├── .gitignore ├── README.md ├── abstract.tex ├── chapter2 ├── chapter2.tex └── exercises │ ├── ex_2.1.tex │ ├── ex_2.10.tex │ ├── ex_2.11.tex │ ├── ex_2.12.tex │ ├── ex_2.13.tex │ ├── ex_2.14.tex │ ├── ex_2.15.tex │ ├── ex_2.17.tex │ ├── ex_2.2.tex │ ├── ex_2.20.tex │ ├── ex_2.22.tex │ ├── ex_2.3.tex │ ├── ex_2.4.tex │ ├── ex_2.5.tex │ ├── ex_2.6.tex │ ├── ex_2.7.tex │ ├── ex_2.8.tex │ └── ex_2.9.tex ├── chapter3 ├── chapter3.tex └── exercises │ ├── ex_3.1.tex │ ├── ex_3.10.tex │ ├── ex_3.14.tex │ ├── ex_3.2.tex │ ├── ex_3.3.tex │ ├── ex_3.4.tex │ ├── ex_3.5.tex │ ├── ex_3.6.tex │ ├── ex_3.7.tex │ ├── ex_3.8.tex │ └── ex_3.9.tex ├── chapter4 ├── chapter4.tex └── exercises │ ├── ex_4.1.tex │ ├── ex_4.11.tex │ ├── ex_4.14.tex │ ├── ex_4.2.tex │ ├── ex_4.20.tex │ ├── ex_4.21.tex │ ├── ex_4.3.tex │ ├── ex_4.4.tex │ ├── ex_4.5.tex │ ├── ex_4.6.tex │ ├── ex_4.7.tex │ └── ex_4.8.tex ├── chapter5 ├── chapter5.tex └── exercises │ ├── ex_5.1.tex │ ├── ex_5.10.tex │ ├── ex_5.11.tex │ ├── ex_5.17.tex │ ├── ex_5.2.tex │ ├── ex_5.22.tex │ ├── ex_5.25.tex │ ├── ex_5.3.tex │ ├── ex_5.4.tex │ ├── ex_5.5.tex │ ├── ex_5.6.tex │ ├── ex_5.7.tex │ ├── ex_5.8.tex │ └── ex_5.9.tex ├── chapter6 ├── chapter6.tex └── exercises │ ├── ex_6.1.tex │ ├── ex_6.10.tex │ ├── ex_6.11.tex │ ├── ex_6.15.tex │ ├── ex_6.2.tex │ ├── ex_6.3.tex │ ├── ex_6.4.tex │ ├── ex_6.5.tex │ ├── ex_6.7.tex │ └── ex_6.8.tex ├── chapter7 ├── chapter7.tex └── exercises │ ├── ex_7.1.tex │ ├── ex_7.16.tex │ ├── ex_7.2.tex │ ├── ex_7.20.tex │ ├── ex_7.4.tex │ ├── ex_7.5.tex │ ├── ex_7.6.tex │ ├── ex_7.7.tex │ └── ex_7.9.tex ├── chapter8 ├── chapter8.tex └── exercises │ └── ex_8.7.tex ├── chapter9 ├── chapter9.tex └── exercises │ ├── ex_9.14.tex │ ├── ex_9.15.tex │ ├── ex_9.19.tex │ ├── ex_9.27.tex │ └── ex_9.6.tex ├── config.tex ├── cover.tex ├── license └── CC_BY-NC-SA.jpg ├── main.tex ├── notation.tex ├── style.tex └── table.tex /.gitignore: -------------------------------------------------------------------------------- 1 | ## Core latex/pdflatex auxiliary files: 2 | *.aux 3 | *.lof 4 | *.log 5 | *.lot 6 | *.fls 7 | *.out 8 | *.toc 9 | 10 | ## Intermediate documents: 11 | *.dvi 12 | *-converted-to.* 13 | # these rules might exclude image files for figures etc. 14 | # *.ps 15 | # *.eps 16 | # *.pdf 17 | 18 | ## Bibliography auxiliary files (bibtex/biblatex/biber): 19 | *.bbl 20 | *.bcf 21 | *.blg 22 | *-blx.aux 23 | *-blx.bib 24 | *.brf 25 | *.run.xml 26 | 27 | ## Build tool auxiliary files: 28 | *.fdb_latexmk 29 | *.synctex 30 | *.synctex.gz 31 | *.synctex.gz(busy) 32 | *.pdfsync 33 | 34 | ## Auxiliary and intermediate files from other packages: 35 | 36 | # algorithms 37 | *.alg 38 | *.loa 39 | 40 | # achemso 41 | acs-*.bib 42 | 43 | # amsthm 44 | *.thm 45 | 46 | # beamer 47 | *.nav 48 | *.snm 49 | *.vrb 50 | 51 | #(e)ledmac/(e)ledpar 52 | *.end 53 | *.[1-9] 54 | *.[1-9][0-9] 55 | *.[1-9][0-9][0-9] 56 | *.[1-9]R 57 | *.[1-9][0-9]R 58 | *.[1-9][0-9][0-9]R 59 | *.eledsec[1-9] 60 | *.eledsec[1-9]R 61 | *.eledsec[1-9][0-9] 62 | *.eledsec[1-9][0-9]R 63 | *.eledsec[1-9][0-9][0-9] 64 | *.eledsec[1-9][0-9][0-9]R 65 | 66 | # glossaries 67 | *.acn 68 | *.acr 69 | *.glg 70 | *.glo 71 | *.gls 72 | 73 | # gnuplottex 74 | *-gnuplottex-* 75 | 76 | # hyperref 77 | *.brf 78 | 79 | # knitr 80 | *-concordance.tex 81 | *.tikz 82 | *-tikzDictionary 83 | 84 | # listings 85 | *.lol 86 | 87 | # makeidx 88 | *.idx 89 | *.ilg 90 | *.ind 91 | *.ist 92 | 93 | # minitoc 94 | *.maf 95 | *.mtc 96 | *.mtc0 97 | 98 | # minted 99 | _minted* 100 | *.pyg 101 | 102 | # morewrites 103 | *.mw 104 | 105 | # nomencl 106 | *.nlo 107 | 108 | # sagetex 109 | *.sagetex.sage 110 | *.sagetex.py 111 | *.sagetex.scmd 112 | 113 | # sympy 114 | *.sout 115 | *.sympy 116 | sympy-plots-for-*.tex/ 117 | 118 | # todonotes 119 | *.tdo 120 | 121 | # xindy 122 | *.xdy 123 | 124 | # WinEdt 125 | *.bak 126 | *.sav 127 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | ### Document 2 | [PDF](https://db.tt/NbUTWDS6) 3 | 4 | ### License 5 | Attribution-NonCommercial-ShareAlike 4.0 International (CC BY-NC-SA 4.0) 6 | -------------------------------------------------------------------------------- /abstract.tex: -------------------------------------------------------------------------------- 1 | \graphicspath{ {license/} } 2 | 3 | \newpage 4 | \phantomsection 5 | \addcontentsline{toc}{chapter}{\textbf{Abstract}} 6 | \begin{abstract} 7 | 8 | These solutions were done as a student who is fresh out of university. 9 | Now I'm studying the book when doing my mandatory military service. 10 | 11 | \vspace{2ex} 12 | 13 | The solutions might be brief. However, I state them as clearly as possible I can. 14 | For the computation part, I might omit details because it is tedious but simple. 15 | 16 | \vspace{2ex} 17 | 18 | This manual now only contains some selected solutions which I can afford. 19 | If there is something vague or incredible, it is possible that it doesn't make sense since it is wrong. 20 | If you have any suggestions or corrections to solutions, please direct to email "chengmao.lee@gmail.com". 21 | Your intelligence will be highly appreciated. 22 | 23 | \vspace{2ex} 24 | 25 | Hope this manual could share some constructive ideas or help you that given important hints to solve problems by yourselves, or just verify whether the answers are consistent or not. 26 | 27 | \vspace{2ex} 28 | 29 | Any way, I hope everybody has good luck and fun in solving problems! 