) For every $\vep>0$, let $\delta=\vep/M$. Then for every $\langle x_n 264 | \rangle_{i=1}^n$ and $\langle y_n\rangle_{i=1}^n$ with $|x_n-y_n|\le\delta$, 265 | \[ 266 | \sum_{i=1}^n|f(x_n)-f(y_n)|\le M\sum_{i=1}^n|x_n-y_n| \le \vep, 267 | \] 268 | as $f$ satisfies the Lipschitz condition.\par 269 | (b) Suppose that $f$ is absolute continuous and $|f\hp|$ is bounded by $M$. 270 | Then for every $x$ and $y$ in the interval, 271 | \[ 272 | |f(x)-f(y)|=\left|\int_x^y f\hp(t)\rd t\right| \le M|x-y|. 273 | \] 274 | Hence, $f$ satisfies the Lipschitz condition. The converse part has been 275 | proved in (a).\par 276 | (c) It is wrong. A counterexample is $f(x)=\chi_{[0,1]}$, $x\in(-1,1)$ 277 | \end{proof} 278 | 279 | \paragraph{21.} 280 | \begin{proof} 281 | $\,$\par 282 | (a) Suppose that $O=\bigcup_{n=1}^\infty(c_n,d_n)$ where $(c_n,d_n)$ are 283 | disjoint. Since $g$ is continuous and increasing, $g\inv(c_n,d_n)$ is still 284 | an open interval, denoting it by $(a_n,b_n)$, and $(a_n,b_n)$ are also 285 | disjoint. Meanwhile, $d_n-c_n=f(a_n)-f(b_n)=\int_{a_n}^{b_n}g\hp $. Hence, 286 | \[ 287 | mO = m\left(\bigcup_{n=1}^\infty(c_n,d_n)\right) = 288 | \sum_{n=1}^\infty(d_n-c_n) = 289 | \sum_{n=1}^\infty \int_{a_n}^{b_n}g\hp = 290 | \int_{g\inv[O]}g\hp. 291 | \]\par 292 | (b) Without loss of generality, we assume that $d\notin E$. For every $\vep 293 | >0$, there exists an open set $O\supset E$ with $mO<\vep$. By Part (a), 294 | \[ 295 | \int_{g\inv[O]\cap H}g\hp = \int_{g\inv[O]}g\hp = mO < \vep. 296 | \] 297 | Since the choice of $\vep$ is arbitrary, $\int_{g\inv[O]\cap H}g\hp = 0$. 298 | Since $g\hp > 0$ on $H$, $g\inv[O]\cap H$ has measure zero.\par 299 | (c) Since $E$ is measurable, so is $g\inv[E]$. Meanwhile, by Theorem 3, 300 | $g\hp$ is measurable, hence $H$ is also measurable. Therefore, $F$ is 301 | measurable.\par 302 | We may assume without loss of generality that $c,d\notin E$. By Proposition 303 | 3.15, there exists some $G\in G_\delta$ such that $E\subset G\subset(c,d)$ 304 | and $m(G\setminus E)=0$. Since $g$ is increasing, $g\inv[G]\cap H = F\cup 305 | (g\inv[G\setminus E]\cap H)$ and by (b), $g[G\setminus E]\cap H$ is of 306 | measure zero. Therefore, $\int_F g\hp=\int_{g\inv[G]\cap H}g\hp$. Namely, it 307 | suffices to show the result for $G\in G_\delta$. \par 308 | Suppose that $G=\bigcap_{n=1}^\infty O_n$ where each $O_n\subset(c,d)$ is 309 | open and $mO_1<\infty$. Without loss of generality, we may assume that 310 | $\langle O_n\rangle$ is decreasing. Then $mG=\lim_{n\to\infty}mO_n$. By (a), 311 | \[ 312 | mO_n = \int_{g\inv[O_n]}g\hp = \int_a^b\chi_{O_n}(g(x))g\hp(x)\rd x. 313 | \] 314 | As $\chi_{O_n}(g(x))g\hp(x)$ is bounded by $|g\hp|$, 315 | \[ 316 | \lim_{n\to\infty}\int_a^b\chi_{O_n}(g(x))g\hp(x)\rd x = 317 | \int_a^b\chi_G(g(x))g\hp(x)\rd x. 318 | \] 319 | Hence, $mG=\int_{g\inv[G]\cap H}g\hp$, completing the proof.\par 320 | (d) By Problem 3.25, $f\circ g$ is measurable. And since $g\hp$ is 321 | measurable by Theorem 3, $(f\circ g)g\hp$ is also measurable. \par 322 | Let $\langle\varphi_n\rangle$ be an increasing sequence of nonnegative 323 | simple functions which converges to $f$, the existence of which is 324 | guaranteed by Problem 4.4. By the monotone convergence theorem, $\int_c^d f 325 | =\lim\int_c^d\varphi_n$.\par 326 | For each $n$, suppose that $\varphi_n(y)=\sum_{k=1}^ma_k^{(n)}(y) 327 | \chi_{E_k^{(n)}}(y)$. Then 328 | \[ 329 | \int_c^d\varphi_n = \sum_{k=1}^ma_k^{(n)}mE_{k}^{(n)}= 330 | \sum_{k=1}^ma_k^{(n)}\int_a^b\chi_{E_k^{(n)}}(g(x))g\hp(x)\rd x= 331 | \int_a^b\varphi_n(g(x))g\hp(x)\rd x, 332 | \] 333 | where the second equality comes from (c). Since $g$ is increasing, 334 | $\langle\varphi_n(g(x))g\hp(x)\rangle$ is an increasing sequence. Hence, 335 | \[ 336 | \int_a^b f(g(x))g\hp(x)\rd x = 337 | \lim_{n\to\infty}\int_a^b\varphi_n(g(x))g\hp(x)\rd x. 338 | \] 339 | Thus, 340 | \[ 341 | \int_c^d f(y)\rd y = 342 | \lim_{n\to\infty}\int_c^d\varphi_n(y)\rd y = 343 | \lim_{n\to\infty}\int_a^b\varphi_n(g(x))g\hp(x)\rd x = 344 | \int_a^b f(g(x))g\hp(x)\rd x. 345 | \] 346 | \end{proof} 347 | 348 | % end 349 | 350 | \subsection{Convex Functions} 351 | \paragraph{23.} 352 | \begin{proof} 353 | $\,$\par 354 | (a) Suppose that $x_0\in(a,b)$ and $y(x)=m(x-x_0)+\varphi(x_0)$ is a 355 | supporting line. As $[a,b)$ is finite, $\varphi\ge\min\{\varphi(a),y(a), 356 | y(b)\}$. \par 357 | (b) If $\varphi$ is monotone, then the limits exists. If $\varphi$ is not 358 | monotone, then since $D^+\varphi$ is nondecreasing, there exists some $[c,d] 359 | \subset(a,b)$ such that $D^+\varphi\le 0$ on $(a,c)$ and $D^+\varphi\ge 0$ 360 | on $(d,b)$. Namely, $\varphi$ is monotone on the $(a,c)$ and $(d,b)$. 361 | Therefore, the limits also exist.\par 362 | Consider a finite interval near the finite endpoint. By (a), the limit can 363 | not be $-\infty$ as $\varphi$ is bounded from below.\par 364 | (c) If $x$ and $y$ are in the interior of $I$, the inequality holds by 365 | definition. By the continuity of $\varphi$, the statement holds for all $x,y 366 | \in I$. 367 | \end{proof} 368 | 369 | \paragraph{24.} 370 | \begin{proof} 371 | Note that the existence of $\varphi^{\prime\prime}$ implies $\varphi$ is 372 | continuously differentiable. Suppose that $\varphi$ is convex on $(a,b)$. 373 | Then $D^+\varphi$ is nondecreasing by Proposition 17, hence $\varphi^{\prime 374 | \prime}(x)\ge 0$ for each $x\in(a,b)$. And the converse of the statement 375 | follows from Proposition 18 immediately. 376 | \end{proof} 377 | 378 | \paragraph{25.} 379 | \begin{proof} 380 | $\,$\par 381 | (a) $\varphi^{\prime\prime}(t)=b^2p(p-1)(a+bt)^{p-2}$ which $\ge 0$ on 382 | $[0,\infty)$ if $p\ge 1$ and $\le 0$ if $0S+T\}=0$. Thus, $S+T\ge \|f+g\|_\infty$ by the definition of 8 | $\esssup$. 9 | \end{proof} 10 | 11 | \paragraph{2.