├── chapter_1 ├── 1_8.tex ├── 1_5.tex ├── 1_6.tex ├── 1_4.tex ├── FA_chapter_1.tex ├── 1_4_comments.txt ├── 1_12.tex ├── 1_10.tex ├── 1_14.tex ├── 1_17.tex ├── 1_9.tex ├── 1_2.tex ├── 1_3.tex ├── 1_7.tex ├── 1_16.tex └── 1_1.tex ├── FA_DM.pdf ├── run_on_mac.sh ├── CONTRIBUTING.md ├── chapter_3 ├── FA_chapter_3.tex ├── 3_3.tex └── 3_11.tex ├── .github └── ISSUE_TEMPLATE │ └── mistakes-tracking.md ├── FA_mainmatter.tex ├── chapter_4 ├── FA_chapter_4.tex ├── 4_13.tex ├── 4_1.tex └── 4_15.tex ├── chapter_6 ├── FA_chapter_6.tex ├── 6_1.tex ├── 6_9.tex ├── 6_6.tex └── 6_1_Mat.tex ├── HOWTO.md ├── chapter_2 ├── 2_3 │ ├── 2_3_0_labels.txt │ ├── 2_3_0_lemma.tex │ ├── 2_3.tex │ ├── 2_3_4_example_2.tex │ ├── 2_3_2_uniform_bound.tex │ ├── 2_3_3_example_1.tex │ └── 2_3_1_radon_measures.tex ├── FA_chapter_2.tex ├── 2_10.tex ├── 2_12.tex ├── 2_16.tex ├── 2_15.tex ├── 2_9.tex └── 2_6.tex ├── bibliography.bib ├── LICENSE ├── .gitignore ├── README.md ├── FA_DM.tex └── notations.tex /chapter_1/1_8.tex: -------------------------------------------------------------------------------- 1 | % 2 | -------------------------------------------------------------------------------- /FA_DM.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/gitcordier/FunctionalAnalysis/HEAD/FA_DM.pdf -------------------------------------------------------------------------------- /run_on_mac.sh: -------------------------------------------------------------------------------- 1 | file_name="FA_DM" 2 | pdf_reader="skim" 3 | 4 | xelatex $file_name 5 | open -a $pdf_reader "${file_name}.pdf" 6 | 7 | -------------------------------------------------------------------------------- /CONTRIBUTING.md: -------------------------------------------------------------------------------- 1 | If you find a mistake, please contact me. This encompasses logical mistakes, of course, but also English grammar mistakes or simply mispellings. Beyond this, any suggestion is welcome:) 2 | -------------------------------------------------------------------------------- /chapter_3/FA_chapter_3.tex: -------------------------------------------------------------------------------- 1 | % 2 | \setcounter{section}{2} 3 | % 4 | \section{Exercise 3. } 5 | \input{\ROOT/chapter_3/3_3.tex} 6 | \newpage 7 | % 8 | \setcounter{section}{10} 9 | % 10 | \section{Exercise 11. Meagerness of the polar} 11 | \input{\ROOT/chapter_3/3_11.tex} 12 | -------------------------------------------------------------------------------- /.github/ISSUE_TEMPLATE/mistakes-tracking.md: -------------------------------------------------------------------------------- 1 | --- 2 | name: Mistakes Tracking 3 | about: Tracking mistakes 4 | title: LOGIC 5 | labels: '' 6 | assignees: '' 7 | 8 | --- 9 | 10 | If you find a mistake, please contact me. This encompasses logical mistakes, of course, but also English grammar mistakes or simply mispellings. Beyond this, any suggestion is welcome:) 11 | -------------------------------------------------------------------------------- /FA_mainmatter.tex: -------------------------------------------------------------------------------- 1 | \setcounter{chapter}{0} 2 | \chapter{Topological Vector Spaces} 3 | \input{\ROOT/chapter_1/FA_chapter_1.tex} 4 | \chapter{Completeness} 5 | \input{\ROOT/chapter_2/FA_chapter_2.tex} 6 | \chapter{Convexity} 7 | \input{\ROOT/chapter_3/FA_chapter_3.tex} 8 | \chapter{Banach Spaces} 9 | \input{\ROOT/chapter_4/FA_chapter_4.tex} 10 | % 11 | \setcounter{chapter}{5} 12 | % 13 | \chapter{Distributions} 14 | \input{\ROOT/chapter_6/FA_chapter_6.tex} -------------------------------------------------------------------------------- /chapter_4/FA_chapter_4.tex: -------------------------------------------------------------------------------- 1 | % 2 | Throughout this set of exercises, $X$ and $Y$ denote Banach spaces, unless the contrary is explicitly stated. 3 | \section{Exercise 1. Basic results} 4 | \input{\ROOT/chapter_4/4_1.tex} 5 | \newpage 6 | % 7 | \setcounter{section}{12} 8 | % 9 | 10 | \section{Exercise 13. Operator compactness in a Hilbert space} 11 | \input{\ROOT/chapter_4/4_13.tex} 12 | \newpage 13 | % 14 | \setcounter{section}{14} 15 | % 16 | \section{Exercise 15. Hilbert-Schmidt operators} 17 | \input{\ROOT/chapter_4/4_15.tex} 18 | \newpage 19 | -------------------------------------------------------------------------------- /chapter_6/FA_chapter_6.tex: -------------------------------------------------------------------------------- 1 | % 2 | \section{Exercise 1. Test functions are almost polynomial} 3 | \input{\ROOT/chapter_6/6_1.tex} 4 | \input{\ROOT/chapter_6/6_1_Mat.tex} 5 | \setcounter{section}{5} 6 | \section{Exercise 6. Around the supports of some distributions} 7 | \input{\ROOT/chapter_6/6_6.tex} 8 | \setcounter{section}{8} 9 | \section{Exercise 9. Convergence in $\mathscr{D}(\Omega)\,$ vs. convergence in $\mathscr{D}'(\Omega)\,$} 10 | \input{\ROOT/chapter_6/6_9.tex} 11 | %\setcounter{section}{16} 12 | %\section{Exercise 17. } 13 | %\input{\ROOT/chapter_6/6_17.tex} 14 | 15 | -------------------------------------------------------------------------------- /HOWTO.md: -------------------------------------------------------------------------------- 1 | # Compiling with bibliography 2 | 3 | xelatex FA_DM 4 | bibtex FA_DM 5 | xelatec FA_DM 6 | xelatex FA_DM 7 | 8 | # Simple compilation 9 | Simply run 10 | 11 | xelatex FA_DM 12 | 13 | Mac users may want to run 14 | 15 | xelatex FA_DM 16 | open -a skim FA_DM.pdf" 17 | 18 | Or, alternatively, the run_on_mac.sh script 19 | 20 | # HTML output 21 | run 22 | 23 | hevea FA_DM 24 | hevea FA_DM 25 | # Second run is optional: We only need it to get/update cross-references 26 | 27 | You will to have to comment/uncomment in the FA_DM.tex before, 28 | see lines 22, …, 25 in FA_DM.tex. Hint: grep -r "HEVEA" . *.tex 29 | 30 | If it fails, then run 31 | 32 | rm *.aux 33 | 34 | before you try again. 35 | 36 | 37 | -------------------------------------------------------------------------------- /chapter_1/1_5.tex: -------------------------------------------------------------------------------- 1 | \textit{% 2 | Consider the definition of ``bounded set'' given in Section 1.6. % 3 | Would the content of this definition be altered if it were required merely % 4 | required that to every neighbourhood $V$ of 0 corresponds }some{ % 5 | \textit{% 6 | $t>0$ such that $E\subset tV$? 7 | } 8 | \begin{proof} 9 | The answer is: No. To prove it, start from (a) of Section 1.14: % 10 | $V$ contains $W$, a balanced neighbourhood of $0$. % 11 | % 12 | Assume that $E$ is bounded in this weaker sense\ie % 13 | there exists a positive $t$ that satisfies % 14 | % 15 | \begin{align}% 16 | E\subset tW. 17 | \end{align} 18 | % 19 | Thus, 20 | % 21 | \begin{align} 22 | E\subset tW \subset sW \subset sV \quad (s>t), 23 | \end{align} 24 | % 25 | since $W$ is balanced. We so reach the definition given in Section 1.6: % 26 | The two ones are equivalent. 27 | \end{proof} 28 | % END 29 | -------------------------------------------------------------------------------- /chapter_1/1_6.tex: -------------------------------------------------------------------------------- 1 | \textit{ 2 | Prove that a set $E$ in a topological vector space is bounded if and only if % 3 | every countable subset of E is bounded. 4 | } 5 | \begin{proof} 6 | It is clear that every subset of a bounded set is bounded. % 7 | Conversely, assume that $E$ is not bounded then pick $V$ % 8 | a neighbourhood of the origin: % 9 | % 10 | No counting number $n=1, 2, \dots$, verifies % 11 | % 12 | $E\subset nV$ (see Exercise 1 in Chapter 1). % 13 | % 14 | In other words, there exists a sequence % 15 | % 16 | $\{x_1, \dots, x_n, \dots\} \subset E$ % 17 | % 18 | such that % 19 | % 20 | \begin{align} 21 | x_n \notin nV. 22 | \end{align} 23 | % 24 | As a consequence, $x_n /n $ fails to converge to $0$ % 25 | as $n$ tends to $\infty $. % 26 | In contrast, $1/n$ succeeds. % 27 | It then follows from Section 1.30 that % 28 | % 29 | $\{x_1, \dots, x_n, \dots\}$ % 30 | % 31 | is not bounded. So ends the proof. 32 | \end{proof} 33 | % END 34 | -------------------------------------------------------------------------------- /chapter_2/2_3/2_3_0_labels.txt: -------------------------------------------------------------------------------- 1 | 2_03_0 2 | \label{2.3 Lemma} 3 | 4 | 2_03_1 5 | Lemma 6 | \label{2.3. Mean value inequality.} 7 | \label{2.3. Mean Value (1).} 8 | \label{2.3. Mean Value (2).} 9 | 10 | 2_03_2 11 | \label{2.3 g bounded operator (1).} 12 | \label{2.3 g bounded operator (2).} 13 | \label{2.3. Convergence in D_K, not vague.} 14 | \label{2.3. Norm in the dual equals norm L1 (1).} 15 | \label{2.3. Norm in the dual equals norm L1 (2).} 16 | \label{2.3. Optimal bound.} 17 | \label{2.3. Uniform bound for the supremum norm.} 18 | 19 | 2_03_3 20 | \label{2.3. Mean value inequality (concrete).} 21 | \label{2.3. Equality from integration by parts.} 22 | \label{2.3. Convergence on D_K (1).} 23 | \label{2.3. Bound with p and M (theoritical).} 24 | 25 | 2_03_4 26 | \label{2.3. Equality from integration by parts.} 27 | \label{2.3. Bound with p and M (M_n).} 28 | \label{2.3. Bound with p and M (concrete).} 29 | \label{2.3. (f_n) converges to Dirac (1).} 30 | -------------------------------------------------------------------------------- /chapter_2/FA_chapter_2.tex: -------------------------------------------------------------------------------- 1 | % 2 | \setcounter{section}{2} 3 | % 4 | 5 | \section{Exercise 3. An equicontinous sequence of measures} 6 | \input{\ROOT/chapter_2/2_3/2_3.tex} 7 | \newpage 8 | % 9 | \setcounter{section}{5} 10 | % 11 | \section{Exercise 6. Fourier series may diverge at $0$} 12 | \input{\ROOT/chapter_2/2_6.tex} 13 | \newpage 14 | % 15 | \setcounter{section}{8} 16 | % 17 | \section{Exercise 9. Boundedness without closedness} 18 | \input{\ROOT/chapter_2/2_9.tex} 19 | \newpage 20 | % 21 | \section{Exercise 10. Continuousness of bilinear mappings} 22 | \input{\ROOT/chapter_2/2_10.tex} 23 | \newpage 24 | % 25 | \setcounter{section}{11} 26 | % 27 | \section{Exercise 12. A bilinear mapping that is not continuous} 28 | \input{\ROOT/chapter_2/2_12.tex} 29 | \newpage 30 | % 31 | \setcounter{section}{14} 32 | % 33 | \section{Exercise 15. Baire's cut} 34 | \input{\ROOT/chapter_2/2_15.tex} 35 | \newpage 36 | % 37 | \section{Exercise 16. An elementary closed graph theorem} 38 | \input{\ROOT/chapter_2/2_16.tex} -------------------------------------------------------------------------------- /bibliography.bib: -------------------------------------------------------------------------------- 1 | @book{BabyRudin, 2 | Author = {Rudin, Walter}, 3 | Keywords = {Analysis, Rudin}, 4 | Publisher = {McGraw-Hill}, 5 | Title = {Principles of Mathematical Analysis}, 6 | Year = {1976}} 7 | @book{BigRudin, 8 | Author = {Rudin, Walter}, 9 | Keywords = {Analysis, Rudin}, 10 | Publisher = {McGraw-Hill}, 11 | Title = {Real and Complex Analysis}, 12 | Year = {1986}} 13 | @book{FA, 14 | Author = {Rudin, Walter}, 15 | Keywords = {Analysis, Rudin}, 16 | Publisher = {McGraw-Hill}, 17 | Title = {Functional Analysis}, 18 | Year = {1991}} 19 | @book{AnalyseIII, 20 | Author = {Schwartz, Laurent}, 21 | Keywords = {analysis, integral calculus, Schwartz}, 22 | Publisher = {Hermann}, 23 | Title = {Analyse}, 24 | Volume = {III (in French)}, 25 | Year = {1997}, 26 | Language={French}} 27 | @book{SpecifyingSystems, 28 | Author = {Lamport, Leslie}, 29 | Keywords = {TLA, formal methods, logic}, 30 | Publisher = {Addison-Wesley}, 31 | Title = {Specifying Systems, The TLA+ Language and Tools for Hardware and Software Engineers}, 32 | Year = {2002}, 33 | Language={English}} 34 | -------------------------------------------------------------------------------- /LICENSE: -------------------------------------------------------------------------------- 1 | MIT License 2 | 3 | Copyright (c) 2020 gitcordier 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 | -------------------------------------------------------------------------------- /chapter_1/1_4.tex: -------------------------------------------------------------------------------- 1 | \textit{Let be % 2 | % 3 | $B = \{(z_{1}, z_{2}) \in \C^2: |z_{1}| \leq |z_{2}| \}$. % 4 | % 5 | Show that $B$ is balanced but that its interior is not. 6 | } 7 | % 8 | \begin{proof} 9 | It is obvious that the nonempty set $B$ contains the origin $(0,0)$. % 10 | Additionally, its interior $B^\circ$ is nonempty as well. % 11 | Indeed, the following set % 12 | % 13 | \begin{align} 14 | \{(z_1, z_2) \in \C^2: |1- z_1| + |2- z_2] < 1/2 \} \subset B % 15 | \end{align} 16 | % 17 | is a neighborhood of $(1, 2) \in B$. % 18 | Moreover, $B$ is balanced, since 19 | \begin{align} 20 | |\alpha z_1| = |\alpha| |z_1| \leq |\alpha| |z_2| = |\alpha z_2| % 21 | \quad (|\alpha| \leq 1) 22 | \end{align} 23 | % 24 | for all $(z_1, z_2)$ in $B$. % 25 | % 26 | Nevertheless, the nonempty set $B^\circ$ is not balanced, what we now % 27 | establish by showing that $(0, 0) \notin B^\circ$. % 28 | % 29 | To do so, assume, to reach a contradiction, % 30 | that the origin has a neighborhood % 31 | % 32 | \begin{align} 33 | U \triangleq \{(z_1, z_2) \in \C^2: |z_1| + |z_2] < r \} \subset B % 34 | \end{align} 35 | % 36 | for some positive $r$. Clearly, $U$ contains $(r/2, 0)$, % 37 | and that special case $(r/2, 0) \in B$ now contradicts the definition of $B$. % 38 | So ends the proof. 39 | \end{proof} 40 | % END 41 | -------------------------------------------------------------------------------- /chapter_6/6_1.tex: -------------------------------------------------------------------------------- 1 | {\it Suppose $f$ is a complex continuous function in $\R^n$, with compact support. \!\!Prove that $\psi P_j\to \!f$ uniformly on $\R^n$, for some $\psi\in \mathscr{D}$ and for some sequence $\{P_j\,\}$ of polynomials.} 2 | \begin{proof} According to 1.16, $\Omega$ is union of a compact sets sequence $\{K_{i\,}\}$ and $\text{supp} (f\,)$ lies in some $K= K_{i\,}$so that $f\,$ is embedded in $\mathscr{D}(\Omega)\,$. We can apply [1.10] to ensure that $\Omega$ encloses a compact set $S=\overline{K +B(\epsilon)}$ for sufficiently small $\epsilon>0\,$.\\ 3 | \\ 4 | One easily checks that the Stone-Weierstraß theorem [5.7] can be applied with the subalgebra $ \{ g\in C(S\,) :\, g \text{ is polynomial}\,\}$ of $C(S\,)\,$. 5 | There so exists a sequence $\{P_j:\, j\in \N\}$ of $\R[X_1,\dotsc,\, X_n]\,$ such that 6 | \begin{align}\label{6_1_1} 7 | \sup_S \lvert\, f-P_{j\,} \rvert \underset{j\infty}{\longrightarrow} 0\quad. 8 | \end{align} 9 | By [6.20], the open set $K +B(\epsilon)$ has a local partition of unity $\{\psi_i\}\subset \mathscr{D}(\Omega)\,$. Moreover, there exists an integer $l$ such that $\psi=\psi_1+\dotsb+\psi_l\,$ equals $1$ on $K\,$. Hence 10 | \begin{align}\label{6_1_2} 11 | \| \, f - \psi P_{j\,}\|_\infty= \, \| \,\psi f - \psi P_{j\,}\|_\infty 12 | = & \, \sup_S \lvert \, \psi f -\! \psi P_{j\,}\rvert \\ 13 | = & \, \sup_S \lvert \, f - \, P_{j\,} \rvert \, \overset{(\ref{6_1_1})}{\underset{j\infty}{\longrightarrow}} 0\quad . 14 | \end{align} 15 | \end{proof} 16 | -------------------------------------------------------------------------------- /chapter_2/2_10.tex: -------------------------------------------------------------------------------- 1 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 2 | % FunctionalAnalysis 3 | % 2_10.tex 4 | % 5 | % encoding: UTF-8 6 | % EOL: LF 7 | % 8 | % format: LaTeX 9 | % indent: spaces (2) 10 | % width: 127 11 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 12 | \textit{% 13 | Prove that a bilinear mapping is continuous if it is continuous at the origin $(0, 0)$. 14 | } 15 | % 16 | \begin{proof} 17 | Let $B: X_1 \times X_2 \to Z$ be a bilinear mapping that is continuous at $(0,0)$, where $X_i$ and $Z$ are topological vector % 18 | spaces. This implies that, for any balanced open set $W$, $X_i$ contains a balanced open set $U_i$ such that % 19 | % 20 | \begin{align} 21 | B(U_1 \times U_2) \subset W. 22 | \end{align} 23 | % 24 | Pick $a_i$ from $ X_i$, and then choose $r_i > 1/2$ large enough so that $a_i \in r_i U_i$. The vector $b_i - a_i$ is now % 25 | restricted to $(r_1 + r_2)^{\minus 1} U_i$. Hence % 26 | % 27 | \begin{align} 28 | B(b_1, b_2) - B(a_1, a_2) & = B(b_1 - a_1, b_2) + B(a_1, b_2) - B(a_1, a_2)\\ 29 | & = B(b_1 - a_1, b_2) + B(a_1, b_2 - a_2)\\ 30 | & = B(b_1 - a_1, b_2 - a_2) + B(b_1 - a_1, a_2) + B(a_1, b_2 - a_2)\\ 31 | & \subset \frac{1}{(r_1 + r_2)^2}W + \frac{r_2}{r_1 + r_2}W + \frac{r_1}{r_1 + r_2}W\\ 32 | & \subset W + W + W 33 | \end{align} 34 | % 35 | We conclude that $B$ is continuous at every $(a_1, a_2)$, since $W$ was arbitrary. 36 | \end{proof} 37 | %END -------------------------------------------------------------------------------- /chapter_1/FA_chapter_1.tex: -------------------------------------------------------------------------------- 1 | % 2 | \section{Exercise 1. Basic results} 3 | \input{\ROOT/chapter_1/1_1.tex} 4 | \newpage 5 | % 6 | \section{Exercise 2. Convex hull} 7 | \input{\ROOT/chapter_1/1_2.tex} 8 | \newpage 9 | % 10 | \section{Exercise 3. Other basic results} 11 | \input{\ROOT/chapter_1/1_3.tex} 12 | \newpage 13 | % 14 | \section{Exercise 4. A nonempty set whose interior is not} 15 | \input{\ROOT/chapter_1/1_4.tex} 16 | \newpage 17 | % 18 | \section{Exercise 5. A first restatement of boundedness} 19 | \input{\ROOT/chapter_1/1_5.tex} 20 | \newpage 21 | % 22 | \section{Exercise 6. A second restatement of boundedness} 23 | \input{\ROOT/chapter_1/1_6.tex} 24 | \newpage 25 | % 26 | \section{Exercise 7. Metrizability and number theory} 27 | \input{\ROOT/chapter_1/1_7.tex} 28 | \newpage 29 | % 30 | \setcounter{section}{8} 31 | % 32 | \section{Exercise 9. Quotient map} 33 | \input{\ROOT/chapter_1/1_9.tex} 34 | \newpage 35 | % 36 | \section{Exercise 10. An open mapping theorem} 37 | \input{\ROOT/chapter_1/1_10.tex} 38 | \newpage 39 | % 40 | \setcounter{section}{11} 41 | % 42 | \begin{section}{Exercise 12. Topology stays, completeness leaves} 43 | \input{\ROOT/chapter_1/1_12.tex} 44 | \end{section} 45 | \newpage 46 | % 47 | \setcounter{section}{13} 48 | % 49 | \section{Exercise 14. $\mathscr{D}_K$ equipped with other seminorms} 50 | \input{\ROOT/chapter_1/1_14.tex} 51 | \newpage 52 | % 53 | \setcounter{section}{15} 54 | % 55 | \section{Exercise 16. Uniqueness of topology for test functions} 56 | \input{\ROOT/chapter_1/1_16.tex} 57 | \newpage 58 | % 59 | \section{Exercise 17. Derivation in some non normed space} 60 | \input{\ROOT/chapter_1/1_17.tex} 61 | -------------------------------------------------------------------------------- /chapter_6/6_9.tex: -------------------------------------------------------------------------------- 1 | \renewcommand{\labelenumi}{(\alph{enumi})} 2 | {\it 3 | \begin{enumerate} 4 | \item Prove that a set $E\subset \mathscr{D}(\Omega)$ is bounded if and only if 5 | \begin{align*} 6 | \sup \{\lvert\Lambda \phi\rvert:\, \phi \in E\,\}< \infty 7 | \end{align*} 8 | for every $\Lambda \in \mathscr{D}(\Omega)$. 9 | \item Suppose $\{\phi_j\}$ is a sequence in $\mathscr{D}(\Omega)$ such that $\{\Lambda \phi_j\}$ is a bounded sequence of numbers, for every $\Lambda \in \mathscr{D}'(\Omega)$. Prove that some subsequence of $\{\phi_j\}$ converges, in the topology of $\mathscr{D}(\Omega)$. 10 | \item Suppose $\{\Lambda _j\}$ is a sequence in $\mathscr{D}'(\Omega)$ such that $\{\Lambda_{j\,} \phi\}$ is bounded, for every $\phi \in \mathscr{D}(\Omega)$. Prove that some subsequence of $\{\Lambda _j\}$ converges in $\mathscr{D}'(\Omega)$ and that the convergence is uniform on every bounded subset of $\mathscr{D}(\Omega)$. Hint: By the Banach-Steinhaus theorem, the restrictions of the $\Lambda_j$ to $\mathscr{D}_K$ are equicontinuous. Apply Ascoli's theorem. 11 | \end{enumerate}} 12 | \begin{proof} 13 | %: (a) 14 | Since $\mathscr{D}(\Omega)$ locally convex space (see (b) of [6.4]), [3.18] states that $E\,$ is bounded if and only if it is weakly bounded. That is (a). \\ 15 | \\ 16 | %: (b) 17 | To prove (b), we first use (a) to conclude that $E= \{\phi_j:\, j\in \N\}$ is bounded: so is $\overline{E}$. % 18 | By (c) of [6.5], there exists some $ \mathscr{D}_K$ that contains $\overline{E}$. % 19 | Since $ \mathscr{D}_K$ has the Heine-Borel property (see [1.46]), $\overline{E}$ is $\tau_K$-compact. % 20 | Apply [A4] with the metrizable space $\mathscr{D}_K$ (see [1.46]) to conclude that $\overline{E}$ has a $\tau_K$ limit point. % 21 | It then follows from (b) of [6.5] that (b) holds. 22 | \end{proof} -------------------------------------------------------------------------------- /chapter_1/1_4_comments.txt: -------------------------------------------------------------------------------- 1 | The following permutation 2 | \begin{align} 3 | \sigma:&\, \C^2 \to \C^2\\ 4 | & (z,z')\mapsto (z',z) 5 | \end{align} 6 | is an automorphism\ie a homeomorphism [\cf (a) of 1.21], of the vector space $\C^2$. The key ingredient is that $\sigma$ is open. Moreover, the invariant points under $\sigma$ form the diagonal 7 | \eq{D\Def \{(z,z)\in\C^2 \}\quad. }\endeq 8 | As the graph of 9 | \begin{align*} 10 | I:& \, \C \to \C\nonumber\quad , \\ 11 | &\, \, z\mapsto z \nonumber 12 | \end{align*} 13 | $D$ is a (closed) proper subspace of $\C^2$. Thus, 14 | \eq{D°=\emptyset\quad . }\endeq 15 | Remark that 16 | \eq{\label{1_4_1} 17 | S\Def \{(z_{1},z_{2})_n :\, n=1,2,3,\dots \}\subseteq B \donc\, \limsup_{n\infty}\, \lvert z_{1,n} \rvert \< \liminf_{n\infty}\, \lvert z_{2,n}\rvert \quad . }\endeq 18 | For if $S$ converges (\ref{1_4_1}) forces $S$ to do so in $B$, which is then closed. 19 | \eq{\label{eq_1_4_2} 20 | \C^2 \setminus B= \{(z_{1},z_{2})\in \C^2:\, \lvert z_{1}\rvert > \lvert z_{2}\rvert \} 21 | }\endeq 22 | is therefore open. So is 23 | \eq{E\Def \sigma(\C^2 \setminus B)= \{(z_{1},z_{2})\in \C^2:\, \lvert z_{1}\rvert < \lvert z_{2}\rvert \}\quad , }\endeq 24 | since $\sigma$ is open. 25 | \eq{B=E \uplus D \quad . }\endeq 26 | Let $(z_1,z_2$ be in $B°$ If such a point would lie in $D$, there would be $V$ a neighbourhood of $(0,0)$ such that 27 | \eq{(z_1,z_2) + V \subseteq B \quad ,}\endeq 28 | $(z_1,z_2) + V )\cap E$ is open. So is $(z_1,z_2) + V )\cap D$. contradicting 29 | \eq{B°\subseteq B\setminus \Delta \C = B_<}\endeq 30 | Let $(z_1,z_2)$ be any element of $B°$. Since $(\Delta\C)°=\emptyset$, $(z_1,z_2)$ is in $B_<$. 31 | \eq{([z] \in \C^\ast / N \cong \R_+ }\endeq 32 | Since $\C^2$ is separated, there exists a neighbourhood of the origin such that 33 | \eq{C(z_1)+V\cup C(z_2)+ V\quad, }\endeq 34 | so that 35 | \eq{z_1 + V \cup z_2 + V}\endeq -------------------------------------------------------------------------------- /chapter_2/2_3/2_3_0_lemma.tex: -------------------------------------------------------------------------------- 1 | If 2 | % 3 | $\phi\in\D_{[a, b]}$, then 4 | % 5 | \begin{align}\label{2.3. mean value inequality.} 6 | \norma{\infty}{D^k \phi} \leq \norma{\infty}{D^p \phi} \left(\frac{b-a}{2}\right)^{p-k} . 7 | \end{align} 8 | % 9 | for all $k \leq p$ in $\N$. 10 | \begin{proof} 11 | Choose $a < x_0 \leq (a+b)/2$ first. By the mean value theorem, there exists $x_1 \in \openinterval{a}{x_0}$ such that % 12 | % 13 | \begin{align} 14 | \phi(x_0) = \phi(x_0) - \phi(a)= D\phi(x_{1})(x_{0} - a). 15 | \end{align} 16 | % 17 | Repeating the same reasoning for $D\phi, D^2 \phi, \dots, D^p \phi \in \D_{[a, b]}$ yields % 18 | % 19 | \begin{align} 20 | \phi(x_0) & = D^0 \phi(x_0) \\ 21 | & = D^1\phi(x_{1})(x_0 - a) \\ 22 | & = D^2\phi(x_{2})(x_1 - a)(x_0 - a) \\ 23 | & \nonumber \mspace{10mu} \vdots \\ 24 | & = D^p\phi(x_{p})(x_{p-1} - a) \cdots (x_0-a), 25 | \end{align} 26 | % 27 | for some $x_1, \dots ,x_p \in \openinterval{a}{x_0}$. Hence % 28 | % 29 | \begin{align}\label{2.3. inequality} 30 | \magnitude{\phi(x_0)} \leq \norma{\infty}{D^p \phi} \left(\frac{b-a}{2}\right)^p . 31 | \end{align} 32 | % 33 | Similarly, if $b > x_0 \geq (a+b)/2$ (interchanging the roles of $a$ and $b$), the same inequality (\ref{2.3. inequality}) holds. % 34 | So, % 35 | % 36 | \begin{align}\label{2.3. mean value inequality, case k=0} 37 | \magnitude{\phi(x_0)} \leq \norma{\infty}{D^p \phi} \left(\frac{b-a}{2}\right)^p \quad ( a < x_0 < b ), 38 | \end{align} 39 | % 40 | which establishes \eqref{2.3. mean value inequality.