30 | 31 | \vspace{2ex} 32 | 33 | \vspace{10cm} 34 | 35 | \includegraphics[]{CC_BY-NC-SA} \\ 36 | $\mathbf{License:\ CC\ BY\text{-}NC\text{-}SA}$ 37 | 38 | \thispagestyle{plain} 39 | \end{abstract} 40 | -------------------------------------------------------------------------------- /chapter2/chapter2.tex: -------------------------------------------------------------------------------- 1 | % === Chapter 2 === 2 | \setcounter{chapter}{1} 3 | \chapter{Basic Topology} 4 | 5 | \input{exercises/ex_2.1} 6 | 7 | \vspace{12pt} 8 | 9 | \input{exercises/ex_2.2} 10 | 11 | \vspace{12pt} 12 | 13 | \input{exercises/ex_2.3} 14 | 15 | \vspace{12pt} 16 | 17 | \input{exercises/ex_2.4} 18 | 19 | \vspace{12pt} 20 | 21 | \input{exercises/ex_2.5} 22 | 23 | \vspace{12pt} 24 | 25 | \input{exercises/ex_2.6} 26 | 27 | \vspace{12pt} 28 | 29 | \input{exercises/ex_2.7} 30 | 31 | \vspace{12pt} 32 | 33 | \input{exercises/ex_2.8} 34 | 35 | \vspace{12pt} 36 | 37 | \input{exercises/ex_2.9} 38 | 39 | \vspace{12pt} 40 | 41 | \input{exercises/ex_2.10} 42 | 43 | \vspace{12pt} 44 | 45 | \input{exercises/ex_2.11} 46 | 47 | \vspace{12pt} 48 | 49 | \input{exercises/ex_2.12} 50 | 51 | \vspace{12pt} 52 | 53 | \input{exercises/ex_2.13} 54 | 55 | \vspace{12pt} 56 | 57 | \input{exercises/ex_2.14} 58 | 59 | \vspace{12pt} 60 | 61 | \input{exercises/ex_2.15} 62 | 63 | \vspace{12pt} 64 | 65 | \setcounter{Exercise}{16} 66 | \input{exercises/ex_2.17} 67 | 68 | \vspace{12pt} 69 | 70 | \setcounter{Exercise}{19} 71 | \input{exercises/ex_2.20} 72 | 73 | \vspace{12pt} 74 | 75 | \setcounter{Exercise}{21} 76 | \input{exercises/ex_2.22} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.1.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.1 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | By definition of empty set, we know 5 | $$ 6 | \forall x: x\notin \emptyset. 7 | $$ 8 | Then by logical axiom PL1 in the Hilbert system, we see that 9 | $$ 10 | \forall x: x\notin A \implies x\notin \emptyset 11 | $$ 12 | for any set $A$. 13 | This is equivalent to 14 | $$ 15 | \forall x: x\in \emptyset \implies x\in A. 16 | $$ 17 | Hence $\emptyset \subseteq A$. 18 | 19 | We conclude the empty set is a subset of every set. 20 | \end{proof} 21 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.10.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.10 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{Prove that this is a metric.} 5 | \begin{proof} 6 | It suffices to prove three properties as following one by one. 7 | \begin{itemize} 8 | \item \textbf{Property $1$ : Positive Definite} 9 | 10 | For $p=q$, $d(p,q) = d(p,p) = 0$. For $p\neq q$, $d(p,q) = 1 > 0$. 11 | 12 | Hence $d(p,q)\geq 0$ with equality if and only if $p=q$. 13 | 14 | \item \textbf{Property $2$ : Symmetric} 15 | 16 | For $p=q$, $d(p,q) = 0 = d(q,p)$. For $p\neq q$, $d(p,q) = 1 = d(q,p)$. 17 | 18 | Hence $d(p,q) = d(q,p)$. 19 | 20 | \item \textbf{Property $3$ : Triangle Inequality} 21 | 22 | Let $r\in X$, then we consider four cases as following. 23 | 24 | \begin{enumerate} 25 | \item $p=q=r$ 26 | $$ 27 | d(p,q) = 0 \leq 0 + 0 = d(p,r) + d(r,q). 28 | $$ 29 | 30 | \item $p=q$ and $q\neq r$ 31 | $$ 32 | d(p,q) = 0 \leq 1 + 1 = d(p,r) + d(r,q). 33 | $$ 34 | 35 | \item $p=r$ and $q\neq r$ 36 | $$ 37 | d(p,q) = 1 \leq 0 + 1 = d(p,r) + d(r,q). 38 | $$ 39 | 40 | \item $p\neq q$ and $q\neq r$ and $p \neq r$ 41 | $$ 42 | d(p,q) = 1 \leq 1 + 1 = d(p,r) + d(r,q). 43 | $$ 44 | \end{enumerate} 45 | 46 | Hence $d(p,q) \leq d(p,r) + d(r,q)$ always. 47 | \end{itemize} 48 | 49 | As a result, we conclude this is a metric. 50 | \end{proof} 51 | 52 | \item $\mathbf{Which\ subsets\ of\ X\ are\ closed?}$ 53 | \begin{answer} 54 | All subsets of $X$. 55 | \end{answer} 56 | \begin{proof} 57 | Let $E$ be an arbitrary subset of $X$ and $p$ be an arbitrary point of $E$. 58 | 59 | Pick $r = \frac{1}{2}$, then the neighborhood $N_r(p)$ only contains $p$ itself, then $p\notin E'$. By the arbitrary choice of $p$, then $E' = \emptyset$. This implies that $E'\subseteq E$ vacuously, so $E$ is closed. 60 | 61 | By the arbitrary choice of $E$, we conclude all subsets of $X$ are closed. 62 | \end{proof} 63 | 64 | \item $\mathbf{Which\ subsets\ of\ X\ are\ open?}$ 65 | \begin{answer} 66 | All subsets of $X$. 67 | \end{answer} 68 | \begin{proof} 69 | Let $E$ be an arbitrary subset of $X$. Since all subsets are closed from the previous result, then we have $E$ is closed; hence we know $E^c$ is open. Notice that $E^c \subseteq X$. 70 | 71 | By the arbitrary choice of $E$, we conclude all subsets of $X$ are open. 72 | \end{proof} 73 | 74 | \item $\mathbf{Which\ subsets\ of\ X\ are\ compact?}$ 75 | \begin{answer} 76 | All finite subsets of $X$. 77 | \end{answer} 78 | \begin{proof} 79 | Let $E$ be an arbitrary subsets of $X$, then $E$ is either finite or infinite. 80 | 81 | If $E$ is finite, suppose $E = \{p_i: 1\leq i\leq n\}$ where $n$ is finite. We can pick an arbitrary open cover $\{G_{\alpha}\}$ of $E$ so that there are indices $\alpha_{1}, \cdots, \alpha_{n}$ such that $p_i\in G_{\alpha_i}$ for each $\alpha_i\in \alpha$. Hence 82 | $$ 83 | E \subset \bigcup_{i=1}^{n}G_{\alpha_i} \subset G_{\alpha}. 84 | $$ 85 | This implies there is a finite subcover of $E$. Moreover, by the arbitrary choice of an open cover, we know every open cover of $E$ contains a finite subcover. By definition, $E$ is compact. 86 | 87 | Otherwise, if $E$ is infinite, since $E$ is open from the previous result, then $E$ might includes subsets of single points. This implies we can select an infinite open cover of $E$ such that each cover is a single element of $E$. Since an open subcover contains an element, we cannot select finite open covers to contains infinitely many elements. So $E$ is not compact under this situation. 88 | 89 | By the arbitrary choice of $E$, we conclude all finite subset of $X$ are compact. 