} 12 | \begin{proof} 13 | Put $S=\|f\|_\infty$. Since $S\ge |f|$ a.e., 14 | \[ 15 | \|f\|_p= 16 | \left\{\int_0^1|f|^p\right\}^{1/p} \le 17 | \left\{\int_0^1S^p\right\}^{1/p} = S. 18 | \] 19 | Therefore, $\uplim_{p\to\infty}\|f\|_p \le S$. For the converse part, let 20 | $\vep$ be any positive number. Then the measure $\delta$ of $E=\{t:\,|f(t)|> 21 | S-\vep\}$ is positive. Hence, 22 | \[ 23 | \left\{\int_0^1|f|^p\right\}^{1/p} \ge 24 | \left\{\int_E|f|^p\right\}^{1/p} \ge 25 | \delta^{1/p}(S-\vep) \to 26 | S-\vep\quad\text{as $p\to\infty$.} 27 | \] 28 | Hence, $\lowlim_{p\to\infty}\ge S$, completing the proof. 29 | \end{proof} 30 | 31 | \paragraph{3.} 32 | \begin{proof} 33 | \[ 34 | \|f+g\|_1 = \int|f+g| \le \int|f|+\int|g| = \|f\|_1+\|g\|_1. 35 | \] 36 | \end{proof} 37 | 38 | \paragraph{4.} 39 | \begin{proof} 40 | For every $M>\|g\|_\infty$, $|g|\le M$ a.e. Hence, 41 | \[ 42 | \int|fg|\le M\int|f| = \|f\|_1 M. 43 | \] 44 | Since the choice of $M$ is arbitrary, $\int|fg|\le \|f\|_1\|g\|_\infty$. 45 | \end{proof} 46 | % end 47 | 48 | \subsection{The Minkowski and Hölder Inequalities} 49 | \paragraph{8} 50 | \begin{proof} 51 | $\,$\par 52 | (a) The logarithm function is concave, so 53 | \[ 54 | \log(a^p/p+b^q/q) \ge \frac{1}{p}\log a^p + \frac{1}{q}\log b^q = \log ab. 55 | \] 56 | Taking $\exp$ on the both sides yields the inequality. The equality holds 57 | iff $a^p=b^q$. \par 58 | (b) The case where $p=\infty$ has been proved in Problem 4 and the case 59 | where $\|f\|_p=0$ or $\|g\|=0$ is straightforward. Hence, we assume that 60 | $10$, there exists some $N$ such that for all $n>N$, $\|f_n-f\|< 94 | \vep$. Hence, for every $n,m>N$, by Minkowski inequality, 95 | \[ 96 | \|f_n-f_m\| \le \|f_n-f\|+\|f-f_m\| < 2\vep. 97 | \] 98 | Hence, $\langle f_n\rangle$ is a Cauchy sequence. 99 | \end{proof} 100 | 101 | \paragraph{10.} 102 | \begin{proof} 103 | Suppose $f_n\to f$. Then $M_n=\|f_n-f\|_\infty=\esssup|f_n-f|\to 0$. Let 104 | $E_n=\{x:\,|f_n(x)-f(x)|>M_m\}$, each of which is with measure zero. And 105 | therefore $E=\bigcup_{n=1}^\infty E_n$ is with measure zero. Note that 106 | $\tilde{E}=\{x:\,|f_n(x)-f(x)|0$, there exists some $N$ such that for every $n>N$ and $x 110 | \in\tilde{E}$, $|f_n(x)-f(x)|<\vep$. Since $mE=0$, this implies $\|f_n-f 111 | \|_\infty=\esssup|f_n(x)-f(x)|<\vep$. Hence, $f_n\to f$ in $L^\infty$. 112 | \end{proof} 113 | 114 | \paragraph{11.} 115 | \begin{proof} 116 | Let $\langle f_n\rangle\subset L^\infty$ be absolutely summable. Put $M_n= 117 | \|f_n\|_\infty$ and $A_n=\{t:\,|f_n(t)|>M_n\}$. By the definition of $\| 118 | \cdot\|_\infty$, $mA_n=0$. Hence, $A=\bigcup_{n=1}^\infty A_n$ is of measure 119 | zero. \par 120 | Note that $|f_n(x)|\le M_n$ for every $n$ and $x\in E\setminus A$. Thus, by 121 | the Weierstrass M-test, $\sum_{n=1}^\infty f_n$ converges uniformly. Hence, 122 | on $E\setminus A$, $\sup|\sum_{n=1}^\infty f_n - \sum_{n=1}^N f_n|\to 0$ as 123 | $N\to\infty$. Since $mA=0$, this implies the summability of $\langle f_n 124 | \rangle$. 125 | \end{proof} 126 | 127 | \paragraph{13.} 128 | \begin{proof} 129 | Suppose $\langle f_n\rangle\subset C$ be absolutely summable. Since for 130 | every $x$, $0\le|f_n(x)|\le\|f_n\|$, $\langle f_n\rangle$ is uniformly 131 | convergent on $[0,1]$. Put $s=\sum_{n=1}^\infty f_n$. Since each $f_n$ is 132 | continuous, so is $s$. Therefore, $s\in C$. \par 133 | For every $\vep>0$, there exists some $N$ such that for every $n>N$ and $x 134 | \in[0,1]$, $\left|s(x)-\sum_{k=1}^nf_k(x)\right|<\vep$. Hence, $\|s- 135 | \sum_{k=1}^nf_k\|<\vep$. Thus, $\langle f_n\rangle$ is summable and 136 | therefore $C$ is a Banach space. 137 | \end{proof} 138 | 139 | \paragraph{16.} 140 | \begin{proof} 141 | Since $\|f_n-f\| \ge |\|f_n\|-\|f\||$, $f_n\to f$ in $L^p$ implies $\|f_n\| 142 | \to \|f\|$. For the converse part, note that $2^p(|f_n|^p+|f|^p)-|f_n-f|^p 143 | \ge 0$ and for almost every $x$, 144 | \[ 145 | 2^p(|f_n|^p+|f|^p)-|f_n-f|^p \to 2^{p+1}|f|^p. 146 | \] 147 | By Fatou's Lemma, 148 | \begin{align*} 149 | 2^{p+1}\|f\|^p = 2^{p+1}\int|f|^p 150 | &\le \lowlim\int\{2^p(|f_n|^p+|f|^p)-|f_n-f|^p\} \\ 151 | &= 2^{p+1}\|f\|^p - \uplim\|f_n-f\|^p. 152 | \end{align*} 153 | Hence, $\uplim\|f_n-f\|^p \le 0$. Since clear that $\lowlim\|f_n-f\|^p\ge0$, 154 | $\lim\|f_n-f\|=0$, i.e., $f_n \to f$ in $L^p$. 155 | \end{proof} 156 | 157 | \paragraph{17.} 158 | I assume that $1/p+1/q=1$. 159 | \begin{proof} 160 | Since $g\in L^p$, $|g|^q$ is integrable on $E=[0,1]$ and therefore for every 161 | $\vep>0$, there exists some $\delta$ such that for every $A\subset E$ with 162 | $mA<\delta$, $\int_A|g|^q<\vep$. Meanwhile, since $f_n(x)\to f(x)$ for 163 | almost every $x$, by Egoroff's Theorem, there exists some $A\subset E$ with 164 | $mA<\delta$ such that $f_ng$ converges to $fg$ uniformly on $E\setminus A$. 165 | \par 166 | From the uniform convergence we conclude 167 | \begin{equation} 168 | \label{eq:6.17.1} 169 | \int_{E\setminus A} fg = \lim_{n\to\infty}\int_{E\setminus A}f_ng. 170 | \end{equation} 171 | Meanwhile, by Hölder inequality, 172 | \begin{align*} 173 | \left|\int_A (f-f_n)g\right| \le \int_A|(f-f_n)g| 174 | &\le\left\{\int_A|f_n-f|^p\right\}^{1/p}\left\{\int_A|g|^q\right\}^{1/q} 175 | \le M\vep^{1/q}. 176 | \end{align*} 177 | Hence, \eqref{eq:6.17.1} can be extended to $E$.\par 178 | For $p=1$, this is not true. $f_n=n\chi_{[0,1/n]}$ and $g=\chi_{[0,1]}$ 179 | gives a counterexample. 180 | \end{proof} 181 | 182 | \paragraph{18.} 183 | \begin{proof} 184 | By Minkowski inequality, 185 | \[ 186 | \|g_nf_n-gf\| = \|g_n(f_n-f)+(g_n-g)f\| \le \|g_n(f_n-f)\|+\|(g_n-g)f\|. 