} when $k=0$. % 41 | Finally, applying \eqref{2.3. mean value inequality, case k=0} to $D^k \phi $ in place of $\phi$ gives % 42 | % 43 | \begin{align} 44 | \norma{\infty}{D^k \phi} \leq \norma{\infty}{D^p \phi} \left(\frac{b-a}{2}\right)^{p-k} 45 | \end{align} 46 | % 47 | for all $0 \leq k \leq p$. % 48 | \end{proof} -------------------------------------------------------------------------------- /chapter_2/2_3/2_3.tex: -------------------------------------------------------------------------------- 1 | % 2 | \textit{ 3 | Put $K=[-1,1]$; define $\D_{K}$ as in section 1.46 4 | (with $\R$ in place of $\R^{n}$). 5 | Supose $\{f_{n}\}$ is a sequence of Lebesgue integrable functions such that 6 | % 7 | $\Lambda\phi 8 | = 9 | \underset{n \to \infty}{\lim} \int_{\minus 1}^1 f_{n}(t)\phi(t)dt$ 10 | % 11 | exists for every $\phi\in\D_{K}$. 12 | Show that $\Lambda$ is a continuous linear functional on $\D_{K}$. 13 | Show that there is a positive integer $p$ and a number $M<\infty$ such that 14 | \begin{align} 15 | \left\lvert 16 | \int_{\minus 1}^1 f_n (t)\phi (t) dt\ 17 | \right\rvert 18 | \leq 19 | M \norma{\infty}{D^{p}} 20 | \nonumber 21 | \end{align} 22 | for all $n$. 23 | For example, if $f_{n}(t)=n^{3}t$ on $[\minus 1/n, 1/n]$ and $0$ elsewhere, 24 | show that this can be done with $p=1$. 25 | Construct an example where it can be done with $p=2$ but not with $p=1$.} 26 | % 27 | % 28 | \renewcommand{\labelenumi}{(\roman{enumi})}% 29 | % 30 | \newline\newline\noindent 31 | We will also consider the case $p=0$. Since all supports of % 32 | % 33 | $\phi, \phi', \phi'', \dots, $ are in $K$, % 34 | % 35 | we make a specialization of the mean value theorem: % 36 | % 37 | \paragraph{Lemma}\label{2.3 Lemma} 38 | \input{\ROOT/chapter_2/2_3/2_3_0_lemma.tex} 39 | %: PROOF OF THE STATEMENT ----------------------------------------------------% 40 | %/ FIRST PART ----------------------------------------------------------------% 41 | % 42 | \begin{proof} 43 | \input{\ROOT/chapter_2/2_3/2_3_1_radon_measures.tex} 44 | % SECOND PART ----------------------------------------------------------------- 45 | \input{\ROOT/chapter_2/2_3/2_3_2_uniform_bound.tex} 46 | \input{\ROOT/chapter_2/2_3/2_3_3_example_1.tex} 47 | \input{\ROOT/chapter_2/2_3/2_3_4_example_2.tex} 48 | \end{proof} 49 | %\renewcommand{\labelenumi}{\alph{enumi}.}% 50 | \renewcommand{\labelenumi}{$(\textit{\alph{enumi}})$}% -------------------------------------------------------------------------------- /chapter_2/2_12.tex: -------------------------------------------------------------------------------- 1 | \textit{% 2 | Let X be the normed space of all real polynomials in one variable, with % 3 | % 4 | \begin{align*} 5 | \norm{f}=\int_0^1\lvert f(t)\rvert\ dt. 6 | \end{align*} 7 | % 8 | Put % 9 | $B(f, g)=\int _0^1 f(t)g(t) dt $, % 10 | % 11 | and show that $B$ is a bilinear continuous functional on $X\times X$ % 12 | which is separately but not continuous. 13 | } 14 | \begin{proof} Let $f$ denote the first variable, $g$ the second one. % 15 | Remark that % 16 | % 17 | \begin{align}\label{magnitude of B(f, g)} 18 | \magnitude{B(f, g)} 19 | & < \norm{f} \cdot \max_{[0,1]} \magnitude{g} ; 20 | \end{align} 21 | % 22 | which is sufficient (\citeresultFA{1.18}) to assert that any % 23 | % 24 | $f \mapsto B(f, g)$ 25 | % 26 | is continuous. The continuity of all % 27 | % 28 | $g \mapsto B(f,g)$ % 29 | % 30 | follows (Put $C(g, f) = B(f, g)$ and proceed as above). % 31 | Suppose, to reach a contradiction, that $B$ is continuous. % 32 | There so exists a positive $M$ such that, 33 | % 34 | \begin{align}\label{magnitude of B(f, g) H_c} 35 | \magnitude{B(f, g)} < M \norm{f}\norm{g}. 36 | \end{align} 37 | % 38 | Put % 39 | % 40 | \begin{align}\label{definition of f_n} 41 | f_n(x)\Def 2 \sqrt{n}\cdot x^{n}\in \R[x] \quad\quad (\counting{n}), 42 | \end{align} 43 | % 44 | so that 45 | % 46 | \begin{align}\label{norm of f_n} 47 | \norm{f_n} = \frac{2 \sqrt{n}}{n+1} \tendsto{n}{\infty}0. 48 | \end{align} 49 | % 50 | On the other hand, 51 | % 52 | \begin{align}\label{norm of B(f_n, g_n)} 53 | B(f_n, f_n)= \frac{4 n}{2n+1} > 1. 54 | \end{align} 55 | % 56 | Finally, we combine % 57 | % 58 | (\ref{norm of B(f_n, g_n)}) and % 59 | (\ref{magnitude of B(f, g) H_c}) with % 60 | (\ref{norm of f_n}) % 61 | % 62 | and so obtain 63 | \begin{align} 64 | 1 < B(f_n, f_n) < M \norm{f_n}^2 \tendsto{n}{\infty} 0. 65 | \end{align} 66 | Our continuousness assumption is then contradicted. So ends the proof. 67 | \end{proof} 68 | -------------------------------------------------------------------------------- /chapter_2/2_16.tex: -------------------------------------------------------------------------------- 1 | \textit{% 2 | Suppose that $X$ and $K$ are metric spaces, % 3 | that $K$ is compact, and that the graph of % 4 | % 5 | $f:X\to K$ % 6 | % 7 | is a closed subset of $X\times K$. % 8 | Prove that $f$ is continuous % 9 | % 10 | (This is an analogue of Theorem 2.15 but much easier.) % 11 | % 12 | Show that compactness of $K$ cannot be omitted from the hypothese, % 13 | even when $X$ is compact. 14 | } 15 | % 16 | \begin{proof}% 17 | Choose a sequence $\set{x_n}{\counting{n}}$ whose limit is an arbitrary $a$. % 18 | By compactness of $K$, the graph $G$ of $f$ contains a subsequence % 19 | % 20 | $\singleton{ 21 | (x_{\rho(n)}, f(x_{\rho(n)})) 22 | }$ 23 | % 24 | of % 25 | % 26 | $\singleton{ 27 | (x_n, f(x_n)) 28 | }$ 29 | % 30 | that converges to some $(a, b)$ of $X\times K$. % 31 | $G$ is closed; therefore, $\singleton{(x_{\rho(n)}, f(x_{\rho(n)}))}$ 32 | converges in $G$. So, $b=f(a)$; % 33 | which establishes that $f$ is sequentially continuous. 34 | % 35 | Since $X$ is metrizable, $f$ is also continuous, see \citeresultFA{[A6]}. % 36 | So ends the proof. 37 | % 38 | \newline\newline\noindent % 39 | % 40 | To show that compactness cannot be omitted from the hypotheses, % 41 | we showcase the following counterexample, % 42 | % 43 | \begin{align} 44 | f: [0, \infty) & \to [0, \infty)\\ 45 | x & \mapsto\begin{cases} 46 | 1/x & (x > 0) \\\nonumber 47 | 0 & (x = 0). 48 | \end{cases} 49 | \end{align} 50 | % 51 | Clearly, $f$ has a discontinuity at $0$. % 52 | Nevertheless the graph $G$ of $f$ is closed. % 53 | To see that, first remark that % 54 | % 55 | \begin{align} 56 | G = \set{(x, 1/x)}{x > 0} \cup \singleton{(0, 0)}. 57 | \end{align} 58 | % 59 | Next, let % 60 | % 61 | $\singleton{(x_n, 1/x_n)}$ % 62 | % 63 | be a sequence in % 64 | % 65 | $G_+ = \set{(x, 1/x)}{x > 0}$ % 66 | % 67 | that converges to $(a, b)$. % 68 | To be more specific: % 69 | $a=0$ contradicts the boundedness of $\singleton{(x_n,1/x_n)}$: % 70 | $a$ is necessarily positive and $b = 1/a$, % 71 | since $x\mapsto 1/x$ is continuous on $R_+$. 72 | This establishes that $(a, b) \in G_+$, hence the closedness $G_+$. % 73 | Finally, we conclude that $G$ is closed, as a finite union of closed sets. 74 | % 75 | \end{proof} 76 | % 77 | % END 78 | -------------------------------------------------------------------------------- /chapter_1/1_12.tex: -------------------------------------------------------------------------------- 1 | \textit{Suppose % 2 | % 3 | $d_1(x,y) = |x-y|, d_2(x,y) = |\phi(x) - \phi(y)|$, % 4 | % 5 | where % 6 | % 7 | $\phi(x)={x}/{(1+\lvert x \rvert )}$. % 8 | % 9 | Prove that $d_1$ and $d_2$ are metrics on $\R$ which induce the same % 10 | topology, although $d_1$ is complete and $d_2$ is not. 11 | } 12 | % 13 | \begin{proof}% 14 | First, each $d_i\, (i=1, 2)$ induces a topology $\tau_i$ % 15 | whose open balls are all % 16 | % 17 | \begin{align}\label{1_12_2} 18 | B_i(a,r) \triangleq \{x \in \R: d_i (a, x) < r \} 19 | \quad (a \in \R, r > 0 ). 20 | \end{align} 21 | % 22 | Next, remark that the monotonically increasing mapping % 23 | $\phi: \R \to ]\minus 1, 1[$ is odd and that % 24 | % 25 | \begin{align}\label{1_12_1} 26 | \phi(x) \tendsto{x}{\infty} 1. 27 | \end{align} 28 | % 29 | $\phi$ is therefore a $\tau_1$-homeomorphsim of $\R$ onto $]\minus 1, 1[$. % 30 | % 31 | A first consequence is that, at fixed $a \in \R$, given any positive scalar % 32 | $\epsilon$, the $\tau_1$-continuousness of $\phi$ supplies an open ball % 33 | $B_1(a,\eta)$ on which $|\phi(a)-\phi|<\epsilon$. % 34 | In terms of balls $B_i$, this reads as follows, % 35 | % 36 | \begin{align} 37 | B_1(a,\eta) \subset B_2(a,\epsilon). 38 | \end{align} 39 | % 40 | The second consequence is that the $\tau_1$-continuousness of % 41 | $\phi^{\, \minus 1}$ yields similar inclusions % 42 | % 43 | \begin{align} \label{1_12_6} 44 | \quad B_2(a, \epsilon') \subset B_1 (a, \eta') 45 | \end{align} 46 | % 47 | provided $\eta'>0$. At arbitrary $\epsilon$, the special case $\eta' = \eta$ % 48 | is the concatenation % 49 | % 50 | \begin{align} 51 | B_2(a, \epsilon') \subset B_1(a,\eta) \subset B_2(a,\epsilon); 52 | \end{align} 53 | % 54 | which proves that $\tau_1 =\tau_2$. % 55 | % 56 | Finally, all inequalities $n < i < j$ over $\N$ together yield % 57 | % 58 | \begin{align} 59 | d_2(i, j)=|\phi(i)-\phi(j)| \tendsto{n}{\infty} 0. 60 | \quad \end{align} 61 | % 62 | The sequence $n=0, 1, 2, \dots$ is therefore $\tau_2$-Cauchy. We will % 63 | nevertheless establish that it $\tau_2$-diverges. To do so, we start by % 64 | offering the $\tau_2$-converge to some $\lambda$: % 65 | The triangle inequality immediately dismiss that assumption, as follows, % 66 | % 67 | \begin{align} 68 | d_2(0, \lambda) \geq 69 | d_2(0, n) - d_2(\lambda, n) = 70 | \phi(n) -d_2(\lambda, n) \tendsto{n}{\infty} 1. 71 | \end{align} 72 | % 73 | We then conclude that $d_2$ fails to be complete. 74 | \end{proof} 75 | % END 76 | -------------------------------------------------------------------------------- /chapter_1/1_10.tex: -------------------------------------------------------------------------------- 1 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 2 | % FunctionalAnalysis 3 | % 1_10.tex 4 | % 5 | % encoding: UTF-8 6 | % EOL: LF 7 | % 8 | % format: LaTeX 9 | % indent: spaces (2) 10 | % width: 127 11 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 12 | \textit{Suppose that X and Y are topological vector spaces, $\dim Y < \infty$, $\Lambda : X \to Y$ is linear, and % 13 | $\Lambda(X) = Y$. 14 | % 15 | \begin{enumerate} 16 | \item{ 17 | Prove that $\Lambda$ is an open mapping.} 18 | \item{ 19 | Assume, in addition, that the null space of $\Lambda$ is closed, 20 | and prove that $\Lambda$ is continuous. 21 | } 22 | \end{enumerate} 23 | % 24 | } 25 | % 26 | \begin{proof} 27 | Discard the trivial case $\Lambda = 0$ and assume that $\dim Y > 0$. From now on, $Y$ has a basis $B= \singleton{e, e', \dots}$ % 28 | and $W$ is an arbitrary neighborhood of $0$. Since addition is continuous, there exists a balanced open $V\subset Y$ such that % 29 | % 30 | \begin{align} 31 | \label{definition of v} 32 | \sum_{e} V \subset W. 33 | \end{align} 34 | % 35 | Moreover, the surjectivity of $\Lambda$ allows us to pick $x_e$ with $\Lambda x_e = e$. This implies that % 36 | every $y = \sum_e y_e e $ ($y_e \in \C$) can be expressed as % 37 | % 38 | \begin{align}\label{1_10_sum} 39 | y = \sum_{e} y_e \Lambda x_e. 40 | \end{align} 41 | % 42 | Note that $\set{x_e}{e\in B}$ is bounded (as a finite set). Hence % 43 | % 44 | \begin{align} 45 | \set{x_e}{e \in B} \subset sV 46 | \end{align} 47 | % 48 | for some $s > 0$. Combining this with (\ref{1_10_sum}) yields % 49 | % 50 | \begin{align} 51 | \label{y in sum of lambda V} 52 | y \in \sum_e s y_e \Lambda (V). 53 | \end{align} 54 | % 55 | Using (\ref{definition of v}) and the balancedness of $\Lambda(V)$, we conclude that 56 | % 57 | \begin{align} 58 | y \in \sum_e \Lambda (V) \subset \Lambda(W) \quad (\norma{\infty}{y} < 1/s). 59 | \end{align} 60 | % 61 | This proves (a) in the case where $Y = \C^n$ and $B$ is the standard basis. The general case now follows from % 62 | [\ref{notations: vector spaces: finite-dimensional vector spaces}].\\\\ 63 | % 64 | To prove (b), assume that the null space $\{\Lambda = 0\}$ is closed and let $f, \pi$ be as in Exercise 1.9, % 65 | $\{\Lambda = 0\}$ playing the role of $N$. Since $\Lambda$ is onto, the first isomorphism theorem (see Exercise 1.9) asserts % 66 | that $f$ is an isomorphism of $X/N$ onto $Y$. We now conclude with the help of % 67 | [\ref{notations: vector spaces: finite-dimensional vector spaces}] that $f$ is a homeomorphism of $X/N$ onto $Y$. % 68 | We have thus established that $f$ is continuous: So is $\Lambda = f\circ \pi$. 69 | \end{proof} 70 | % END 71 | -------------------------------------------------------------------------------- /chapter_1/1_14.tex: -------------------------------------------------------------------------------- 1 | \textit{Put $K =[0, 1]$ and define $\D_K$ as in Section 1.46. % 2 | Show that the following three families of seminorms % 3 | (where $n = 0, 1, 2, \dots$) define the same topology on $\D_K$. 4 | If $D = d/dx$: 5 | % 6 | \begin{enumerate} 7 | \item{ 8 | $\| D^n f \|_\infty = \sup\set{\left| D^n f(x)\right|}{\infty< x< \infty}$ 9 | } 10 | \item{ 11 | $\| D^n f \|_1 =\int_0^1 \left|D^n f(\varit{x}) \right| d\varit{x}$ 12 | } 13 | \item{ 14 | $\| D^n f \|_2 = \left\{ 15 | \int_0^1 | D^n f(\varit{x}) |^2 d\varit{x} 16 | \right\}^{1/2}.$ 17 | } 18 | \end{enumerate} 19 | % 20 | } 21 | % 22 | \begin{proof}% 23 | Let us equipp $\D_K$ with the inner product % 24 | % 25 | $\bra{f}\ket{g} = \int_0^1 f \, \bar{g}$so that % 26 | $\bra{f}\ket{f} = \| f \|_2^2$. % 27 | % 28 | The following % 29 | % 30 | \begin{align} 31 | \int_0^1 1\,| D^n f| \leq \| 1 \|_2 \| D^n f \|_2 32 | \end{align} 33 | % 34 | is then a Cauchy-Schwarz inequality, see Theorem 12.2 of \cite{FA}. % 35 | We so obtain % 36 | % 37 | \begin{align}\label{inequalities 1} 38 | \| D^n f \|_1 39 | \leq 40 | \| D^n f \|_2 41 | \leq 42 | \| D^n f \|_\infty 43 | <\infty 44 | \end{align} 45 | % 46 | since $K$ has length $1$. % 47 | % 48 | Obviously, the support of $D^n f$ lies in $K$, hence the below equality % 49 | % 50 | \begin{align}\label{inequalities 2} 51 | |D^n f(x)| 52 | = 53 | \left|\int_{0}^x D^{n+1}f\right| 54 | \leq 55 | \int_{0}^x |D^{n+1}f| 56 | \leq 57 | \|D^{n+1}f \|_1. 58 | \end{align} 59 | % 60 | Take the supremum over all $|D^n f(x)|$: Combining (\ref{inequalities 1}) % 61 | with (\ref{inequalities 2}) now reads as follows, % 62 | % 63 | \begin{align}\label{inequalities 3} 64 | \|D^n f\|_1 65 | \leq 66 | \|D^{n} f\|_2 67 | \leq 68 | \|D^{n} f\|_\infty 69 | \leq 70 | \|D^{n+1}f\|_1 71 | \leq 72 | \cdots < \infty. 73 | \end{align} 74 | % 75 | Finally, put % 76 | % 77 | \begin{align} 78 | V^{(i)}_n & \triangleq \{f\in \D_K: \|f \|_i < 2^{\,\minus n}\}, \\ 79 | \mathscr{B}^{(i)} & \triangleq \{V^{(i)}_n: n = 0, 1, 2, \dots\}, 80 | \end{align} 81 | % 82 | so that (\ref{inequalities 3}) is mirrored by neighborhood inclusions, % 83 | provided $i=1, 2, \infty$: 84 | % 85 | \begin{align}\label{inclusions} 86 | V^{(1)}_n 87 | \supseteq 88 | V^{(2)}_n 89 | \supseteq 90 | V^{(\infty)}_n 91 | \supseteq 92 | V^{(1)}_{n+1} 93 | \supseteq 94 | \cdots . 95 | \end{align} 96 | % 97 | Their subchains $V^{(i)}_n\supseteq V^{(i)}_{n+1}$ turn $\mathscr{B}^{(i)}$ % 98 | into a local base of a topology $\tau_i$. 99 | The whole chain (\ref{inclusions}) then forces % 100 | % 101 | \begin{align} 102 | \tau_1 \subset \tau_2 \subset \tau_\infty \subset \tau_1; 103 | \end{align} 104 | % 105 | which achieves the proof. 106 | \end{proof} 107 | % END 108 | 109 | -------------------------------------------------------------------------------- /chapter_1/1_17.tex: -------------------------------------------------------------------------------- 1 | \textit{In the setting of Section 1.46, prove that 2 | % 3 | $f \mapsto D^{\alpha}f$ 4 | % 5 | is a continuous mapping of 6 | % 7 | $C^{\infty}\left(\Omega\right)$ into 8 | $C^{\infty}\left(\Omega\right)$ and also of 9 | $\D_{K}$ into 10 | $\D_{K}$, for every multi-index $\alpha$. 11 | % 12 | } 13 | \begin{proof} 14 | In both cases, $D^\alpha$ is a linear mapping. 15 | It is then sufficient to establish continuousness at the origin. 16 | % 17 | We begin with the $C^\infty\left(\Omega\right)$ case. \\ 18 | \\ 19 | Let $U$ be an aribtray neighborhood of the origin. 20 | There so exists $N$ such that $U$ contains 21 | % 22 | \begin{align} 23 | V_{N}= \left\{ 24 | \phi \in C^\infty\left(\Omega\right): 25 | \max\set{ 26 | | D^\beta\phi (x) | 27 | }{ 28 | \magnitude{ \beta } \leq N, x\in K_N 29 | } 30 | < 1/N 31 | \right\}. 32 | \end{align} 33 | % 34 | Now pick $g$ in $V_{N+\magnitude{\alpha}}$so that 35 | % 36 | \begin{align} 37 | \max 38 | \set{ 39 | \magnitude{ D^\gamma g\left(x\right) } 40 | }{ 41 | \magnitude{ \gamma | \leq N+| \alpha }, 42 | x\in K_{N} 43 | } 44 | < \frac{1}{N+\magnitude{\alpha}}. 45 | \end{align} 46 | % 47 | (the fact that $K_N\subset K_{N+\magnitude{\alpha}}$ was tacitely used). 48 | % 49 | The special case $\gamma = \beta + \alpha$ yields 50 | %‡ 51 | \begin{align} 52 | \max 53 | \set{ 54 | | D^\beta D^\alpha g(x)| 55 | }{ 56 | \magnitude{\beta} \leq N, 57 | x\in K_{N} 58 | } 59 | < \frac{1}{N}. 60 | \end{align} 61 | % 62 | We have just proved that 63 | % 64 | \begin{align}\label{1.17. inclusion} 65 | g \in V_{N + \magnitude{\alpha}} 66 | \then 67 | D^\alpha g \in V_{N}, 68 | \quad 69 | \ie 70 | \quad 71 | D^\alpha \left(V_{N+\magnitude{\alpha}}\right) \subset V_N, 72 | \end{align} 73 | % 74 | which establishes the continuity of % 75 | % 76 | $D^{\alpha}: C^\infty \left(\Omega\right)\to C^\infty \left(\Omega\right)$. % 77 | % 78 | \\\\ 79 | To prove the continuousness of the restriction 80 | % 81 | $D^\alpha \lvert_{\D_K}: \D_K \to \D_K$, % 82 | % 83 | % 84 | we first remark that the collection of the 85 | % 86 | ${V_N \cap \D_K}$ % 87 | % 88 | is a local base of the subspace topology of $\D_K$. 89 | % 90 | % 91 | $V_{N+\magnitude{\alpha}} \cap \D_K$ % 92 | % 93 | is then a neighborhood of $0$ in this topology. % 94 | Furthermore, 95 | % 96 | \begin{align} 97 | % 98 | D^\alpha \lvert_{\D_K} \left(V_{N+\magnitude{\alpha}} \cap \D_K\right) 99 | % 100 | & = 101 | D^\alpha\left(V_{N+\magnitude{\alpha}} \cap \D_K\right) \\ 102 | % 103 | & \subset 104 | D^\alpha\left(V_{N+\magnitude{\alpha}}\right) 105 | \cap 106 | D^\alpha\left(\D_K\right) \\ 107 | % 108 | & \subset 109 | V_N 110 | \cap 111 | \D_K 112 | % 113 | \quad (\text{see }(\ref{1.17. inclusion})) 114 | % 115 | % 116 | \end{align} 117 | % 118 | So ends the proof. 119 | \end{proof} 120 | -------------------------------------------------------------------------------- /chapter_2/2_15.tex: -------------------------------------------------------------------------------- 1 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 2 | % FunctionalAnalysis 3 | % 2_15.tex 4 | % 5 | % encoding: UTF-8 6 | % EOL: LF 7 | % 8 | % format: LaTeX 9 | % indent: spaces (2) 10 | % width: 127 11 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 12 | \textit{Suppose $X$ is an F-space and $Y$ is a subspace of $X$ whose complement is of the first category. Prove that $Y=X$. }% 13 | Hint: \textit{$Y$ must intersect $x+Y$ for every $x\in X$. }% 14 | % 15 | \begin{proof}% 16 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 17 | % A. Assumptions. 18 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 19 | Let $\singleton{E_n}$ be a sequence of nowhere dense sets such that $X\setminus Y = \bigcup_{n=1}^\infty E_n$. Note that % 20 | every $V_n = X \setminus \closure{E_n}$ is dense in $X$ . Now pick an arbitrary $x \in X$. Since the translation by $x$ is a % 21 | self-homeomorphism, $x + V_n$ is open and dense as well. Alternatively, observe that $X = x + X \subset \closure{x + V_n}$, % 22 | which is a special case of \citeresultFA{1.3 (b)}. % 23 | % 24 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 25 | % B. Where Baire's theorem is involved. 26 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 27 | Next, apply Baire's theorem twice to establish that % 28 | % 29 | \begin{enumerate} 30 | \item every intersection $W_n = V_n \cap [x+V_n]$ is dense in $X$, 31 | \item so is the nonempty intersection $\bigcap_{n=1}^\infty W_n$. 32 | \end{enumerate} 33 | % 34 | Moreover, the intersection $\bigcap_{n=1}^\infty V_n$ does not cut any $E_n$. To sum up, % 35 | % 36 | \begin{align}\label{2_15_1} 37 | w \in \bigcap_{n=1}^\infty W_n \subset \bigcap_{n=1}^\infty V_n \subset Y 38 | \end{align} 39 | % 40 | for some $w = w(x)$. % 41 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 42 | % C. Key ingredient! 43 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 44 | Furthermore, $w$ also lies in every $x + V_n$, by (a). Hence % 45 | % 46 | \begin{align}\label{2_15_2} 47 | w - x \in \bigcap_{n=1}^\infty V_n \subset Y. 48 | \end{align} 49 | % 50 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 51 | % D. Put B and C together then conclude. 52 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 53 | Finally, (\ref{2_15_1}) and (\ref{2_15_2}) together yield % 54 | % 55 | \begin{align} 56 | x = w - (w - x) \in Y - Y = Y, 57 | \end{align} 58 | % 59 | as $Y$ is a subgroup of $X$. This establishes that % 60 | % 61 | \begin{align} 62 | X \subset Y, 63 | \end{align} 64 | % 65 | since $x$ was arbitrary. 66 | \end{proof} 67 | % 68 | % END 69 | -------------------------------------------------------------------------------- /chapter_2/2_9.tex: -------------------------------------------------------------------------------- 1 | \textit{ 2 | Suppose $X,Y,Z$ are Banach spaces and 3 | % 4 | \begin{align*} 5 | B:X\times Y \to Z 6 | \end{align*} 7 | % 8 | is bilinear and continuous. Prove that there exists $M<\infty$ such that 9 | % 10 | \begin{align*} 11 | \norm{B(x, y)} 12 | \leq 13 | M \norm{x}\norm{x} \quad (x\in X, y\in Y). 14 | \end{align*} 15 | % 16 | Is completeness needed here?} 17 | \begin{proof}% 18 | % % % % % % % 19 | % FIRST PART % 20 | % % % % % % % 21 | The answer is: No. To prove this, we only assume that % 22 | % 23 | $\mathit{X}$, $\mathit{Y}$, $\mathit{Z}$ % 24 | % 25 | are normed spaces. % 26 | % 27 | Since $B$ is continous at the origin, there exists a positive $r$ % 28 | such that % 29 | % 30 | \begin{align}\label{bound for B} 31 | \norm{x} + \norm{y} < r \Rightarrow % 32 | \norm{B(x, y)} < 1. 33 | \end{align} 34 | % 35 | % 36 | Given nonzero $\mathit{x}, \mathit{y}$, let $s$ range over $]0, r[$so that the folllowing bound % 37 | % 38 | \begin{align} 39 | \norm{B(x, y)} = \frac{4\norm{x}\norm{y}}{s^2} % 40 | \left\| 41 | B \left(\frac{s}{2\norm{x}}x, \frac{s}{2\norm{y}}y \right) 42 | \right\| 43 | \overset{(\ref{bound for B})}{<} \frac{4 \norm{x} \norm{y}}{s^2} 44 | %\quad (x\neq 0, y \neq 0) 45 | \end{align} 46 | is effective. % 47 | It is now obvious that %% 48 | % 49 | \begin{align} 50 | B(x, y) \leq % 51 | \frac{\!\!\!4\!}{\,s^{\,2}}\norm{x}\norm{y} 52 | \tendsto{s}{r} 53 | \frac{\!\!\!4\!}{\,r^{\,2}}\norm{x}\norm{y} 54 | \quad ((x, y) \in X \times Y); 55 | \end{align} 56 | % 57 | which achieves the proof.\\\\ % 58 | % % % % % % % % 59 | % SECOND PART % 60 | % % % % % % % % 61 | As a concrete example, choose % 62 | % 63 | $X = Y = Z = C_c(\R)$, % 64 | % 65 | topologized by the supremum norm. % 66 | % 67 | $C_c(\R)$ is not complete % 68 | % 69 | (see 5.4.4 of \cite{AnalyseIII}), % 70 | % 71 | nevertheless the bilinear product % 72 | % 73 | \begin{align} 74 | \begin{aligned} 75 | B: &&& C_c(\R)^{2} & \to &&& C_c(\R) \\\nonumber 76 | &&& (f, g) & \mapsto &&& f \cdot g 77 | \end{aligned} 78 | \end{align} 79 | % 80 | is bounded (since % 81 | $\norma{\infty}{f\cdot g} \leq \norma{\infty}{f}\cdot \norma{\infty}{g}$% 82 | ), % 83 | and continuous. To show this, pick a positive scalar % 84 | $\epsilon $ smaller than $1$, provided any $(f, g)$. Next, define % 85 | % 86 | \begin{align} 87 | r \triangleq \frac{\epsilon}{1 + \norma{\infty}{f} + \norma{\infty}{g}} < 1. % 88 | \end{align} 89 | % 90 | We now restrict $(u, v)$ to a particular neighborhood of $(f, g)$. % 91 | More specifically, 92 | \begin{align} 93 | \norma{\infty}{f-u} + \norma{\infty}{g-v} < r. % 94 | \end{align} 95 | % 96 | Next, remark that % 97 | % 98 | $\norma{\infty}{u} \leq r + \norma{\infty}{f}$ % 99 | % 100 | and so obtain (bear in mind that $r < 1$)% 101 | % 102 | \begin{align} 103 | \norma{\infty}{fg - uv} 104 | & = 105 | \norma{\infty}{(f-u) \cdot g + u \cdot (g-v)} \\ 106 | & \leq 107 | \norma{\infty}{f-u} \cdot \norma{\infty}{g} + 108 | \norma{\infty}{u} \cdot \norma{\infty}{g-v} \\ 109 | & < r \cdot \norma{\infty}{g} + (r + \norma{\infty}{f}) \cdot r \\ 110 | & < r \cdot (r + \norma{\infty}{f} + \norma{\infty}{g}) \\ 111 | & < \epsilon. 