90 | \end{proof} 91 | \end{itemize} 92 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.11.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.11 === 2 | \begin{Exercise} 3 | \begin{answer} 4 | $d_2$, $d_5$ $\mathbf{are}$ metrics. 5 | $d_1$, $d_3$, $d_4$ $\mathbf{are\ not}$ metrics. 6 | \end{answer} 7 | \begin{solution} 8 | In order to determine whether these functions are metric or not, the verification is similar to Exercise 2.10. Here we only show something nontrivial. Notice that we suppose $z\in\mathbb{R}^1$. 9 | 10 | \begin{itemize} 11 | \item \textbf{$d_1$ violates the Triangle Inequality property.} 12 | 13 | Pick $x=4$, $y=0$, $z=1$. Then 14 | \begin{alignat*}{7} 15 | \quad&& d_1(x,y) &\leq d_1(x,z) + d_1(z,y) \\ 16 | \implies&& (x-y)^2 &\leq (x-z)^2 + (z-y)^2 \\ 17 | \implies&& 16 &\leq 9 + 1. 18 | \end{alignat*} 19 | This is a contradiction. 20 | 21 | \item \textbf{$d_2$ obeys the Triangle Inequality property.} 22 | 23 | \begin{alignat*}{7} 24 | \quad&& |x-y| &\leq |x-z| + |z-y| \\ 25 | \implies&& |x-y| &\leq |x-z| + |z-y| + 2\sqrt{|x-z| |z-y|} \\ 26 | \implies&& \sqrt{|x-y|}^2 &\leq \left( \sqrt{|x-z|} + \sqrt{|z-y|} \right)^2 \\ 27 | \implies&& \sqrt{|x-y|} &\leq \sqrt{|x-z|} + \sqrt{|z-y|} \\ 28 | \implies&& d_2(x,y) &\leq d_2(x,z) + d_2(z,y). 29 | \end{alignat*} 30 | This implies what we need to show. 31 | 32 | \item \textbf{$d_3$ violates the Positive Definite property.} 33 | 34 | Pick $x=1$, $y=-1$, then $d_3(x,y) = 0$ with $x\neq y$. 35 | 36 | \item \textbf{$d_4$ violates the Symmetry property.} 37 | 38 | Pick $x=0$, $y=1$, then 39 | $$ 40 | d_4(x,y) = |x-2y| = 2 \neq 1 = |y-2x| = d_4(y,x). 41 | $$ 42 | 43 | \item \textbf{$d_5$ obeys the Triangle Inequality property.} 44 | 45 | Consider 46 | $$ 47 | |x-y| \leq |x-z| + |z-y|. 48 | $$ 49 | Add $|x-z| |x-y| + |x-z| |z-y| + |x-z| |z-y| + |x-z| |x-y| |z-y|$ on both sides, then 50 | $$ 51 | |x-y|(1+|x-z|)(1+|z-y|) \leq |x-z|(1+|x-y|)(1+|z-y|) + |z-y| + |x-z| |z-y|. 52 | $$ 53 | Add $|z-y| |x-y| + |z-y| |x-y| |x-z|$ on the right side, then 54 | \begin{align*} 55 | |x-y|(1+|x-z|)(1+|z-y|) 56 | \leq& |x-z|(1+|x-y|)(1+|z-y|) \\ 57 | \quad& +|z-y|(1+|x-y|)(1+|x-z|). 58 | \end{align*} 59 | This implies 60 | $$ 61 | \frac{|x-y|}{1+|x-y|} \leq \frac{|x-z|}{1+|x-z|} + \frac{|z-y|}{1+|z-y|}. 62 | $$ 63 | We obtain $d_5(x,y) \leq d_5(x,z) + d_5(z,y)$ as promised. 64 | \end{itemize} 65 | \end{solution} 66 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.12.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.12 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Let $\{G_{\alpha}\}$ be arbitrary over cover of $K$, then $K\subset \bigcup_{\alpha}G_{\alpha}$. 5 | 6 | There is an open set $G_0 \in \{G_{\alpha}\}$ such that $0\in G_0$. Since $G_0$ is open, then $0\in (G_0)^{\circ}$. It follows that there is a neighborhood $N_r(0)$ such that $N_r(0)\subset G_0$ with $r>0$. Notice that $N_r(0)$ contains points $\frac{1}{n}\in K$ such that $\frac{1}{n}\frac{1}{r}$. 7 | 8 | For which points are not contained in $N_r(0)$; that is, let $m := \lfloor \frac{1}{r} \rfloor$, then points $\frac{1}{i}$ for $i=1,2,\cdots,m$ are not contained in this neighborhood. We can pick $G_i\in \{G_{\alpha}\}$ containing the point $\frac{1}{i}$ for $1\leq i \leq m$ so that 9 | $$ 10 | K\subset G_0\cup \left( \bigcup_{i=1}^{m}G_i \right) \subset \{G_{\alpha}\}. 11 | $$ 12 | Notice that $m$ is finite, then we know there is a finite open subcover $G_0\cup \left( \bigcup_{i=1}^{m}G_i \right)$ of $G_{\alpha}$. 13 | 14 | By the arbitrary choice of $G_{\alpha}$, we know every open cover of $K$ contains a finite subcover; hence we conclude $K$ is compact. 15 | \end{proof} 16 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.13.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.13 === 2 | \begin{Exercise} 3 | \begin{answer} 4 | $\{0\}\cup \{\frac{1}{n} + \frac{1}{m}:n,m\in\mathbb{N}\}$ 5 | \end{answer} 6 | \begin{solution} 7 | Let $A = \{0\}\cup \{\frac{1}{n} + \frac{1}{m}:n,m\in\mathbb{N}\}$. 8 | Then $A' = \{0\}\cup \{\frac{1}{n}:n\in\mathbb{N}\}$ which is countable. 9 | Notice that $A\cup A' = \overline{A}$ is bounded and closed so is compact. 10 | 11 | \end{solution} 12 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.14.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.14 === 2 | \begin{Exercise} 3 | \begin{answer} 4 | $\left\{ \left(\frac{1}{n+1},1\right) \right\}_{n\in\mathbb{N}}$ 5 | \end{answer} 6 | \begin{proof} 7 | Consider the open cover $\left\{ \left(\frac{1}{n+1},1\right) \right\}_{n\in\mathbb{N}}$. If $r\in(0,1)$, then let $n\in\mathbb{N}$ be such that $\frac{1}{n+1}\max_{i\in I}\{i+1\}$, then $\frac{1}{n+1}\in(0,1)$; however, $\frac{1}{n+1} < \frac{1}{i+1}$ for all $i\in I$. Hence 10 | $$ 11 | \frac{1}{n+1} \notin \bigcup_{i\in I}\left(\frac{1}{i+1},1\right). 12 | $$ 13 | This means this open cover has no finite subcover. 14 | \end{proof} 15 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.15.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.15 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{The term is replaced by "\textit{closed}".} 5 | \end{itemize} 6 | \begin{proof} 7 | Let $C_n := \left\{m\geq n:m\in\mathbb{N}\right\}$ which is closed for each $n\in\mathbb{N}$ since $C_n$ has no limit points; therefore every limit point of $C_n$ is in $C_n$ vacuously. Consider any finite subcollection $\{C_i\}_{i\in I}$ of $\{C_n\}$ where $I\subset \mathbb{N}$ and $|I|$ is finite. 8 | 9 | Suppose to contrary that 10 | $$ 11 | \bigcap_{n=1}^{\infty} C_n \neq \emptyset. 12 | $$ 13 | Then this implies 14 | $$ 15 | \bigcap_{i\in I} C_i \neq \emptyset. 