187 | \] 188 | Fix $\vep>0$. Since $f,g_n,g\in L^p$, $|g_n-g|^p|f|^p$ is integrable and 189 | therefore there exists some $\delta>0$ such that for all subsets with 190 | measure $<\delta$, the integral of over it $<\vep$. Meanwhile, since $g_n\to 191 | g$ a.e., by Egoroff's Theorem, there exists some $A\subset E=[0,1]$ with $mA 192 | <\delta$ such that $g_n\to g$ uniformly on $E\setminus A$ and therefore 193 | there exists some $N_1>0$ such that for all $n>N_1$, $|g_n(x)-g(x)|^p<\vep$ 194 | for $x\in E\setminus A$. Thus, for every $n>N_1$, 195 | \begin{align*} 196 | \|(g_n-g)f\| 197 | &=\left\{\int_{E\setminus A}|g_n-g|^p|f|^p\right\}^{1/p} + 198 | \left\{\int_A|g_n-g|^p|f|^p\right\}^{1/p} \\ 199 | & \le \sqrt[p]{\vep}\|f\| + \sqrt[p]{\vep} \le (\|f\|+1)\vep. 200 | \end{align*} 201 | Since $|g_n|\le M$, $\|g_n(f_n-f)\|\le M\|f_n-f\|$. And since $f_n\to f$ in 202 | $L^p$, there exists some $N_2>0$ such that for all $n>N_2$, $\|f_n-f\|< 203 | \vep$. Put $N=\max(N_1,N_2)$, then for every $n>N$, 204 | \[ 205 | \|g_nf_n-gf\| \le (\|f\|+1+M)\vep. 206 | \] 207 | Hence, $g_nf_n\to gf$ in $L^p$. 208 | \end{proof} 209 | % end 210 | 211 | \subsection{Approximation in $L^p$} 212 | \paragraph{19.} 213 | \begin{proof} 214 | Since $\|T_\Delta f\|\le \|T_\Delta|f|\|$ and $\|f\|=\||f|\|$, we may assume 215 | without loss of generality that $f\ge 0$. For $p>1$, by Jensen's inequality, 216 | \begin{align*} 217 | \|T_\Delta f\|_p^p 218 | &= \sum_{k=1}^m\int_{\xi_{k-1}}^{\xi_k} 219 | \left(\frac{1}{\xi_k-\xi_{k-1}}\int_{\xi_{k=1}}^{\xi_k}f\right)^p \\ 220 | &\le \sum_{k=1}^m\int_{\xi_{k-1}}^{\xi_k} 221 | \frac{1}{\xi_k-\xi_{k-1}}\int_{\xi_{k-1}}^{\xi_k}f^p \\ 222 | &= \sum_{k=1}^m\int_{\xi_{k-1}}^{\xi_k}f^p \\ 223 | &= \int_0^1f^p = \|f\|_p^p. 224 | \end{align*} 225 | \end{proof} 226 | % end 227 | -------------------------------------------------------------------------------- /real_analysis_3rd/ch7_metric_spaces.tex: -------------------------------------------------------------------------------- 1 | \section{Metric Spaces} 2 | \setcounter{subsection}{6} 3 | \subsection{Compact Metric Space} 4 | \paragraph{27.} 5 | \begin{proof} 6 | If $\rho(F, K) > 0$, then clear that $F\cap K = \varnothing$. For the reverse 7 | direction, consider the function $h(x) = \rho(x, F)= \inf_{y\in F}\rho(x,y)$. 8 | Clear that for $x\in K$, $h(x) \le \rho(K, F)$. 9 | 10 | First, we show that $h$ is continuous. Let $x$ be fixed. For every 11 | $x\hp \in X$ 12 | \[ 13 | h(x) \le \rho(x, y) \le \rho(x, x\hp) + \rho(x\hp, y), 14 | \quad\forall y\in F. 15 | \] 16 | Take infimum on the right hand side and we get 17 | \[ 18 | h(x) \le \rho(x, x\hp) + h(x\hp) 19 | \quad\Rightarrow\quad 20 | h(x) - h(x\hp) \le \rho(x, x\hp). 21 | \] 22 | Similarly, we have $h(x\hp) - h(x) \le \rho(x, x\hp)$. Thus, $h$ is 23 | continuous. 24 | 25 | Since $K$ is compact and $h$ is continuous, $h$ attains its infimum $c$ 26 | at some point $x_0 \in K$. Assume, to obtain a contradiction, that $c = 0$. 27 | Then, for every $\vep > 0$, there is a $y \in F$ s.t. $\rho(x_0, y) < \vep$. 28 | Namely, $x_0$ is a cluster point of $F$. Since $F$ is closed, $x_0 \in F$, 29 | which contradicts with $F\cap K=\varnothing$. Thus, $c > 0$ and therefore 30 | $\rho(F, K) > 0$. 31 | \end{proof} 32 | 33 | \paragraph{28.} 34 | \begin{proof} 35 | $\,$\par 36 | (a) Let $\vep > 0$ be fixed. Since $f$ is uniformly continuous, there exists 37 | some $\delta > 0$ s.t. for every $x_1,x_2\in X$ with $\rho(x_1, x_2) < 38 | \delta$, $\rho(f(x_1), f(x_2)) < \vep$. Let $B\subset X$ be a ball with 39 | radius $\delta/2$. Then $f(B) \subset Y$ is contained by some ball of 40 | radius $\vep$. Hence, we can cover $Y$ with finitely many balls of radius 41 | $\vep$ as long as we can cover $X$ with finitely many balls of radius 42 | $\delta / 2$ and this can be done since $X$ is totally bounded. 43 | 44 | (b) Clear that $X = (0, 1)$ is totally bounded while $Y = (1, \infty)$ is 45 | not. The function $1/x$ a continuous function that maps $X$ onto $Y$. Hence, 46 | the result does not hold. 47 | \end{proof} 48 | 49 | 50 | \paragraph{29.} 51 | \begin{proof} 52 | $\,$\par 53 | (a) First, by the definition of open cover and open sets, $\varphi(x) > 0$. 54 | Meanwhile, since $X$ is compact, it is bounded and, therefore, 55 | $\varphi(x) < \infty$. 56 | 57 | (b) Let $x$ be fixed and $r$ be such that there exists some $O\in \mcal{U}$ 58 | with $B_{x, r}\subset O$. Let $s = r - \rho(x, y)$. If $s \le 0$, then 59 | clear that $s \le \varphi(y)$. If $s > 0$, then we have $B_{y, s} 60 | \subset B_{x, r} \subset O$. Hence, $s \le \varphi(y)$. Thus, 61 | $r - \rho(x, y) \le \varphi(y)$. Take supremum on the left hand side and we 62 | get $\varphi(x) - \rho(x, y) \le \varphi(y)$. 63 | 64 | (c) It follows immediately from (b). 65 | 66 | (d) Let $(x_n)$ be a sequence s.t. $\varphi(x_n) \to \vep$. Since $X$ is 67 | sequentially compact, $(x_n)$ has a convergent subsequence. For the sake of 68 | convenience, we assume the subsequence is $(x_n)$ itself. Suppose $x_n 69 | \to x$. Since $\varphi$ is continuous, $\varphi(x) = \lim\varphi(x_n) = 70 | \vep$. Thus, by (a), $\vep > 0$. 71 | 72 | (e) Let $\delta$ be any positive number that is less than $\vep$. For every 73 | $x$, since $\delta < \vep = \inf\varphi$, $\delta < \varphi(x)$ and, 74 | therefore, $B_{x, \delta} \subset O$ for some $O \in \mcal{U}$. 75 | \end{proof} 76 | 77 | \subsection{Baire Category} 78 | \paragraph{31.(a)} 79 | \begin{proof} 80 | Suppose that $F$ is nowhere dense. Since $F$ is closed, $F^c$ is dense. 81 | Therefore, for every point $x\in F$, every neighborhood of $x$ contains a 82 | point of $F^c$. Thus, $F$ contains to open set. For the reverse direction, 83 | since $F$ contains no open set, every neighborhood of every $x\in X$ contains 84 | a point in $F^c$, which implies that $F^c$ is dense. 85 | \end{proof} 86 | 87 | \paragraph{34.} 88 | \begin{proof} 89 | For the first part, it suffices to show that $\partial E$ is nowhere dense 90 | for any set $E \subset X$. Since $\partial E$ is closed and contains no open 91 | set, by 31.(a), it is nowhere dense. 92 | 93 | For the second part, assume, to obtain a contradiction, that $F$ is not 94 | nowhere dense. Then by 31.(a), it contains an nonempty open set $O$. By 95 | 32.(a), $O$ is also meager, which contradicts the Baire category theorem. 96 | \end{proof} 97 | 98 | \paragraph{35.