112 | \end{align} 113 | % 114 | Since $\epsilon$ was arbitrary, it is now established that $B$ continuous % 115 | at every $(f, g)$. 116 | \end{proof} 117 | % END 118 | -------------------------------------------------------------------------------- /chapter_1/1_9.tex: -------------------------------------------------------------------------------- 1 | \textit{Suppose 2 | \begin{enumerate} 3 | \item $X$ and $Y$ are topological vector spaces, 4 | \item $\Lambda: X\to Y$ is linear. 5 | \item $N$ is a closed subspace of $X$, 6 | \item$\varit{\pi}: X\to X/N$ is the quotient map, and 7 | \item $\Lambda \varit{x}=0$ for every $\varit{x}\in N$. 8 | \end{enumerate} 9 | Prove that there is a unique $f:X/N\to Y$ which satisfies 10 | % 11 | $\Lambda = \mathit{f\circ \pi}$, 12 | % 13 | that is, 14 | % 15 | $\Lambda \varit{x}= \mathit{f(\pi (x))}$ for all $\mathit{x\in X}$. 16 | % 17 | Prove that $\varit{f}$ is linear and that $\Lambda$ is continuous % 18 | % 19 | if and only if % 20 | % 21 | $\varit{f} $ is continuous. 22 | % 23 | Also, $\Lambda$ is open if and only if $\varit{f}$ is open. 24 | } 25 | % 26 | \begin{proof} Bear in mind that 27 | % 28 | $\pi$ continously maps $X$ onto the topological (Hausdorff) space $X/N$, 29 | since $N$ is closed (see \citeresultFA{1.41}). 30 | % 31 | Moreover, the equation 32 | % 33 | $\Lambda = f \circ \pi$ 34 | % 35 | has necessarily a unique solution, which is the binary relation 36 | % 37 | \begin{align}\label{1.9. definition of f.} 38 | f \Def\set{(\pi x, \Lambda x)}{x \in X} \subset X/N \times Y. 39 | \end{align} 40 | % 41 | To ensure that $f$ is actually a mapping, simply remark that 42 | the linearity of $\Lambda$ implies 43 | % 44 | \begin{align} 45 | %\forall ( x, x') \in X^2: 46 | % 47 | \Lambda x \neq \Lambda x' \then \pi x' \neq \pi x'. 48 | \end{align} 49 | % 50 | It straightforwardly derives from (\ref{1.9. definition of f.}) that 51 | $f$ inherits linearity from $\pi$ and $\Lambda$.\\ 52 | \\ 53 | {\bf Remark.} The special case 54 | % 55 | $N = \singleton{\Lambda = 0}$ \ie $\Lambda x= 0$ \iif $x\in N$ (\cf (e)), % 56 | % 57 | is the first isomorphism theorem in the topological spaces context. 58 | To see this, remark that this strenghtening of (e) yields 59 | % 60 | \begin{align} 61 | f(\pi x) = 0 62 | \citethen{\ref{1.9. definition of f.}} 63 | \Lambda x = 0 64 | \then 65 | x \in N 66 | \then 67 | \pi x = N 68 | \end{align} 69 | and so conclude that $f$ is also one-to-one. 70 | % 71 | %We have thus proved 72 | % 73 | %the first isomorphism theorem in the topological spaces context. % 74 | % 75 | \\\\ 76 | Now assume $f$ to be continuous. Then so is 77 | % 78 | $\Lambda = f\circ \pi $, 79 | % 80 | by \citeresultFA{1.41 (a)}. 81 | % 82 | Conversely, 83 | % 84 | if $\Lambda$ is continuous, then for each neighborhood $V$ of $0_Y$ 85 | there exists a neighborhood $U$ of $0_X$ such that 86 | % 87 | \begin{align} 88 | \Lambda(U) = f\left(\pi(U)\right) 89 | \subset 90 | V. 91 | \end{align} 92 | % 93 | Since $\pi$ is open (\citeresultFA{1.41 (a)}), $\pi(U)$ is a neighborhood of 94 | % 95 | $N=0_{X/N}$: 96 | %s 97 | This is sufficient to establish that the linear mapping $f$ is continuous. 98 | % 99 | If $f$ is open, so is $\Lambda = f\circ \pi$, by \citeresultFA{1.41 (a)}. 100 | % 101 | To prove the converse, remark that 102 | % 103 | every neighborhood $W$ of $0_{X/N}$ satisfies % 104 | % 105 | \begin{align} 106 | W = \pi(V) 107 | \end{align} 108 | % 109 | for some neighborhood $V$ of $0_X$. So, 110 | % 111 | \begin{align} 112 | f(W) = f \left(\pi(V)\right) = \Lambda(V). 113 | \end{align} 114 | % 115 | As a consequence, 116 | % 117 | if $\Lambda$ is open, then $f(W)$ is a neighborhood of $0_Y$. % 118 | % 119 | So ends the proof. 120 | \end{proof} 121 | 122 | 123 | -------------------------------------------------------------------------------- /chapter_2/2_3/2_3_4_example_2.tex: -------------------------------------------------------------------------------- 1 | In order to construct the second requested example, we give % 2 | % 3 | ${f_n}$ % 4 | % 5 | a \textit{derivative}% 6 | % 7 | \footnote{ 8 | See \citeresultFA{6.1} 9 | for a further discussion. 10 | } % 11 | ${f_n}'$, as follows % 12 | % 13 | \begin{align} 14 | {f_n}': \D_K &\to \C\\ 15 | \phi &\mapsto \minus \bra{f_n}\ket{\phi'}. \nonumber 16 | \end{align} 17 | % 18 | It has been proved that every $\bra{f_n}$ is continuous. % 19 | So is % 20 | % 21 | \begin{align} 22 | D : \D_K & \to \D_K \\ 23 | \phi & \mapsto \phi'; \nonumber 24 | \end{align} 25 | see Exercise 1.17. % 26 | ${f_n}'$ is therefore continuous. Now apply % 27 | % 28 | (\ref{2.3. (f_n) converges to Dirac (1).}) % 29 | % 30 | with $\phi'$ and so obtain % 31 | % 32 | \begin{align} 33 | \minus \bra{f_n}\ket{\phi'} &\tendsto{n}{\infty} \frac{4}{3} \phi''(0) 34 | \quad(\phi \in \D_K), \nonumber 35 | \end{align} 36 | % 37 | \ie % 38 | % 39 | \begin{align} 40 | {f_n}' &\tendsto{n}{\infty} \frac{4}{3} \delta''. 41 | \end{align} 42 | % 43 | It follows from % 44 | % 45 | (\ref{2.3. Bound with p and M (concrete).}) % 46 | % 47 | that, 48 | % 49 | \begin{align} 50 | \lvert \bra{f_n}\ket{\phi'} \rvert 51 | \leq 52 | \frac{4}{3} \norma{\infty}{ \phi''} \quad (\counting{n}). 53 | \end{align} 54 | % 55 | It is therefore possible to uniformly bound 56 | % 57 | ${f_n}'$ % 58 | % 59 | with respect to a norm % 60 | % 61 | $\norma{\infty}{D^p\cdot}$, % 62 | % 63 | namely $\norma{\infty}{D^2\cdot}$. 64 | % 65 | Then arises a question: % 66 | % 67 | Is $2$ the smallest $p$? 68 | % 69 | The answer is: Yes. 70 | % 71 | To show this, we first assume, to reach a contradiction, that % 72 | % 73 | there exists a positive constant $M$ such that 74 | % 75 | \begin{align} 76 | \lvert \langle f_n \lvert \phi' \rangle \rvert 77 | \leq 78 | M \norma{\infty}{\phi'} 79 | \quad (\counting{n}). 80 | \end{align} 81 | % 82 | Define % 83 | % 84 | \begin{align} 85 | \phi_{j}(x) = \int_{\minus 1}^x \phi_j. 86 | \end{align} 87 | % 88 | The oddness of $\phi_j$ forces % 89 | % 90 | $\phi_{j}$ to vanish outside $[\minus 1, 1]$: % 91 | % 92 | $\phi_{j}$ is therefore in $\D_K$. So, under our assumption, 93 | % 94 | \begin{align} 95 | \lvert 96 | \langle{f_n}\lvert \phi_j' \rangle 97 | \rvert 98 | \leq 99 | M \norma{\infty}{\phi_{j}'} 100 | \quad (\counting{n}); 101 | \end{align} 102 | % 103 | which is % 104 | % 105 | \begin{align} 106 | \lvert 107 | \langle{f_n}\lvert{\phi_j}\rangle 108 | \rvert 109 | \leq 110 | M 111 | \quad (\counting{n}). 112 | \end{align} 113 | % 114 | We have thus reached a contradiction (again with the sequence % 115 | % 116 | $\singleton{\langle f_n\lvert \phi_{\rho(n)} \rangle}$) % 117 | % 118 | and so conclude that there is no constant $M$ such that % 119 | % 120 | \begin{align} 121 | \lvert 122 | \langle \lvert f_n \phi'\rangle 123 | \rvert 124 | \leq 125 | M \norma{\infty}{\phi'} 126 | \quad (\counting{n}). 127 | \end{align} 128 | % 129 | Finally, assume, to reach a contradicton, that % 130 | there exists a constant $M$ such that 131 | % 132 | \begin{align} 133 | \lvert \langle f_n \lvert \phi' \rangle \rvert 134 | \leq 135 | M \norma{\infty}{\phi}. 136 | \end{align} 137 | % 138 | The mean value theorem (see (\ref{2.3. Mean value inequality.})) asserts that % 139 | % 140 | \begin{align} 141 | \lvert \langle f_n \lvert \phi' \rangle \rvert 142 | \leq 143 | M \norma{\infty}{\phi} 144 | \leq 145 | M \norma{\infty}{\phi'}; 146 | \end{align} 147 | % 148 | which is, again, a desired contradiction. So ends the proof. 149 | 150 | -------------------------------------------------------------------------------- /chapter_2/2_6.tex: -------------------------------------------------------------------------------- 1 | \textit{ 2 | Define the Fourier coefficient $\hat{f}(n)$ of a function % 3 | % 4 | $f\in L^2(T)$ ($T$ is the unit circle) % 5 | % 6 | by 7 | % 8 | \begin{align*} 9 | \hat{f}(n) 10 | = 11 | \frac{1}{2\pi} \int_{\minus \pi}^{\pi} % 12 | % 13 | f(e^{i\theta}) e^{\minus i n\theta} d\theta 14 | % 15 | \end{align*} 16 | % 17 | for all $n\in \Z$ (the integers). Put 18 | % 19 | \begin{align*} 20 | \Lambda_n f =\sum_{k=\minus n }^{n} \hat{f}(k). 21 | \end{align*} 22 | % 23 | Prove that % 24 | $\set{f \in {L}^2(T)}{\underset{n\infty}{\lim}\,\Lambda_n f \text{ exists}} $ % 25 | % 26 | is a dense subspace of $L^2(T)$ of the first category. % 27 | } 28 | % 29 | \begin{proof}% 30 | Let $f(\theta)$ stand for $f(e^{i\theta})$so that % 31 | % 32 | $L^2(T)$ is identified with a closed subset of $L^2([\minus \pi, \pi])$, % 33 | % 34 | hence the inner product % 35 | % 36 | \begin{align}\label{2.06. Inner product.} 37 | \hat{f}(n) 38 | = 39 | (f,e_n) 40 | = 41 | \frac{1}{2\pi} \int_{\minus \pi}^\pi f(\theta)e^{\minus in\theta}d\theta. 42 | \end{align} 43 | % 44 | We believe it is customary to write % 45 | \begin{align}\label{2.06. Rewriting of Lambda_n.} 46 | \Lambda_n (f) = (f, e_{\minus n}) + \cdots + (f, e_n). 47 | \end{align} 48 | Moreover, a well known (and easy to prove) result is % 49 | % 50 | \begin{align}\label{2.06. Orthonormality.} 51 | (e_{n}, e_{n'}) = [n=n'], \text{\ie } % 52 | \set{e_n}{n\in\Z} \text{ is an orthormal subset of } L^2(T). 53 | \end{align} 54 | % 55 | For the sake of brevity, we assume the isometric ($\equiv$) % 56 | identification % 57 | % 58 | $L^2\equiv (L^2)^\ast$. % 59 | % 60 | So, % 61 | % 62 | \begin{align}\label{2.06. Norm of Lambda_n.} 63 | \norm{\Lambda_n}^2 64 | \citeq{\ref{2.06. Rewriting of Lambda_n.}} 65 | \norm{e_{\minus n} + \cdots + e_n}^2 66 | \citeq{\ref{2.06. Orthonormality.}} 67 | \norm{e_{\minus n }}^2 + \cdots + \norm{e_{n}}^2 68 | \citeq{\ref{2.06. Orthonormality.}} 69 | 2n+1. 70 | \end{align} 71 | % 72 | %$\Lambda_n$ is therefore a bounded linear functional, of norm $\sqrt{2n+1}$. % 73 | We now assume, to reach a contradiction, that % 74 | % 75 | \begin{align} 76 | B \Def \set{f \in {L}^2(T)}{ 77 | %{\underset{n\infty} 78 | \sup\set{ 79 | \magnitude{\Lambda_n \,f}}{ 80 | \counting{n} 81 | } < \infty 82 | } % 83 | \end{align} 84 | % 85 | is of the second category. % 86 | So, the Banach-Steinhaus theorem \citeresultFA{2.5} asserts that the sequence % 87 | % 88 | $\singleton{\Lambda_n}$ % 89 | % 90 | is norm-bounded; which is a desired contradiction, since % 91 | % 92 | \begin{align} 93 | \norm{\Lambda_n} \citeq{ 94 | \ref{2.06. Norm of Lambda_n.} 95 | } \sqrt{2n + 1} \tendsto{n}{\infty}\infty. 96 | % 97 | \end{align} 98 | % 99 | We have just established that $B$ is actually of the first category; % 100 | and so is its subset % 101 | % 102 | L=$\set{f\in {L}^2(T)}{\lim_{n\longrightarrow\infty}\Lambda_n f \text{ exists}}$. % 103 | %. 104 | We now prove that $L$ is nevertheless dense in $L^2(T)$. % 105 | To do so, we let $P$ be $\text{span}\set{e_k}{k\in Z} $, % 106 | the collection of the trignometric polynomials % 107 | % 108 | $p(\theta)= \sum \lambda_k e^{ik\theta}$: % 109 | % 110 | Combining % 111 | % 112 | (\ref{2.06. Rewriting of Lambda_n.}) with % 113 | (\ref{2.06. Orthonormality.}) % 114 | % 115 | shows that % 116 | $\Lambda_n(p)= \sum \lambda_k$ for almost all $n$. % 117 | Thus, 118 | % 119 | \begin{align}\label{2.06. Basic inclusions.} 120 | P \subset L \subset L^2(T). 121 | \end{align} 122 | % 123 | We know from the Fejér theorem (the Lebesgue variant) that % 124 | $P$ is dense in $L^2(T)$. We then conclude, with the help of % 125 | % 126 | (\ref{2.06. Basic inclusions.}), % 127 | % 128 | that % 129 | % 130 | \begin{align} 131 | L^2(T) = \overline{P} = \overline{L}. 132 | \end{align} 133 | % 134 | %It is now clear that $\overline{L} = L^2(T)$. 135 | So ends the proof 136 | \end{proof} 137 | 138 | 139 | -------------------------------------------------------------------------------- /chapter_1/1_2.tex: -------------------------------------------------------------------------------- 1 | \textit{The convex hull of a set $A$ in a vector space $X$ is the set of all % 2 | convex combinations of members of $A$, that is the set of all sums % 3 | % 4 | $t_1 x_1 +\cdots +t_n x_n$ % 5 | % 6 | in which $x_i \in A,\, t_i \geq 0$, $\sum t_i = 1$; $n$ is arbitrary. 7 | % 8 | Prove that the convex hull of a set $A$ is convex and that is the intersection 9 | of all convex sets that contain $A$.} 10 | % 11 | \begin{proof} The convex hull of a set $S$ will be denoted by $\co{S}$. % 12 | Remark that $S \supseteq eq \co{S}$ % 13 | (to see that, take $t_1 = 1$ for each $x_1$ in $S$) and that % 14 | $\co{A} \supseteq eq \co{B}$ where $A \supseteq eq B$ (obvious). \\ 15 | \\ 16 | Our proof will directly derive from % 17 | % 18 | (i) $\Rightarrow$ (iv) % 19 | % 20 | in the following lemma, 21 | \renewcommand{\labelenumi}{(\roman{enumi})} 22 | \begin{quote} 23 | \textit{Let $S$ be a subset of a vector space $X$: Its convex hull $\co{S}$ % 24 | is convex and the following statements % 25 | \begin{enumerate} 26 | \item $S$ is convex; % 27 | \item{ 28 | % 29 | $s_1 S + \dots + s_n S = (s_1 + \cdots + s_n) S$ % 30 | % 31 | for all positive scalar variables $\mathit{s_1}, \dots, \mathit{s_n}$; % 32 | } 33 | \item{ 34 | % 35 | $\mathit{t_1} S + \dots + \mathit{t_n} S = S$ % 36 | % 37 | for all positive scalar variables $\mathit{s_1}, \dots, \mathit{s_n}$ % 38 | such that % 39 | % 40 | $s_1 + \cdots + s_n = 1$; % 41 | } 42 | \item $\co{S} = S$ 43 | \end{enumerate} 44 | are equivalent. % 45 | } 46 | \end{quote} 47 | % 48 | From now on, we skip the trivial case $S=\emptyset$ % 49 | then only consider nonempty sets. % 50 | To prove the first part, let $\mathit{a}$, $\mathit{b}$ % 51 | range over $\co{S}$so that % 52 | % 53 | $a = t_1 x_1 + \cdots + t_n x_n$ and % 54 | $b = t_{n+1} x_{n+1} + \cdots + t_{n+p} x_{n+p}$ % 55 | % 56 | for some $(\mathit{t_i}, \mathit{x_i})$. % 57 | % 58 | Every sum $sa + (1-s)b $ ($0\leq s \leq 1$) is then in the convex hull of % 59 | $\{\mathit{x_1}, \dots, \mathit{x_{n+p}}\}$, since % 60 | % 61 | \begin{align} 62 | sa + (1-s)b = \sum_{i=1}^n st_i x_i + \sum_{i=n+1}^{n+p} (1-s)t_i x_i 63 | \end{align} 64 | % 65 | and 66 | \begin{align} 67 | \sum_{i=1}^n st_i + \sum_{i=n+1}^{n+p} (1-s)t_i &= 68 | s\sum_{i=1}^n t_i +(1-s) \sum_{i=n+1}^{n+p} t_i = 1. 69 | \end{align} 70 | % 71 | In terms of sets $S$, this reads as follows, % 72 | % 73 | \begin{align} 74 | s\co{S} + (1-s)\co{S} \subset \co{S}; 75 | \end{align} 76 | % 77 | which was our fist goal. % 78 | We now aim at the equivalence % 79 | % 80 | (i) $\Rightarrow$ $\cdots$ $\Rightarrow$ (iv) $\Rightarrow$ (i): % 81 | % 82 | An easy proof by induction makes the implication (i) $\Rightarrow$ (ii) % 83 | directly come from (d) of the above exercise 1, chapter 1. % 84 | % 85 | (iii) is a special case of (ii), 86 | and the implication (iii) $\Rightarrow$ (iv) derives from the definition of % 87 | the convex hull. % 88 | % 89 | We now close the chain with (iv) $\Rightarrow$ (i), % 90 | by remarking that $S$ is convex whether $S = \co{S}$. % 91 | % 92 | The lemma being proved, let us establish the second part. \\ 93 | \\ 94 | To do so, we start from the convexity of $\co{A}$ then set % 95 | % 96 | $F = \{\co{A}\}$. % 97 | % 98 | We may enrich $F$ as follows, % 99 | \begin{align} 100 | B \in F \Rightarrow B \text{ is convex and contains }A. 101 | \end{align} 102 | Note that our initial predicate % 103 | ``[$F$ only encompasses] \textit{all convex sets that contain A}", % 104 | is now the special case % 105 | % 106 | \begin{align} 107 | B \in F \Leftrightarrow B \text{ is convex and contains }A. 108 | \end{align} 109 | % 110 | In any case, the key ingredient is that $\co{A} \in F$ implies % 111 | % 112 | \begin{align} 113 | \co{A} \supseteq \bigcap_{B \in F} B. 114 | \end{align} 115 | % 116 | Conversely, the next formula % 117 | % 118 | \begin{align} 119 | \co{A} \subset \co{B} \overset{(i) \Rightarrow (iv)}{=} B \quad (B \in F) 120 | \end{align} 121 | % 122 | is valid and implies % 123 | \begin{align} 124 | \co{A} \subset \bigcap_{B \in F} B. 125 | \end{align} 126 | % 127 | So ends the proof 128 | \end{proof} 129 | % END 130 | -------------------------------------------------------------------------------- /.gitignore: -------------------------------------------------------------------------------- 1 | ## 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Editors: 243 | # WinEdt 244 | *.bak 245 | *.sav 246 | 247 | # Texpad 248 | .texpadtmp 249 | 250 | # LyX 251 | *.lyx~ 252 | 253 | # Kile 254 | *.backup 255 | 256 | # gummi 257 | .*.swp 258 | 259 | # KBibTeX 260 | *~[0-9]* 261 | 262 | # TeXnicCenter 263 | *.tps 264 | 265 | # auto folder when using emacs and auctex 266 | ./auto/* 267 | *.el 268 | 269 | # expex forward references with \gathertags 270 | *-tags.tex 271 | 272 | # standalone packages 273 | *.sta 274 | 275 | # Makeindex log files 276 | *.lpz 277 | 278 | # Files that are not ready / not relevant 279 | dyadic* 280 | Annex* 281 | 282 | # Misc 283 | misc* 284 | 285 | *misc* 286 | 287 | # MacOS files 288 | .DS_Store 289 | 290 | # Passwords 291 | *.ini 292 | 293 | # Excel files 294 | *.xlsx -------------------------------------------------------------------------------- /chapter_6/6_6.tex: -------------------------------------------------------------------------------- 1 | % 2 | \renewcommand{\labelenumi}{(\alph{enumi})} 3 | {\it 4 | \begin{enumerate} 5 | \item Suppose that $c_m=\exp\{\minus (m!)!\}$, $m=0,\, 1,\, 2,\, \dots\, $. Does the series 6 | \begin{align*}{ 7 | \sum_{m=0}^\infty c_m (D^m\phi)(0) 8 | }\end{align*} 9 | converges for every $\phi\in C^{\,\infty} (R)$? 10 | \item Let $\Omega$ be open in $\R^n$, suppose $\Lambda_i\in \mathscr{D}'(\Omega)$, and suppose that all $\Lambda_i$ have their supports in some fixed compact $K\subset \Omega$. Prove that the sequence $\{\Lambda_i\}$ cannot converge in $\mathscr{D}'(\Omega)$ unless the orders of the $\Lambda_j$ are bounded. Hint: Use the Banach-Steinhaus theorem. 11 | \item Can the assumption about the supports be dropped in (b)? 12 | \end{enumerate}} 13 | \begin{proof} The answer is: no. Let us establish this assertion. Assume, to reach a contradiction, that the above series converges for every smooth $\phi:\, \R\to\,\C\,$. \\ 14 | \\ 15 | The sequence $\{c_m\, (D^m\phi)(0)\}$ so converges to $0$. Nevertheless, it is proved in [1.46] that $C^{\,\infty}(\Omega)$ is not locally bounded. In other words, it is always possible to excavate a $\phi\,$ for which the magnitude of the $m$-th derivative at $0$ is as large as we please\footnote{indeed [1.46] provides sufficient tools for constructive proof of this, see the $\phi_j-\check{\phi}_j$ involved in (\ref{2_3_phicheck}).}\eg greater than $1/c_m$. 16 | A desired contradiction is then reached. We now prove (b), again by contradiction.\\ 17 | \\ 18 | To do so we assume $\{\Lambda_j\}$ to converge to some $\Lambda$ of $\mathscr{D}'(\Omega)$ and we let $Q$ run through the compact sets of $\Omega\,$. Next, we define 19 | \begin{align}{\label{6_6_1} 20 | S(T,\, Q\,)\Def \{N\in \N, \, \exists C\in \R_+:\, \lvert T\phi\, \rvert \leq C\, \| \phi \|_N \,\text{ for all }\phi \text{ of } \mathscr{D}_Q \}\quad (T\in \mathscr{D}(\Omega))\quad . 21 | }\end{align} 22 | Such subset of $\N$ has a minimum $\omega(T,\, Q)$. The following value 23 | \begin{align}{\label{6_6_2} 24 | \omega (T\,) \Def \max\{ \omega(T,\, Q\,): \, Q\subset \Omega\, , \,\, Q \text{ compact }\}\leq \infty 25 | }\end{align} 26 | is then the order of $T$. Assume, to reach a contradiction, that, after passage to a subsequence, 27 | \begin{align}{\label{6_6_3} 28 | \omega (\Lambda_{j\,},\, Q_j\,) =j\quad (j=1,\, 2,\, 3,\, \dots) 29 | }\end{align} 30 | for some compact $Q=Q_j\,$. By (a) of [6.24], $Q_j$ cuts $\text{supp}\Lambda_j\,$, say in $p_j\,$. Since $K$ encloses $\text{supp}\Lambda_j\,$, $\{p_j\}$ tends, after passage to a subsequence, to some $p$ of $K\,$. 31 | Choose a positive scalar $r$ so that 32 | \begin{align}{\label{6_6_5} 33 | \overline{B}(p,\, r)\Def \{ x\in \R^n:\, \lvert x-p\,\rvert\leq r\}\subset \Omega \quad . 34 | }\end{align} 35 | Such closed ball $\overline{B}(p,\, r)$ is a compact subset of $\Omega$. By (b) of [6.5] (which refers to [1.46]) $\mathscr{D}_{\overline{B}(p,\, r)}$ is then a Fréchet space. It now follows from [2.6] that $\{\Lambda_j\}$ is equicontinuous on $\mathscr{D}_{\overline{B}(p,\, r)}\,$. There so exists\footnote{For more details, see Exercise 2.3.} a nonnegative integer $N$ such that 36 | \begin{align}{\label{6_6_6} 37 | \lvert \Lambda \phi \,\rvert \leq C\, \| \phi \|_N \quad (\phi \in \mathscr{D}_{\overline{B}(p,\, r)}) 38 | }\end{align} 39 | for some positive constant $C$. On the other hand, $\overline{B}(p,\, r)$ contains almost all the $p_j$'s. Hence 40 | \begin{align}{\label{6_6_7} 41 | \lvert \Lambda_N\,\phi \,\rvert > C\, \| \phi \|_N 42 | }\end{align} 43 | for some $\phi$ of $\mathscr{D}_{\overline{B}(p,r)}\,$. (b) is then established.\\ 44 | \\ 45 | To prove (c), we introduce a sequence $\{x_m:\, m\in \Z\}$ of $ \Omega$ that has no limit point. Let $\{ \alpha_m:\, m\in \Z\}$ be in $\N$ and so define\footnote{As $\Omega=\R\,$, the case $\alpha_m= m$ is the ``counterpart" of the series of (a) and the case $(x_m,\, \alpha_m)= (m,\, 0)$ is the \textsl{Dirac comb}.} 46 | \begin{align}{ 47 | \Lambda:\, \mathscr{D}(\Omega) \to & \, \C \qquad\qquad\qquad\qquad .\\ 48 | \phi \mapsto & \, \sum_{m=\minus \infty}^\infty (D^{\,\alpha_m}\phi)(x_m) \nonumber 49 | }\end{align} 50 | $\Lambda$ belongs to $\mathscr{D}'(\Omega)$, since $\{x_m\}$ has no limit point. Next, we easily check that 51 | \begin{align}{ 52 | \Lambda_j:\, \mathscr{D}(\Omega) \to & \, \C \qquad\qquad\qquad\qquad(j\in \N)\\ 53 | \phi \mapsto & \, \sum_{\lvert m\rvert \leq j} (D^{\,\alpha_m}\phi)(x_m) \nonumber 54 | }\end{align} 55 | is also a distribution and that $\{\Lambda_j\}$ tends to $\Lambda$ in $\mathscr{D}'(\Omega)$. Nevertheless, no $\Lambda_j$'s can have common support because $\{x_m\}$ has no limit point. Our assumption can therefore be dropped.\end{proof} 56 | 57 | 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 | 69 | 70 | -------------------------------------------------------------------------------- /chapter_2/2_3/2_3_2_uniform_bound.tex: -------------------------------------------------------------------------------- 1 | From now on, unless the contrary is explicitely stated, % 2 | we asume that $\phi$ only denotes an element of $C_K^\infty(\R)$. 3 | Let $f_n$ be a Lebesgue integrable function such that % 4 | % 5 | \begin{align}\label{2.3. Convergence on D_K (1).} 6 | \Lambda \phi = \underset{n \to \infty}{\lim} 7 | \int_{K} f_n\phi 8 | % 9 | \quad (\phi \in C_K^\infty(\R)). 10 | \end{align} 11 | % 12 | for some linear form $\Lambda$. % 13 | % 14 | Since $\phi$ vanishes outside $K$, we can suppose without loss of generality % 15 | that the support of $f_n$ lies in $K$. So, % 16 | % 17 | (\ref{2.3. Convergence on D_K (1).}) % 18 | % 19 | can be restated as follows, 20 | % 21 | \begin{align} 22 | \Lambda \phi = \underset{n \to \infty}{\lim} 23 | Lambda_n\phi 24 | % 25 | \quad (\phi \in C_K^\infty(\R)). 26 | \end{align} 27 | % 28 | Let $K_1, K_2, \dots, $ be compact sets that satisfy the conditions 29 | specified in \citeresultFA{1.44}. % 30 | $\D_K$ is $C_K^\infty(\R)$ topologized by the related seminorms % 31 | % 32 | $p_1, p_2, \dots$, see \citeresultFA{1.46, 6.2} and Exercise 1.16. 33 | % 34 | We know that $K\subset K_m$ for some index $m$ % 35 | (see Lemma 2 of Exercise 1.16): From now on, we only consider the indices 36 | $N \geq m$so that% 37 | % 38 | \renewcommand{\labelenumi}{(\alph{enumi})}% 39 | % 40 | \begin{enumerate} 41 | \item{ 42 | $p_N(\phi) = \norma{N}{\phi} \Def \max 43 | \set{\lvert D^{\alpha}\phi(x) \rvert}{\alpha \leq N, x \in \R}$, % 44 | for $\phi \in \D_K$; 45 | } 46 | \item{ 47 | The collection of the sets % 48 | $V_N = \set{ \phi \in \D_K}{ \norma{N}{\phi} < 2^{\minus N}}$ % 49 | is a (decreasing) local base of $\tau_K$, the subspace topology of $\D_K$; % 50 | see \citeresultFA{6.2} for a more complete discussion. 51 | } 52 | \end{enumerate} 53 | % 54 | \renewcommand{\labelenumi}{(\roman{enumi})} 55 | % 56 | % 57 | Let us specialize (\ref{2.3 g bounded operator (1).