16 | $$ 17 | Now pick $r = \max_{i\in I}\{i\}$ where $r\in\mathbb{N}$, then $r\in C_i$ for all $i\in I$ and so $r\in \bigcap_{i\in I}C_i$. We can choose $k\in\mathbb{N}$ such that $k>r$, then $r\notin C_k$ and so 18 | $$ 19 | r \notin \bigcap_{n=1}^{\infty}C_n. 20 | $$ 21 | 22 | We conclude the statement is false. 23 | \end{proof} 24 | 25 | \begin{itemize} 26 | \item \textbf{The term is replaced by "\textit{bounded}".} 27 | \end{itemize} 28 | \begin{proof} 29 | Let $B_n := \left(0,\frac{1}{n}\right)$ which is bounded for each $n\in\mathbb{N}$. Consider any finite subcollection $\{B_i\}_{i\in I}$ of $\{B_n\}$ where $I\subset \mathbb{N}$ and $|I|$ is finite. 30 | 31 | Suppose to contrary that 32 | $$ 33 | \bigcap_{n=1}^{\infty} B_n \neq \emptyset. 34 | $$ 35 | Then this implies 36 | $$ 37 | \bigcap_{i\in I} B_i \neq \emptyset. 38 | $$ 39 | Now pick $n = \min_{i\in I}\{i\}$. By Archimedean property, there is $r\in\mathbb{R}$ such that $n r > 1$. This means $r>\frac{1}{n}$ and hence $r\notin\left(0, \frac{1}{n} \right)$ implies 40 | $$ 41 | r\notin \bigcap_{n=1}^{\infty} B_n. 42 | $$ 43 | So this contradicts the supposition. 44 | 45 | We conclude the statement is false. 46 | \end{proof} 47 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.17.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.17 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{Is $E$ countable?} 5 | \end{itemize} 6 | \begin{answer} 7 | No. 8 | \end{answer} 9 | \begin{proof} 10 | Suppose to contrary that $E$ is countable. Let $A$ be a countable subset of $E$, and $A$ consist of the sequences $s_1, s_2, \cdots$. We construct a sequence $s$ as follows. 11 | 12 | If $n$-th digit of $s_n$ is $4$, then we set $n$-th digit of $s$ as $7$, and vice versa. We know $s$ differs from every number of $A$ in at least one place; hence $s\notin A$. However $s\in E$ so that $A\subset E$. 13 | 14 | We have shown that every countable subset of $E$ is a proper subset $E$, then $E\subset E$ is a contradiction; hence we conclude $E$ is uncountable. 15 | \end{proof} 16 | 17 | \begin{itemize} 18 | \item \textbf{Is $E$ dense in $[0,1]$?} 19 | \end{itemize} 20 | \begin{answer} 21 | No. 22 | \end{answer} 23 | \begin{proof} 24 | Let $x := 0.22$. Pick $r = 0.1 > 0$, then we know $N_r(x) = (0.21, 0.23)$. For any element $y\in N_r(x)$, it implies $y\notin E$, then $x\notin E'$. And $x\notin E$ trivially; however $x\in [0,1]$. So we find an element (e.g. $0.22$) in $[0,1]$ is neither a point of $E$ nor a limit point of $E$. 25 | 26 | As a result, we conclude $E$ is not dense. 27 | \end{proof} 28 | 29 | \begin{itemize} 30 | \item \textbf{Is $E$ compact?} 31 | \end{itemize} 32 | \begin{answer} 33 | Yes. 34 | \end{answer} 35 | \begin{proof} 36 | Notice that $E\subset \mathbb{R}^1$, then $E$ is compact if and only if $E$ is bounded and closed. 37 | 38 | Since $E\subset[0,1]$, then for any $e\in E$, we know $|e| \leq 1$ and hence $E$ is bounded by $1$. 39 | 40 | Now we claim $E$ is closed. This means every limit point of $E$ is a member of $E$, so is equivalent to that every member which is not a member of $E$ is not a limit point of $E$. 41 | 42 | Let $x\in[0,1]\setminus E$. Since $x\notin E$, in particular, we let 43 | $$ 44 | x := 0.a_1 a_2 a_3 \cdots 45 | $$ 46 | where $a_i\notin \{4,7\}$ for some $i$. Now let $k$ be any index such that $a_k\notin\{4,7\}$. Pick $r = \frac{1}{10^{k+1}}$, then we consider three cases. 47 | 48 | \begin{enumerate} 49 | \item $0 < a_k < 9$. 50 | 51 | The ($k$+$1$)-st decimal place will be $a_k$ for every element of $N_r(x)$ and so no element of $N_r(x)$ is a member of $E$. 52 | 53 | \item $a_k = 0$. 54 | 55 | The ($k$+$1$)-st decimal place will be $9$, $0$, or $1$ for every element of $N_r(x)$ and so no element of $N_r(x)$ is a member of $E$. 56 | 57 | \item $a_k = 9$. 58 | 59 | The ($k$+$1$)-st decimal place will be $8$, $9$, or $0$ for every element of $N_r(x)$ and so no element of $N_r(x)$ is a member of $E$. 60 | \end{enumerate} 61 | Hence $N_r(x)$ contains no points in $E$. This means $x\notin E'$. 62 | 63 | By the arbitrary choice of $x$, every member which is not a member of $E$ is not a limit point of $E$. It follows that $E$ is compact. 64 | \end{proof} 65 | 66 | \begin{itemize} 67 | \item \textbf{Is $E$ perfect?} 68 | \end{itemize} 69 | \begin{answer} 70 | Yes. 71 | \end{answer} 72 | \begin{proof} 73 | We have known $E$ is closed in the previous proof. Now we claim every point of $E$ is a limit point of $E$. 74 | 75 | Let $x\in E$ arbitrarily. And we change the $k$-th digit of $x$ from $4$ to $7$ or from $7$ to $4$; moreover, we denote this number $x_k$. 76 | 77 | Since $d(x, x_k)=3\times 10^{-k}$, by Archimedean property, there is $k\in\mathbb{Z}$ such that $3\times 10^{-k} < r$. This means we can find such $x_k\in E$ in the neighborhood $N_r(x)$. By the arbitrary choice of $x$, we obtain every neighborhood of every point in $E$ contains another point in $E$. So every point in $E$ is a limit point. 78 | 79 | It follows that $E$ is perfect. 80 | \end{proof} 81 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.2.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.2 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Continue with the hint, we let $A_N$ be the set of finitely many equations with 5 | $$ 6 | n+|a_0|+|a_1|+\cdots+|a_n| = N. 7 | $$ 8 | Then the set of all algebraic numbers is 9 | $$ 10 | \bigcup_{N=1}^{\infty}A_N. 11 | $$ 12 | 13 | Since $A_N$ is a countable set for any $N\in\mathbb{N}$, by Theorem 2.12, we know $\bigcup_{N=1}^{\infty}A_N$ is countable which means the set of all algebraic numbers is countable. 