} 99 | \begin{proof} 100 | Let $E$ be a subset set of the complete metric space $X$. If $E$ is residual, 101 | then $E^c = \bigcup_n K_n$ where $K_n$ are nowhere dense sets. Hence, 102 | \[ 103 | E = \bigcap K_n^c \supset \bigcap (\cl K_n)^c. 104 | \] 105 | Note that $(\cl K_n)^c$ is a $G_\delta$. Meanwhile, since each $(\cl K_n)^c$ 106 | is a dense open set and $X$ is complete, By the theorem of Baire, 107 | $\bigcap (\cl K_n)^c$ is also dense. The proof of the reverse direction is 108 | similar. 109 | \end{proof} 110 | 111 | \paragraph{39.} 112 | \begin{proof} 113 | Let $E_m$ has the same meaning as in the proof of Theorem 32. Let $O := 114 | \bigcup_m E_m^\circ$. By Prop. 31, $O$ is a dense residual open set. For 115 | every $x\in O$, $x \in E_m^\circ$ for some $m$. Since $E_m^\circ$ is open, 116 | there is a neighborhood $U$ of $x$ which is contained by $E_m^\circ$. Thus, 117 | $\mcal{F}$ is uniformly bounded by $m$ in $U$. 118 | \end{proof} 119 | 120 | 121 | \setcounter{subsection}{9} 122 | \subsection{The Ascoli-Arzel\'{a} Theorem} 123 | \paragraph{47.} 124 | \begin{proof} 125 | Let $x$ be an arbitrary fixed point in $X$ and $(x_n)\subset X$ a sequence 126 | that converges to $x$. Clear that $K = \{x\}\cup(x_n)$ is (sequently) 127 | compact. Hence, $f_n$ converges to $f$ uniformly on $K$. Thus, $f$ is also 128 | continuous on $K$ and therefore at $x$. Namely, $f$ is continuous. 129 | \end{proof} 130 | 131 | \paragraph{49.} 132 | \begin{proof} 133 | Let $x$ be an arbitrary point in $X$. For every $\vep > 0$, since $\mcal{F}$ 134 | is equicontinuous, there is an open neighborhood $O$ of $x$ s.t. 135 | $\sigma(f(x), f(y)) < \vep$ for every $y\in O$ and $f\in\mcal{F}$. Now, for 136 | every $f^+\in \mcal{F}^+$, suppose $f_n \to f$. We have 137 | \[ 138 | \sigma(f^+(x), f^+(y)) 139 | = \sigma\left(\lim_n f_n(x), \lim_m f_m(y)\right) 140 | = \lim_n\lim_m \sigma(f_n(x), f_m(y)) \le \vep, 141 | \] 142 | where the second equality comes from the continuity of $\sigma$ and the last 143 | inequality comes from the equicontinuity of $\mcal{F}$. 144 | \end{proof} 145 | 146 | \paragraph{50.} I assume that the norm on $C[0, 1]$ is the sup norm. 147 | \begin{proof} 148 | Let $\mcal{F} := \{f\in C[0, 1]\mid \|f\|_\alpha \le 1 \}$. First, we show 149 | that $\mcal{F}$ is equicontinuous. By the definition of $\|\cdot\|_\alpha$, 150 | $f\in\mcal{F}$ only if it is bounded by $1$ and $|f(x) - f(y)| \le 151 | |x - y|^\alpha$ for all $x, y \in [0, 1]$. For every $x\in X$ and $\vep > 0$, 152 | consider the $\sqrt[\alpha]{\vep}$-ball $B$ centered at $x$. For every 153 | $y \in B$, $|f(x) - f(y)| \le |x - y|^\alpha < \vep$. Hence, $\mcal{F}$ is 154 | equicontinuous. Let $(f_n)$ be a sequence in $\mcal{F}$. Since $f_n$ is 155 | bounded by $1$, by Corollary 41, $(f_n)$ contains a convergent subsequence. 156 | Thus, $\mcal{F}$ is a (sequently) compact subset of $C[0, 1]$. 157 | \end{proof} 158 | 159 | 160 | 161 | 162 | 163 | 164 | 165 | 166 | -------------------------------------------------------------------------------- /real_analysis_3rd/ch8_topological_spaces.tex: -------------------------------------------------------------------------------- 1 | \section{Topological Spaces} 2 | \subsection{Fundamental Notions} 3 | \paragraph{3.} 4 | \begin{proof} 5 | If $A$ is open, then clear that $x \in A \subset A$. For the converse, let 6 | $E = \bigcup_{x\in A}O_x$ where $x\in O_x \subset A$. Since $O_x\subset A$, 7 | $E \subset A$. Meanwhile, for every $x\in A$, $x \in O_x\subset E$. Hence, 8 | $A = E$. Thus, $A$ is open since $E$ is the union of open sets. 9 | \end{proof} 10 | 11 | \paragraph{7.} 12 | \begin{proof} 13 | $\,$\par 14 | (a) We argue by contradiction. Assume $x \in F^c$. Since $F$ is closed, $F^c$ 15 | is open and therefore there is neighborhood $O$ of $x$ s.t. $O \cap F 16 | \ne \varnothing$, which contradicts the fact that $x$ is a cluster point. 17 | Thus, $x \in F$. 18 | 19 | (b) Let $y := f(x)$ and $y_n := f(x_n)$. Let $O$ be an arbitrary 20 | neighborhood of $y$. Since $f$ is continuous, there is a neighborhood $U$ of 21 | $x$ s.t. $f(U) \subset O$. Since $x = \lim x_n$, there is an integer $N$ s.t. 22 | $x_n \in U$ for every $n > N$. Hence, $y_n = f(x_n) \in f(U) \subset O$ for 23 | all $n > N$. Thus, $y_n \to y$. 24 | 25 | (c) The previous argument, \textit{mutatis mutandis}, yields the result. 26 | \end{proof} 27 | 28 | \paragraph{10.} 29 | \begin{proof} 30 | $\,$\par 31 | (a) Suppose that both $A_1$ and $A_2$ are open. 32 | Let $f_1 := f|_{A_1}$, $x\in A$ and $y = f(x)$. We may assume without 33 | loss of generality that $x \in A_1$. For every neighborhood $O$ of $y$, since 34 | $f_1$ is continuous, there is a neighborhood $A_1\cap U$ s.t. $f(A_1\cap U) 35 | \subset O$ where $U$ is an open set of $X$. Since $A_1\cap U$ is still open, 36 | $f$ is continuous at $x$. Thus, $f$ is continuous. 37 | 38 | (b) Let $f(x) = 0$ on $A_1 = (-1, 0)$ and $f(x) = 1$ on $A_2 = [0, 1]$. 39 | \end{proof} 40 | 41 | \subsection{Bases and Contability} 42 | \paragraph{11.} 43 | \begin{proof} 44 | $\,$\par 45 | (a) Suppose for every $B\in\mcal{B}$ containing $x$, there is a $y\in 46 | B\cap E$. Let $O$ be an arbitrary open set containing $x$. By the definition 47 | of bases, there is a base set $B$ s.t. $x \in B \subset \mcal{B}$. Thus, 48 | $O\cap E \supset B\cap E \ne \varnothing$. Namely, $x \in \cl E$. The 49 | reversed direction follows immediately from the definition of cluster points. 50 | 51 | (b) If there is sequence in $E$ that converges to $x$, clear that $x\in 52 | \cl E$. Now we show the reverse. Let $\mcal{B}_x = (B_k)$ be a countable base 53 | at $x$ and $S_n = \bigcap_{k=1}^n B_k$. Choose $x_n \in S_n$. For every 54 | open set $O$ containing $x$, there exists a $B_m$ s.t. $x \in B_m 55 | \subset O$. Hence, for every $n > m$, $x_n \in S_n\subset S_m 56 | \subset S_m \subset O$. Thus, $x_n \to x$. 57 | 58 | (c) It follows immediately from (b). 59 | \end{proof} 60 | 61 | \paragraph{13.