}) with % 58 | % 59 | $u=f_n$ and $ \phi \in V_m$ % 60 | % 61 | then conclude that $\bra{f_n}$ is bounded by $\norma{L^1}{f_n}$ on $V_m$: % 62 | Every linear functional $\bra{f_n}$ is therefore $\tau_K$-continuous, see % 63 | % 64 | \citeresultFA{1.18}. \newline\newline\noindent 65 | % 66 | To sum it up: % 67 | % 68 | \begin{enumerate} 69 | \item{$\D_K$, equipped the topology $\tau_K$, is a Fréchet space % 70 | (see \citeresultFA{section 1.46})}; 71 | % 72 | \item{Every linear functional $\bra{f_n}$ is continuous % 73 | with respect to this topology;} 74 | % 75 | \item{ 76 | $Lambda_n\phi \tendsto{n}{\infty} \Lambda \phi$ for all $\phi$, 77 | \ie 78 | $ \Lambda-\bra{f_n} \tendsto{n}{\infty}0$. 79 | } 80 | \end{enumerate} 81 | % 82 | With the help of \citeresultFA{[2.6] and [2.8]}, we conclude that % 83 | % 84 | $\Lambda$ is continuous 85 | % 86 | and that the sequence 87 | % 88 | $\singleton{\bra{f_n}}$ % 89 | % 90 | is equicontinuous. 91 | % 92 | So is the sequence % 93 | % 94 | $\singleton{\Lambda - \bra{f_n}}$, % 95 | % 96 | since addition is continuous. 97 | % 98 | There so exists $i, j$ such that, for all $n$, 99 | % 100 | \begin{align} 101 | \magnitude{\Lambda \phi} < 1/2 & \quad \text{if }\phi\in V_i, \\ 102 | \magnitude{\Lambda \phi - \lambda_n \phi} < 1/2 & \quad \text{if } \phi\in V_j. 103 | \end{align} 104 | % 105 | Choose $p\geq i, j $ so that $V_p \subset V_i \cap V_j$: % 106 | The latter inequalities imply that % 107 | % 108 | \begin{align} 109 | \magnitude{\Lambda \phi} & \leq \magnitude{\Lambda\phi - \Lambda_n \phi} + \magnitude{\Lambda \phi} < 1 \quad\text{if } \phi\in V_p. % 110 | \end{align} 111 | % 112 | Note that, at arbitrary $\epsilon > 0$, the mapping $\phi_\epsilon = 2^{\minus p}\phi / (\norma{p}{\phi} + \epsilon)$ is a member of $V_p$. % 113 | It is now clear that each below statement is true and implies the following one % 114 | % 115 | \begin{align} 116 | \magnitude{\Lambda_n \phi_\epsilon} & < 1 \\ 117 | \magnitude{\Lambda_n \phi} & < 2^p \left(\norma{p}{\phi} + \epsilon\right)\\ 118 | \magnitude{\Lambda_n \phi} & \leq 2^p \norma{p}{\phi} \\ 119 | \magnitude{\Lambda_n \phi} & \leq 2^p \left(\norma{\infty}{\phi} + \norma{\infty}{D \phi} + \cdots + \norma{\infty}{D^p \phi}\right). 120 | \end{align} 121 | % 122 | Finally, we apply \eqref{2.3. mean value inequality.} with $\phi, D\phi, \dots, D^p \phi$ to obtain % 123 | % 124 | \begin{align}\label{2.3. Bound with p and M (theoritical).} 125 | \magnitude{\Lambda_n \phi} \leq 2^p (p+1) \norma{\infty}{D^p \phi}. 126 | \end{align} 127 | % 128 | The first part is so proved, with some $p$ and some $ M = M(p)$. % 129 | \newline\newline\noindent 130 | % 131 | % END -------------------------------------------------------------------------------- /chapter_3/3_3.tex: -------------------------------------------------------------------------------- 1 | \textit{% 2 | Suppose X is a real vector space (without topology). % 3 | Call a point $x_0\in A \subset X$ an internal point of $A$ if % 4 | $A- x_0$ is an absorbing set. % 5 | % 6 | %\renewcommand{\labelenumi}{(\alph{enumi})} 7 | % 8 | \begin{enumerate} 9 | \item{% 10 | Suppose $A$ and $B$ are disjoint convex sets in $X$, % 11 | and $A$ has an internal point. % 12 | Prove that there is a nonconstant linear functional $\Lambda$ such that % 13 | $\Lambda(A)\cap \Lambda(B)$ contains at most one point. % 14 | (The proof is similar to that of Theorem 3.4) % 15 | }% 16 | \item{% 17 | Show (with $X=\R^2$, for example) that it may not possible to have % 18 | $\Lambda (A)$ and $\Lambda (B)$ disjoint, under the hypotheses of (a). % 19 | } 20 | \end{enumerate} 21 | } 22 | % 23 | \begin{proof}% 24 | Take $A$ and $B$ as in (a); the trivial case $B=\emptyset$ is discarded. % 25 | Since $A-x_0$ is absorbing, so is its convex superset % 26 | % 27 | $C= A-B - x_0 + b_0$ $(b_0 \in B)$. % 28 | % 29 | Note that $C$ contains the origin. % 30 | Let $p$ be the Minkowski functional of $C$. Since $A$ and $B$ are disjoint, % 31 | $b_0-x_0$ is not in $C$, hence $p(b_0-x_0) \geq1$. % 32 | We now proceed as in the proof of the Hahn-Banach theorem \citeresultFA{3.4} % 33 | to establish the existence of a linear functional % 34 | % 35 | $\Lambda: X\to \R$ such that % 36 | % 37 | \begin{align} 38 | \Lambda \leq p % 39 | \end{align} 40 | % 41 | and % 42 | % 43 | \begin{align} 44 | \Lambda(b_0-x_0) =1. 45 | \end{align}% 46 | % 47 | Then % 48 | % 49 | \begin{align} 50 | \Lambda a -\Lambda b + 1 =\Lambda (a-b+ b_0-x_0) \leq p (a-b+ b_0-x_0) \leq 1\quad (a\in A, b \in B). 51 | \end{align} 52 | % 53 | Hence % 54 | % 55 | \begin{align}\label{3_3_3} 56 | \Lambda a \leq \Lambda b . 57 | \end{align} 58 | % 59 | We now prove that $\Lambda (A) \cap \Lambda (B) $ contains at most one point. % 60 | Suppose, to reach a contradiction, that this intersection contains % 61 | % 62 | $y_1$ and $y_2$. % 63 | % 64 | There so exists % 65 | % 66 | $(a_i, b_i)$ %% 67 | % 68 | in $A\times B$ ($i=1, 2$) such that % 69 | % 70 | \begin{align}\label{3_3_4} 71 | \Lambda a_i= \Lambda b_i = y_i . 72 | \end{align} 73 | % 74 | Assume without loss of generality that $y_1< y_2$. Then, % 75 | %\begin{align} 76 | %y_1=\Lambda b_1 < \Lambda \left( \frac{1}{2} a_1\right ) + \Lambda \left( \frac{1}{2} a_2\right ) = \frac{1}{2} (y_1+y_2) \quad . 77 | %\end{align} 78 | % 79 | \begin{align}\label{3_3_5} 80 | 2\cdot y_1= \Lambda b_1+ \Lambda b_1 < \Lambda ( a_1 + a_2) = (y_1+y_2) \quad . 81 | \end{align} 82 | % 83 | Remark that $a_3= \frac{1}{2} (a_1+ a_2)$ lies in the convex set $A$. % 84 | This implies % 85 | % 86 | \begin{align} 87 | \Lambda b_1 \overset{(\ref{3_3_5})}{<} \Lambda a_3 \overset{(\ref{3_3_3})}{\leq} \Lambda b_1\quad ; 88 | \end{align} 89 | % 90 | which is a desired contradiction. (a) is so proved and we now deal with (b). % 91 | % 92 | \newline\newline\noindent 93 | % 94 | From now on, the space $X$ is $\R^2$. Fetch % 95 | % 96 | \begin{align} 97 | S_1 & \Def \set{(x, y )\in \R^2}{x\leq 0, y \geq 0}, \\ 98 | S_2 & \Def \set{(x, y )\in \R^2}{x>0, y > 0}, \\ 99 | A & \Def S_1\cup S_2, \\ 100 | B & \Def X\setminus A. 101 | \end{align} 102 | % 103 | Pick $(x_i, y_i)$ in $S_i$. % 104 | Let $t$ range over the unit interval, and so obtain % 105 | % 106 | \begin{align} 107 | t\cdot \left(\begin{array}{c}x_1 \\y_1\end{array}\right)+ 108 | (1-t) \cdot \left(\begin{array}{c}x_2 \\ y_2\end{array}\right)= 109 | \left(\begin{array}{c}t\cdot x_1+(1-t)\cdot x_2 \\t\cdot y_1 + (1-t)\cdot y_2\end{array}\right) 110 | \in \R\times \R_+\subset A. 111 | \end{align} 112 | % 113 | Thus, every segment that has an extremity in $S_1$ and the other one in $S_2$ % 114 | lies in $A$. % 115 | Moreover, each $S_i$ is convex. We can now conclude that $A$ is so. % 116 | The convexity of $B$ is proved in the same manner. Furthermore, % 117 | $A$ hosts a non degenerate triangle\ie $A^{\circ}$ is nonempty\footnote{% 118 | % 119 | For a immediate proof of this, remark that a triangle boundary is % 120 | compact/closed and apply [1.10] or 2.5 of \cite{BigRudin}. 121 | }: % 122 | $A$ contains an internal point. % 123 | % 124 | \newline\newline\noindent 125 | % 126 | Let $L$ be a vector line of $\R^2$. % 127 | In other words, $L$ is the null space of a linear functional % 128 | % 129 | $\Lambda: \R^2\to \R$ % 130 | % 131 | (to see this, take some nonzero $u$ in $L^\bot$ and set % 132 | % 133 | $\Lambda x= (x,u)$ % 134 | % 135 | for all $x$ in $\R^2$). One easily checks that both $A$ and $B$ cut $L$. % 136 | Hence % 137 | % 138 | \begin{align} 139 | \Lambda (L)=\{0\}\subset \Lambda (A)\cap \Lambda (B)\neq\emptyset\quad. 140 | \end{align} 141 | % 142 | So ends the proof. 143 | \end{proof} 144 | -------------------------------------------------------------------------------- /chapter_4/4_13.tex: -------------------------------------------------------------------------------- 1 | % 2 | { \CMUCS 3 | \begin{enumerate} 4 | \renewcommand{\labelenumi}{(\alph{enumi})} 5 | \item Suppose $T\in \mathscr{B}(X,Y), T_n \in \mathscr{B}(X,Y)\, $ for $n=1,\, 2, \, 3,\, \dots$, each $T_n\,$ has finite-dimensional range, and $\lim \| T-T_n\|=0\, $. Prove that $T$ is compact. 6 | \item Assume $Y$ is a Hilbert space, and prove the converse of (a): Every compact $T\in \mathscr{B}(X,Y)$ can be approximated in the operator norm by operators with finite-dimensional ranges. Hint: In a Hilbert space there are linear projections of norm 1 onto any closed subspace. (See theorems 5.16, 12.4.) 7 | \end{enumerate} 8 | } 9 | %: a 10 | \begin{proof} Since each $T_n$ is compact, (a) follows from (c) of [4.18]. Besides, we take the opportunity to alternatively prove that the compact operators subspace is norm closed.\\ 11 | \\ 12 | Reset every $T_n$ as a compact operator. Let $\{x_{\,0}^{\,i}:\, i\in \N\}$ be in $U$ the open unit ball of $X$. Since $T_1$ is compact, $\{x^i_0\}$ contains a subsequence $\{x^{\,i}_{\,1}:\, i\in \N\}$ such that $\{T_1x^{i}_{\,1}\}$ converges to a point $y_{\,1}$ of $Y$. The same reasoning can be recursively applied to $T_{n}$ and $\{x^{\,i}_{\, n-1}\}\subset U$ so that $\{T_{n\,}x^{\,i}_{\, n}\}$ tends to some $y_{n}$ of $Y$, as $\{x^{\,i}_{\, n}\}$ is a subsequence of $\{x^{\,i}_{\,n-1}\}$. Then 13 | \begin{align}\label{4_13_a_0} 14 | T_{n\,} x^{i}_p \,\underset{i\to \infty}{\longrightarrow}\,y_n\quad (p\> n=1,\, 2,\, 3,\, \dotsb )\quad . 15 | \end{align} 16 | Applied with $\{x^{\,i}_{\,n}:\, (n,i\,)\in\N^2\}$, a Cantor's diagonal process therefore provides a subsequence $\{\tilde{x}_{\, j}:\, j\in \N\} $ such that 17 | \begin{align}\label{4_13_a_1} 18 | &T_{j\,} \tilde{x}_k \underset{k\to \infty}{\longrightarrow}y_{j}\quad ;\\ 19 | &T_{j\,} \tilde{x}_{\, j} \underset{j\to \infty}{\longrightarrow} y_{j}\quad . 20 | \end{align} 21 | %Since $\|T-T_k\|\underset{k\to \infty}{\longrightarrow}0$, (\ref{4_13_a_1}) implies that 22 | %\begin{align}\label{4_13_a_2} 23 | %T \tilde{x}_k -y_k \underset{k\to \infty}{\longrightarrow}0\quad. 24 | %\end{align} 25 | We now easily obtain 26 | \begin{align}\label{4_13_a_3} 27 | \| T_{j\,} \tilde{x}_{j\,} -T_{\, k} \tilde{x}_{\, k}\| \leq 28 | \| T_{j\,} \tilde{x}_{j\,}- y_{j\,}\| + 29 | \| y_{j\,} -T_{j\,} \tilde{x}_{k\,}\| + 30 | \| T_{j\,} - T_{ k\,}\| \underset{k> j\to \infty}{\longrightarrow}0\quad . 31 | \end{align} 32 | $\{T_{ j\,} \tilde{x}_{\, j} \}$ is then a Cauchy sequence. So is $\{T \,\tilde{x}_{\, j}\}$, since $\| T-T_{j}\,\| \to 0$. On the other hand, $Y$ is complete: (a) is then proved and we now establish the counterpart in a Hilbert space.\\ 33 | \\ 34 | %: b 35 | Fix $\epsilon$ as a positive scalar. Since $T$ is compact, $Y$ contains a finite set $C$ such that 36 | \begin{align}\label{4_13_b_1} 37 | T(U\,)\subset \bigcup_{c\in C} B(c,\, \epsilon)\quad . 38 | \end{align} 39 | As a Hilbert space, $Y$ contains a \textsl{maximal orthonormal set} (or \textsl{Hilbert basis}) $M$. This implies that $\text{span}(M)$ is dense in $Y$; \cf 4.18 \& 4.22 of \cite{BigRudin}. The finiteness of $C$ forces $M$ to enclose a finite set $S$ so that 40 | \begin{align}\label{4_13_b_2} 41 | \forall c\in C, \, \exists s(c\,) \in \text{span} (S\,):\, \|c - s(c\,)\| <\epsilon\quad . 42 | \end{align} 43 | Let $x$ be in $U$. It follows from (\ref{4_13_b_1}) that 44 | \begin{align}\label{4_13_b_3} 45 | \|Tx - c_x \| < \epsilon 46 | \end{align} 47 | for some $c_x$ of $C$. We now combine (\ref{4_13_b_2}) and (\ref{4_13_b_3}) to obtain 48 | \begin{align} 49 | \|Tx - s(c_x) \| \leq \| Tx - c_x \| + \| c_x - s(c_x)\| < 2\epsilon 50 | \end{align} 51 | As a finite-dimensional subspace, $\text{span}(S\,)$ is closed (see footnote 4, Exercise 1.10). We so obtain 52 | \begin{align} 53 | Y=\text{span}(S\,)\oplus \text{span}(S\,)^\bot \quad , 54 | \end{align} 55 | by [12.4]. There so exists a unique projection projection $\pi=\pi(\epsilon)$ of $Y$ onto itself (see [5.6] for the definition) such that 56 | \begin{align} 57 | \pi (Y) = \text{span}(S\,)\, ,\, \, (I-\pi)(Y\,) = \text{span}(S\,)^\bot\quad . 58 | \end{align} 59 | It is easily checked that $\pi$ has norm $1$. Moreover, 60 | \begin{align} 61 | \pi s = s \quad (s\in\text{span}(S\,))\quad . 62 | \end{align} 63 | Thus, 64 | \begin{align} 65 | (I-\pi) (Tx\,)= (I-\pi ) (Tx - s(c_x)) \quad (x\in U\,)\quad . 66 | \end{align} 67 | Then, 68 | \begin{align} 69 | \|( I- \pi )(T x) \| \leq \| I- \pi \| \, \| Tx -s(c_x)\| < 4 \epsilon \quad (x\in U)\quad 70 | \end{align} 71 | (the fact that $\pi$ has norm $1$ is hidden in the right side inequality). We have just so proved that 72 | \begin{align} 73 | \| T-\pi\circ T\, \| \in \underset{\epsilon \sim 0}{O}(\epsilon) \quad . 74 | \end{align} 75 | That is particularly true whether $\epsilon=\epsilon_0,\, \epsilon_1,\, \epsilon_2,\, \dots ,\, \epsilon_n \underset{n\to \infty}{\longrightarrow} 0\,$. Let so $T_n$ be $ \pi(\epsilon_n) \circ T\,$ and conclude that these (compact) operators approximate $T$ in the desired fashion\ie 76 | \begin{align} 77 | \| T-T_n \| \underset{n \to \infty}{\longrightarrow} 0\quad . 78 | \end{align} 79 | \end{proof} 80 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # Forewords 2 | 3 | A long time ago, I solved a (tiny…) bunch of problems from the Walter Rudin's 4 | *Functional Analysis*. 5 | 6 | I am carefully rewriting my solutions, aiming at the cleanest possible result. 7 | 8 | # Requirements 9 | 10 | - TeX 11 | - XeLaTeX 12 | - packages 13 | - AMS packages: 14 | - amsmath 15 | - physics 16 | - amssymb 17 | - mathrsfs 18 | - amsthm 19 | - mathspec 20 | - cite 21 | - geometry 22 | - xltxtra 23 | - xunicode 24 | - xetex 25 | - Optional: Relevant fonts: 26 | - CMU Serif 27 | - CMU Typewriter Text Light 28 | 29 | You may want to install the latest 30 | [TexLive](https://www.tug.org) 31 | 32 | # Legal issues 33 | 34 | - I do not broadcast nor sell any copy of *Functional Analysis* 35 | - I do not make available any content from this book, 36 | **excepted** problems statements 37 | 38 | Feel free to get copies of this great book by yourself. 39 | *Functional Analysis*' ISBNs are 40 | 41 | - ISBN-10: 0070542368 42 | - ISBN-13: 978-0070542365 43 | 44 | ## Files 45 | 46 | - FA_DM.pdfOutput from [Xelatex](https://www.tug.org) compilation.You can also get an html output from 47 | [Hevea](http://hevea.inria.fr) compilation, see HOWTO. 48 | - FA_DM.tex 49 | - FA_mainmatter.tex 50 | - FA_chapter_1.tex 51 | - chapter_1/ 52 | - 1_1.tex 53 | Basic results that straightforwardly follow from the axioms 54 | as given as in section 1.4. 55 | - 1_2.tex 56 | The convex hull of a set A is convex and that is the intersection of 57 | all convex set(s) that contain A. 58 | - 1_3.tex 59 | - 1_4.tex 60 | - 1_5.tex 61 | - 1_6.tex 62 | - 1_7.texI choose to start with this because it is a lovely result, 63 | since it connects a topological result 64 | (*to be metrizable or not to be*) with number theory. 65 | - 1_9.texContinuousness, openness of a linear mapping. 66 | - 1_10.texContinuousness, openess of a linear mapping onto a finite dimensional 67 | space.Not trivial, since the domain may be infinite-dimensional. 68 | - 1_14.texAlternative ways to the define topology of the test functions space D_K, 69 | in the special case K=[0, 1]. 70 | - 1_16.texThis is about showing that a function test topology is independent 71 | from the "supremum seminorms" we consider. 72 | It is then more than an exercise, it should be regarded as a very part of 73 | the textbook (sections 1.44, 1.46). 74 | - 1_17.tex 75 | Given a multi-index $\alpha$, the differential operators $D^\alpha$ is 76 | continuous in the test functions topology. 77 | - chapter_2/ 78 | - 2_3/ 79 | - 2_3.texIn $D_K$, some Lebesgue integrable functions converge to $\delta'$, 80 | which is not a Radon measure. 81 | Their weak derivatives converge to $\delta''$. 82 | - 2_3_0_labels.texReferences 83 | - 2_3_0_lemma.tex.Specialization of mean value theorem. 84 | - 2_3_1_radon_measures.texStart by looking into $C_0(R)^\ast$. 85 | - 2_3_2_uniform_bound.tex. 86 | $D_K$ topology allows equicontinuity. 87 | - 2_3_3_example_1.texConvergence of Lebesgue integrable functions to $\delta'$ in $D_K$. 88 | - 2_3_4_example_2.texTheir weak derivatives converge to $\delta''$. 89 | - TODO: Add proof that the Dirac derivative is not a Radon measure, 90 | to complete the figure. 91 | - 2_6.texThe Banach-Steinhaus theorem applied to $L^2(T)$ the $L2$ functions of the 92 | unit circle of $C$: 93 | The series made on the Fourier coefficients may diverge. 94 | Nevertheless, convergence holds in a dense space 95 | (by the the Fejér theorem, for instance). 96 | TODO: (?) Add some comparison with the Carleson's theorem. 97 | - 2_9.texGiven normed spaces (X, Y, Z), any continuous bilinear mapping 98 | $B: (x, y) \in X\times Y \mapsto B(x, y) \in Z$ is bounded. 99 | Thoses spaces need not be complete. An easy example is given by 100 | $B(f, g)= fg$ where $f$ and $g$ keep in $C_c(R)$. 101 | - 2_10.texA bilinear mapping that is continuous at the origin is continuous. 102 | Actually, 2.09 contains all the relevant material. 103 | In the more general topological vector space context proof, 104 | the norm is replaced by Minkowski functionals on balanced open sets. 105 | - 2_12.texA bilinear mapping that is separately continuous, but not continuous. 106 | - 2_15.texIn a F-space X, the complement C of a subgroup Y is not 107 | of the first category, unless X=Y. 108 | To sum it up: If Y is a proper subspace, 109 | then Y is of the first category BUT its complement C 110 | is of the second category, as X is. 111 | - 2_16.tex 112 | A simpler version of the closed graph theorem. 113 | Roughly speaking, compactness replaces completeness. Compactness cannot 114 | be dropped: A counterexample is given. 115 | - chapter 3/ 116 | - 3_4.tex 117 | - 3_11.tex 118 | Meagerness of the polar (in the infinite dimensional case) 119 | of the neighborhoods of the origin: 120 | Hahn-Banach theorem and polar. We only involve the weak star 121 | -closedness of the polar, not its weak star-compactness! 122 | - chapter 4/ 123 | - 4_1.tex 124 | - 4_13.tex 125 | - 4_15.tex 126 | - chapter 6/ 127 | - 6_1.tex 128 | - 6_9.tex 129 | - 6_9.tex 130 | - LICENSE 131 | - HOWTO 132 | -------------------------------------------------------------------------------- /chapter_6/6_1_Mat.tex: -------------------------------------------------------------------------------- 1 | We will actually prove more by showing that $\mathscr{D}(\Omega)$ is separable for each nonempty open subset $\Omega\,$ of $\R^n\,$. 2 | \begin{proof} 3 | The following is split in three parts. The first one is about the above requested result: % 4 | That was our first part. We now go further by proving the separability of $\mathscr{D}(\Omega)\,$. To do so, we keep $(\alpha,\, j\,)$ in $\N^n\times \N\,$. Remark that $S$ encloses $\text{supp}(D^{\, \alpha}f\,) $: according to the first part, there exists a sequence $\{ P_{\alpha,\, j}:\, j\in \N \} \subset\R [X_1,\dotsc ,\, X_n]$ such that 5 | \begin{align}\label{6_1_4} 6 | \| D^{\, \alpha} f- \psi P_{\alpha,\, j \,} \|_\infty \underset{j\infty}{\longrightarrow} 0 \quad . 7 | \end{align} 8 | Now let $m$ range over $\{1,\, 2,\,3,\, \dots\}$ and set $W_{m,\, j\,}$ in $\mathscr{D}(\Omega)$ as follows 9 | \begin{align}\label{6_1_5a} 10 | D^{\,\minus \alpha} \phi \in \mathscr{D}(\Omega):\, D^{\,\alpha} D^{\,\minus \alpha} \phi= \phi\quad . 11 | \end{align} 12 | %\begin{align}\label{6_1_5c} 13 | %D^{\,\minus \alpha} \phi(x\,) \Def 14 | %\underset{ \alpha_1 \text{ time(s)} }{ 15 | % \underbrace{ \int_{\minus \infty}^{x_1} \dotsc\int_{\minus\infty}^{x_1}} 16 | % } 17 | % \,\,\dotsc\, 18 | % \underset{ \alpha_n \text{ time(s)} }{ 19 | % \underbrace{ \int_{\minus \infty}^{x_n} \dotsc\int_{\minus\infty}^{x_n}} 20 | % } 21 | % \,\, \phi \quad (x\in \R^n)\quad . 22 | % \end{align} 23 | \begin{align}\label{6_1_5} 24 | W_{m,\, j\,}(x\,)\Def D^{\,\minus (m,\dotsc,\,m)} ( \psi P_{(m,\dotsc, \, m),\, j\,}) 25 | \end{align} 26 | By (\ref{6_1_4}), there exists a natural number $k(\!m)$ such that 27 | \begin{align}\label{6_1_6} 28 | \|D^{(m,\dotsc,\, m)} (f -W_{m,\, j}) \|_\infty < 1/m \quad (j\geq k(\!m) )\quad . 29 | \end{align} 30 | Assume without loss of generality that $S$ has diameter $1$so that (\ref{6_1_6}) yields 31 | \begin{align}\label{6_1_7} 32 | \|D^{\, \lambda } (f -W_{m,\, k(\!m)} ) \|_\infty < 1/m \quad ( \lvert \lambda \rvert \leq m )\quad , 33 | %\max \{ \lvert D^{\, \lambda}(f- W_{m,\, k(\!m) })(x\,)\rvert :\, x\in S,\, \lvert \lambda\rvert \leq m\,\} < 1/m \quad 34 | \end{align} 35 | by the mean value theorem. In other words $(\text{remark that }\text{supp} (f -W_{m,\, k(\!m)}) \subset S\,)$, 36 | \begin{align}\label{6_1_8} 37 | f- W_{m,\, k(\!m)} \in U_m \Def \{ \phi \in \mathscr{D}_S :\, \| \phi \|_m< 1/m \} \supseteq U_{m+1}\supseteq \dotsb \quad( m=1,\, 2,\, 3,\, \dots) \quad . 38 | \end{align} 39 | Pick $W$ in $\beta$ (see (b) of [6.3]): $W\cap\mathscr{D}_S$ contains a neighbourhood of $0$. Hence $W$ contains some $U_{m\,}$, for $m$ sufficiently large. Thus 40 | \begin{align}\label{6_1_9} 41 | W_{m,\, k(\!m) } \underset{m\infty} {\longrightarrow} f \quad (\text{in } \mathscr{D}(\Omega))\quad . 42 | \end{align} 43 | We have so established that the $W_{m,\, k(\!m)}$'s family is dense in $\mathscr{D}(\Omega)$. We now aim to disclose a countable set $\tilde{W}\,$ that has the same property.\\ 44 | \\ 45 | Choose $\delta $ in $\R_+$ and fetch any $W_{m,\, k(\!m)}$. Let $X$ be $(X_1,\dotsc,\, X_n)$ and express $P_{(m,\dotsc,\, m\,),\, k(\!m)}$ as 46 | \begin{align}\label{6_1_10} 47 | P(X\,)=\sum_{\lvert \gamma \rvert \leq d} p_\gamma \cdot X^{\,\gamma} \quad . 48 | \end{align} 49 | Since $\bar{\Q}=\R$, $\Q[X\,]$ hosts some ${\Q(X\,)=\sum_{\lvert \gamma \rvert \leq d} q_\gamma \cdot X^{\,\gamma}}$ such that $\lvert p_\gamma - q_\gamma\rvert < \delta\,$ for all $\gamma$. Thus, 50 | \begin{align}\label{6_1_11} 51 | \lvert P(x\,)-Q(x\,) \,\rvert \leq \sum_{\lvert \gamma \rvert \leq d} \lvert p_\gamma-q_\gamma\rvert\, \lvert x\,\rvert^{\lvert \gamma\rvert} \leq \delta \sum_{l \leq d} \binom{l+n-1}{n-1}\, \| \, x\, \|^{l}_\infty \quad (x\in \R^n)\quad . 52 | \end{align} 53 | Since $S$ is bounded, we so obtain 54 | \begin{align}\label{6_1_12} 55 | \|\psi (P- Q) \|_\infty \in O(\delta) \quad . 56 | \end{align} 57 | Now define $\tilde{W}_m\,$ in terms of $Q$ as $W_{m,\, k(\!m)}$ was defined in terms of $P$, and consider the integrations made in (\ref{6_1_5}): each $D^{\,\lambda} \tilde{W}_m\,\, (\lvert \lambda \rvert\leq m)$ can be obtained from some of them. So (\ref{6_1_12}) yields 58 | \begin{align}\label{6_1_13} 59 | \| D^{\,\lambda} (W_{m,\, k(\!m)} - \tilde{W}_m ) \|_\infty \in O(\delta)\quad (\lvert \lambda \rvert\leq m) \quad . 60 | \end{align} 61 | To be more specific, these $\lambda$'s only exist in finite amount, so the big O can be assumed to be the same for all them. Since $\delta$ was arbitrary, combining (\ref{6_1_9}) with (\ref{6_1_13}) establishes the density of the all $\tilde{W}_m$'s family $\tilde{W}$.\\ 62 | \\ 63 | Furthermore, each member of $\tilde{W}$ is only made of two ingredients: $\psi$ and a polynomial of $\Q[X]$. The mapping $\psi$ is attached to some $K_i$ and $\Q[X]$ inherits countableness from $\Q$. Note that the ``integrations packs" of (\ref{6_1_5}) only exist in countable amount. % 64 | Our $\tilde{W}$ is then countable. % 65 | \end{proof} 66 | 67 | 68 | 69 | 70 | 71 | 72 | 73 | 74 | 75 | 76 | 77 | 78 | -------------------------------------------------------------------------------- /chapter_1/1_3.tex: -------------------------------------------------------------------------------- 1 | \renewcommand{\labelenumi}{(\alph{enumi})} 2 | \textit{ 3 | Let be X as topological vector space. All sets mentioned below are understood to be the subsets of X. Prove the following statements: 4 | \begin{enumerate} 5 | \item The convex hull of every open set is open. 6 | \item If X is locally convex then the convex hull of every bounded set is bounded. 7 | \item If A and B are bounded, so is A+B. 8 | \item If A and B are compact, so is A+B. 9 | \item If A is compact and B is closed, then A+B is closed. 10 | \item The sum of two closed sets may fail to be closed. 