14 | \end{proof} 15 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.20.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.20 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{Are closures of connected sets always connected?} 5 | \end{itemize} 6 | \begin{answer} 7 | Yes. 8 | \end{answer} 9 | \begin{proof} 10 | Let $E := A\cup B$ be a connected set in a metric $X$, then $A\cap\overline{B}\neq\emptyset$ and $\overline{A}\cap B\neq\emptyset$. Notice that $\overline{E} = \overline{A\cup B} = \overline{A}\cup\overline{B}$. If $\overline{E}$ is connected, it suffices to prove $\overline{A}\cap \overline{B}\neq\emptyset$. 11 | 12 | We claim $\overline{A}\cap\overline{B}\neq\emptyset$. Since $A\cap\overline{B}\neq\emptyset$ and $\overline{A}\cap B\neq\emptyset$, then there is an element $p$ such that $p\in A\cap\overline{B}$ or $p\in \overline{A}\cap B$. In either of two cases, $p\in (A\cup A')\cap(B\cup B')$. This implies $p\in \overline{A}\cap\overline{B}$; hence $\overline{A}\cap\overline{B}\neq\emptyset$. 13 | 14 | By the arbitrary choice of $E$ which is a connected set, we conclude every closure of connected set is also connected. 15 | \end{proof} 16 | 17 | \begin{itemize} 18 | \item \textbf{Are interiors of connected sets always connected?} 19 | \end{itemize} 20 | \begin{answer} 21 | No. 22 | \end{answer} 23 | \begin{proof} 24 | Consider $E = \left\{(x,y):-1\leq x \leq 1, -|x|\leq y \leq |x|\right\} \subset \mathbb{R}^2$. Notice that $E$ is connected. Then we know 25 | $$ 26 | E^{\circ} = \left\{(x,y):-10\right\}$; hence $E^{\circ}$ is not connected. 29 | 30 | We have shown a counter-example, so we complete the proof. 31 | \end{proof} 32 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.22.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.22 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Follow the hint. It is clear that $\mathbb{Q}^k\subset\mathbb{R}^k$. By Theorem 2.13 and its corollary, we know $\mathbb{Q}^k$ is countable. Now we need to prove $\mathbb{Q}^K$ is dense in $\mathbb{R}^k$. 5 | 6 | Let $a = (a_1,\cdots, a_k)$ be an arbitrary point in $\mathbb{R}^k$. Pick a radius $r > 0$ arbitrarily. Let $b = (b_1,\cdots, b_k)$ where $b_i\in\mathbb{Q}^k$ such that 7 | $$ 8 | a_i < b_i < a_i+\frac{r}{\sqrt{k}} 9 | $$ 10 | for each $i$. Then we know 11 | $$ 12 | d(a,b) 13 | = \|a - b \| 14 | = \sqrt{\sum_{i=1}^{k} (a_i-b_i)^2} 15 | < \sqrt{\sum_{i=1}^{k} \left( \frac{r}{\sqrt{k}} \right)^2} \ 16 | = \sqrt{k\cdot\frac{r^2}{k}} 17 | = |r| 18 | = r. 19 | $$ 20 | This means there is $b\in\mathbb{Q}^k$ in the neighborhood $N_r(a)$. Since $r$ was arbitrary, then $a$ is a limit point of $\mathbb{Q}^k$. 21 | 22 | By the arbitrary choice of $a$, we know every point in $\mathbb{R}^k$ is a limit of $\mathbb{Q}^k$. So $\mathbb{Q}^k$ is dense. 23 | 24 | Since $\mathbb{R}^k$ contains a countable dense subset $\mathbb{Q}^k$, we conclude $\mathbb{R}^k$ is separable. 25 | \end{proof} 26 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.3.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.3 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Suppose to contrary that all real numbers are algebraic, then the set of all real numbers called $\mathbb{R}$ is countable by Exercise 2.2. 5 | However, this contradicts that $\mathbb{R}$ is uncountable; hence there exist real numbers which are not algebraic. 6 | \end{proof} 7 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.4.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.4 === 2 | \begin{Exercise} 3 | \begin{answer} 4 | No. 5 | \end{answer} 6 | \begin{proof} 7 | Suppose to contrary that the set of all irrational numbers is countable. 8 | By the corollary of Theorem 2.13, we know the set of all rational numbers is countable. 9 | Then the union of all irrational numbers and all rational numbers is $\mathbb{R}$ which is uncountable. 10 | This contradicts the supposition. 11 | \end{proof} 12 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.5.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.5 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Let $E := \left\{ 1+\frac{1}{n} : n\in \{2,3,\cdots\} \right\}$. 5 | \begin{itemize} 6 | \item \textbf{Claim only $1$ is a limit point.} 7 | 8 | By Archimedean Property, we can pick radius $r$ such that $\frac{1}{n} < r$. 9 | Then $p_n = 1+\frac{1}{n}$ is a member of $E$ and $d(1, p_n) = \frac{1}{n} < r$. 10 | Hence $1$ is a limit point. 11 | 12 | On the other hand, consider a point $x$ such that $x\notin E\cup \{1\}$. 13 | We set 14 | $$ 15 | \delta = \min\{d(x,1), d(x,2), d(x, p_n), d(x, p_{n+1})\}, 16 | $$ 17 | and $r = \frac{\delta}{2}$. 18 | Then there is no point of $E$ in $N_r(x)$. 19 | \end{itemize} 20 | 21 | Moreover, we set $F := \left\{ 2+\frac{1}{n} : n\in \{2,3,\cdots\} \right\}$. 22 | And $G := \left\{ 3+\frac{1}{n} : n\in \{2,3,\cdots\} \right\}$. 23 | A similar argument establishes only $2$ and $3$ are limit points. 24 | 25 | Now we set $K := E\cup F\cup G$, then $K$ is bounded by $4$. 26 | Hence we construct $K$ which is a bounded set of real numbers with exactly three limit points $1,2$ and $3$ as promised. 27 | \end{proof} 28 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.6.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.6 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{Prove $E'$ is closed.} 5 | \begin{proof} 6 | By definition, it suffice to prove $(E')' \subseteq E'$. 7 | 8 | Let $x\in (E')'$ arbitrarily. Now fix $x$ and pick arbitrarily small radius $r$ such that $s := \frac{r}{2}$ and $t := \frac{r}{4}$. 