} 62 | \begin{proof} 63 | By the construction of $\mcal{B}$ and Prop. 5, $\mcal{B}$ is a base for some 64 | topology on $X$. Let $\mcal{T}$ be a topology containing $\mcal{C}$. Since 65 | $X \in \mcal{T}$ and $\mcal{T}$ is closed under finite intersection, 66 | $\mcal{T} \supset \mcal{B}$. Thus, $\mcal{T}$ contains the topology generated 67 | by $\mcal{B}$. Since the choice of $\mcal{T}$ is arbitrary, we conclude that 68 | $\mcal{B}$ is a base for the weakest topology containing $\mcal{C}$. 69 | \end{proof} 70 | 71 | 72 | \paragraph{16.} 73 | \begin{proof} 74 | Let $\mcal{B}$ be a countable base for the topology on $X$ and 75 | $\mcal{U}$ an open cover of $X$. For each $B\in\mcal{B}$, if there is 76 | some $U \in \mcal{U}$ with $B \subset U$, then pick that $U$. This yields 77 | a at most countable subset $\mcal{V}$ of $\mcal{U}$. Now we show that 78 | $\mcal{V}$ covers $X$. For every $x\in X$, since $\mcal{U}$ covers $X$, there 79 | is a $U\in\mcal{U}$ s.t. $x\in U$. Meanwhile, since $\mcal{B}$ is a base, 80 | there is a $B \in \mcal{B}$ s.t. $x\in B \subset U$. Hence, by our 81 | construction, there is a $U\hp \in \mcal{U}$ containing $x$ is picked. Thus, 82 | $x$ is covered by $\mcal{V}$. 83 | \end{proof} 84 | 85 | \subsection{The Separation Axioms and Continuous Real-Valued Functions} 86 | \paragraph{18.} 87 | \begin{proof} 88 | $\,$\par 89 | (a) If $x \ne y$. then $d(x, y) > 0$. It suffices to choose the open balls 90 | of radius $d(x, y) / 2$ centered at $x$ and $y$ respectively. 91 | 92 | (b) Let $O_1 = \{x\mid \rho(x, F_1) < \rho(x, F_2) \}$ and $O_2 = \{ 93 | x\mid \rho(x, F_1) > \rho(x, F_2)\}$. Clear that $O_1$ and $O_2$ are 94 | disjoint. Since $F_1$ and $F_2$ are closed and disjoint, $\rho(F_1, F_2) 95 | > 0$. Hence, for every $x\in F_1$, $\rho(x, F_1) = 0$ and $\rho(x, F_2) 96 | > 0$. Therefore, $F_1 \subset O_1$. Similarly, $F_2 \subset O_2$. Meanwhile, 97 | for every $x\in O_1$, the open ball centered at $x$ and of radius 98 | $(\rho(x, F_2) - \rho(x, F_1))/3$ is contained in $O_1$. Therefore, $O_1$ is 99 | open and similarly for $O_2$. Thus, $X$ is normal. 100 | \end{proof} 101 | 102 | \paragraph{20.} 103 | \begin{proof} 104 | If $f$ is continuous, then clear that $\{x\mid f(x) < a\} = 105 | f\inv((-\infty, a))$ and $\{x\mid f(x) > a\} = f\inv((a, \infty))$ are open. 106 | Now we show the reverse. For every open interval $(a, b)$, $f\inv((a, b)) 107 | = f\inv((a, \infty))\cap f\inv((-\infty, a))$ is open. Meanwhile, every open 108 | set $O$ in $\mathbb{R}$ is a union $\bigcup I_\alpha$of open intervals. Thus, 109 | $f\inv(O) = \bigcup f\inv(I_\alpha)$ is open. Namely, $f$ is continuous. 110 | The second results follows immediately from $\{f \ge a\}$ is closed iff 111 | $\{f < a\}$ is open. 112 | \end{proof} 113 | 114 | \paragraph{23.} 115 | It seems that in (a), the Hausdorff condition is not necessary. 116 | \begin{proof} 117 | $\,$\par 118 | (a) First, suppose that $X$ is normal. Let $G = O^c$. $F$ and $G$ 119 | are disjoint closed set and, therefore, there are disjoint open sets $U$ and 120 | $V$ s.t. $F\subset U$ and $G\subset V$. To show $\cl U\subset O$, note that 121 | $O = G^c \supset V^c \supset U$, since $U$ and $V$ are disjoint. Since $V^c$ 122 | is closed, $V^c \supset \cl U$. Thus, $F\subset U$ and $\cl U \subset O$. 123 | 124 | For the reverse, let $F_1$ and $F_2$ be two disjoint closed sets. Then 125 | $F_1^c$ is an open set containing $F_2$ and therefore there exists an open 126 | $U_2$ s.t. $F_2 \subset U_2$ and $\cl U_2\subset F_1^c$. Note that $\cl U_2$ 127 | and $F_1$ are again disjoint closed sets. Hence, similarly, we can find an 128 | open $U_1$ s.t. $F_1\subset U_1$ and $\cl U_1 \subset (\cl U_2)^c$. Thus, 129 | $X$ is normal. 130 | 131 | (b) We index the sequence by $n$ instead of $r$. Suppose that $N$ is the 132 | smallest integer s.t. $r = p2^{-N} < 1$. By (a), we may find a open set 133 | $U_{N}$ s.t. $F\subset U_N$ and $\cl U_N\subset O$. Now, $U_N$ is again an 134 | open set containing $F$ and therefore we can find $U_{N+1}$ s.t. 135 | $F\subset U_{N+1}$ and $\cl U_{N+1} \subset U_N$. Proceed iteratively and 136 | we get the required sequence. 137 | 138 | (c) Clear that $0 \le f \le 1$, $\equiv 0$ on $F$ and $f\equiv 1$ on $O^c$. 139 | Hence, it suffices to show the continuity. For every $x \in X$ and $\vep> 0$, 140 | choose $r_1, r_2 \in \mathbb{Q}$ such that 141 | \[ 142 | f(x) - \vep < r_1 < f(x) < r_2 < f(x) + \vep. 143 | \] 144 | Let $U = U_{r_2} \setminus \cl U_{r_1}$. Clear that $U$ is open. Meanwhile, 145 | \[ 146 | f(x) < r_2 \quad\Rightarrow\quad 147 | \inf\{r\mid x \in U_r\} < r_2 \quad\Rightarrow\quad 148 | \exists r < r_2 \text{ s.t. } x \in U_r. 149 | \] 150 | Hence, $x \in U_{r_2}$. If $x \in \cl U_{r_1}$, then $x \in U_r$ for all 151 | $r > r_1$ since $U_r \supset \cl U_{r_1}$ and, therefore, $f(x) \le r_1$. 152 | Contradiction. Thus, $x \notin U_{r_1}$. Hence, $U$ is an open neighborhood 153 | of $x$. For every $y \in U$, clear that $f(x) < r_2$. Also, $y \notin 154 | \cl U_{r_1}$ implies that $f(y) \ge r_1$. Hence, $f(U)\subset (f(x) - \vep, 155 | f(x) + \vep)$. Thus, $f$ is continuous. 156 | 157 | (d) If $X$ is normal, then clear that the function described in (c) satisfies 158 | the requirements. For the reverse, let $O_1 = \{x\mid f(x) < 1/2\}$ and 159 | $O_2 = \{x \mid f(x) > 1/2\}$. Clear that $O_1$ and $O_2$ are disjoint and 160 | since $f$ is continuous, they are open. Meanwhile, as $f\equiv 0$ on $A$ 161 | and $f\equiv 1$ on $B$, $O_1$ contains $A$ and $O_2$ contains $B$. Thus, 162 | $X$ is normal. 163 | \end{proof} 164 | 165 | \paragraph{24.} 166 | The function in (f) should be $g = \varphi k / (1 - |\varphi k|)$. 167 | \begin{proof} 168 | $\,$\par 169 | (a) Obvious. 170 | 171 | (b) Clear that $B$ and $C$ are disjoint. Since $f$ is continuous, so is 172 | $h$ and therefore $B$ and $C$ are closed. By Urysohn's lemma, there exists 173 | continuous $0 \le h_{1, B}, h_{1, C} \le 1$ s.t. $h_{1, B} \equiv 1$ (resp. 174 | $h_{1, C} \equiv 1$) on $B$ (resp. $C$) and vanishes on $C$ (resp. $B$). Let 175 | $h_1 = (-h_{1, B} + h_{1, C}) / 3$. Then $h_1 \equiv -1/3$ on $B$ and 176 | $h_1 \equiv 1/3$ on $C$. Meanwhile, $h_1$ is continuous and $|h_1| \le 1/3$. 