11 | \end{enumerate} 12 | } 13 | %: (a) 14 | \begin{proof} 15 | \renewcommand{\labelenumi}{(\alph{enumi})} 16 | \begin{enumerate} 17 | \item% 18 | Pick an open set $A$ then let the variables $\mathit{V}_i$ % 19 | ($i=1, 2, \dots$) run through all open subsets of $A$so that % 20 | % 21 | \begin{align} 22 | \co{A} \subset 23 | \bigcup_{t_i} \, % 24 | (t_1V_1 + \cdots + t_i V_i + \cdots) 25 | \subset \co{A} 26 | \end{align} 27 | % 28 | given all convex combinations % 29 | % 30 | $t_1V_1 + \cdots + t_i V_i + \cdots $. % 31 | % 32 | We know from Section 1.7 of \cite{FA} that those sums are open; % 33 | which achieves the proof. % 34 | % 35 | %:(b) 36 | \item Provided a bounded set $E$, % 37 | pick $V$ a neighbourhood of $0$: By (b) of Section 1.14 in \cite{FA}, % 38 | $V$ contains a convex neighbourhood of $0$, say $W$. % 39 | % 40 | There so exists a positive scalar $s$ such that 41 | % 42 | \begin{align} 43 | E \subset tW \subset tV \quad (t>s); 44 | \end{align} 45 | % 46 | which yields % 47 | % 48 | \begin{align} 49 | \co{E} \subset \co{tW} = t\co{W} = t W \subset tV. 50 | \end{align} 51 | % 52 | So ends the proof. % 53 | % 54 | %:(c) 55 | \item At fixed $V$, neighbourhood of the origin, % 56 | we combine the continuousness of $+$ with Section 1.14 of \cite{FA} % 57 | to conclude that there exists $U$ a balanced neighborhood of the origin % 58 | such that % 59 | % 60 | \begin{align} 61 | U+U\subset V. 62 | \end{align} 63 | % 64 | Moreover, by the very definition of boundedness, % 65 | $A \subset r U$ for some positive scalar $r$. % 66 | Similarly, $B \subset s U$ for some positive $s$. % 67 | % 68 | Finally, 69 | % 70 | \begin{align} 71 | A+B \subset rU + sU \subset tU + tU \subset tV \quad (t > r, s), 72 | \end{align} 73 | % 74 | since $U$ is balanced. So ends the proof. % 75 | % 76 | %:(d) 77 | \item First, $A$ and $B$ are compact: So is $A\times B$. % 78 | Next, $+$ maps continuously $A\times B$ onto $A+B$. % 79 | In conclusion, $A+B$ is compact. % 80 | % 81 | %:(e) 82 | \item From now on, we assume that neither $A$ nor $B$ is empty, % 83 | since otherwise the result is trivial. % 84 | Now pick $c\in X$ outside $A+B$: % 85 | The result will be established by showing that $c$ is not in the closure % 86 | of $A+B$. \\ 87 | \\ 88 | To do so, we let the variable $\mathit{a}$ range over $A$: % 89 | Every set $a+B$ is closed as well, see Section 1.7 of \cite{FA}. % 90 | % 91 | Trivially, $a+B \neq c$: By Section 1.10 of \cite{FA}, % 92 | there so exists $V=V(a)$ a neighborhood of the origin such that % 93 | % 94 | \begin{align}\label{separation} 95 | (a+B + V) \cap (c+V) = \emptyset. 96 | \end{align} 97 | % 98 | Moreover, there are finitely many $a+V$, say $a_1 + V_1, a_2 + V_2, \dots$, % 99 | whose union $U$ contains the compact set $A$. Therefore, % 100 | % 101 | \begin{align}\label{U + B encloses A + B} 102 | A+B \subset U + B. 103 | \end{align} 104 | % 105 | Now define % 106 | \begin{align} 107 | W \triangleq V_1 \cap V_2 \cap \cdots, 108 | \end{align} 109 | % 110 | so that 111 | % 112 | \begin{align} 113 | (a_i + B + V_i) \cap (c + W) \overset{(\ref{separation})}{=} \emptyset % 114 | \quad (i = 1, 2, \dots). 115 | \end{align} 116 | % 117 | As a conclusion, $c$ is not in the closure of $U+B$. % 118 | Finally, (\ref{U + B encloses A + B}) asserts that % 119 | $c$ is not in $\overline{A+B}$ either; which achieves the proof. \\ 120 | \\ 121 | \textbf{Corollary}: If $B$ is the closure of a set $S$, then % 122 | % 123 | \begin{align} 124 | A+B \subset \overline{A+S} \subset \overline{A+B} = A + B 125 | \end{align} 126 | % 127 | by (b) of Section 1.13 of \cite{FA} (since $A$ is closed; % 128 | see Section 1.12, from the same source). % 129 | The special case $A = \{x\}$, $B=X$ % 130 | will occur in the proof of Exercise 15 in chapter 2. % 131 | % 132 | %:(f) 133 | \item The last proof will consist in exibhiting a counterexample. % 134 | To do so, let $f$ be any continuous mapping of the real line such that % 135 | \renewcommand{\labelenumii}{(\roman{enumii})} 136 | \begin{enumerate} 137 | \item $f(x) + f(\minus x) \neq 0$ \quad ($x \in \R$); 138 | \item $f$ vanishes at infinity. 139 | \end{enumerate} 140 | For instance, we may combine (ii) with $f$ even and $f>0$ by setting % 141 | % 142 | $f(x) = 2^{\minus |x|}$, % 143 | $f(x) = e^{\minus x^2}$, % 144 | $f(x) = 1/(1+|x|)$, \dots, % 145 | % 146 | and so on. \\ 147 | \\ 148 | As a continous function, $f$ has closed graph $G$, see [2.14] of \cite{FA}. % 149 | % 150 | Moreover, (i) implies that the origin % 151 | % 152 | $(0, 0) \neq \left(x-x, f(x)+ f(\minus x)\right)$ % 153 | % 154 | is not in $G+G$. % 155 | % 156 | On the other hand, 157 | % 158 | \begin{align} 159 | \{ \left(0, f(n) + f(\minus n)\right): n=1, 2, \dots\} \subset G + G. 160 | \end{align} 161 | % 162 | 163 | % 164 | Now the key ingredient is that % 165 | % 166 | \begin{align} 167 | \left(0, f(n)+f(\minus n)\right) \overset{(ii)}{\tendsto{n}{\infty}} (0, 0). 168 | \end{align} 169 | % 170 | We have so constructed a sequence in $G+G$ that converges outside $G+G$. % 171 | So ends the proof. 172 | \end{enumerate} 173 | \end{proof} 174 | %END 175 | -------------------------------------------------------------------------------- /chapter_3/3_11.tex: -------------------------------------------------------------------------------- 1 | \textit{\noindent 2 | Let $X$ be an infinite-dimensional Fréchet space. % 3 | Prove that $X^\ast$, with its $\weakstar$topology, % 4 | is of the first category in itself. 5 | } 6 | \newline\newline\noindent 7 | This is actually a consequence of the lemma below, % 8 | which we prove first. % 9 | The proof that $X^\ast$ is of the first category in itself comes right after, % 10 | as a corollary.% 11 | % 12 | \paragraph{Lemma.}{\it % 13 | f $\varit{X}$ is an infinite dimensional topological vector space whose dual % 14 | % 15 | $\mathit{X^{\,\ast}}$ % 16 | % 17 | separates points on $X$, then the polar 18 | % 19 | \begin{align} 20 | \varit{K}_A\Def \{ \Lambda \in \mathit{X^\ast}:\, \magnitude{\Lambda} \leq \mathrm{1} \text{ on } A\} 21 | \end{align} 22 | % 23 | of any absorbing subset $\varit{A}$ is a $\varit{\weakstar}$closed set that has empty interior. 24 | } 25 | % 26 | \begin{proof}% 27 | Let $x$ range over $X$. The linear form % 28 | % 29 | $\Lambda \mapsto \Lambda x$ % 30 | % 31 | is $\weakstar$continuous, see \citeresultFA{3.14}. % 32 | Therefore, % 33 | % 34 | $P_x= \set{\Lambda \in X^\ast}{\magnitude{\Lambda x} \leq 1}$ % 35 | % 36 | is $\weakstar$ closed: % 37 | %In particular, every $P_a$ ($a \in A$) is $\weakstar$closed: % 38 | As the intersection of $\set{P_a}{a\in A}$, % 39 | $K_A$ is also a $\weakstar$closed set. % 40 | We now prove the second half of the statement. % % 41 | % 42 | \newline\newline\noindent 43 | % 44 | From now on, $X$ is assumed to be endowed with its weak topology: % 45 | $X$ is then locally convex, but its dual space is still % 46 | % 47 | $X^\ast$ (see \citeresultFA{3.11}). % 48 | % 49 | Put % 50 | % 51 | \begin{align} 52 | W_{F, x} \Def \bigcap_{x\in F} \set{\Lambda \in X^\ast}{\magnitude{\Lambda x} < r_x} \quad\quad (r_x > 0) 53 | \end{align} 54 | % 55 | where $F$ runs through the nonempty finite subsets of $X$. % 56 | % 57 | Clearly, the collection of all such $W$ is a local base of $X^\ast$. % 58 | Pick one of those $W$ and remark that the following subspace % 59 | % 60 | \begin{align} 61 | M \Def \text{span}(F) 62 | \end{align} 63 | % 64 | is finite dimensional. % 65 | Assume, to reach a contradiction, that $A\subset M$. % 66 | So, every $x$ lies in $t_xM=M$ for some $t_x>0$, since $A$ is absorbing. % 67 | As a consequence, $X$ is the finite dimensional space $M$, % 68 | which is a desired contradiction. 69 | % 70 | We have just established that $A\not\subset M$: % 71 | Now pick $\varit{a}$ in $A\setminus M$ and so conclude that %f 72 | % 73 | \begin{align} 74 | b\Def \frac{a}{t_a}\in A 75 | \end{align} 76 | % 77 | Remark that $b\notin M$ (otherwise, $a = t_a b \in t_a M=M$ would hold) % 78 | and that $M$, as a finite dimensional space, is closed % 79 | (see \citeresultFA{1.21 (b)} for a proof): % 80 | By the Hahn-Banach theorem \citeresultFA{3.5}, % 81 | there exists $\Lambda_a$ in $X^\ast$ such that 82 | % 83 | \begin{align}\label{3.11. Polar_4.} 84 | \Lambda_a b > 2 85 | \end{align} 86 | % 87 | and 88 | % 89 | \begin{align}\label{3.11. Polar_4bis.} 90 | \Lambda_a (M) = \singleton{0}. 91 | \end{align} 92 | % 93 | The latter equality implies that $\Lambda_a$ vanishes on $F$; % 94 | hence $\Lambda_a$ is an element of $W$. % 95 | On the other hand, given an arbitrary $\Lambda \in K_A$, % 96 | the following inequalities % 97 | % 98 | \begin{align} 99 | \magnitude{\Lambda_a b + \Lambda b} 100 | \geq 101 | 2 - \magnitude{\Lambda b} 102 | > 103 | 1. 104 | \end{align} 105 | % 106 | show that $\Lambda + \Lambda_a$ is not in $K_A$. % 107 | % 108 | We have thus proved that 109 | % 110 | \begin{align} 111 | \Lambda + W\not\subset K_A. 112 | \end{align} 113 | % 114 | Since $W$ and $\Lambda$ are both arbitrary, this achieves the proof. % 115 | \end{proof} 116 | % 117 | %\newline\newline 118 | \noindent 119 | We now give a proof of the original statement. % 120 | % 121 | \paragraph{Corollary.}% 122 | {\it If $\varit{X}$ is an infinite-dimensional Fréchet space, % 123 | then $\mathit{X^\ast}$ is meager in itself. 124 | } 125 | % 126 | \begin{proof}% 127 | From now on, $X^\ast$ is only endowed with its $\weakstar$topology. % 128 | Let $d$ be an invariant distance that is compatible with the topology of $X$, % 129 | so that the following sets 130 | % 131 | \begin{align} 132 | B_n \Def \set{x\in X}{d(0, x) < 1/n}\quad\quad (\counting{n}) 133 | \end{align} 134 | % 135 | form a local base of $X$. % 136 | % 137 | If $\Lambda$ is in $X^\ast$, then % 138 | % 139 | \begin{align} 140 | \magnitude{\Lambda} \leq m \text{ on } B_n 141 | \end{align} 142 | % 143 | for some $(n, m) \in \singleton{1, 2, 3, \dots}^2$, see \citeresultFA{1.18}. % 144 | % 145 | Hence, $X^\ast$ is the countable union of all % 146 | % 147 | \begin{align}\label{3.11. Countable union.} 148 | m\cdot K_n \quad\quad (\counting{m,n}), 149 | \end{align} 150 | % 151 | where $K_n$ is the polar of $B_n$. % 152 | Clearly, showing that every $m\cdot K_n$ is nowhere dense % 153 | is now sufficient. % 154 | To do so, we use the fact that $X^\ast$ separates points; % 155 | see \citeresultFA{3.4}. % 156 | As a consequence, the above lemma implies % 157 | % 158 | \begin{align} 159 | \left({\overline{K}_n}\right)^\circ = \left({{K}_n}\right)^\circ=\emptyset. 160 | \end{align} 161 | % 162 | Since the multiplication by $m$ is a homeomorphism (see \citeresultFA{1.7}), % 163 | this is equivalent to % 164 | % 165 | \begin{align}\label{3.11. Nowhere dense.} 166 | \left(\,{\overline{m\cdot K_n}}\, \right)^\circ 167 | = 168 | %\left(\, m \cdot \overline{{{K}_n}} \, \right)^\circ 169 | % = 170 | %\left(m \cdot K_n \right)^\circ 171 | % = 172 | m\cdot \left({{K}_n}\right)^\circ 173 | = 174 | \emptyset. 175 | \end{align} 176 | % 177 | So ends the proof.% 178 | \end{proof} -------------------------------------------------------------------------------- /chapter_2/2_3/2_3_3_example_1.tex: -------------------------------------------------------------------------------- 1 | We now come back to the special case $f_n = g_n$ (see the first part). % 2 | From now on, $f_n(x) = n^3 x \Iverson{\minus 1/n \leq x \leq 1/n}$. % 3 | Actually, we will prove that 4 | % 5 | \renewcommand{\labelenumi}{(\alph{enumi})} 6 | % 7 | \begin{enumerate} 8 | \item{$\int_K f_n \phi \tendsto{n}{\infty} \Lambda \phi \quad (\phi\in\D_{K})$,} 9 | \item{A uniform bound % 10 | % 11 | $\magnitude{\Lambda_n} \leq M \norma{\infty}{D^p }$ % 12 | % 13 | exists for all $f_n$, with $p=1$ as the smallest possible $p$. 14 | } 15 | \end{enumerate} 16 | % 17 | \renewcommand{\labelenumi}{(\roman{enumi})} 18 | % 19 | % 20 | Bear in mind that % 21 | % 22 | $K \subset K_m$ % 23 | % 24 | and shift the $K_N$'s indices %by $\minus m$ 25 | so that % 26 | % 27 | $K_{m+1}$ becomes $K_1$, $K_{m+2}$ becomes $K_2$, and so on. % 28 | % 29 | The resulting topology $\tau_K$ remains unchanged (see Exercise 1.16). % 30 | % 31 | First, we set % 32 | % 33 | \begin{align} 34 | \Mu_{n}(\zeta) 35 | & 36 | \Def \max_{[\minus 1/n , 1/n]} \magnitude{\zeta}\\ 37 | %\Def \max\set{\magnitude{\phi(x)}}{x \in [\minus 1/n, 1/n]}, \\ 38 | % 39 | \Delta_{n}(\zeta) & \Def \max_{[\minus 1/n , 1/n]} \magnitude{\zeta - \zeta(0)} 40 | %\Delta_{n}(\phi) 41 | %& 42 | %\Def \max\set{\magnitude{\phi(x) - \phi(0)}}{x \in [\minus 1/n, 1/n]}. 43 | \end{align} 44 | % 45 | for all $\zeta \in \D_K$. Next, we easily establish the following mean value inequality, 46 | % 47 | \begin{align}\label{2.3. Mean value inequality (concrete).} 48 | \magnitude{ \phi(1/n) - \phi(\minus 1/n)} 49 | \leq 50 | \frac{2}{n} \Mu_{n}(\phi'). 51 | % 52 | \end{align} 53 | % 54 | Independently, an integration by parts shows that % 55 | % 56 | \begin{align} 57 | \Lambda_n\phi = 58 | %& = 59 | %\left[ \frac{n^3 t^2}{2} \phi(t) \right]^{1/n}_{\minus 1/n} 60 | %- \frac{n^3}{2} \int_{\minus 1/n}^{1/n} t^{2} \phi'(t)dt\\ 61 | % 62 | \label{2.3. Equality from integration by parts.} 63 | \frac{n}{2}\left(\phi(1/n) - \phi(\minus 1/n)\right) 64 | - \frac{n^3}{2} \int_{\minus 1/n}^{1/n} t^{2} \phi'(t)dt. 65 | % 66 | \end{align} 67 | Combining % 68 | % 69 | (\ref{2.3. Mean value inequality (concrete).}) % 70 | % 71 | with % 72 | (\ref{2.3. Equality from integration by parts.}) 73 | % 74 | yields % 75 | % 76 | \begin{align} 77 | \magnitude{\Lambda_n\phi} 78 | & \leq 79 | \frac{n}{2}\magnitude{\phi(1/n) - \phi(\minus 1/n)} 80 | + 81 | \frac{n^3}{2}\int_{\minus1/n}^{1/n} t^{2}\lvert\phi'(t)\rvert dt 82 | \\ 83 | % 84 | & \leq 85 | \Mu_{n}(\phi') 86 | + 87 | \frac{n^3}{2} \Mu_{n}(\phi') \int_{\minus1/n}^{1/n} t^{2} dt \\ 88 | % 89 | \label{2.3. Bound with p and M (M_n).} 90 | & \leq 91 | \frac{4}{3} \Mu_{n}(\phi') \\ 92 | \label{2.3. Bound with p and M (concrete).} 93 | & \leq 94 | \frac{4}{3} \norma{\infty}{\phi'}. 95 | % 96 | \end{align} 97 | % 98 | Futhermore, % 99 | % 100 | (\ref{2.3. Bound with p and M (M_n).}) % 101 | % 102 | gives a hint about the convergence of $f_n$: Since % 103 | % 104 | $\Mu_{n}(\phi')$ tends to $\magnitude{\phi'(0)}$, % 105 | % 106 | we may expect that % 107 | % 108 | $f_n$ tends to $\frac{4}{3}\phi'(0)$. % 109 | % 110 | This is actually true: A straightforward computation shows that % 111 | % 112 | \begin{align} 113 | \Lambda_n\phi - \frac{4}{3}\phi'(0) 114 | \citeq{\ref{2.3. Equality from integration by parts.}} 115 | \frac{\phi(1/n) - \phi(\minus 1/n)}{1/n - (\minus 1/n)} 116 | - \phi'(0) 117 | - \frac{n^3}{2} \int_{\minus 1/n}^{1/n} (\phi' - \phi'(0)) t^{2} dt. 118 | \end{align} 119 | % 120 | Hence, 121 | % 122 | \begin{align} 123 | \magnitude{\Lambda_n\phi -\frac{4}{3} \phi'(0)} 124 | & \leq 125 | \magnitude{ 126 | \frac{\phi(1/n) - \phi(\minus 1/n)}{1/n - (\minus 1/n)} 127 | - \phi'(0) 128 | } 129 | + \frac{1}{3}\Delta_{n}(\phi') 130 | % 131 | \tendsto{n}{\infty} 0. 132 | \end{align} 133 | % 134 | We have just proved that 135 | % 136 | \begin{align}\label{2.3. (f_n) converges to Dirac (1).} 137 | \Lambda_n\phi \tendsto{n}{\infty} %\Lambda\phi \Def 138 | \frac{4}{3}\phi'(0) 139 | % 140 | \quad (\phi\in \mathscr{D}_K). 141 | \end{align} 142 | % 143 | In other words, 144 | % 145 | \begin{align} 146 | \Lambda_n \tendsto{n}{\infty}\minus \frac{4}{3}\delta' , 147 | \end{align} 148 | % 149 | where $\delta$ is the \textit{Dirac measure} and 150 | $\delta', \delta'', \dots, $ its \textit{derivatives}; % 151 | % 152 | see \citeresultFA{6.1 and 6.9}. 153 | % 154 | \newline\newline\noindent 155 | It follows from the previous part that % 156 | % 157 | $\minus \frac{4}{3}\delta'$ is $\tau_K$-continuous, % 158 | % 159 | and from %% 160 | % 161 | (\ref{2.3. Bound with p and M (concrete).}) % 162 | % 163 | that %% 164 | % 165 | \begin{align} 166 | \lvert \Lambda_n\phi \rvert \leq \frac{4}{3} \norma{\infty}{\phi'} 167 | \quad (\counting{n}) 168 | \end{align} 169 | % 170 | (which is a concrete version of % 171 | % 172 | (\ref{2.3. Bound with p and M (theoritical).}) ). % 173 | % 174 | Furthermore, we have already spotlighted a sequence % 175 | % 176 | \begin{align} 177 | \set{ 178 | \langle{f_n} \lvert \phi_{\rho(n)} \rangle 179 | }{ 180 | \norma{\infty}{\phi_{\rho(n)}} = 1; \counting{n} 181 | } 182 | \end{align} 183 | % 184 | that is not bounded. We then restate % 185 | % 186 | (\ref{2.3. Uniform bound for the supremum norm.}) % 187 | % 188 | in a more precise fashion: % 189 | There is no constant $M$ such that % 190 | % 191 | \begin{align} 192 | \lvert \langle f_n \lvert \phi \rangle \rvert \leq 193 | M 194 | \norma{\infty}{\phi} 195 | \quad (\phi\in C_K^\infty(\R)). 196 | \end{align} 197 | % 198 | The previous bound of $\bra{f_n}$ % 199 | % 200 | - see (\ref{2.3. Bound with p and M (concrete).}), 201 | % 202 | is therefore the best possible one\ie % 203 | % 204 | % 205 | $p=1$ is the smallest possible $p$ and, given $p=1$, % 206 | % 207 | $M=\frac{4}{3}$ is the smallest possible $M$ % 208 | (to see that, compare % 209 | (\ref{2.3. Bound with p and M (M_n).}) with 210 | (\ref{2.3. (f_n) converges to Dirac (1).})% 211 | ); which is (b). % 212 | % 213 | \newline\newline\noindent 214 | % 215 | %\newline\newline\noindent 216 | %END 217 | -------------------------------------------------------------------------------- /chapter_4/4_1.tex: -------------------------------------------------------------------------------- 1 | % 2 | \textit{ 3 | Let $\phi$ be the embedding of $X$ into $X^{\,\ast\ast}$ decribed in Section 4.5. Let $\tau$ be the weak topology of $X$, and let $\sigma$ be the weak$^{\,\ast}$- topology of $X^{\,\ast\ast}$- the one induced by $X^{\,\ast}$. 4 | \begin{enumerate} 5 | \item Prove that $\phi$ is a homeomorphism of $(X,\, \tau)$ onto a dense subspace of $(X^{\,\ast\ast},\, \sigma)$. 6 | \item If $B$ is the closed unit ball of $X$, prove that $\phi(B\,)$ is $\sigma$-dense in the closed unit ball of $X^{\,\ast\ast}$. (Use the Hahn-Banach separation theorem.) 7 | \item Use (a), (b), and the Banach-Alaoglu theorem to prove that $X$ is reflexive if and only if $B$ is weakly compact. 8 | \item Deduce from (c) that every norm-closed subspace of a reflexive space is reflexive. 9 | \item If $X$ is reflexive and $Y$ is a closed subspace of $X$, prove that $X/Y$ is reflexive. 10 | \item Prove that $X$ is reflexive if and only $X^{\,\ast} $ if reflexive. \\ Suggestion: One half follows from (c); for the other half, apply (d) to the subspace $\phi(X\,)$ of $X^{\,\ast\ast}$. 11 | \end{enumerate} 12 | } 13 | % 14 | \begin{proof} 15 | Let $\psi$ be the isometric embedding of $X^\ast$ into $X^{\ast\ast\ast}$. % 16 | The dual space of $(X^{\ast\ast},\sigma)$ is then $\psi(X^\ast)$. \\ 17 | \\ 18 | % 19 | It is sufficient to prove that 20 | \begin{align} 21 | \phi^{\minus 1}: \phi(X) \to & X \\ 22 | \phi (x) \mapsto & x 23 | \end{align} 24 | % 25 | is a homeomorphism (with respect to $\tau$ and $\sigma$). We first consider 26 | \begin{align} 27 | V \Def & \set{ 28 | x^{\ast\ast} \in X^{\ast\ast} }{ 29 | \magnitude{ \bra{x^{\ast\ast} }\ket{\psi x^\ast}} < r 30 | } & (x^{\ast} \in X^{\ast}, r > 0); \\ 31 | U \Def & \set{ 32 | x \in X }{ 33 | \magnitude{ \bra{x}\ket{x^\ast} } < r 34 | } & (x^\ast \in X^\ast, r > 0 ). 35 | \end{align} 36 | and remark that the so defined $V\,$'s (respectively $U\,$'s) shape a local subbase $\mathscr{S}_\sigma$ (respectively $\mathscr{S}_\tau$) of $\sigma$ (respectively $\tau$). We now observe that 37 | \begin{align} 38 | U=\phi^{\minus 1} \left(V \cap \phi(X\,) \right) = \phi^{\minus 1} (V) \cap X \quad ( V\in \mathscr{S}_\sigma\,,\,\,U\in \mathscr{S}_\tau) \quad , 39 | \end{align} 40 | since $\phi^{\minus 1}$ is one-to-one. This remains true whether we enrich each subbase $\mathscr{S}$ with all finite intersections of its own elements, for the same reason. It then follows from the very definition of a local base of a weak / weak$^\ast$-topology that $\phi^{\minus 1}$ and its inverse $\phi$ are continuous.\\ 41 | \\ 42 | The second part of (a) is a special case of [3.5] and is so proved. First, it is evident that 43 | \begin{align} 44 | \overline{\phi(X\,)}_\sigma\subset X^{\,\ast\ast}\quad . 45 | \end{align} 46 | and we now assume- to reach a contradiction- that $(X^{\,\ast\ast},\, \sigma)$ contains a point $z^{\,\ast\ast}$ outside the $\sigma$-closure of $\phi(X\,)$. By [3.5], there so exists $y^{\,\ast}$ in $X^{\,\ast}$ such that 47 | \begin{align} 48 | \label{4_1_6} \langle \phi x,\, \psi y^{\,\ast} \rangle=&\langle y^{\,\ast} ,\,\phi x \rangle= \langle x,\, y^{\,\ast}\rangle= 0\quad (x\in X\,)\quad ; \\ 49 | \label{4_1_7} \langle z^{\,\ast\ast},\, \psi y^{\,\ast} \rangle=&\,1 50 | \end{align} 51 | (\ref{4_1_6}) forces $y^{\,\ast}$ to be a the zero of $X^{\,\ast}$. The functional $\psi y^{\,\ast}$ is then the zero of $X^{\,\ast\ast\ast}$: (\ref{4_1_7}) is contradicted. Statement (a) is so proved; we next deal with (b).\\ 52 | \\ 53 | %: (b) 54 | The unit ball $B^{\,\ast\ast}$ of $X^{\,\ast\ast}$ is weak$^\ast$-closed, by (c) of [4.3]. On the other hand, 55 | \begin{align} 56 | \phi(B\,)\subset B^{\,\ast\ast}\quad , 57 | \end{align} 58 | since $\phi$ is isometric. Hence 59 | \begin{align} 60 | \overline{\phi(B\,)}_\sigma\subset \overline{ (B^{\,\ast\ast})}_\sigma= B^{\,\ast\ast}\quad . 61 | \end{align} 62 | Now suppose, to reach a contradiction, that $B^{\,\ast\ast}\setminus \overline{\phi(B\,)}_\sigma$ contains a vector $z^{\,\ast\ast}$. By [3.7], there exists $y^{\,\ast}$ in $X^{\,\ast}$ such that 63 | \begin{align} 64 | \label{4_1_10} \lvert \psi y^{\,\ast} \rvert \leq &1 \quad \text{on}\,\,\overline{\phi(B\,)}_\sigma \quad ; \\ 65 | \label{4_1_11} \langle z^{\,\ast\ast},\, \psi y^{\,\ast} \rangle > &1\quad . 66 | \end{align} 67 | It follows from (\ref{4_1_10}) that 68 | \begin{align} 69 | \lvert \, \psi y^{\,\ast}\, \rvert \leq 1 \text{ on }\phi(B\,)\, , \, \, \ie \lvert\, y^{\,\ast} \, \rvert \leq 1 \text{ on } B\quad . 70 | \end{align} 71 | We have so proved that 72 | \begin{align} 73 | y^{\,\ast} \in B^\ast \quad . 74 | \end{align} 75 | Since $ z^{\,\ast\ast}$ lies in $B^{\,\ast\ast}$, it is now clear that 76 | \begin{align} 77 | \lvert \langle z^{\,\ast\ast},\, \psi y^{\,\ast} \rangle\rvert \leq 1\quad; 78 | \end{align} 79 | what it contradicts (\ref{4_1_11}), and thus proves (b). We now aim at (c).\\ 80 | \\ 81 | %: c 82 | It follows from (a) that 83 | \begin{align}\label{4_15} 84 | B\text{ is weakly compact if and only if }\phi(B\,) \text{ is weak}^\ast\text{-compact. } 85 | \end{align} 86 | If $B$ is weakly compact, then $\phi(B\,)$ is weak$^\ast$-closed. So, 87 | \begin{align} 88 | \phi(B\,)= \overline{\phi(B\,)}_\sigma\overset{(b)}{=} B^{\,\ast\ast}\quad . 89 | \end{align} 90 | $\phi$ is therefore onto\ie $X$ is reflexive.\\ 91 | Conversely, keep $\phi$ as onto: one easily checks that $\phi(B\,)=B^{\,\ast\ast}$. The image $\phi (B\,)$ is then weak$^\ast$-compact by (c) of [4.3]. The conclusion now follows from (\ref{4_15}).\\ 92 | \\ 93 | %: (d) 94 | Next, let $X$ be a reflexive space $X$, whose closed unit ball is $B$. Let $Y$ be a norm-closed subspace of $X$: $Y$ is then weakly closed (\cf [3.12]). On the other hand, it follows from (c) that $B$ is weakly compact. We now conclude that the closed unit ball $B\cap Y$ of $Y$ is weakly compact. We again use (c) to conclude that $Y$ is reflexive. (d) is therefore established. Now proceed to (e).\\ 95 | \\ 96 | %: (e) 97 | Let $\equiv$ stand for ``isometrically isomorphic" and apply twice [4.9] to obtain, first 98 | \begin{align}\label{4_17} 99 | (X/Y\,)^\ast \equiv Y^\bot \quad , 100 | \end{align} 101 | next, 102 | \begin{align}\label{4_18} 103 | (X/Y\,)^{\ast\ast} \equiv (Y^\bot)^\ast 104 | \equiv X^{\,\ast\ast} / (Y^\bot)^\bot 105 | \equiv X/Y\quad . 106 | \end{align} 107 | Combining (\ref{4_17}) with (\ref{4_18}) makes (e) to hold.\\ 108 | \\ 109 | It remains to prove (f). To do so, we state the following trivial lemma (L) 110 | \begin{quotation} 111 | \noindent { \CMUCS Given a reflexive Banach space $Z$, the weak$^{\,\ast}$-topology of $Z^{\,\ast}$ is its weak one.} 112 | \end{quotation} 113 | %Its proof is nothing else but a tiny rewriting: if $Z^{\ast\ast}\equiv Z$, then $ Z^\ast_{\text{w}^\ast}=Z^\ast_{\text{w}}$.