9 | 10 | Since $x\in (E')'$, there is a point $y\in E'$ in the neighborhood $N_s(x)$. And since $y\in E'$, there is also a point $z\in E$ in the neighborhood $N_t(y)$. 11 | 12 | Notice that $y\neq z$ by the choice of radius $s$ and $t$. 13 | 14 | Consider 15 | $$ 16 | d(x,z) 17 | \leq d(x,y) + d(y,z) 18 | < s + t 19 | = \frac{r}{2} + \frac{r}{4} 20 | < r. 21 | $$ 22 | Hence for arbitrary radius $r$, there is a point $z\in E$ in the neighborhood $N_r(x)$, then we see that $x\in E'$. Moreover, $x$ is arbitrary, we know $(E')' \subseteq E'$ ; hence we conclude $E'$ is closed. 23 | \end{proof} 24 | 25 | \item \textbf{Prove $\overline{E}$ and $E$ have the same limit points.} 26 | 27 | \begin{proof} 28 | We will claim $(\overline{E})' \subseteq E'$ and $E' \subseteq (\overline{E})'$ to obtain $(\overline{E})' = E'$ which is asked to prove. 29 | 30 | Let $x\in (\overline{E})'$ and arbitrary radius $r > 0$, then there is a point $y\in \overline{E}$ in the neighborhood $N_r(x)$. 31 | 32 | Since $\overline{E} = E\cup E'$, then if $y\in E$, this makes $x\in E'$ hold trivially; otherwise, if $y\in E'$, then $x\in E'$ holds from the previous proof. Hence we know $(\overline{E})' \subseteq E'$. 33 | 34 | On the other hand, if $x\in E'$, then every neighborhood of $x$ contains an element of $E$. This implies every neighborhood of $x$ contains an element of $\overline{E}$ trivially. Hence we know $E' \subseteq (\overline{E})'$. 35 | 36 | Finally, we conclude $(\overline{E})' = E'$ which means $\overline{E}$ and $E$ have the same limit points. 37 | \end{proof} 38 | 39 | \item \textbf{Do $E$ and $E'$ always have the same limit points?} 40 | \begin{answer} 41 | No. 42 | \end{answer} 43 | \begin{proof} 44 | We give a counter-example. 45 | Let $E := \{0\} \cup \left\{ \frac{1}{n}:n\in\mathbb{N}\setminus\{1\} \right\}$, then $E' = \{0\}$ and $(E')' = \emptyset$. Hence $E' \neq (E')'$ which means $E$ and $E'$ don't have the same limit points. 46 | \end{proof} 47 | \end{itemize} 48 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.7.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.7 === 2 | \begin{Exercise} 3 | \begin{lemma}\label{lemma:ex_2.7a} 4 | Let $E,F,G$ be subsets of a metric space. 5 | If $E=F\cup G$, then $\overline{E} \subseteq \overline{F}\cup\overline{G}$. 6 | \end{lemma} 7 | \begin{proof} 8 | Let $x\in\overline{E}$, then $x\in E$ or $x\in E'$. 9 | 10 | If $x\in E$, then $x\in F$ or $x\in G$ imply $x\in\overline{F}$ or $x\in\overline{G}$, respectively. 11 | In either of two cases, we know $x\in \overline{F}\cup\overline{G}$. 12 | 13 | Otherwise, if $x\in E'$, then $x\in F'$ or $x\in G'$. 14 | Suppose to contrary that $x\notin F'\cap G'$. 15 | Pick arbitrarily small radius $s$ and $t$ such that there is no point of $F$ in the neighborhood $N_s(x)$, and there is no point of $G$ in the neighborhood $N_t(x)$. 16 | Let $r := \min\{s,t\}$, since $N_r(x)\subseteq N_s(x)$ and $N_r(x)\subseteq N_t(x)$, then $N_r(x)$ contains no points of $F\cup G = E$. 17 | This contradicts the supposition that $x\in E'$. 18 | Hence $x\in F'$ or $x\in G'$. 19 | These imply $x\in \overline{F}$ or $x\in \overline{G}$, respectively. 20 | In either of two cases, we know $x\in \overline{F}\cup \overline{G}$. 21 | 22 | Finally, no matter $x\in E$ or $x\in E'$, we know $x\in \overline{F}\cup \overline{G}$ always holds; hence we conclude $\overline{E} \subseteq \overline{F}\cup \overline{G}$. 23 | \end{proof} 24 | 25 | \begin{lemma}\label{lemma:ex_2.7b} 26 | Let $E,F,G$ be subsets of a metric space. 27 | If $E=F\cup G$, then $\overline{E} \supseteq \overline{F}\cup\overline{G}$. 28 | \end{lemma} 29 | \begin{proof} 30 | Let $x\in \overline{F}\cup \overline{G}$, then $x\in \overline{F}$ or $x\in \overline{G}$. 31 | In particular, $x\in F$ or $x\in F'$ or $x\in G$ or $x\in G'$. 32 | 33 | If $x\in F$ or $x\in G$, then $x\in F\cup G$; hence $x\in \overline{F\cup G}$. This implies $x\in \overline{E}$. 34 | 35 | Otherwise, if $x\in F'$ or $x\in G'$, then every neighborhood of $x$ contains an element of $F$ or $G$. 36 | Moreover, every element of $F$ or $G$ is an element of $F\cup G = E$. 37 | By definition, we see that $x\in E'$. 38 | Since $E' \subseteq E\cup E' = \overline{E}$, we have $x\in \overline{E}$. 39 | 40 | Finally, in either of cases, we always know $x\in \overline{E}$; hence we conclude $\overline{E} \supseteq \overline{F}\cup\overline{G}$. 41 | \end{proof} 42 | \begin{enumerate}[a)] 43 | \item 44 | \begin{proof} 45 | By Lemma \ref{lemma:ex_2.7a} and Lemma \ref{lemma:ex_2.7b}, we know $\overline{E} = \overline{F}\cup\overline{G}$. 46 | 47 | For $n=1$, if $B_1 = A_1$, then $\overline{B_1} = \overline{A_1}$ holds. 48 | 49 | Suppose for $n=k$, 50 | $$ 51 | B_k = \bigcup_{i=1}^{k}A_i \implies \overline{B_k} = \bigcup_{i=1}^{k}\overline{A_i}. 52 | $$ 53 | Then for $n=k+1$, if $B_{k+1} = \bigcup_{i=1}^{k+1}A_i$, then we have 54 | \begin{align*} 55 | \overline{B_{k+1}} 56 | &= \overline{\cup_{i=1}^{k+1}A_i} \\ 57 | &= \overline{\left( \cup_{i=1}^{k}A_i \right) \cup A_{k+1}} \\ 58 | &= \overline{\cup_{i=1}^{k}A_i} \cup \overline{A_{k+1}} \\ 59 | &= \overline{B_k} \cup \overline{A_{k+1}} \\ 60 | &= \left( \cup_{i=1}^{k}\overline{A_i} \right) \cup \overline{A_{k+1}} \\ 61 | &= \cup_{i=1}^{k+1}\overline{A_i} 62 | \end{align*} 63 | also holds. 64 | 65 | By induction, we conclude the formula holds for all $n\in\mathbb{N}$. 66 | \end{proof} 67 | 68 | \item 69 | \begin{proof} 70 | From Lemma \ref{lemma:ex_2.7a}, when we pick $r$ which is the minimum radius among all the candidates, $r$ exists if and only if the amount of candidates is finite. However, since $n=\infty$ which means candidates are infinite, then $r$ doesn't exist. 