177 | Let $x \in A$. If $h(x) \le -1/3$, then $|h(x) - h_1(x)| = |h(x) + 1/3| < 178 | 2/3$. Similarly for $x$ with $h(x) \ge 1/3$. If $-1/3 < h(x) < 1/3$, then 179 | $|h(x) - h_1(x)| < |h(x)| + |h_1(x)| \le 2/3$. Thus, $|h - h_1| < 2/3$. 180 | 181 | (c) Suppose that we have constructed $h_n$. Let $s_n = \sum_{i=1}^n h_i$. Let 182 | \[ 183 | B_n = \{x \mid h(x) - s(x) < -2^n/3^{n+1}\} 184 | \quad\text{and}\quad 185 | C_n = \{x \mid h(x) - s(x) > -2^n/3^{n+1}\}. 186 | \] 187 | The previous argument, \textit{mutatis mutandis}, yields a continuous 188 | function $h_{n+1}$ s.t. $|h_{n+1}| < 2^n/3^{n+1}$ and $|h - s_n - h_{n+1}| 189 | = |h - s_{n+1}| < 2^{n+1} / 3^{n+1}$ for all $x \in A$. 190 | 191 | (d) By the Weierstrass $M$-test, $h_n$ is uniformly summable. Hence, 192 | $k = \sum_{n=1}^\infty h_n$ is continuous as each $h_n$ is. Clear that $|k| 193 | \le 1$. Moreover, by the estimation in (c), $h = k$ on $A$. 194 | 195 | (e) By Urysohn's lemma, there is a continuous function $\varphi$ on $X$ s.t. 196 | $\varphi \equiv 1$ on $A$ and $\varphi \equiv 0$ on $\{x\mid k(x) = 1\}$. 197 | 198 | (f) Let $g = \varphi k / (1 - |\varphi k|)$. By (e), $g$ is well-defined on 199 | entire $X$. Also, $g$ is continuous and for $x \in A$ 200 | \[ 201 | g(x) = \frac{\varphi(x) k(x)}{1 - |\varphi(x) k (x)|} 202 | = \frac{1\times h(x)}{1 - |1\times h(x)|} 203 | = \frac{\frac{f}{1 + |f|}}{1 - \frac{|f|}{1 + |f|}} 204 | = f. 205 | \] 206 | \end{proof} 207 | 208 | \paragraph{26.} 209 | \begin{proof} 210 | Let $\mcal{J}$ be the topology generated by $\mcal{F}$. Since every 211 | $f \in \mcal{F}$ is continuous, $\mcal{J} \subset \mcal{T}$. Let $O \in 212 | \mcal{T}$. For every $x \in O$, there is a continuous function $f\in\mcal{F}$ 213 | s.t. $f(x) = 1$ and vanishes on $O^c$. Namely, 214 | \[ 215 | U_x := f\inv (\mathbb{R}\setminus\{0\}) \in \mcal{J} 216 | \] 217 | is an open set with $x \in U_x \subset O$. Clear that $O = \bigcup_x U_x$. 218 | Thus, $O \in \mcal{J}$. Namely, $\mcal{T} = \mcal{J}$. 219 | \end{proof} 220 | 221 | 222 | \subsection{Connectedness} 223 | \paragraph{32.} 224 | \begin{proof} 225 | Assume that $G$ is not connected and let $(O_1, O_2)$ be a separation of $G$. 226 | Since each $G_\alpha$ is connected, $G_\alpha$ is contained by exactly 227 | one of $O_1$ and $O_2$ since otherwise $(G_\alpha\cap O_1, G_\alpha\cap O_2)$ 228 | would be a separation of $G_\alpha$. Thus, there are two of $\{G_\alpha\}$ 229 | s.t. one is contained by $O_1$ and the other contained by $O_2$ and hence 230 | they have no point in common. Contradiction. 231 | \end{proof} 232 | 233 | \paragraph{33.} 234 | \begin{proof} 235 | Assume that $B$ is not connected and let $(O_1, O_2)$ be a separation of $B$. 236 | Since $A$ is connected, either $A\cap O_1$ or $A\cap O_2$ is empty. Assume 237 | without loss of generality that $A\cap O_2 = \varnothing$. Then, $O_2 238 | \subset \partial A$ and therefore has empty interior. Contradiction. 239 | \end{proof} 240 | 241 | \paragraph{35.} 242 | \begin{proof} 243 | $\,$\par 244 | (a) If $X$ is not connected, then we choose two points $x, y$ from each set 245 | of a separation. Clear that they can not be connected by some arc since the 246 | image of $[0, 1]$ under a continuous map is still connected. 247 | 248 | (b) Assume that $X$ is not connected and let $(O_1, O_2)$ be a separation. 249 | Since each of $X_1=\{(x, y) \mid x = 0, -1\le y \le 1\}$ and $X_2\{(x, y) 250 | \mid y = \sin x, 0 < x \le 1\}$ is connected, they are contained in, say, 251 | $O_1$ and $O_2$ respectively. Clear that any neighborhood of $(0, 0)$ 252 | contains points in $X_2$ and therefore $O_1\cap O_2 \ne \varnothing$. 253 | Contradiction. 254 | 255 | Now we show that $X$ is not arcwise connected. (TODO) 256 | 257 | (c) Let $x \in G$ and $H$ be the points of $G$ that can be connected to $x$ 258 | by a polygonal arc. For every $y \in H$, since $G$ is open, there is an open 259 | ball $B$ centered at $y$ with $B \subset G$. Clear that we can connect every 260 | $z \in B$ by the arc connecting $x$ and $y$, and the segment from $y$ to $z$. 261 | Thus, $H$ is open. Now, we show that $K = G\setminus H$ is open. Let $y$ be 262 | a point in $K$ and $B$ a small open ball centered at $y$ that is contained in 263 | $G$. Assume, to obtain a contradiction, that $B \cap H \ne \varnothing$. Then 264 | the previous argument, \textit{mutatis mutandis}, show that $y$ can be 265 | connected to $x$. Contradiction. Thus, $H$ is both open and closed in $G$. 266 | Since $G$ is connected, $H = G$. Thus, $G$ is arcwise connected. 267 | \end{proof} 268 | 269 | 270 | 271 | 272 | 273 | 274 | 275 | 276 | 277 | 278 | 279 | 280 | 281 | 282 | 283 | 284 | -------------------------------------------------------------------------------- /real_analysis_3rd/ch9_compact_and_locally_compact_spaces.tex: -------------------------------------------------------------------------------- 1 | \section{Compact and Locally Compact Spaces} 2 | \subsection{Compact Spaces} 3 | 4 | \paragraph{2.} 5 | I further assume that $X$ is Hausdorff. 6 | \begin{proof} 7 | Assume, to obtain a contradiction, that $F_n = K_n\setminus O \ne\varnothing$ 8 | for every $n$. Since $F_n \supset F_{n+1}$, $\{F_n\}$ is collection of closed 9 | subsets of compact set $K_1$ with finite intersection property. Hence, 10 | $\bigcap F_n$ is nonempty, contradicting with $\bigcap K_n \subset O$. 11 | \end{proof} 12 | 13 | \paragraph{3.} 14 | \begin{proof} 15 | Let $F$ be a closed set and $x \notin F$. Since $X$ is Hausdorff, for every 16 | $y \in F$, there are two disjoint open sets $U_y$ and $O_y$ s.t. $y \in U_y$ 17 | and $x \in O_y$. Since $X$ is compact, so is $F$. Note that $\{U_y\}$ is an 18 | open cover for $F$. Hence, it has a finite subcover $\{U_{y_i}\}_{i=1}^n$. 