\\ 114 | Assume first that $X$ is reflexive. Since $B^{\,\ast}$ is weak$^\ast$ compact, by (c) of [4.3], (L) implies that $B^{\,\ast}$ is also weakly compact. Then (c) turns $X^{\,\ast}$ into a reflexive space. \\ 115 | \\ 116 | Conversely, let $X^{\,\ast}$ be reflexive. What we have just proved that makes $X^{\,\ast\ast}$ reflexive. On the other hand, $\phi(X\,)$ is a norm-closed subspace of $X^{\,\ast\ast}$; \cf [4.5]. Hence $\phi(X\,)$ is reflexive, by (d). It now follows from (c) that $B^{\,\ast\ast} \cap \phi(X\,)$ is weakly compact\ie weak$^\ast$-compact (to see this, apply (L) with $Z=X^{\,\ast}$). \\ 117 | \\ 118 | By (a), $B$ is therefore weakly compact\ie $X$ is reflexive, see (c). So ends the proof. 119 | \end{proof} 120 | 121 | 122 | 123 | 124 | 125 | 126 | 127 | 128 | 129 | 130 | 131 | 132 | 133 | 134 | 135 | 136 | -------------------------------------------------------------------------------- /chapter_2/2_3/2_3_1_radon_measures.tex: -------------------------------------------------------------------------------- 1 | We first consider % 2 | % 3 | $C_0(\R)$ % 4 | % 5 | topologized by the supremum norm. % 6 | Given a Lebesgue integrable function $u$, we put % 7 | \begin{align} 8 | \bra{u}\ket{\phi} 9 | \Def 10 | \int_\R u\phi 11 | \quad(\phi \in C_0(\R)). 12 | \end{align} 13 | % 14 | The following inequalities % 15 | % 16 | \begin{align}\label{2.3 g bounded operator (1).} 17 | % 18 | \left\lvert\bra{u}\ket{\phi}\right\rvert 19 | \leq 20 | \int_\R \left\lvert u\phi \right\rvert 21 | \leq 22 | \norma{L^1}{u} 23 | \quad(\norma{\infty}{\phi} \leq 1) 24 | \end{align} 25 | % 26 | imply that every linear functional 27 | % 28 | \begin{align} 29 | \bra{u}: C_0(\R) &\to \C \\ 30 | \phi &\mapsto \bra{u}\ket{\phi} \nonumber 31 | \end{align} 32 | % 33 | is bounded on the open unit ball. It is therefore continuous; % 34 | % 35 | see \citeresultFA{1.18}. % 36 | % 37 | Conversely, ${u}$ can be identified with % 38 | % 39 | $\bra{u}$, % 40 | % 41 | since $u$ is determined (a.e) by the integrals % 42 | % 43 | $\bra{u}\ket{\phi}$. % 44 | % 45 | In the Banach spaces terminology, 46 | % 47 | $u$ is then (identified with) a linear \textit{bounded}\hspace{2pt}\footnote{ 48 | % 49 | see \citeresultFA{1.32, 4.1} 50 | } % 51 | % % 52 | operator $\bra{u}$, of norm % 53 | % 54 | \begin{align}\label{2.3 g bounded operator (2).} 55 | \sup\set{ 56 | \magnitude{\bra{u}\ket{\phi}} 57 | }{ 58 | \norma{\infty}{\phi} = 1 59 | } 60 | = \norma{L^1}{u}. 61 | \end{align} 62 | % 63 | Note that, in the latter equality, $\leq \norma{L^1}{u}$ follows from % 64 | % 65 | (\ref{2.3 g bounded operator (1).}), % 66 | % 67 | as the converse follows from the Stone-Weierstrass theorem\footnote{ 68 | % 69 | See 7.26 of \cite{BabyRudin}. 70 | }. % 71 | % 72 | We now consider the special cases % 73 | % 74 | $u = g_n$ % 75 | %($\counting{n}$), % 76 | % 77 | , where $g_n$ is % 78 | % 79 | \begin{align} 80 | g_n : \R & \to \R\\ 81 | x & \mapsto \begin{cases} 82 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 83 | n^3 x & \left(x \in \left[\minus \frac{1}{n}, \frac{1}{n}\right]\right)\\ 84 | 0 & \left(x \notin \left[\minus \frac{1}{n}, \frac{1}{n}\right]\right). 85 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 86 | \end{cases}\nonumber 87 | \end{align} 88 | % 89 | First, remark that $g_n(x)$ $\longrightarrow 0$, % 90 | as the sequence $\singleton{g_n}$ fails to converge in $C_0(\R)$ % 91 | % 92 | (since $g_n(1/n)= n^2 \geq 1$), % 93 | % 94 | and also in $L^1$ % 95 | % 96 | (since $\int_\R \lvert g_n \rvert = n^2 \longrightarrow{\infty}$) 97 | . % 98 | Nevertheless, we will show that % 99 | % 100 | the $\bra{g_n}$ % 101 | % 102 | converge pointwise\footnote{ 103 | % 104 | See \citeresultFA{3.14} for a definition of the related topology. % 105 | % 106 | } on $\D_K$ % 107 | \ie % 108 | there exists a $\tau_K$-continuous linear form $\Lambda$ such that % 109 | % 110 | \begin{align}\label{2.3. Convergence in D_K, not vague.} 111 | \bra{g_n}\ket{\phi} \tendsto{n}{\infty} \Lambda \phi, % 112 | \end{align} 113 | % 114 | where $\phi$ ranges over $\D_K$. % 115 | % 116 | We now prove (\ref{2.3 g bounded operator (2).}) in the special cases % 117 | % 118 | $u = g_n$. % 119 | % 120 | To do so, we fetch % 121 | % 122 | $\phi^{+}_{1}, \dots, \phi^{+}_{j}, \dots,$ from $C_K^\infty(\R)$. % 123 | % 124 | More specifically, % 125 | % 126 | \begin{enumerate} 127 | \item{ 128 | $\phi^{+}_{j} = 1$ on $[e^{\minus j}, 1-e^{\minus j}]$; 129 | }% 130 | \item{ 131 | $\phi^{+}_{j} = 0$ on $\R\setminus [-1, 1]$;% 132 | } 133 | \item{ 134 | $0 \leq \phi^{+}_{j} \leq 1$ on $\R$; % 135 | } 136 | \end{enumerate} 137 | % 138 | see \citeresultFA{[1.46]} for a possible construction of those $\phi^{+}_{j}$. 139 | % 140 | Let % 141 | % 142 | $\phi^{-}_{1}, \dots, \phi^{-}_{j}, \dots,$ % 143 | % 144 | mirror the $\phi^{+}_{j}$, in the sense that % 145 | % 146 | $\phi^{-}_{j}(x) = \phi^{+}_{j}(\minus x)$, % 147 | % 148 | so that % 149 | % 150 | \begin{enumerate}\addtocounter{enumi}{3} 151 | \item{ 152 | $\phi_{j} \Def \phi^{+}_{j} - \phi^{-}_{j}$ is odd, as $g_n$ is; 153 | } 154 | \item{ 155 | every $\phi_{j}$ is in $C_K^\infty(\R)$; 156 | } 157 | \item{ 158 | The sequence $\singleton{\phi_j}$ converges (pointwise) to % 159 | $1_{[0, 1]} - 1_{[\minus 1, 0]}$, and $\norma{\infty}{\phi_{j}} = 1$. % 160 | } 161 | \end{enumerate} 162 | % 163 | Thus, with the help of the Lebesgue's convergence theorem, 164 | % 165 | \begin{align}\label{2.3. Norm in the dual equals norm L1 (1).} 166 | \langle{g_n}\lvert{\phi_j}\rangle 167 | = 168 | 2 \int_{0}^1 g_n(t) \phi^{+}_{j}(t) d t 169 | \tendsto{j}{\infty} 170 | 2 \int_{0}^1 g_n(t) d t 171 | = \norma{L^1}{g_n} = n. 172 | \end{align} 173 | % 174 | Finally, % 175 | % 176 | \begin{align}\label{2.3. Norm in the dual equals norm L1 (2).} 177 | \norma{L^1}{g_n} 178 | \citeleq{\ref{2.3. Norm in the dual equals norm L1 (1).}} 179 | \sup\set{ 180 | \magnitude{\bra{g_n}\ket{\phi}} 181 | }{ 182 | \norma{\infty}{\phi} = 1 183 | } 184 | \citeleq{\ref{2.3 g bounded operator (2).}} 185 | \norma{L^1}{g_n}; 186 | \end{align} % 187 | % 188 | which is the desired result. % 189 | % 190 | So, in terms of boundedness constants: % 191 | Given $n$, there exists $C_{n} < \infty$ such that 192 | % 193 | \begin{align}\label{2.3. Optimal bound.} 194 | \left\lvert 195 | \bra{g_n}\ket{\phi} 196 | \right\rvert 197 | \leq 198 | C_{n} \quad (\norma{\infty}{\phi} = 1); 199 | \end{align} 200 | % 201 | see (\ref{2.3 g bounded operator (1).}). 202 | Furthermore, % 203 | % 204 | $\norma{L^1}{g_n}$ is actually the best\ie lowest, possible $C_{n}$, see % 205 | % 206 | (\ref{2.3. Norm in the dual equals norm L1 (2).}). % 207 | % 208 | But, on the other hand, % 209 | % 210 | (\ref{2.3. Norm in the dual equals norm L1 (1).}) % 211 | % 212 | shows that there exists a subsequence %% 213 | % 214 | $\singleton{\langle{g_n}\lvert{\phi_{\rho(n)}}\rangle}$ % 215 | % 216 | such that % 217 | % 218 | $\langle{g_n}\lvert{\phi_{\rho(n)}}\rangle$ % 219 | % 220 | is greater than, say, $n -0.01$, as $\norma{\infty}{\phi_{\rho(n)}} = 1$. % 221 | % 222 | Consequently, there is no bound $M$ such that %% 223 | % 224 | \begin{align}\label{2.3. Uniform bound for the supremum norm.} 225 | \left\lvert 226 | \bra{g_n}\ket{\phi} 227 | \right\rvert 228 | \leq M 229 | \quad (\norma{\infty}{\phi} = 1; \counting{n}). 230 | \end{align} 231 | % 232 | In other words, the $g_n$ have no \textit{uniform bound} in ${L^1}$, % 233 | \ie % 234 | the collection of all continous linear mappings % 235 | % 236 | $\bra{g_n}$ % 237 | % 238 | is not equicontinous % 239 | % 240 | (see discussion in \citeresultFA{2.6}). % 241 | % 242 | As a consequence, % 243 | % 244 | the $\bra{g_n}$ % 245 | % 246 | do not converge pointwise (or ``vaguely'', in Radon measure context): % 247 | A vague (\ie pointwise) convergence would be (by definition) % 248 | % 249 | \begin{align} 250 | \bra{g_n}\ket{\phi} \tendsto{n}{\infty} \Lambda \phi % 251 | \quad (\phi \in C_0(\R)) % 252 | \end{align} 253 | % 254 | for some $\Lambda\in C_0(\R)^\ast$, which would make % 255 | % 256 | (\ref{2.3. Uniform bound for the supremum norm.}) 257 | % 258 | hold, see \citeresultFA{2.6, 2.8}. % 259 | % 260 | This by no means says that the % 261 | % 262 | $\bra{g_n}$ % 263 | % 264 | do not converge pointwise, in a relevant space, to some $\Lambda$ (see % 265 | % 266 | (\ref{2.3. Convergence in D_K, not vague.}). % 267 | % 268 | \newline\newline\noindent 269 | % END 270 | -------------------------------------------------------------------------------- /chapter_4/4_15.tex: -------------------------------------------------------------------------------- 1 | % 2 | { \CMUCS Suppose $\mu$ is a finite (or $\sigma$-finite) positive measure on a measure space $\Omega$, $\mu\times\mu$ is the corresponding product measure on $\Omega\times\Omega$, and $K\in L^2(\mu\times\mu)$. Define 3 | \begin{align*} 4 | (Tf\,)(s\,)=\int_\Omega K(s,t\,) f\,(t\,) d\mu(t\,)\quad [\,f\in L^2(\mu)\,]. 5 | \end{align*} 6 | \begin{enumerate} 7 | \renewcommand{\labelenumi}{(\alph{enumi})} 8 | \item Prove that $T\in \mathscr{B}(L^2(\mu))$ and that 9 | \begin{align*} 10 | \| T\, \|^2 \leq \int_\Omega\int_\Omega \lvert K (s,t\,)\rvert ^2 d\mu (s\,) d\mu(t\,). 11 | \end{align*} 12 | \item Suppose $a_i$, $b_i$ are members of $L^2(\mu)$, for $1\leq i\leq n$, put $K_1=\sum a_i(s\,)b_i(t\,)$ and define $T_1$ in terms of $K_1$ a $T$ was defined in terms of $K$. Prove that $\dim \mathscr{R}(T_1)\leq n$. 13 | \item Deduce that $T$ is a compact operator in $L^2(\mu)$. Hint: Use exercise 13. 14 | \item Suppose $\lambda\in \C,\, \lambda\neq 0$. Prove: Either the equation 15 | \begin{align*}Tf-\lambda f=g\end{align*} 16 | has a unique solution $f\in L^2(\mu)$ for every $g\in L^2(\mu)$ or there are infinitely many solutions for some $g$ and none for others. (This is known as the \textsl{Fredholm alternative}.). 17 | \item Describe the adjoint of $T$. 18 | \end{enumerate} 19 | } 20 | %: a 21 | \begin{proof} Let $X$ (respectively $P\,$) be the Banach space $L^2(\mu)$ (respectively $L^2(\mu\times \mu)\,$). A consequence of the Radon-Nikodym theorem (\cf 6.16 of \cite{BigRudin}\,) is that there exists a group isomorphism $\rho:\, X\to \, X^{\,\ast},\, f\mapsto f^{\,\,\ast}$ such that 22 | \begin{align} 23 | \langle u ,\, f^{\,\,\ast} \rangle =\int_\Omega u\cdot f \, \d\mu \quad (u\in X,\, f\in X)\quad . 24 | \end{align} 25 | Define a.e $K_s,\, K_t:\, \Omega\to \C$ by setting 26 | \begin{align} 27 | K_s(t\,)\Def K_t(s\,)\Def K(s,\, t\,)\, \, \text{ a.e}\quad \left((s,t\,)\in \Omega\right)\quad . 28 | \end{align} 29 | $T$ is clearly linear. Moreover, 30 | \begin{align} 31 | \lvert (T f\,)(s\,) \rvert = \lvert \langle K_s,\, f^{\,\,\ast} \rangle \rvert \leq \| K_s\|_X\quad(\|\,f\,\,\|_X < 1) \quad 32 | \end{align} 33 | (the latter inequality is a Cauchy-Schwarz one). Now apply the Fubini's theorem with $\lvert K\,\rvert^2$ to obtain 34 | \begin{align} 35 | \|Tf\,\,\|^2_X \leq \int_\Omega \| K_s\|^2_X \, \,\d \mu (s\,) =\| K\,\|_P^2 \,< \infty \quad(\|\,f\,\,\|_X < 1)\quad . 36 | \end{align} 37 | (a) is then proved. \\ 38 | \\ 39 | %: b 40 | To show (b), remark that 41 | \begin{align} 42 | \int_\Omega a_i (s\,) \cdot b_i \cdot f \,\,\d\mu \,\,\in \C\cdot a_i(s\,)\,\,\,\text{a.e}\quad \quad (f\in X, \, s\in \Omega) \quad. 43 | \end{align} 44 | It is now clear that $T$ maps any $f$ of $X$ into $\C \cdot a_1+\dotsb+ \C\cdot a_n$. We so conclude that $\dim R(T_1)\leq n $. \\ 45 | \\ 46 | %: c 47 | We now aim at (c). The current part refers to Exercise 4.13. $X$ is also a Hilbert space and so contains a Hilbert basis $M$. Define a.e 48 | \begin{align} 49 | a_b: \Omega \to &\,\C \\ 50 | s \mapsto &\,(K_s,\, b) \nonumber 51 | \end{align} 52 | whenever $b$ ranges $M$. Hence, 53 | \begin{align} 54 | K_s= \sum_{b\in M} a_b(s\,) \cdot b \, \text{ a.e} \quad (s\in \Omega) \quad. 55 | \end{align} 56 | Provided any positive scalar $\epsilon$, there so exists a finite subset $S=S(\epsilon)$ of $M$ such that 57 | \begin{align} 58 | \| K_s - \sum_{b\in S} a_b(s\,) \cdot b\,\|_X < \epsilon \quad (s\in \Omega)\quad . 59 | \end{align} 60 | Remark that $\underset{b\in S}{\sum} a_b \cdot b$ matches the definition of $K_1$; \cf(b): from now on, 61 | \begin{align} 62 | K_1\Def \sum_{b\in S} a_b \cdot b\quad . 63 | \end{align} 64 | It follows from (b) that 65 | \begin{align} 66 | \dim R(K_1) < \infty \quad. 67 | \end{align} 68 | Now turn back to (a), with $K-K_1$ playing the role of $K$, and so obtain 69 | \begin{align} 70 | \|T-T_1\| < \epsilon \mu(\Omega)\leq \infty \quad . 71 | \end{align} 72 | For if $\mu$ is finite, use (a) of Exercise 4.13 to conclude that $T$ is compact. Assume henceforth that $\mu$ is not (necessarily) finite and pick $\delta$ in $\R_+$. The simple functions (with finite measure support\,) form a dense family of an $L^p$ space ($1\leq p<\infty$); \cf 3.13 of \cite{BigRudin}. It then exists a simple function $K_\delta$ of $L^2(\mu\times \mu)$ such that 73 | \begin{align} 74 | (\mu\times\mu)\left(\{K_\delta\neq 0\}\right) <\infty \,, \,\, \| K-K_\delta \|_P <\delta\quad . 75 | \end{align} 76 | Define an operator $T_\delta$ in terms of $K_\delta$ as $T$ was defined in terms of $K$, and proceed as in (a) with $T-T_\delta$ instead of $T$. Then 77 | \begin{align}\label{4_15_13} 78 | \|T-T_\delta\| < \delta\quad . 79 | \end{align} 80 | The key ingredient is that $K_\delta$ can be identified with an element of the finite measure space $L^2(\{K_\delta\neq 0\},\mu\times\mu)\,$. What we have attempted to approximate $T$ by $T_1$ can therefore be reiterated (with $K_\delta$ playing the role of $K$) to achieve an approximation $T_{\delta,1}$ of $T_\delta$ so that 81 | \begin{align}\label{4_15_14} 82 | \|T_{\delta}-T_{\delta,1}\| < \epsilon\quad . 83 | \end{align} 84 | It now follows from (\ref{4_15_13}) and (\ref{4_15_14}) that 85 | \begin{align} 86 | \|T-T_{\delta,1}\|\leq \|T-T_{\delta}\| +\|T_{\delta}-T_{\delta,1}\| < \epsilon+\delta\quad . 87 | \end{align} 88 | Since $\epsilon$ and $\delta$ were arbitrary, the $\sigma$-finite case is proved. We now establish (d).\\ 89 | \\ 90 | %: d 91 | Provided $g$ of $X$, let $E_g$ be the following equation on $X$ 92 | \begin{align}Tf-\lambda f =g \quad ,\end{align} 93 | whose solution set is denoted by $S_g\,$. Note that $S_0$ is $\ker (T-\lambda)$ and discard the trivial case $S_0=X\,$\footnote{\eg$ X=L^2(\{0\},\, \delta)\,$so that $I=\lambda^{\minus 1} T $ is compact. }: each $f$ of $X $ lies in $S_{\,Tf-\lambda f\,} $, as some $Tf-\lambda f\,$'s are nonzero. Some $S_g$'s are then nonempty. Remark that 94 | \begin{align}\label{4_15_17} 95 | S_g= f + S_0 \quad (f\in S_g) \quad 96 | \end{align} 97 | in such case. Furthermore, the equality $\beta= \alpha $ of [4.25] yields 98 | \begin{align} 99 | (T-\lambda I\,)(X)\neq X , \text{\ie} S_0 \neq\{0\} \quad . 100 | \end{align} 101 | So if $T-\lambda I$ is not onto, not only some $S_g$'s are empty, but also $S_0\neq\{0\}$. Every nonempty $S_g$ (such sets always exist, see above) is then infinite, by (\ref{4_15_17}).\\ 102 | Otherwise, $T-\lambda I$ is bijective and every equation $E_g$ has then a unique solution $f$. The Fredholm alternative is so proved. \\ 103 | \\ 104 | Our last step is the description of $T^{\,\ast}$. Let $S:\, X\to X$ be such that 105 | \begin{align} 106 | (Sf\,)(t\,)\Def \int_\Omega K_t \cdot f \,\,\, \text{a.e}\quad \quad (\,f\in X,\, t\in \Omega) 107 | \end{align} 108 | Proceed as in (a), with $S$ instead of $T$: $S$ lies in $\mathscr{B}(X)$. Next, we claim that 109 | \begin{align} 110 | \langle u,\, T^{\,\ast} f^{\,\,\ast} \rangle = &\, \langle Tu,\,f^{\,\,\ast} \rangle\\ 111 | =& \int_{\Omega} (T u ) \cdot f \,\, \,\d\mu \\ 112 | \label{int_Fubini}=& \int_{\Omega^2} K\cdot f \cdot u \,\,\, \d(\mu\times\mu) \\ 113 | =& \int_{\Omega} (S f\, ) \cdot u \,\,\, \d\mu\, \\ 114 | =&\, \langle u,\, (S f\, )^\ast \rangle \quad , 115 | \end{align} 116 | whenever $u$ and $f$ run through the closed unit ball of $X$. Since $\|T\, \|$, $\| T^{\,\ast} \|$ are equal and finite, only exactness of (\ref{int_Fubini}) is possibly in doubt; the justification below dissipates it. In conclusion, 117 | \begin{align} 118 | T^{\,\ast} = \rho S \rho^{\minus 1} \quad . 119 | \end{align} 120 | Informally, 121 | \begin{align} 122 | T^{\,\ast} = S\quad . 123 | \end{align} 124 | \\ 125 | \underline{Justification of (\ref{int_Fubini})}. The current proof shall be complete once we have justified (\ref{int_Fubini}). To do so, keep $u$ and $f$ as above. Let us introduce 126 | \begin{align} 127 | A(s\,)\Def \int_\Omega \lvert K_s(t\,) \cdot u (t\,)\rvert \,\, \d\mu(t\,) \, \,\,\text{a.e} \quad (s\in \Omega)\quad , 128 | \end{align} 129 | to make hold the following Cauchy-Schwarz inequality 130 | \begin{align} 131 | A(s\,)\leq \| K_s\|_X \quad (s\in \Omega)\quad . 132 | \end{align} 133 | Thus, 134 | \begin{align} 135 | \int_{\Omega^2} \lvert K(s,\,t\,) \, u(t\,)\, f\,(s\,) \rvert \,\d\mu(s\,)\d\mu(t\,) 136 | = & \int_\Omega \lvert \,f\,(s\,) \rvert \,A(s\,) \, \d\mu(s\,) \\ 137 | \leq & \int_\Omega \lvert \,f\,(s\,) \rvert \, \| K_s\|_X\,\d\mu(s\,) \\ 138 | \label{4_15_complement_2} \leq & \left[ \int_\Omega \| K_s\|_X^2\,\,\d\mu(s\,) \right]^{\frac{1}{2}} 139 | = \| K\,\|_P < \infty \quad . 140 | \end{align} 141 | The inequality in (\ref{4_15_complement_2}) is a Cauchy-Schwarz one, the following equality follows from the Fubini's theorem. This achieves the proof.\end{proof} 142 | 143 | 144 | 145 | 146 | 147 | 148 | 149 | 150 | -------------------------------------------------------------------------------- /chapter_1/1_7.tex: -------------------------------------------------------------------------------- 1 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 2 | % FunctionalAnalysis 3 | % 1_7.tex 4 | % 5 | % encoding: UTF-8 6 | % EOL: LF 7 | % 8 | % format: LaTeX 9 | % indent: spaces (2) 10 | % width: 127 11 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 12 | \textit{% 13 | Let $X$ be the vector space of all complex functions on the unit interval $[0, 1]$, topologized by the family of seminorms % 14 | % 15 | \begin{align} 16 | p_x(f)= \magnitude{f(x)} \quad (0\leq x\leq 1).\nonumber 17 | \end{align} 18 | % 19 | This topology is called the topology of pointwise convergence. Justify this terminology. \\ 20 | \\ 21 | Show that there is a sequence % 22 | $\singleton{f_n}$ in $X$ such that (a) $\singleton{f_n}$ converges to $0$ as $n \to \infty$, but (b) if $\singleton{\gamma_n}$ % 23 | is any sequence of scalars such that $\gamma_n \to \infty$ then $\singleton{\gamma_n f_n}$ does not converge to $0$. % 24 | (Use the fact that the collection of all complex sequences converging to $0$ has the same cardinality as $[0, 1]$.) % 25 | This shows that metrizability cannot be omitted in (b) of Theorem 1.28. 26 | } 27 | % 28 | 29 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 30 | % First part: Justification of the terminology 31 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 32 | \subsection{Justifying the terminology} 33 | \begin{proof} 34 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 35 | % A. tau-convergence => pointwise convergence. 36 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 37 | The family of seminorms $p_x$ is separating: The collection $\mathscr{B}$ of all finite intersections of the sets % 38 | % 39 | \begin{align} 40 | V(x,k) \triangleq \singleton{p_x < 2^{\minus k}} 41 | \quad (x \in [0, 1], k=1, 2, 3, \dots) 42 | \end{align} 43 | % 44 | is therefore a local base for a topology $\tau$ on $X$, see Section 1.37 of \cite{FA}. Thus, given $\singleton{g_n} \subset X$, 45 | the following inequalities, expressed in Iverson's notation, hold: % 46 | % 47 | \begin{align} 48 | % 49 | \label{Inequality boolean series} 50 | % 51 | \sum_{n=1}^\infty \Iverson{g_n \notin \cap_{i=1}^m U_i} 52 | \leq 53 | \sum_{n=1}^\infty \sum_{i=1}^m \Iverson{g_n \notin U_i} 54 | = 55 | \sum_{i=1}^m \sum_{n=1}^\infty \Iverson{g_n \notin U_i}. 56 | \end{align} 57 | % 58 | Now assume that $\singleton{f_n}$ $\tau$-converges to some $f$. By definition, % 59 | % 60 | \begin{align} 61 | % 62 | \sum_{n=1}^\infty \Iverson{f_n - f \notin W} < \infty \quad (W \in \mathscr{B}). 63 | \end{align} 64 | % 65 | The special case $W = V(x, k)$ implies that, for a fixed $k$, $\magnitude{f_n(x)- f(x)} < 2^{\minus k}$ for all but % 66 | finitely many $n$. In other words, $\singleton{f_n(x)}$ converges to $f(x)$. % 67 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 68 | % B. tau-divergence => pointwise divergence. 69 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 70 | Conversely, assume that $\singleton{f_n}$ does not $\tau$-converge. This implies that, for every $f \in X$, there exist % 71 | finitely many $V(x_1, k_1), \dots, V(x_m, k_m)$ such that % 72 | % 73 | \begin{align} 74 | % 75 | \label{Divergence} 76 | \sum_{n=1}^\infty \Iverson{f_n - f \notin \cap_{i=1}^m V(x_i, k_i)} = \infty. 77 | \end{align} 78 | % 79 | In (\ref{Inequality boolean series}), the special case $g_n = f_n - f$ and $U_i = V(x_i, k_i)$ then gives% 80 | % 81 | \begin{align} 82 | \sum_{n=1}^\infty \Iverson{f_n - f \notin \cap_{i=1}^m V(x_i, k_i)} \citeleq{\ref{Inequality boolean series}} 83 | \sum_{i=1}^m \sum_{n=1}^\infty \Iverson{f_n - f \notin V(x_i, k_i)} 84 | = \infty. 85 | \end{align} 86 | % 87 | It is now obvious that % 88 | % 89 | \begin{align} 90 | \sum_{n=1}^\infty \bigl[f_n - f \notin V(x_i, k_i)\bigr] = \infty 91 | \end{align} 92 | % 93 | for some $i$, which shows that $\singleton{f_n(x_i)}$ does not converge to $f(x_i)$. Thus, $\tau$-convergence coincides with % 94 | pointwise convergence on $X$. 95 | \end{proof} 96 | % 97 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 98 | % SECOND PART 99 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 100 | % 101 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 102 | % First proof: Use the given hint 103 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 104 | \subsection{Proof (with the given hint)} 105 | We now prove the second part by constructing a specific sequence $\singleton{f_n}$ that satisfies both (a) and (b). % 106 | Indeed, the hint suggests that there exists a bijection % 107 | % 108 | \begin{align} 109 | \phi: \bigl\{(\theta_n): \theta_n \xrightarrow{n\infty} 0\bigr\} & \to [0, 1] \\ 110 | (\theta_1, \dots, \theta_n, \dots) & \mapsto x. \nonumber 111 | \end{align} 112 | % 113 | \begin{proof} 114 | We set % 115 | % 116 | \begin{align} 117 | f_n(x) & \Def \theta_n \quad \bigl(x = \phi(\theta_1, \dots, \theta_n, \dots)\bigr) 118 | \end{align} 119 | % 120 | so that $\singleton{f_n}$ tends pointwise to $0$. Note that, with this construction, the special case % 121 | % 122 | \begin{align} 123 | x_\gamma % 124 | \Def \phi\Bigl(1 / \sqrt{1 + \magnitude{\gamma_1}}, \dots, 1 / \sqrt{1 + \magnitude{\gamma_n}}, \dots\Bigr) % 125 | \end{align} 126 | % 127 | outputs % 128 | % 129 | \begin{align} 130 | \gamma_n f_n(x_\gamma) = \gamma_n / \sqrt{1 + \magnitude{\gamma_n}} % 131 | \tendsto{n}{\infty} \infty, 132 | \end{align} 133 | % 134 | given a sequence $\singleton{\gamma_n}$ diverging to $\infty$. This proves (b), since % 135 | $\singleton{\gamma_n f_n(x_\gamma)}$ diverges. 136 | \end{proof} 137 | % 138 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 139 | % Second proof: No hint, the hard way. 140 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 141 | \subsection{Proof with binary expansions (no hint)} 142 | We rely on the following assertion: Every irrational number has a binary expansion that is not eventually periodic. % 143 | % 144 | \begin{proof} 145 | More precisely, there exists a bijective sum % 146 | % 147 | %\def\bin{\text{\fw sum}} 148 | \def\bin{\sigma} 149 | \begin{align} 150 | \bin: \bigl\{\beta \in \{0, 1\}^{\N_+}: \beta \text{ is not eventually periodic}\bigr\} & \to [0, 1] \setminus \Q \\ 151 | (\beta_1, \dots, \beta_n, \dots) & \mapsto \sum_{k=1}^\infty \beta_k 2^{\minus k}. \nonumber 152 | \end{align} 153 | % 154 | A suitable $\singleton{f_n}$ can be defined as follows: % 155 | % 156 | \begin{align} 157 | \label{definition of f_n(alpha)} f_n(x) \Def \begin{cases} 158 | 2^{\minus (\beta_1 + \cdots + \beta_n)} & \bigr(x = \bin(\beta_1, \dots, \beta_n, \dots) \notin \Q) \\ 159 | 0 & (x \in \Q). 