71 | 72 | On the other hand, from Lemma \ref{lemma:ex_2.7b}, it doesn't matter $n$ is finite or infinite; hence a similar argument from part a) establishes 73 | $$ 74 | \overline{B} \supseteq \bigcup_{i=1}^{\infty}\overline{A_i}. 75 | $$ 76 | 77 | \begin{itemize} 78 | \item \textbf{Claim this inclusion can be proper.} 79 | \end{itemize} 80 | 81 | For example, from Exercise 2.5, we let $A_i = \left\{ 1+\frac{1}{i+1} \right\}$ for all $i\in\mathbb{N}$ and $B=\cup_{i=1}^{\infty} A_i$. It follows that $\cup_{i=1}^{\infty}\overline{A_i} = \emptyset$ and $\overline{B} = \{1\}$; hence $\overline{B} \neq \bigcup_{i=1}^{\infty}\overline{A_i}$. 82 | 83 | We conclude 84 | $$ 85 | \overline{B} \supset \bigcup_{i=1}^{\infty}\overline{A_i} 86 | $$ 87 | might occur, so this claim holds; furthermore, we complete the proof. 88 | \end{proof} 89 | \end{enumerate} 90 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.8.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.8 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{Is every point of every open set $E \subset \mathbb{R}^2$ a limit point of $E$?} 5 | \begin{answer} 6 | Yes. 7 | \end{answer} 8 | \begin{proof} 9 | Let $x = (a, b)\in E$ arbitrarily. 10 | Since $E$ is open, then $x\in E^{\circ}$; hence there is a neighborhood $N_r(x)$ where $r>0$ such that $N_r(x)\subset E$. 11 | 12 | Fix $x$, consider the neighborhood $N_s(x)$ for any radius $s>0$. 13 | 14 | We can pick $y = \left( a+\frac{1}{2}\min\{r,s\}, b \right)$, then $d(x,y) = \frac{1}{2}\min\{r,s\} < r$. 15 | This means there is a point $y\in N_r(x) \subset E$ in the neighborhood $N_s(x)$. 16 | So every neighborhood of $x$ contains a point $y\neq x$ such that $y\in E$; hence $x\in E'$. 17 | 18 | Because $x$ was arbitrary, we conclude $E \subseteq E'$ for any open set $E\subset \mathbb{R}^2$. 19 | \end{proof} 20 | 21 | \item \textbf{Is every point of every closed set $E \subset \mathbb{R}^2$ a limit point of E?} 22 | \begin{answer} 23 | No. 24 | \end{answer} 25 | \begin{proof} 26 | For example, let $E := \left\{ (0,0) \right\}$, then $E' = \emptyset$; hence $(0,0)\notin E'$. 27 | 28 | Generally speaking, every finite subset $E$ of $\mathbb{R}^2$ is closed, however $E' = \emptyset$. 29 | \end{proof} 30 | \end{itemize} 31 | \end{Exercise} -------------------------------------------------------------------------------- /chapter2/exercises/ex_2.9.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 2.9 === 2 | \begin{Exercise} 3 | \begin{enumerate}[a)] 4 | \item 5 | \begin{proof} 6 | Let $x\in E^{\circ}$ arbitrarily, then there is a neighborhood $N_r(x)\subset E$ with $r>0$. 7 | 8 | Let $y\in N_r(x)$ arbitrarily, then we have $d(x,y)0$. It follows that $N_s(y)\subset E$. 10 | 11 | Let $z\in N_s(y)$ arbitrarily, since 12 | $$ 13 | d(x,z) \leq d(x,y)+d(y,z) < d(x,y) + s = r, 14 | $$ 15 | there is a neighborhood $N_s(y)\subset E$ with $s>0$. 16 | Because $z$ was arbitrary, then we have $y\in E^{\circ}$. 17 | 18 | Moreover, $y$ was arbitrary, then $N_r(x)\subset E^{\circ}$ which means $x\in (E^{\circ})^{\circ}$; also, by the choice of $x$ arbitrarily, hence $E^{\circ}$ is open. 19 | \end{proof} 20 | 21 | \item 22 | \begin{proof} 23 | $(\Longrightarrow)$ 24 | Since $E$ is open, then every point of $E$ is an interior point of $E$; hence $E^{\circ} = E$. 25 | 26 | $(\Longleftarrow)$ 27 | Since $E = E^{\circ}$, and we know $E^{\circ}$ is always open by the Exercise 2.9a, then $E$ is open. 28 | \end{proof} 29 | 30 | \item 31 | \begin{proof} 32 | Let $x\in G$ arbitrarily. 33 | Since $G$ is open, we know $x\in G^{\circ}$. 34 | Then there is a neighborhood $N_r(x)\subset G$ with $r>0$. 35 | Because $G\subset E$ by hypothesis, $N_r(x)\subset E$ which means $x\in E^{\circ}$. 36 | By the arbitrary choice of $x$, we conclude $G\subset E^{\circ}$. 37 | \end{proof} 38 | 39 | \item 40 | \begin{proof} 41 | We need to prove $(E^{\circ})^c = \overline{E^c}$ which means $(E^{\circ})^c \subseteq \overline{E^c}$ and $(E^{\circ})^c \supseteq \overline{E^c}$. 42 | 43 | \begin{itemize} 44 | \item \textbf{Claim $(E^{\circ})^c \subseteq \overline{E^c}$.} 45 | 46 | Let $x\in(E^{\circ})^c$ arbitrarily, then $x\notin E^{\circ}$. 47 | This implies every neighborhood of $x$ contains no elements in $E$; hence every neighborhood of $x$ contains some elements $y\in E^c$. 48 | 49 | If $x=y$ for at least one of these neighborhoods, then we know $x\in E^c$ because $x=y$ and $y\in E^c$. 50 | 51 | Otherwise, if $x\neq y$ for all neighborhoods, then by definition of a limit point, we know $x\in (E^c)'$. 52 | 53 | In either of two cases, $x$ is always a member of $E^c \cup (E^c)' = \overline{E^c}$. 54 | 55 | By the arbitrary choice of $x$, we conclude $(E^{\circ})^c \subseteq \overline{E^c}$. 56 | 57 | \item \textbf{Claim $(E^{\circ})^c \supseteq \overline{E^c}$.} 58 | 59 | Let $x\in\overline{E^c}$ arbitrarily, then $x\in E^c$ or $x\in (E^c)'$. 60 | 61 | If $x\in E^c$, then $x\notin E$. 62 | This implies that every neighborhood of $x$ is not contained in $E$, so $x\notin E^{\circ}$; hence $x\in (E^{\circ})^c$. 63 | 64 | Otherwise, if $x\in (E^c)'$, then every neighborhood $N_r(x)$ contains a point $y\in E^c$ with $r>0$. 65 | Since 66 | $$ 67 | d(x,x) = 0 < d(x,y) < r, 68 | $$ 69 | we know $x\in(E^{\circ})^c$. 70 | 71 | In either of two cases, $x\in(E^{\circ})^c$ always; also, by arbitrary choice of $x$, we conclude $(E^{\circ})^c \supseteq \overline{E^c}$. 72 | \end{itemize} 73 | Finally, we together two claims to obtain $(E^{\circ})^c = \overline{E^c}$. 74 | \end{proof} 75 | 76 | \item 77 | \begin{answer} 78 | No. 79 | \end{answer} 80 | \begin{proof} 81 | Let $E := (0,1) \cup (1,2) \subset \mathbb{R}$, then $\overline{E} = [0,2]$; hence $E^{\circ} = E$ and $\overline{E}^{\circ} = (0,2)$. 