19 | Let $U = \bigcup U_{y_i}$ and $O = \bigcap O_{y_i}$. Clear that they are 20 | disjoint open sets s.t. $F \subset U$ and $x \in O$. 21 | \end{proof} 22 | 23 | \paragraph{6.} 24 | \begin{proof} 25 | Let $\vep > 0$ be fixed. Since $\mcal{F}$ is equicontinuous, for every 26 | $x \in X$, there is a neighborhood $O_x$ s.t. for every $x\hp \in X$, 27 | $\sigma(f(x), f(x\hp)) < \vep$. Clear that $\{O_x\}$ is an open cover and 28 | since $X$ is compact, it has a finite subcover $\{O_{x_i}\}_{i=1}^m$. 29 | 30 | For $x_i$, since $f_n(x_i) \to f(x_i)$, there is an integer $N_i$ s.t. for 31 | every $n > N$, $\sigma(f_n(x_i), f(x_i)) < \vep$. Since $\sigma(f_n(x_i), 32 | f_n(x)) < \vep$ holds for all $n$, $\sigma(f(x_i), f(x)) \le \vep$. 33 | Hence, for every $x \in O_{x_i}$ and $n > N$, 34 | \[ 35 | \sigma(f_n(x), f(x)) 36 | \le \sigma(f_n(x), f_n(x_i)) + \sigma(f_n(x_i), f(x_i)) 37 | + \sigma(f(x_i), f(x)) 38 | < 3\vep. 39 | \] 40 | Let $N = \max N_{x_i}$ and we get the desired result. 41 | \end{proof} 42 | 43 | \subsection{Countable Compactness and the Bolzano-Weierstrass Property} 44 | \paragraph{9.} 45 | \begin{proof} 46 | $\,$\par 47 | (a) It follows immediately from the definition and Problem 8.20. 48 | 49 | (b) For every $\alpha \in \mathbb{R}$, 50 | \[ 51 | f + g < \alpha 52 | \quad\text{iff}\quad 53 | f < \alpha - g 54 | \quad\text{iff}\quad 55 | \exists q\in \mathbb{Q} \text{ s.t. } f < q,\, q < \alpha - g. 56 | \] 57 | Hence, 58 | \[ 59 | \{f + g < \alpha\} 60 | = \bigcup_{q \in \mathbb{Q}} \{f < q\} \cap \{g < \alpha - q\}, 61 | \] 62 | which is open. Thus, $f + g$ is also upper semicontinuous. 63 | 64 | (c) Since $(f_n)$ is a decreasing sequence, we can write $f(x) = \inf_n 65 | f_n(x)$. Hence, for every $\alpha \in \mathbb{R}$, $f < \alpha$ iff there 66 | exists some $n$ s.t. $f_n(x) < \alpha$. Hence, 67 | \[ 68 | \{f < \alpha\} = \bigcup_n \{f_n < \alpha\}, 69 | \] 70 | which is open. Thus, $f$ is also upper semicontinuous. 71 | 72 | (d) Note that $(f_n - f)$ is a decreasing sequence of upper semicontinuous 73 | functions that converges to $0$. Hence, by Dini's theorem, the convergence is 74 | uniform. 75 | 76 | (e) Suppose that $x \in \{f < \alpha\}$. Let $\vep$ be a positive real 77 | number. Since $f_n \to f$ uniformly, there is an integer $n$ s.t. 78 | $|f(y) - f_n(y)| < \vep$ for all $y \in X$. Meanwhile, since $f_n$ is upper 79 | semicontinuous, there is a $\delta > 0$ s.t. for every $y$ in the 80 | $\delta$-ball $B$ centered at $x$, $f_n(y) < f_n(x) + \vep$. Hence, for every 81 | $y \in B$, 82 | \begin{align*} 83 | f(y) = f(y) - f_n(y) + f_n(y) - f_n(x) + f_n(x) - f(x) + f(x) 84 | \le 3\vep + f(x). 85 | \end{align*} 86 | Thus, for sufficiently small $\vep > 0$, we have $B \subset \{f < \alpha\}$. 87 | Namely, $\{f < \alpha\}$ is open whence $f$ is upper semicontinuous. 88 | \end{proof} 89 | 90 | \paragraph{10.} 91 | \begin{proof} 92 | $\,$\par 93 | (i. $\Rightarrow$ iii.) Let $f$ be a bounded continuous real-valued function 94 | and $M := \sup f < \infty$. Let $F_n = \{f \ge M - 1/n\}$. Since $f$ is 95 | continuous, $F_n$ is closed. Note that $(F_n)$ is a countable family of 96 | closed sets with finite intersection property. Hence, $\bigcap F_n = 97 | \{f \ge M\}$ is nonempty as $X$ is countably compact. Namely, the maximum 98 | can be attained. 99 | 100 | (iii. $\Rightarrow$ ii.) Let $f$ be a continuous function and assume, to 101 | obtain a contradiction, that $f$ is unbounded. Then the function 102 | $-1/(|f| + 1)$ is a continuous bounded function whose maximum can not be 103 | attained. Contradiction. 104 | 105 | (ii. $\Rightarrow$ i.) Assume, to obtain a contradiction, that $X$ does not 106 | have the Bolzano-Weierstrass property, that is, there is a sequence $(x_n)$ 107 | in $X$ that has no cluster point. Then $F := \{x_n\}_{n=1}^\infty$ is closed. 108 | Define $f: F \to \mathbb{R}$ by $f(x_n) = n$. Note that $f$ is continuous on 109 | $F$ and by Tietze's extension theorem, it can be continuously extended to 110 | $X$. However, $f$ is unbounded, contradicting (ii.). Thus $X$ has the 111 | Bolzano-Weierstrass and, therefore, is countably compact. 112 | \end{proof} 113 | 114 | \subsection{Products of Compact Spaces} 115 | \paragraph{13.} 116 | \begin{proof} 117 | Let $E$ be a closed and bounded set in $\mathbb{R}^n$. Then it is contained 118 | in some closed cube $K = \prod_{i=1}^n[a_i, b_i]$. By Tychonoff's theorem, 119 | $K$ is compact. Thus, $K$, a closed subset of a compact set, is also compact. 120 | \end{proof} 121 | 122 | \paragraph{15.} 123 | \begin{proof} 124 | Let $X = \prod_{n=1}^\infty$ be the product of sequentially compact spaces 125 | $(X_n)$. Let $(x_n)$ be a sequence in $X$. Since $X_1$ is sequentially 126 | compact, we may choose a subsequence $x_n^1$ of $x_n$ s.t. the first 127 | coordinate of $(x_n^1)$ converges to some $x^1$. Similarly, from $(x_n^1)$ we 128 | may choose a subsequence $x_n^2$ whose second coordinate converges to some 129 | $x^2$. Proceed inductively and we get a sequence of sequence. Finally, 130 | consider the sequence $(x_n^n)$. Since each coordinate converges and we are 131 | dealing with the product topology, $x_n^n$ converges to $(x_1, x_2, \dots)$. 132 | \end{proof} 133 | 134 | 135 | \subsection{Locally Compact Spaces} 136 | \paragraph{18.} 137 | \begin{proof} 138 | For every $x \in K$, since $X$ is locally compact, there is an open set 139 | $O_x$ with $\cl O_x$ compact. Note that $\{O_x\}_{x\in K}$ is an open cover 140 | for the compact set $K$. Hence, it has a finite subcover 141 | $\{O_{x_i}\}_{i=1}^n$. Then, $O = \bigcup_{i=1}^n O_{x_i}$ is an open set 142 | containing $K$ whose closure is compact. 143 | \end{proof} 144 | 145 | \paragraph{19.