160 | \end{cases} 161 | \end{align} 162 | % 163 | Indeed, we note that every bit stream $\bin^{\minus 1}(x)$ has infinitely many $1$'s, which implies that % 164 | $f_n(x) \xrightarrow{n\infty} 0$. Next, pick an arbitrary $\gamma_n \to \infty$. Thus, for any positive integer $k$, % 165 | $\gamma_n > 4^k$ for all sufficiently large $n$, say $n > N_k$. We select $n_k > N_k$ so large that % 166 | % 167 | \begin{align} 168 | \label{definition of n_k} 169 | n_{k+1} - n_k > k. 170 | \end{align} 171 | % 172 | The crucial point is that the sequence $1_{\{n_1, n_2, \dots \}}$ is not eventually periodic. Moreover, the particular choice % 173 | % 174 | \begin{align} 175 | \beta^{\gamma} \Def 1_{\{n_1, n_2, \dots \}} 176 | \end{align} 177 | % 178 | implies % 179 | % 180 | \begin{align} \label{p sum of bits} 181 | \beta^\gamma_1 + \dots + \beta^\gamma_{n_1} + \dots + \beta^\gamma_{n_k} = k. 182 | \end{align} 183 | % 184 | Finally, (\ref{definition of f_n(alpha)}) and (\ref{p sum of bits}) together yield % 185 | % 186 | \begin{align} 187 | \gamma_{n_k} f_{n_{k}}(\bin(\beta^\gamma)) 188 | = {\gamma_{n_k}}/{2^k} 189 | %> 4^k / 2^k 190 | > 2^k 191 | \tendsto{k}{\infty}\infty. 192 | \end{align} 193 | % 194 | In conclusion, every sequence of scalars $\gamma_n$ such that $\gamma_n \to \infty$ contains a subsequence% 195 | $\singleton{\gamma_{n_k}}$ that causes $\singleton{\gamma_{n_k}f_{n_k}}$ to diverge. This is (b). 196 | \end{proof} 197 | % 198 | % END 199 | -------------------------------------------------------------------------------- /FA_DM.tex: -------------------------------------------------------------------------------- 1 | % a4paper = ISO 216 standard. 2 | % Don't forget to revert to default if you want the US Letter instead. 3 | \documentclass[a4paper,10pt,titlepage,openany, leqno]{book} 4 | 5 | %: MATH PACKAGE USED 6 | %\usepackage{graphicx} 7 | 8 | \usepackage{amsmath} 9 | \usepackage{ 10 | physics, 11 | amssymb, amsfonts, 12 | mathrsfs, 13 | amsthm, 14 | mathspec, 15 | fouridx, 16 | %%stix 17 | %%tensor 18 | theorem, 19 | tikz 20 | } 21 | \usetikzlibrary{arrows,snakes,backgrounds,patterns,matrix,shapes,fit,calc,shadows,plotmarks} 22 | 23 | %\usetikzlibrary{shapes.geometric} 24 | %\usetikzlibrary{arrows.meta,arrows} 25 | %\usepackage{tlatex} 26 | % TEXs LOGOS! 27 | %\usepackage{dtklogos} 28 | 29 | %\usepackage[bottom=3cm,top=2cm]{geometry} 30 | % Metric system, sorry;) 31 | \usepackage[left=1.9cm,right=1.32cm,top=1.9cm,bottom=3.67cm]{geometry} 32 | %: COLOR 33 | \usepackage{array} 34 | %\usepackage[dvipsnames]{xcolor} 35 | %\usepackage[]{color} 36 | %\usepackage{framed} 37 | \usepackage{colortbl} 38 | 39 | %: FOR XELATEX. SKIP IF ERROR MESSAGE 40 | \usepackage{xltxtra,xunicode} 41 | %: for HEVEA- Uncomment before using XeTeX 42 | %\usepackage[utf8]{inputenc} 43 | \usepackage[xetex]{hyperref} 44 | %\usepackage{hyperref} 45 | %\usepackage{listings} 46 | \hypersetup{% 47 | pdfborder = {0 0 0}, 48 | colorlinks, 49 | citecolor=red, 50 | filecolor=green, 51 | linkcolor=black, 52 | urlcolor=blue%cyan!50!black!90 53 | } 54 | \defaultfontfeatures{Mapping=tex-text} 55 | 56 | % FONTS: PERSONNAL CHOICE- CAN BE CHANGED! 57 | \newfontfamily{\CMUCS}{CMU Classical Serif} 58 | \newfontfamily{\CMU}{CMU Serif} 59 | \newfontfamily{\CMUSS}{CMU Sans Serif} 60 | \newfontfamily{\fw}{CMU Typewriter Text Light} 61 | %\newfontfamily{\Arabic}{Al Bayan} 62 | % On mac OS: 63 | %\newfontfamily{\Didot}{Didot} 64 | % Alternatively, for all Unix systems: 65 | %\newfontfamily{\Didot}{Theano Didot} 66 | %\usepackage{mathspec} 67 | \def\mainfont{CMU Serif} 68 | \setmainfont{\mainfont} 69 | 70 | \setmathfont(Latin)[Uppercase=Regular,Lowercase=Regular]{\mainfont} 71 | \setmathfont(Greek)[Uppercase=Regular,Lowercase=Regular]{\mainfont} 72 | \setmathrm{\mainfont} 73 | %\setmathbb{\mainfont} 74 | %\setmathit{\mainfont} 75 | \setmathtt{CMU Typewriter Text Light} 76 | %\setmathbf{CMU Serif} 77 | 78 | 79 | %: TYPOGRAPHIC CONVENTIONS, XETEX 80 | \usepackage{polyglossia} 81 | %\selectbackgroundlanguage[variant=usmax]{english} 82 | \setdefaultlanguage[variant=usmax]{english} 83 | 84 | % 85 | %\usepackage{glossaries} 86 | %\makeglossaries 87 | 88 | 89 | %\newglossaryentry{maths} 90 | %{ 91 | % name=mathematics, 92 | % description={Mathematics is what mathematicians do} 93 | %} 94 | %: ----------------------------PRIMITIVES------------------------------------- 95 | \newcommand{\insmall}[1]{\text{\small{#1}}} 96 | % Some acceptable minus sign. 97 | \def\minus{\insmall{-}} 98 | 99 | % The field C and the usual subsets R, Q, Z, N. 100 | \newcommand\usualSet[1]{{\mathbf #1}} 101 | \def\C{\usualSet{C}} 102 | \def\R{\usualSet{R}} 103 | \def\Q{\usualSet{Q}} 104 | \def\Z{\usualSet{Z}} 105 | \def\N{\usualSet{N}} 106 | \def\K{\usualSet{K}} 107 | 108 | % "Defined as equals to", alternative to := . 109 | \def\Def{\triangleq} 110 | \def\defiif{\quad \iif \quad} 111 | 112 | % The usual scalar field = C, most of the time. 113 | \def\field{\C} 114 | 115 | % Counting numbers 116 | \newcommand\counting[1]{#1=1, 2, 3, \dots} 117 | % integers (nonnegative) 118 | \newcommand\integers[1]{#1=0, 1, 2, \dots} 119 | 120 | % Indices: upper, lower 121 | \newcommand{\up}[1]{^{(#1)}} 122 | \newcommand{\low}[1]{_{#1}} 123 | \newcommand{\upnw}[2]{\fourIdx{#2}{}{}{}#1} 124 | \newcommand{\downsw}[2]{\fourIdx{}{#2}{}{}#1} 125 | \newcommand{\scriptatleft}[3]{\fourIdx{#2}{#3}{}{}#1} 126 | \newcommand{\diagscript}[3]{\fourIdx{#2}{}{}{#3}#1} 127 | %\tensor*[^x]{V}{_k} 128 | 129 | % function, relation 130 | \newcommand{\function}[1]{\mathtt{#1}} 131 | \newcommand{\relation}[2]{{#1}_{#2}} 132 | \newcommand{\f}[2]{#1(#2)} 133 | \newcommand{\id}[1]{\text{id}_{#1}} 134 | %\newcommand{\DeclareMathOperator}[1]{\text{#1}} 135 | 136 | % Sets 137 | %\renewcommand{\notin}{\tiny{\not\in}} 138 | %\renewcommand{\ni}{\small \ni} 139 | \renewcommand{\subset}{\subseteq} 140 | \renewcommand{\supset}{\supseteq} 141 | \DeclareMathOperator\opcard{card} 142 | \newcommand{\card}[1]{\opcard(#1)} 143 | % 144 | \def\contains{\supseteq} 145 | \def\cuts{\cap} 146 | \DeclareMathOperator\cvxhull{co} 147 | \DeclareMathOperator\Leibnizdiff{d} 148 | \DeclareMathOperator\support{supp} 149 | \newcommand{\co}[1]{\cvxhull(#1)} 150 | \newcommand{\set}[2]{\{#1: #2\}} 151 | \newcommand{\singleton}[1]{\{#1\}} 152 | \newcommand{\interior}[1]{\overset{\circ}{#1}} 153 | \newcommand{\closure}[1]{\overline{#1}} 154 | \newcommand{\supp}[1]{\support{(#1)}} 155 | 156 | % Arithmetics 157 | \newcommand{\ceil}[1]{\lceil #1 \rceilf} 158 | 159 | % Analysis 160 | \newcommand\openinterval[2]{\left]#1, #2\right[} 161 | %\newcommand\magnitude[1]{\left\lvert\, #1 \,\right\rvert} 162 | \newcommand\magnitude[1]{\abs{\,#1\,}} 163 | \newcommand{\norma}[2]{\norm{#2}_{#1}} 164 | \newcommand{\Continuous}{{\mathscr C}} 165 | \newcommand{\diff}[1]{\Leibnizdiff\!#1} 166 | \newcommand{\D}{{\mathscr D}} 167 | %\renewcommand{\norm}[2]{\norm{#2}_{#1}} 168 | \def\weakstar{\text{weak}^\ast\text{-}} 169 | 170 | % Topology 171 | \newcommand{\localbase}[1]{\mathscr #1} 172 | 173 | %Iverson bracket 174 | \newcommand{\Iverson}[1]{\bigl[#1\bigr]} 175 | 176 | % limits 177 | \newcommand{\tendsto}[2]{\xrightarrow[#1 \to #2]{}} 178 | 179 | % Variables 180 | \newcommand{\varit}[1]{\mathit{#1}} 181 | \def\vart{\varit{t}} 182 | %Usual terms 183 | \def\ie{\textit{, i.e., }} 184 | \def\eg{\textit{, e.g., }} 185 | \def\cf{\textit{cf.\,}} 186 | \def\iif{{\bf iff} } 187 | \def\IIF{{\bf iff}} 188 | \def\wlg{{\bf wlg }} % Without loss og generality 189 | 190 | % TODO: Make choices, be consistent 191 | \def\then{\Rightarrow} 192 | \def\THEN{\Rightarrow} 193 | \def\THEREFORE{\Rightarrow} 194 | \def\therefore{\Rightarrow} 195 | \def\since{\Leftarrow} 196 | \def\because{\Leftarrow} 197 | \def\IF{\Leftarrow} 198 | \renewcommand{\iff}{\Leftrightarrow} 199 | 200 | % Misc 201 | % phi is varphi: 202 | \renewcommand{\phi}{\varphi} 203 | % epsilon is varepsilon 204 | \renewcommand{\epsilon}{\varepsilon} 205 | 206 | % Citations 207 | \newcommand{\citehere}[2]{\overset{(#1)}{#2}} 208 | \newcommand{\citeq}[1]{\citehere{#1}{=}} 209 | %\newcomand{\citeineq}[1]{\citehere{#1}{=}} 210 | \newcommand{\citeleq}[1]{\citehere{#1}{\leq}} 211 | \newcommand{\citegeq}[1]{\citehere{#1}{\geq}} 212 | \newcommand{\citeleast}[1]{\citehere{#1}{<}} 213 | \newcommand{\citegreater}[1]{\citehere{#1}{>}} 214 | 215 | \newcommand{\citesubset}[1]{\citehere{#1}{\subset}} 216 | \newcommand{\citesubseteq}[1]{\citehere{#1}{\subseteq}} 217 | \newcommand{\citesupset}[1]{\citehere{#1}{\supset}} 218 | \newcommand{\citesupseteq}[1]{\citehere{#1}{\supseteq}} 219 | 220 | \newcommand{\citethen}[1]{\citehere{#1}{\Rightarrow}} 221 | %\newcommand{\citesince}[1]{\citehere{#1}{\Leftarrow}} 222 | \newcommand{\citeresult}[2]{#1 of #2} 223 | \newcommand{\citeresultFA}[1]{\citeresult{#1}{\cite{FA}}} 224 | 225 | \newcommand{\citein}[1]{\citehere{#1}{\in}} 226 | % Misc 227 | \newcommand{\underbarwithindex}[2]{\underline{#1}\,\!_{#2}} 228 | \newcommand{\dy}[1]{{\function{dyadic}}(#1)} 229 | \def\ddy{{\function{decay}}} 230 | 231 | %--------------END OF PRIMITIVES-------------------------------- 232 | \def\ROOT{./} 233 | \def\TITLE{ 234 | Solutions to some exercises from Walter Rudin's \textit{Functional Analysis} 235 | } 236 | \def\EMAIL{} 237 | \def\AUTHOR{gitcordier} 238 | 239 | \begin{document} 240 | % Changes the "proof" in proof environment. 241 | % source: https://tex.stackexchange.com/questions/8089/changing-style-of-proof 242 | \let\oldproofname=\proofname 243 | \renewcommand{\proofname}{{\rm \small PROOF}} 244 | %\begin{abstract} 245 | % \input{\ROOT/abstract.tex} 246 | %\end{abstract} 247 | \title{\TITLE} 248 | \author{\AUTHOR} 249 | \date{\today} 250 | \maketitle 251 | 252 | % FORMAT ENUMERATION (DEFAULT. OPTIONS: ALPH, ARABIC, ROMAN,…) 253 | \renewcommand{\labelenumi}{$(\textit{\alph{enumi}}\,)$} 254 | % IF LANG = @fr 255 | %\renewcommand{\chaptername}{Chapitre} 256 | % 257 | % CHAPTER NAME : 258 | \frontmatter 259 | \tableofcontents 260 | %\clearpage 261 | 262 | %\printglossary 263 | 264 | \input{\ROOT/notations.tex} 265 | 266 | \mainmatter 267 | %\part{Content} 268 | \renewcommand{\thechapter}{\arabic{chapter}} 269 | \renewcommand{\thesection}{\arabic{section}} 270 | \renewcommand{\thesubsection}{\arabic{subsection}} 271 | \input{\ROOT/FA_mainmatter.tex} 272 | %\backmatter 273 | %\part{Annex} 274 | %\renewcommand\thechapter{\Alph{chapter}} 275 | %\setcounter{chapter}{1} 276 | %\chapter{Additional results} 277 | %\newcounter{annex} 278 | %\setcounter{annex}{1} 279 | %\renewcommand\thesection{\thechapter.\arabic{annex}} 280 | %\section{Number theory} 281 | %\input{\ROOT/Annex_number_theory.tex} 282 | \bibliographystyle{plain} 283 | \bibliography{bibliography}{} 284 | \addcontentsline{toc}{chapter}{Bibliography} 285 | \end{document} 286 | 287 | 288 | %Made on \XeTeX 289 | -------------------------------------------------------------------------------- /chapter_1/1_16.tex: -------------------------------------------------------------------------------- 1 | \textit{ 2 | Prove that the topology of $C(\Omega)$ does not depend on the particular 3 | choice of $\singleton{K_n}$, as long as this sequence satisfies the conditions 4 | specified in section 1.44. Do the same for $C^\infty(\Omega)$ (Section 1.46).} 5 | % 6 | \paragraph{Comment}This is an invariance property: 7 | The function test topology only depends on the existence of the 8 | supremum-seminorms $p_n$, then, eventually, 9 | only on the ambient space itself. 10 | This should be regarded as a very part of the textbook \cite{FA} 11 | % 12 | The proof consists in combining trivial consequences of the local base 13 | definition with a well-known result (for instance, see [2.6] in \cite{BigRudin}) 14 | about intersection of nonempty compact sets. 15 | 16 | \paragraph{Lemma 1} {\it % 17 | Let $X$ be a topological space with a countable local base % 18 | $\mathit{\set{V_n}{\counting{n}}}$. 19 | If 20 | % 21 | $\mathit{\tilde{V}_{n} = V_1 \cuts \cdots \cuts V_n}$, 22 | % 23 | then every subsequence 24 | % 25 | $\mathit{\singleton{\tilde{V}_{\rho(n)}}}$ 26 | % 27 | is a decreasing (\ie 28 | % 29 | $\mathit{\tilde{V}_{\rho(n)} \contains \tilde{V}_{\rho(n+1)}}$) 30 | % 31 | local base of $\varit{X}$. 32 | } 33 | % 34 | \begin{proof} 35 | The decreasing property is trivial. Now remark that 36 | % 37 | $V_n \contains \tilde{V}_{n}$: 38 | % 39 | This shows that 40 | % 41 | $\singleton{\tilde{V}_{n}}$ 42 | % 43 | is a local base of $X$. Then so is 44 | % 45 | $\singleton{\tilde{V}_{\rho(n)}}$, 46 | % 47 | since $\tilde{V}_{n} \contains \tilde{V}_{\rho(n)}$. 48 | \end{proof} 49 | % 50 | \noindent The following special case 51 | % 52 | $V_{n} = \tilde{V}_{n}$ 53 | % 54 | is one of the key ingredients: 55 | %: COROLLARY 1 OF LEMMA 1-----------------------------------------------------% 56 | \paragraph{Corollary 1 (special case $V_{n} = \tilde{V}_{n}$)} 57 | {\it Under the same notations of Lemma 1, if $\mathit{\singleton{V_{n}}}$ % 58 | is a decreasing local base, then so is $\mathit{\singleton{V_{\rho(n)}}}$.} 59 | % 60 | %: COROLLARY 2 OF LEMMA 1-----------------------------------------------------% 61 | \paragraph{Corollary 2}{\it % 62 | If 63 | % 64 | $\mathit{\mathit{\singleton{Q_n}}}$ 65 | % 66 | is a sequence of compact sets that satisfies the conditions specified 67 | in section 1.44, then every subsequence 68 | % 69 | $\mathit{\singleton{Q_{\rho(n)}}}$ 70 | % 71 | also satisfies theses conditions. 72 | % 73 | Furthermore, if $\mathit{\tau_{Q}}$ is the $\mathit{C(\Omega)}$'s 74 | (respectively $\mathit{C^\infty (\Omega)}$'s) topology of the seminorms % 75 | $\mathit{p_{n}}$, 76 | as defined in section 1.44 (respectively 1.46), then the seminorms 77 | % 78 | $\mathit{p_{\rho(n)}}$ 79 | % 80 | define the same topology $\mathit{\tau_{Q}}$. % 81 | } 82 | % 83 | \begin{proof}% 84 | % 85 | Let $X$ be $C(\Omega)$ topologized by the seminorms $p_{n}$ 86 | (the case $X=C^\infty(\Omega)$ is proved the same way). 87 | % 88 | If 89 | % 90 | $V_{n} = \singleton{p_{n} < 1/n}$, 91 | % 92 | then 93 | % 94 | $\singleton{V_{n}}$ 95 | % 96 | is a decreasing local base of $X$. 97 | % 98 | Moreover, 99 | % 100 | \begin{align} 101 | Q_{\rho(n)} 102 | \subset 103 | \interior{Q}_{\rho(n) + 1} 104 | \subset 105 | Q_{\rho(n) + 1} 106 | \subset 107 | Q_{\rho(n+ 1)}. 108 | \end{align} 109 | % 110 | Thus, 111 | % 112 | \begin{align} 113 | Q_{\rho(n)} 114 | \subset 115 | \interior{Q}_{\rho(n+ 1)}. 116 | \end{align} 117 | % 118 | In other words, 119 | % 120 | $Q_{\rho(n)}$ satisfies the conditions specified in section 1.44. 121 | % 122 | % 123 | $\singleton{p_{\rho(n)}}$ 124 | % 125 | then defines a topology $\tau_{Q_\rho}$ for which 126 | % 127 | $\singleton{V_{\rho(n)}}$ 128 | % 129 | is a local base. So, 130 | % 131 | $\tau_{Q_\rho} \subset \tau_{Q}$. 132 | % 133 | Conversely, the above corollary asserts that 134 | % 135 | $\singleton{V_{\rho(n)}}$ 136 | % 137 | is a local base of $\tau_{Q}$, which yields 138 | % 139 | $\tau_{Q}\subset \tau_{Q_\rho}$. 140 | % 141 | \end{proof} 142 | %: LEMMA 2 -------------------------------------------------------------------% 143 | \paragraph{Lemma 2}{ \it \label{1.16 Lemma 2} 144 | If a sequence of compact sets $\mathit{\singleton{Q_n}}$ satisfies the conditions 145 | specified in section 1.44, then every compact set $K$ lies in allmost all 146 | % 147 | $\mathit{Q^{\,\circ}_n}$\ie 148 | % 149 | there exists $\varit{m}$ such that 150 | % 151 | \begin{align}\mathit{ 152 | K \subset 153 | \interior{Q}_m 154 | \subset 155 | \interior{Q}_{m+1} 156 | \subset 157 | \interior{Q}_{m+2} 158 | \subset 159 | \cdots.} 160 | \end{align} 161 | } 162 | % 163 | \begin{proof} 164 | The following definition 165 | % 166 | \begin{align} 167 | C_n \Def K \setminus \interior{Q}_n 168 | \end{align} 169 | % 170 | shapes $\singleton{C_n}$ as a decreasing sequence of compact\footnote{ 171 | See (b) of 2.5 of \cite{BigRudin}. 172 | } 173 | sets. We now suppose (to reach a contradiction) that 174 | % 175 | no $C_n$ is empty 176 | % 177 | and so conclude\footnote{ 178 | In every Hausdorff space, the intersection of a decreasing sequence of % 179 | nomempty compact sets is nonempty. % 180 | This is a corollary of 2.6 of \cite{BigRudin}. 181 | } 182 | that the $C_n$'s intersection contains a point that is not in any $Q^\circ_n$. 183 | On the other hand, the conditions specified in [1.44] force the 184 | % 185 | $Q^\circ_n$'s collection 186 | % 187 | to be an open cover. 188 | % 189 | This contradiction reveals that 190 | % 191 | $C_m = \emptyset$, 192 | \ie 193 | $K \subset Q^\circ_m$, 194 | % 195 | for some $m$. 196 | % 197 | Finally, 198 | % 199 | \begin{align} 200 | K\subset 201 | \interior{Q}_m 202 | \subset 203 | Q_m 204 | \subset 205 | \interior{Q}_{m+1} 206 | \subset 207 | Q_{m +1} 208 | \subset 209 | \interior{Q}_{m+2} 210 | \subset 211 | \cdots. 212 | \end{align} 213 | % 214 | \end{proof} 215 | %: THEOREM -------------------------------------------------------------------% 216 | \noindent We are now in a fair position to establish the following: 217 | %\newpage 218 | \paragraph{Theorem}{\it 219 | The topology of % 220 | % 221 | $\mathit{C(\Omega)}$ % 222 | % 223 | does not depend on the particular choice of % 224 | % 225 | $\mathit{\singleton{K_n}}$, % 226 | % 227 | as long as this sequence satisfies the conditions specified in section 1.44. % 228 | Neither does the topology of % 229 | % 230 | $\mathit{C^{\,\infty} (\Omega)}$, % 231 | % 232 | as long as this sequence satisfies the conditions specified in section 1.44. 233 | } 234 | % 235 | \begin{proof}% 236 | With the second corollary's notations, 237 | % 238 | $\tau_{K} = \tau_{K_\lambda}$, 239 | % 240 | for every subsequence $\singleton{K_{\lambda(n)}}$. 241 | % 242 | Similarly, let 243 | % 244 | $\singleton{L_n}$ 245 | % 246 | be another sequence of compact subsets of $\Omega$ that satisfies 247 | the condition specified in [1.44], 248 | so that 249 | % 250 | $\tau_{L} = \tau_{L_\kappa}$ 251 | % 252 | for every subsequence $\singleton{L_{\kappa(n)}}$. 253 | % 254 | Now apply the above Lemma 2 with $K_i$ ($\counting{i}$) and so conclude that 255 | % 256 | $K_i 257 | \subset 258 | L^\circ_{m_i} 259 | \subset 260 | L^\circ_{m_{i}+1} 261 | \subset 262 | \cdots$ 263 | % 264 | for some $m_i$. In particular, the special case $\kappa_i = m_i + i$ is 265 | % 266 | \begin{align} 267 | % 268 | \label{1_16. K subset interior L} 269 | K_i 270 | \subset 271 | \interior{L}_{\kappa_i}. 272 | \end{align} 273 | % 274 | Let us reiterate the above proof with $K_n$ and $L_n$ in exchanged roles 275 | then similarly find a subsequence $\set{\lambda_j}{\counting{j}}$ such that 276 | % 277 | \begin{align} 278 | % 279 | \label{1_16. L subset interior K} 280 | % 281 | L_j \subset \interior{K}_{\lambda_j} 282 | \end{align} 283 | % 284 | Combine 285 | % 286 | (\ref{1_16. K subset interior L}) with 287 | (\ref{1_16. L subset interior K}) 288 | % 289 | and so obtain 290 | % 291 | \begin{align} 292 | K_1 293 | \subset 294 | \interior{L}_{\kappa_1} 295 | \subset 296 | L_{\kappa_1} 297 | \subset 298 | \interior{K}_{\lambda_{\kappa_1}} 299 | \subset 300 | K_{\lambda_{\kappa_1}} 301 | \subset 302 | \interior{L}_{\kappa_{\lambda_{\kappa_1}}} 303 | \subset 304 | \cdots, 305 | \end{align} 306 | % 307 | which means that the sequence 308 | % 309 | $Q = ( 310 | K_1, 311 | L_{\kappa_1}, 312 | K_{\lambda_{\kappa_1}}, 313 | %L_{\kappa_{\lambda_{\kappa_1}}}, 314 | \dots 315 | )$ 316 | % 317 | satisfies the conditions specified in section 1.44. 318 | It now follows from the corollary 2 that 319 | % 320 | \begin{align} 321 | \tau_{K} 322 | = 323 | \tau_{K_\lambda} 324 | = 325 | \tau_{Q} 326 | = 327 | \tau_{L_\kappa} 328 | = \tau_{L}. 329 | \end{align} 330 | % 331 | So ends the proof 332 | \end{proof} 333 | % END 334 | -------------------------------------------------------------------------------- /chapter_1/1_1.tex: -------------------------------------------------------------------------------- 1 | %:1 2 | \renewcommand{\labelenumi}{(\alph{enumi})} 3 | \textit{Suppose $X$ is a vector space. All sets mentioned below are understood 4 | to be subsets of $X$. Prove the following statements from the axioms 5 | as given as in section 1.4. 6 | \begin{enumerate} 7 | \item{If $x,\,y\in X$ there is a unique $z\in X$ such that $x+z=y$.} 8 | \item{ $0\cdot x=0=\alpha\cdot 0 \quad (\alpha\in\field, x\in X)$.} 9 | \item{ $2A\subset A+A$.} 10 | \item{ $A$ is convex if and only if $(s+t)A=sA+tA$ % 11 | for all positive scalars $s$ and $t$.} 12 | \item{ Every union (and intersection) of balanced sets is balanced.} 13 | \item{ Every intersection of convex sets is convex.} 14 | \item{ If $\Gamma$ is a collection of convex sets that is totally ordered by 15 | set inclusion, then the union of all members of $\Gamma$ is convex.} 16 | \item{ If $A$ and $B$ are convex, so is $A+B$.} 17 | \item{ If $A$ and $B$ are balanced, so is $A+B$.} 18 | \item{ Show that parts (f\,), (g) and (h) hold with subspaces in place of 19 | convex sets.} 20 | \end{enumerate} 21 | } 22 | % 23 | \renewcommand{\labelenumi}{(\alph{enumi})} 24 | \begin{proof} 25 | \begin{enumerate} 26 | %: (a) 27 | \item Such property only depends on the group structure of $X$: Each $x$ in 28 | $X$ has an opposite $\minus x$. Let $x'$ be any opposite of $x$so that 29 | ${x-x=0}=x+x'$. % 30 | Thus, $\minus x +x -x = \minus x + x + x' $, % 31 | which is equivalent to $\minus x = x'$. So is established the uniqueness of % 32 | $\minus x $. % 33 | % 34 | It is now clear that $x+z=y$ \iif $z=\minus x +y$, % 35 | which asserts both the existence and the uniqueness of $z$. 36 | %: (b) 37 | \item Remark that % 38 | % 39 | \begin{align} 40 | 0\cdot x & =(0+0)\cdot x=0\cdot x+0\cdot x \\ 41 | & =(0+0)\cdot x=0 +0\cdot x 42 | \end{align} 43 | % 44 | then conclude from (a) that $0\cdot x=0$. So, % 45 | \begin{align} \label{inverse of x} 46 | 0=0\cdot x=(1-1)\cdot x &=x+(\minus 1)\cdot x 47 | \Rightarrow \minus 1\cdot x= \minus x. 48 | \end{align} 49 | % 50 | Finally, % 51 | % 52 | \begin{align} 53 | \alpha\cdot 0\overset{(\ref{inverse of x})}{=} 54 | \alpha\cdot (x+(\minus 1\cdot x)) 55 | = \alpha \cdot x + \alpha \cdot (\minus 1) \cdot x 56 | = (\alpha-\alpha ) \cdot x =0\cdot x = 0, 57 | \end{align} 58 | % 59 | which proves (b). 60 | % 61 | %: (c) 62 | \item Remark that 63 | % 64 | \begin{align} 65 | 2x =(1+1) x = x + x 66 | \end{align} 67 | % 68 | for every $x$ in $X$, and so conclude that % 69 | % 70 | \begin{align}\label{double lies in sum} 71 | 2A = \{2x: x\in A \} 72 | = \{x + x: x \in A \} 73 | \subset \{ x + y : (x,\,y) \in A^2 \} 74 | = A+A 75 | \end{align} 76 | % 77 | for all subsets $A$ of $X$; which proves (c). % 78 | %: (d) 79 | \item If $A$ is convex, then % 80 | % 81 | \begin{align} 82 | A \subset \frac{s}{s+t} A + \frac{t}{s+t} A \subset A; 83 | \end{align} 84 | % 85 | which is % 86 | % 87 | \begin{align} 88 | sA + tA = (s+t)A. 89 | \end{align} 90 | % 91 | Conversely, the special case $s+t=1$ is % 92 | % 93 | \begin{align} 94 | sA + (1-s)A = A. 95 | \end{align} 96 | % 97 | The latter extends to $s=0$, since % 98 | % 99 | \begin{align} 100 | 0A + A \overset{(b)}{=}\{0\}+A=A. 101 | \end{align} 102 | % 103 | The extension to $s=1$ is analogously established % 104 | (or simply use the fact that $+$ is commutative!). 105 | So ends the proof. % 106 | %: (e) 107 | \item Let $A$ range over $B$ a collection of balanced subsetsso that % 108 | % 109 | \begin{align} 110 | \alpha \bigcap B \subset \alpha A \subset A \subset \bigcup B 111 | \end{align} 112 | % 113 | for all scalars $\alpha$ of magnitude $\leq 1$. % 114 | The inclusion $\alpha \bigcap B \subset A$ establishes the first part. % 115 | Now remark that % 116 | % 117 | \begin{align} 118 | \alpha A \subset \bigcup {B} 119 | \end{align} 120 | % 121 | implies % 122 | % 123 | \begin{align} 124 | \alpha \bigcup {B} \subset \bigcup {B}; 125 | \end{align} 126 | which achieves the proof. % 127 | % 128 | %: (f) 129 | \item Let $A$ range over $C$ a collection of convex subsetsso that % 130 | % 131 | \begin{align} 132 | (s+t) \bigcap C \subset s\bigcap C + t\bigcap C \subset sA + tA 133 | \overset{(d)}{\subset} (s+t)A 134 | \end{align} 135 | % 136 | for all positives scalars $\mathit{s}$, $\mathit{t}$. % 137 | Inclusions at both extremities force % 138 | % 139 | \begin{align} 140 | s\bigcap C + t\bigcap C = (s+t) \bigcap C. 141 | \end{align} 142 | % 143 | We now conclude from (d) that the intersection of $C$ is convex. % 144 | So ends the proof. 145 | %: (g) 146 | \item Skip all trivial cases % 147 | % 148 | $\Gamma = \emptyset$, % 149 | $\{ \emptyset\}$, % 150 | $\{\{x\}\}$, % 151 | $\{\emptyset$, % 152 | $\{x\}\}$ % 153 | % 154 | then pick $x_1, x_2$ in $\bigcup \Gamma$, 155 | so that each $x_i$ ($i=1, 2)$ lies in some $C_i \in \Gamma$. % 156 | % 157 | Since $\Gamma$ is totally ordered by set inclusion, we henceforth assume % 158 | without loss of generality that $C_1$ is a subset of $C_2$. % 159 | % 160 | So, $x_1, x_2$ are now elements of the convex set $C_2$. % 161 | Every convex combination of our $x_i$'s is then in % 162 | $C_2 \subset \bigcup \Gamma$. Hence (g). % 163 | % 164 | %: (h) 165 | \item Simply remark that 166 | % 167 | \begin{align} 168 | s (A+B) + t (A+B) = s A+ t A +s B +t B = (s+t)(A+B) 169 | \end{align} 170 | % 171 | for all positive scalars $\mathit{s}$ and $\mathit{t}$, % 172 | then conclude from (d) that $A + B$ is convex. % 173 | %: (i) 174 | \item Given any $\alpha$ from the closed unit disc, % 175 | % 176 | \begin{align} 177 | \alpha(A+B)=\alpha A+ \alpha B \subset A+B. 178 | \end{align} 179 | % 180 | There is no more to prove: $A+B$ is balanced. % 181 | %: (j) 182 | \item Our proof will be based on the following lemma, % 183 | % 184 | \renewcommand{\labelenumii}{(\roman{enumii})} 185 | \textit{ 186 | \begin{quote} 187 | If $S$ is nonempty, then each of the following three properties 188 | \begin{enumerate} 189 | \item $S$ is a vector subspace of $X$; 190 | \item $S$ is convex balanced such that $S + S = S$; 191 | \item $S$ is convex balanced such that $\lambda S=S\quad (\lambda > 0)$ 192 | \end{enumerate} 193 | implies the other two. 194 | \end{quote} 195 | } 196 | % 197 | To prove the lemma, let $\mathit{S}$ % 198 | run through all nonempty subsets of $X$. % 199 | First, assume that (i) holds: Clearly, every $S$ is convex balanced. % 200 | Moreover, $S+S \subset S$. Conversely, $S = S + \{0\} \subset S + S $; % 201 | which establishes (ii). % 202 | % 203 | Next, assume (only) (ii): A proof by induction shows that % 204 | % 205 | \begin{align}\label{induction nS} 206 | nS = (n-1) S + S = S + S = S \quad (n=1,2,3, \dots) 207 | \end{align} 208 | % 209 | with the help of (b) and (d). % 210 | Pick $\lambda >0$ then choose $n$ so large that $1 < n \lambda < n^2$. % 211 | Thus, % 212 | % 213 | \begin{align} 214 | nS \overset{(\ref{induction nS})}{\subset} S 215 | \subset n\,\lambda S 216 | \subset n^2 S, 217 | \end{align} 218 | % 219 | since $S$ is balanced. % 220 | For instance, set % 221 | % 222 | $n = \lceil{1/\lambda}\rceil + \lceil{\lambda}\rceil$. % 223 | % 224 | Dividing the latter inclusions by $n$ shows that % 225 | % 226 | \begin{align} 227 | S \subset \lambda S \subset nS \overset{(\ref{induction nS})}{\subset} S, 228 | \end{align} 229 | % 230 | which is (iii). Finally, dropping (ii) in favor of (iii) leads to % 231 | % 232 | \begin{align} 233 | \alpha S +\beta S 234 | \overset{(a)}{=} |\alpha | S + | \beta | S 235 | \overset{(d)}{=} (|\alpha | + | \beta| )S 236 | \overset{(iii)}{=} S 237 | % 238 | \quad (|\alpha| + |\beta| > 0); 239 | \end{align} 240 | % 241 | where the equality at the left holds as $S$ is balanced. % 242 | % 243 | Moreover (under the sole assumption that $S$ is balanced), % 244 | this extends to $|\alpha| + |\beta| = 0$, as follows, % 245 | % 246 | \begin{align} 247 | \alpha S + \beta S = 0S + 0S\overset{(b)}{=} \{0\} 248 | \overset{(b)}{=} 0S \subset S. 249 | \end{align} 250 | % 251 | Hence (i), which achieves the lemma's proof. % 252 | We will now offer a straightforward proof of (j). \\ 253 | \\ 254 | Let $V$ be a collection of vector spaces of $X$, % 255 | of intersection $I$ and union $U$. 256 | % 257 | First, remark that every member of $V$ is convex balanced: % 258 | So is $I$ (combine (e) with (f)). % 259 | % 260 | Next, let $\mathit{Y}$ range over $V$so that % 261 | % 262 | \begin{align} 263 | I + I \subset Y + Y \subset Y; 264 | \end{align} 265 | % 266 | which yields 267 | % 268 | \begin{align} 269 | I + I = I 270 | \end{align} 271 | % 272 | (the fact that $I = I + \{0\} \subset I + I$ was tacitely used). % 273 | % 274 | It now follows from the lemma's (ii) $\Rightarrow$ (i) that % 275 | $I$ is a vector subspace of $X$. % 276 | % 277 | Now temporarily assume that $S$ is totally ordered by set inclusion: % 278 | Combining (e) with (g) establishes that $U$ is convex balanced. % 279 | % 280 | To show that $U$ is more specifically a vector subspace, % 281 | we first remark that such total order implies that either % 282 | $Z \subset Y$ or $Y \subset Z$, as $\mathit{Z}$ ranges over $V$. % 283 | A straightforward consequence is that 284 | % 285 | \begin{align} 286 | Y \subset Y + Z \subset Y\cup Z. 287 | \end{align} 288 | % 289 | Another one is that $Y \cup Z$ ranges over $V$ as well. % 290 | Combined with the latter inclusions, this leads to % 291 | % 292 | \begin{align} 293 | U \subset U + U \subset U. 294 | \end{align} 295 | % 296 | It then follows from the lemma's (ii) $\Rightarrow$ (i) that % 297 | $U$ is a vector subspace of $X$. % 298 | % 299 | Finally, let $\mathit{A},\mathit{B}$ run through all vector subspaces of $X$: % 300 | Combining (h) with (i) proves that $A+B$ is convex balanced as well. % 301 | % 302 | Furthermore, % 303 | % 304 | \begin{align} 305 | A + B \overset{(i) \Rightarrow (ii)}{=} (A + A) + (B + B) = (A + B) + (A + B), 306 | \end{align} 307 | % 308 | where the equality at the right holds as $X$ is an abelian group. % 309 | We now conclude from (ii) that any $A+B$ is a vector subspace of $X$. % 310 | % 311 | So ends the proof. % 312 | \end{enumerate} 313 | \end{proof} 314 | % END 315 | -------------------------------------------------------------------------------- /notations.tex: -------------------------------------------------------------------------------- 1 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 2 | % FunctionalAnalysis 3 | % notations.tex 4 | % 5 | % encoding: UTF-8 6 | % EOL: LF 7 | % 8 | % format: LaTeX 9 | % indent: spaces (2) 10 | % width: 127 11 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 12 | \renewcommand{\thesection}{\Roman{section}} 13 | \renewcommand{\thesubsection}{\roman{subsection}} 14 | % 15 | \chapter{Notations and Assumptions}% 16 | %\addcontentsline{toc}{chapter}{Notations and Conventions} 17 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 18 | % LOGIC 19 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 20 | \section{Logic}% 21 | \subsection{Propositional logic operators} 22 | Given propositional variables $\mathit{p}$, $\mathit{q}$, the boolean % 23 | operators $\lnot$, $\lor$, $\land$, $\iff$, $\then$, % 24 | $\Leftarrow$, are assigned boolean \textit{truth values} (``true'' and ``false'') as follows, 25 | \begin{enumerate} 26 | \item[$\lnot$]{% 27 | $\lnot p$ and $p$ have opposite values: 28 | } 29 | \item[$\lor$]{ 30 | The \textit{disjunction} (``or'') $p \lor q$ is true unless $p$ is false and $q$ is false. 31 | } 32 | \item[$\land$]{ 33 | The \textit{conjunction} (``and'') $p \land q$ is true if and only if $p$ is true and $q$ is true. 34 | } 35 | \item[$\iff$]{% 36 | The \textit{logical equivalence} expresses \textit{tautologies}: % 37 | $p \iff q$ is true unless $p$ and $q$ have opposite values. It is easily checked that % 38 | $(p \iff q) \iff \bigl((p \then q) \land (p \IF q) \bigr)$, see the definitions below. 39 | } 40 | \item[$\then$]{% 41 | The logical connection \textit{$p$ implies $q$} is supported by $\then$: $p \then q$ means \textit{if $p$ then $q$} % 42 | and is formally defined as $\lnot p \lor q$. Note that the ``reasoning'' $p \then q $ is always true unless % 43 | $p$ is true and $q$ is false. Moreover, $p \land (p \then q) \then q$ is always true. % 44 | This deductive rule is known as \textit{modus ponens}. 45 | } 46 | \item[$\Leftarrow$]{ $q \IF p$ means that $q$ is implied by $p$ and is defined as $ p\then q $. % 47 | It is commonly read aloud as ``$q$ if $p$'' or ``$q$ is a consequence of $p$''. 48 | } 49 | \end{enumerate} 50 | % 51 | See Section 1.3 and Subsection 16.1.3 of \cite{SpecifyingSystems} for further reading. 52 | % 53 | \subsection{Iverson notation}% 54 | Given a boolean expression $\phi$, % 55 | $\Iverson{\phi}$ returns the truth value of $\phi$, encoded as follows, % 56 | % 57 | \begin{align} \nonumber 58 | \Iverson{\phi}\triangleq 59 | \begin{cases} 60 | 0 & \quad\quad \text{if } \phi \text{ is false;} \\ 61 | 1 & \quad\quad \text{if } \phi \text{ is true.} 62 | \end{cases} 63 | \end{align} 64 | % 65 | For example, $\Iverson{1 > 0} = 1$ but $\Iverson{ \sqrt{2} \in \Q} = 0$. 66 | \section{Special terms} 67 | \subsection{Halmos' $\iif$ and definitions}% 68 | \iif is a shorthand for ``if and only if". Splitting \iif into \textit{if-then} clauses shows that it is just a rewording of % 69 | the logical equivalence $\iff$. All definitions will use the \iif format, which is consistent with the fact that % 70 | every definition expresses a tautology. 71 | % 72 | \subsection{The assignment operator $\Def$}% 73 | Given variables $\varit{a}$ and $\varit{b}$, $\triangleq$ is a specialization of $=$. We say that $a\triangleq b$ \iif % 74 | $a$ and $b$ are assumed to be equal. Usually, $a\triangleq b$ means that $a$ is assigned the previously known value % 75 | $b$ (some authors write $a:=b$) but this is not a limitation. Definitions can be redundant and may overlap. % 76 | % 77 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 78 | % SETS 79 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 80 | \section{Sets} 81 | \subsection{Subsets and supersets}% 82 | $\subset$ and $\supset$ are the standard symbols for set ordering, as follows: % 83 | \begin{align} 84 | X \subset Y & \defiif x \in X \then x\in Y \\ 85 | Y \supset X & \defiif X \subset Y. 86 | \end{align} 87 | No specific symbol is reserved for strict ordering. If necessary, $X \neq Y$ will be explicitly stated. 88 | % 89 | \subsection{Special mappings} 90 | The identity $I$ is the special mapping $I_X = \set{(x,x)}{x\in X}$. Similarly, the projection % 91 | $\pi = \set{((x, y), x)}{x \in X, y \in Y}$ always exists. Note that $I$ is the diagonal of $X^2$. % 92 | % 93 | \subsection{Equinumerosity}% 94 | $X\equiv Y$ means that there exists a bijection $\phi$ that maps $X$ to $Y$, which lets us identify $X$ with $Y$. % 95 | In a metric space context, $X\equiv Y$ means that $\phi$ is isometric. % 96 | % 97 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 98 | % TVS 99 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 100 | \section{Topological vector spaces} 101 | \subsection{Scalar field}% 102 | $\C$ extends $\R$, which implies that a property\eg linearity, that is true on $\C$ is also true on $\R$. The complex case % 103 | is then a \textit{special case} of the real case. This restriction may be significant in some contexts. Nevertheless, the % 104 | standard scalar field is $\C$, which means that considering $\R$ instead of $\C$ makes no difference, unless stated otherwise.% 105 | % 106 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 107 | % Vector spaces 108 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 109 | \subsection{Vector space bases}\label{notations: vector spaces: vector space bases} 110 | Given a vector space $X$ over $\C$ (or, more generally, over a field), a subset $B$ of $X$ is a basis of $X$ \iif % 111 | the sum % 112 | % 113 | \begin{align} 114 | \bigl\{(z_u)_{u\in B}: z_u \in \C, \set{u}{z_u \neq 0} \text{ is finite} \bigr\} & \to X \\\nonumber 115 | (z_u) & \mapsto \sum_{z_u \neq 0 } z_u u 116 | \end{align} 117 | bijectively maps all \textit{finitely supported} $(z_u)$ \textit{onto} $X$. % 118 | % 119 | The axiom of choice (AC) forces % 120 | \begin{enumerate} 121 | \item the existence of such $B$ % 122 | (the proof is similar to the second part of the Hahn-Banach theorem [3.1] of \cite{FA} with $B$ playing the role of 123 | $\Lambda$); % 124 | \item all bases to have the same cardinal, which is called the {\it dimension} of $X$ and is denoted as $\dim(X)$. % 125 | \end{enumerate} 126 | % 127 | We now turn to the finite-dimensional case. Note that the $0$-dimensional case is the degenerate case $B=\emptyset$, % 128 | which is equivalent to $X=\singleton{0}$. Our first step is to study $\C^n$ ($n>0$) the standard $n$-dimensional vector space. 129 | % 130 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 131 | % Finite dimension 132 | % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 133 | \subsection{Finite-dimensional spaces}\label{notations: vector spaces: finite-dimensional vector spaces}% 134 | \subsubsection{The product topology of $\C^n$}% 135 | \label{notations: vector spaces: finite-dimensional vector spaces: the product topology of Cn} 136 | % 137 | $\C^n$ has a standard basis $1_{\singleton{1}}, \dots, 1_{\singleton{n}} \in \{0, 1\}^{\{0, \dots, n-1\}}$ % 138 | so that $z_k$ ($1 \leq k \leq n$) is the $k$-th component of a given % 139 | $(z_1,\dots, z_n) = z_1 (1, 0, \dots, 0) + \cdots + z_n (0, \dots, 0, 1) \in \C^n$. Furthermore, % 140 | $\C^n$ is endowed with the topology generated by all polydiscs % 141 | % 142 | \begin{align} 143 | \prod_{i=1}^{n} \underbrace{\set{z_i \in C}{\magnitude{z_i} < r_i}}_{D_{r_i}} \quad (r_i > 0 ). 144 | \end{align} 145 | % 146 | Equivalently, we may equip $\C^n$ with the Euclidean norm % 147 | % 148 | \begin{align} 149 | \norma{2}{z} \Def \sqrt{\magnitude{z_1}^2 + \cdots + \magnitude{z_n}^2} \quad \left(z = (z_1, \dots, z_n) \in \C^n\right), 150 | \end{align} 151 | % 152 | whose open balls centered at the origin are all % 153 | % 154 | \begin{align} 155 | B_r \Def \set{z\in \C^n}{\norma{2}{z} < r} \quad (r > 0). 156 | \end{align} 157 | % 158 | To show the equivalence, first set $r_i = r/\sqrt{n}$. Hence % 159 | % 160 | \begin{align} 161 | \prod_{i=1}^{n} D_{r_i} \subset B_r. 162 | \end{align} 163 | % 164 | Conversely, put $r = \min\{r_1, \dots, r_n\}$ so that % 165 | % 166 | \begin{align} 167 | B_r \subset \prod_{i=1}^{n} D_{r_i} . 168 | \end{align} 169 | % 170 | \subsubsection{Topology of a finite-dimensional vector space} 171 | It is customary to identify any $n$-dimensional vector space with $\C^n$ equipped with the Euclidean norm, % 172 | see [\ref{notations: vector spaces: finite-dimensional vector spaces: the product topology of Cn}]. % 173 | % 174 | To show this, pick an $n$-dimensional vector space $Y$ and let $f$ be an isomorphism of $\C^n$ onto $Y$. % 175 | For instance, require that $u_k = f(e_k)$, like in [1.20] of \cite{FA}, as $\singleton{u_k}$ is a basis of $Y$, % 176 | see [\ref{notations: vector spaces: vector space bases}]. It follows from [1.21] of \cite{FA} that $f$ is a homeomorphism. 177 | The striking consequence is that 178 | % 179 | \begin{quote} 180 | $\set{f(U)}{U \text{ open in }\C^n}$ {is the only vector space topology for $Y$}. % 181 | \end{quote} 182 | % 183 | Thus, $Y$ is necessarily locally convex and locally bounded \ie normable, see [1.39] of \cite{FA}. Note that % 184 | $\norm{y}= \norma{2}{f^{\minus 1}(y)}$ ($y\in Y$) is an example of a norm. Additionally, $Y$ is locally compact, % 185 | since the closed unit ball of $\C^n$ is compact. Now pick an $n$-dimensional topological vector space $W$, then repeat the % 186 | same reasoning, first with $g: \C^n \to W$, next with $h = g\circ f^{\,\minus 1}$ playing the role of $f$. This establishes % 187 | that the homeomorphism $h$ maps $Y$ onto $W$ and that $W$ is normable as well. It is now clear that the following assertions % 188 | are equivalent in the finite-dimensional context: % 189 | % 190 | \begin{enumerate} 191 | \item $\dim(W) = \dim(Y),$ 192 | \item $W$ and $Y$ are isomorphic to each other,% 193 | \item $W$ and $Y$ are homeomorphic to each other, they are normable. 194 | \end{enumerate} 195 | % 196 | Furthermore, the norms on $W$ and $Y$ are \textit{equivalent}. That is, for any given norm $\norm{\, \cdot \, }$ on $Y$ and % 197 | any given norm $\norma{W}{\,\cdot \,}$ on $W$, there exists a positive constant (termed as ``modulus of continuity'') $C= C_h$ % 198 | such that % 199 | \begin{align} 200 | \norma{W}{w} & \leq C \norm{y} \quad\quad \left((y, w) \in h \right), 201 | \end{align} 202 | % 203 | as $h$ is continuous. The special case $W=Y$ is that all norms on $Y$ are equivalent, in the sense that % 204 | % 205 | \begin{align} 206 | \norma{{\text{copy of }Y}}{h(y)} \leq C \norm{y}. 207 | \end{align} 208 | % 209 | \subsubsection{The standard norms $\norma{1}{\,\cdot\,}$, $\norma{2}{\,\cdot\,}$, $\norma{\infty}{\,\cdot\,}$} When $\C^n$ % 210 | is equipped with standard norms $1, 2, \infty$, the sharp constant\ie the smallest $C_{i, j}$ such that % 211 | % 212 | \begin{align} 213 | \norma{j}{z} \leq C_{i, j} \norma{i}{z} %\quad \left(z \in \C^n \right) 214 | \end{align} 215 | % 216 | is easily derived from definitions - see [1.19] of \cite{FA} - with the noticeable exception of $C_{2, 1} = \sqrt{n}$. % 217 | Indeed, $\norma{1}{z} \leq \sqrt{n} \norma{2}{z}$ is a special Cauchy-Schwarz inequality, see (1) in [12.2] of \cite{FA}. % 218 | The steps of this classical trick are left to the reader. % 219 | % 220 | \section{Measure theory on $\R^n$} 221 | \subsection{Radon measures} 222 | A (positive) Radon measure is a linear functional % 223 | $\Lambda: C_c(\R^n) \to \C, \phi \mapsto \Lambda \phi$ that % 224 | % 225 | \renewcommand{\labelenumi}{(\roman{enumi})} 226 | \begin{enumerate} 227 | \item is positive\ie $\phi \geq 0 \then \Lambda\phi \geq 0$, 228 | \item is continuous in the sense that, for each compact $K \subset \R^n$ there exists a modulus of continuity $M_K$ such that % 229 | \begin{align} 230 | \magnitude{\Lambda\phi} \leq M_K\norma{\infty}{\phi} \quad (\supp{\phi} \subset K ). 231 | \end{align} 232 | \end{enumerate} 233 | \renewcommand{\labelenumi}{(\alph{enumi})} 234 | % 235 | Theorem [2.15] of \cite{BigRudin} shows that (i) actually implies (ii), as (ii) defines Radon measures in a weaker sense, % 236 | see \cite{AnalyseIII}. Moreover, $\Lambda$ is bounded on $C_c(\R^n)$ with respect to the supremum norm \iif (ii) is % 237 | satisfied with all $M_K\leq \norma{\infty}{\Lambda} <\infty$. These bounded linear functionals are at the core of the theory, % 238 | and according to [6.19] of \cite{BigRudin}, each of them is isometrically identified with a specific regular Borel measure % 239 | $\mu$. % 240 | % 241 | \subsection{Lebesgue integration} 242 | Theorem [2.14] of \cite{BigRudin} states that every positive Radon measure $\Lambda$ (bounded or not) is identified with a % 243 | specific positive and regular Borel measure $\beta$. This justifies the introduction of % 244 | % 245 | \begin{align}\label{definition of sum} 246 | \int_{\R^n} \phi \diff{\beta} \Def \Lambda \phi \quad \left(\phi \in C_c(\R^n)\right), 247 | \end{align} 248 | % 249 | where the left side is a Lebesgue integral. The regularity property implies that $\Lambda$ has mass % 250 | % 251 | \begin{align} 252 | \int_{\R^n} 1 \diff{\beta} \Def \sup \bigl\{\Lambda \phi: \phi \in C_c(\R^n), \norma{\infty}{\phi} \leq 1 \bigr\}. 253 | \end{align} 254 | % 255 | When we substitute $f\diff{\beta}$ for $\diff{\beta}$, with $f$ Borel measurable, \eqref{definition of sum} takes the form % 256 | % 257 | \begin{align}\label{Borel integral} 258 | \int_{\R^n} f \phi \diff{\beta} = \Lambda \phi. 259 | \end{align} 260 | % 261 | In particular, if $f$ is positive, 262 | % 263 | \begin{align} 264 | \int_{\R^n} f \diff{\beta} = \sup \bigl\{\Lambda \phi: \phi \in C_c(\R^n), \norma{\infty}{\phi} \leq 1 \bigr\}. 265 | \end{align} 266 | % 267 | Moreover, given $N = \set{f}{f = 0 \text{ almost everywhere with respect to }\beta}$, we see that, % 268 | in \eqref{Borel integral}, density $f$ and measure $\Lambda$ are identified modulo $N$. Algebraically speaking: % 269 | \begin{align} 270 | f_1\beta = f_2\beta \iff f_1-f_2 \in N 271 | \end{align} 272 | % 273 | Thus, we define the following vector spaces % 274 | % 275 | \begin{align} 276 | \mathcal{L}^1 \Def \Biggl\{f: \int_{\R^n}\magnitude{f} \diff{\beta} < \infty\Biggr\} 277 | \subset \Biggl\{f: \int_{\R^n} 1_K \magnitude{f} \diff{\beta} < \infty % 278 | \text{, when $K \subset \R^n$ compact} \Biggr\} 279 | \Def \mathcal{L}^1_{\text{loc}}, 280 | \end{align} 281 | % 282 | where $\mathcal{L}^1$ is equipped with the seminorm 283 | % 284 | \begin{align} 285 | \magnitude{f}_1 \Def \int_{\R^n} \magnitude{f} \diff{\beta} \quad (f \in \mathcal{L}^1). 286 | \end{align} 287 | % 288 | Notably, $N$ is closed in $\mathcal{L}^1$. This implies that $L^1 = \mathcal{L}^1/N$, equipped with the norm % 289 | % 290 | \begin{align} 291 | \norma{1}{f + N} \Def \int_{\R^n} \magnitude{f} \diff{\beta} \quad (f \in \mathcal{L}^1) 292 | \end{align} 293 | % 294 | is a Banach space. % 295 | \subsection{Dirac's impulse, a physicist's detour} 296 | Suppose we wish to record the impact of a particle on a surface. To do so, we introduce a carrier signal $\phi(t)$ that % 297 | vanishes outside the observation interval. A given collision occurs at time $t_Q$, which means the surface absorbs a pulsed % 298 | energy quantum $Q$. Thus, the particle contribution is encoded as the Heaviside step function $H(t) = \Iverson{t \geq 0}$, % 299 | as time $t$ ranges over the real line. This models the fact that energy $H(t)$ only increases by an infinitely abrupt jump % 300 | at reference time $t_Q=0$. In this detour, we heuristically conclude that % 301 | % 302 | \begin{align} 303 | \frac{\diff{H}}{\diff{t}} = \infty \times \Iverson{t=0}. 304 | \end{align} 305 | % 306 | The latter informal derivative $\diff{H}/\diff{t}$ is empirically known as $\delta(t)$ the \textit{Dirac delta function}, 307 | or \textit{Dirac's impulse}. Interpreted as a density, $\delta(t)$ carries unit mass because % 308 | % 309 | \begin{align} 310 | \int_{\minus \infty}^\infty \delta(t) \diff{t} = \int_{\minus \infty}^\infty \diff{H} = H(\infty) - H(\minus \infty) = 1. 311 | \end{align} 312 | % 313 | This expresses that the collision releases the energy quantum. We now apply this heuristic to the carrier signal $\phi$: % 314 | % 315 | \begin{align} 316 | \int_{\minus \infty}^\infty \phi(t) \delta(t)\diff{t} & = \int_{\minus \infty}^\infty \phi \diff{H} & \\ 317 | & = \bigl[ H\phi \bigr]_{-\infty}^{\infty} - \int_{\minus \infty}^\infty H \diff{\phi} & [\text{integration by parts}]\\ 318 | & = -\int_{\minus \infty}^\infty H \diff{\phi} \label{Dirac's impulse as a distribution}&[\text{$\supp{\phi}$ is compact}]\\ 319 | & = \phi(0) & 320 | \end{align} 321 | % 322 | The key point is that the sum \eqref{Dirac's impulse as a distribution} no longer involves any $\diff{H}$. Moreover, % 323 | we obtain $\Delta \phi = \phi(0)$, the filtered carrier signal $\phi \mapsto \phi\vert_{t=0}$. This motivates the following % 324 | definitions, % 325 | % 326 | \begin{align} 327 | H(\phi) & \Def \int_{\minus \infty}^\infty H \phi\diff{t}, \\ 328 | \diff{H}(\phi) & \Def \minus \int_{\minus \infty}^\infty H \diff{\phi} = \phi(0), \\ 329 | \delta(\phi) & \Def \phi(0).\label{Dirac's impulse definition} 330 | \end{align} 331 | % 332 | This definition of $\delta$ conveys the idea of energy increasing by a fixed amount at pulsed time while being % 333 | mathematically sound. Extension of \eqref{Dirac's impulse definition} to all $\phi \in C_c(\R^n)$ is immediate and turns % 334 | $\delta$ into a Radon measure. Notably, its Borel-measure counterpart is the \textit{Dirac measure} % 335 | % 336 | \begin{align} 337 | \delta: E \to \Iverson{0 \in E} 338 | \end{align} 339 | % 340 | restricted to Borel sets in $\R^n$. Dirac's $\delta$ plays a pivotal role in mathematical physics, as it is the convolution % 341 | identity. To see that, we set $\phi_s(t) = \phi(s-t)$ ($s \in \R$), then we interpret the convolution $[\delta \ast \phi](s)$ % 342 | as integration with respect to the Dirac measure. This reads as % 343 | % 344 | \begin{align} 345 | [\delta \ast \phi](s) \Def \int_{\minus \infty}^\infty \phi(s-t) \diff\delta(t) = \int_\R \phi_s \diff{\delta} 346 | \citeq{\ref{definition of sum}} \delta(\phi_s) = \phi(s). 347 | \end{align} 348 | % 349 | % END 350 | % 351 | 352 | --------------------------------------------------------------------------------