} 146 | \begin{proof} 147 | $\,$\par 148 | (a) Since $X$ is a locally compact Hausdorff space and $K$ is compact, there 149 | exists an open set $V$ with compact closure s.t. $V\supset K$. Since 150 | $\cl V$ is compact, it is normal. Therefore, by Urysohn's lemma, there is 151 | a continuous function $f: \cl V \to [0, 1]$ s.t. $f\equiv 1$ on $K$ and 152 | $f\equiv 0$ on $\partial V$. Extend $f$ to $X$ by setting $f \equiv 0$ 153 | outsides $\cl V$. Then $f$ is continuous and $f \equiv 1$ on $K$. Meanwhile, 154 | since $\supp f\subset \cl V$, it is also compact. 155 | \end{proof} 156 | 157 | \paragraph{24.} 158 | Assume that $X$ is also Hausdorff. 159 | \begin{proof} 160 | $\,$\par 161 | (a) Clear that if $F$ is closed, so is $F\cap K$ for each closed compact 162 | $K$. For the reverse, we show that $F^c$ is open. 163 | Let $x \notin F$. Since $X$ is locally compact, there is a neighborhood $U$ 164 | of $x$ whose closure is compact. If $F\cap\cl U = \varnothing$, then we are 165 | done. If $F\cap \cl U \ne \varnothing$, then by the hypothesis, it is closed. 166 | Therefore, $U\setminus(F\cap\cl U)$ is again an open neighborhood of $x$. 167 | Since $X$ is a locally compact Hausdorff space, we can find an open 168 | neighborhood $V$ with $\cl V \subset U\setminus(F\cap\cl U)$. In both cases, 169 | $F^c$ is open. 170 | 171 | (b) Suppose that for each closed compact $K$, $F\cap K$ is closed. For every 172 | $x \in \cl F$, since $X$ is first-countable, there exists a sequence $(x_n) 173 | \subset F$ which converges to $x$. Then $E := \{x\}\cup\{x_n\}_{n=1}^\infty$ 174 | is closed and compact. Thus, by the hypothesis, $F\cap E$ is also closed, 175 | which implies that $x \in F$. Hence, $F$ is closed. 176 | \end{proof} 177 | 178 | \paragraph{26.} 179 | Assume that $X$ is Hausdorff. 180 | \begin{proof} 181 | Let $x$ be an arbitrary point in $X$ and $V$ any neighborhood of $x$. Since 182 | $X$ is a locally compact Hausdorff space, we may choose a open neighborhood 183 | $U_1$ of $x$ whose closure is compact and contained by $V$. Since $O_1$ is 184 | dense, $U_1\cap O_1$ is a nonempty neighborhood of $x$. Then, choose a 185 | neighborhood $U_2$ of $x$ s.t. $\cl U$ is compact and $\cl U \subset 186 | U_1\cap O_1$. Proceed inductively and we get a sequence $(U_n)$ s.t. 187 | $\cl U_n$ is compact and $\cl U_{n+1} \subset O_n\cap U_n$. 188 | 189 | Since $(\cl U_n)$ is a nested sequence of compact sets, $\bigcap \cl U_n$ 190 | is nonempty. Choose $x_* \in \bigcap\cl U_n$. For every $O_n$, $x_* 191 | \in \cl U_{n+1} \subset O_n$. Thus, $x_* \in V\cap\bigcap_{n=1}^\infty O_n$. 192 | Namely, $\bigcap_{n=1}^\infty O_n$ is dense. 193 | \end{proof} 194 | 195 | \paragraph{29.} 196 | \begin{proof} 197 | $\,$\par 198 | (a) Let $F$ be a closed subset of a locally compact space $X$. For every 199 | $x\in F\subset X$, there is a neighborhood $U$ of $x$ whose closure is 200 | compact in $X$. Then, $U\cap F$ is also compact. Thus, $F$ is locally 201 | compact. 202 | 203 | (b) Let $O$ be an open subset of a locally compact Hausdorff space $X$. For 204 | every $x\in O \subset X$, there is a neighborhood $U$ of $x$ whose closure is 205 | compact in $X$ and contained by $O$. Note that $\cl U$ is also compact in 206 | $O$. Thus, $O$ is locally compact. 207 | \end{proof} 208 | 209 | 210 | 211 | 212 | 213 | 214 | 215 | 216 | 217 | 218 | 219 | 220 | 221 | 222 | 223 | 224 | -------------------------------------------------------------------------------- /real_analysis_3rd/def.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt, a4paper]{article} 2 | 3 | \usepackage[margin=1in]{geometry} 4 | \usepackage{ 5 | color, 6 | clrscode, 7 | amssymb, 8 | amsmath, 9 | listings, 10 | xcolor, 11 | supertabular, 12 | multirow, 13 | mathtools, 14 | mathrsfs, 15 | amsthm, 16 | systeme, 17 | amsfonts 18 | } 19 | \definecolor{bgGray}{RGB}{36, 36, 36} 20 | \usepackage[ 21 | colorlinks, 22 | linkcolor=bgGray, 23 | anchorcolor=blue, 24 | citecolor=green 25 | ]{hyperref} 26 | 27 | \newcommand{\vep}{\varepsilon} 28 | 29 | \newenvironment{solution} 30 | {\begin{proof}[Solution]} 31 | {\end{proof}} 32 | \newcommand{\rd}{\mathrm{d}} 33 | \newcommand{\inv}{^{-1}} 34 | \newcommand{\hp}{^\prime} 35 | \newcommand{\mcal}{\mathcal} 36 | \newcommand{\ubar}[1]{\text{\b{$#1$}}} 37 | 38 | 39 | \DeclareMathOperator*\lowlim{\underline{lim}} 40 | \DeclareMathOperator*\uplim{\overline{lim}} 41 | \DeclareMathOperator*\esssup{ess\,sup} 42 | \DeclareMathOperator\sgn{sgn} 43 | \DeclareMathOperator{\cl}{cl} 44 | \DeclareMathOperator{\supp}{supp} 45 | \def\upint{\mathchoice 46 | {\mkern13mu\overline{\vphantom{\intop}\mkern7mu}\mkern-20mu}% 47 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 48 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 49 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 50 | \int} 51 | \def\lowint{\mkern3mu\underline{\vphantom{\intop}\mkern7mu}\mkern-10mu\int} 52 | 53 | -------------------------------------------------------------------------------- /real_analysis_3rd/main.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Engineev/solutions/4e33274fe1ed9e46fd0e6671c57cb589704939bd/real_analysis_3rd/main.pdf -------------------------------------------------------------------------------- /real_analysis_3rd/main.tex: -------------------------------------------------------------------------------- 1 | \input{def.tex} 2 | 3 | \title{Real Analysis} 4 | \author{Yunwei Ren} 5 | \date{} 6 | 7 | \begin{document} 8 | \maketitle 9 | \tableofcontents 10 | 11 | \newpage 12 | \setcounter{section}{2} 13 | \input{ch3_lebesgue_measure.tex} 14 | \newpage 15 | \input{ch4_the_lebesgue_integral.tex} 16 | \newpage 17 | \input{ch5_differentiation_and_integration.tex} 18 | \newpage 19 | \input{ch6_the_classical_banach_spaces.tex} 20 | \newpage 21 | \input{ch7_metric_spaces} 22 | \newpage 23 | \input{ch8_topological_spaces} 24 | \newpage 25 | \input{ch9_compact_and_locally_compact_spaces} 26 | \newpage 27 | \setcounter{section}{10} 28 | \input{ch11_measure_and_integration} 29 | \newpage 30 | \input{ch12_measure_and_outer_measure} 31 | 32 | 33 | \end{document} 34 | --------------------------------------------------------------------------------