82 | This implies $E^{\circ} \neq \overline{E}^{\circ}$ as a counter-example. 83 | \end{proof} 84 | 85 | \item 86 | \begin{answer} 87 | No. 88 | \end{answer} 89 | \begin{proof} 90 | Let $E := \mathbb{Q}$ which is the set of all rational numbers, then $E^{\circ} = \emptyset$; hence $\overline{E} = \mathbb{R}$ and $\overline{E^{\circ}} = \emptyset$. 91 | This implies $\overline{E}\neq \overline{E^{\circ}}$ as a counter-example. 92 | \end{proof} 93 | \end{enumerate} 94 | \end{Exercise} -------------------------------------------------------------------------------- /chapter3/chapter3.tex: -------------------------------------------------------------------------------- 1 | % === Chapter 3 === 2 | \setcounter{chapter}{2} 3 | \chapter{Numerical Sequences and Series} 4 | 5 | \input{exercises/ex_3.1} 6 | 7 | \vspace{12pt} 8 | 9 | \input{exercises/ex_3.2} 10 | 11 | \vspace{12pt} 12 | 13 | \input{exercises/ex_3.3} 14 | 15 | \vspace{12pt} 16 | 17 | \input{exercises/ex_3.4} 18 | 19 | \vspace{12pt} 20 | 21 | \input{exercises/ex_3.5} 22 | 23 | \vspace{12pt} 24 | 25 | \input{exercises/ex_3.6} 26 | 27 | \vspace{12pt} 28 | 29 | \input{exercises/ex_3.7} 30 | 31 | \vspace{12pt} 32 | 33 | \input{exercises/ex_3.8} 34 | 35 | \vspace{12pt} 36 | 37 | \input{exercises/ex_3.9} 38 | 39 | \vspace{12pt} 40 | 41 | \input{exercises/ex_3.10} 42 | 43 | \vspace{12pt} 44 | 45 | \setcounter{Exercise}{13} 46 | \input{exercises/ex_3.14} 47 | 48 | -------------------------------------------------------------------------------- /chapter3/exercises/ex_3.1.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 3.1 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Notice that we assume the metric space is a Euclidean space. 5 | 6 | Suppose $\{s_n\}$ converges to $s$. That is, given $\epsilon>0$, there exists $N>0$ such that 7 | $$ 8 | n\geq N \implies |s_n-s| < \epsilon. 9 | $$ 10 | By Triangle Inequality, we know 11 | $$ 12 | \left| |s_n|-|s| \right| \leq |s_n-s|. 13 | $$ 14 | Hence 15 | $$ 16 | n\geq N \implies \left| |s_n|-|s| \right| < \epsilon. 17 | $$ 18 | This implies $\{|s_n|\}$ converges to $|s|$. 19 | 20 | \begin{itemize} 21 | \item $\mathbf{Is\ the\ converse\ true?}$ 22 | \end{itemize} 23 | \begin{answer} 24 | No. 25 | \end{answer} 26 | \begin{proof} 27 | Let $s_n = (-1)^n$, then $\{|s_n|\} = \{ |(-1)^n| \} = \{ 1 \}$ converges to $1$. However, $\{ s_n \} = \{ (-1)^n \}$ is a sequence $\{-1,1,-1,1,\cdots \}$ which diverges. 28 | \end{proof} 29 | \end{proof} 30 | \end{Exercise} -------------------------------------------------------------------------------- /chapter3/exercises/ex_3.10.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 3.10 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | By hypothesis, we know $a_n\in\mathbb{Z}$ and $|a_n|\geq 1$. 5 | Since 6 | $$ 7 | \alpha := \limsup_{n\to\infty} \sqrt[n]{|a_n|} 8 | \geq \limsup_{n\to\infty} \sqrt[n]{1} 9 | = 1, 10 | $$ 11 | then 12 | $$ 13 | R = \frac{1}{\alpha} 14 | \leq 1. 15 | $$ 16 | This means the radius of convergence is at most $1$. 17 | \end{proof} 18 | \end{Exercise} -------------------------------------------------------------------------------- /chapter3/exercises/ex_3.14.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 3.14 === 2 | \begin{Exercise} 3 | \begin{enumerate}[(a)] 4 | \item 5 | \begin{proof} 6 | Since $\lim_{n\to\infty} s_n = s$, then given $\epsilon>0$, there exists $N>0$ such that 7 | $$ 8 | n\geq N \implies d(s_n,s) < \epsilon. 9 | $$ 10 | This means 11 | $$ 12 | n\geq N \implies s-\epsilon < s_n < s+\epsilon. 13 | $$ 14 | Consider 15 | $$ 16 | \sigma_n = \frac{\sum_{i=0}^{n} s_i}{n+1} = \frac{\sum_{i=0}^{N-1} s_i + \sum_{i=N}^{n} s_i}{n+1}. 17 | $$ 18 | Since 19 | \begin{alignat*}{7} 20 | \quad&& \frac{\sum_{i=0}^{N-1}s_i + (n-N+1)(s-\epsilon)}{n+1} <& \sigma_n &<& \frac{\sum_{i=0}^{N-1}s_i + (n-N+1)(s+\epsilon)}{n+1} \\ 21 | \implies&& \frac{\sum_{i=0}^{N-1}s_i}{n+1}+\frac{n(s-\epsilon)}{n+1}-\frac{(N-1)(s-\epsilon)}{n+1} <& \sigma_n &<& \frac{\sum_{i=0}^{N-1}s_i}{n+1}+\frac{n(s-\epsilon)}{n+1}-\frac{(N-1)(s+\epsilon)}{n+1}. 22 | \end{alignat*} 23 | Taking the limit as $n\to\infty$, then 24 | $$ 25 | s \leq \lim_{n\to\infty} \sigma_n \leq s. 26 | $$ 27 | It follows from the Squeeze Theorem. 28 | Hence we conclude $\lim_{n\to\infty} \sigma_n = s$. 29 | \end{proof} 30 | 31 | \item 32 | \begin{answer} 33 | $s_n = (-1)^n$. 34 | \end{answer} 35 | \begin{proof} 36 | Since 37 | $$ 38 | \frac{0}{n+1} \leq \sigma_n = \frac{\sum_{i=0}^{n} s_n}{n+1} \leq \frac{1}{n+1}. 39 | $$ 40 | Taking the limit as $n\to\infty$, then $\lim_{n\to\infty} \sigma_n = 0$. 41 | However, $\{s_n\}$ diverges. 42 | \end{proof} 43 | 44 | \item 45 | \begin{answer} 46 | It can happen. 47 | \end{answer} 48 | \begin{proof} 49 | Consider 50 | $$ 51 | s_n = \begin{cases} 52 | \left( \frac{1}{3} \right)^n + \sqrt[3]{n} & \mbox{if } n=k^3, k\in\mathbb{Z} \\ 53 | \left( \frac{1}{3} \right)^n & \mbox{otherwise} 54 | \end{cases}. 55 | $$ 56 | Notice that $\limsup_{n\to\infty} s_n = \infty$ and $s_n>0$ for all $n$. 57 | Then 58 | \begin{align*} 59 | \sum_{i=0}^{n}s_n 60 | &= \sum_{i=0}^{n} \left( \frac{1}{3} \right)^i + \sum_{i=0}^{\lfloor \sqrt[3]{n} \rfloor}i \\ 61 | &\leq \sum_{i=0}^{\infty} \left( \frac{1}{3} \right)^i + \sum_{i=0}^{\lfloor \sqrt[3]{n} \rfloor}i \\ 62 | &= \frac{3}{2} + \frac{\sqrt[3]{n}(\sqrt[3]{n}+1)}{2} \\ 63 | &\leq \frac{3+\sqrt[3]{n}(\sqrt[3]{n}+\sqrt[3]{n})}{2} \\ 64 | &= \frac{3+2\sqrt[3]{n^2}}{2}. 65 | \end{align*} 66 | Since 67 | $$ 68 | |\sigma_n| 69 | = \left| \frac{\sum_{i=0}^{n}s_i}{n+1} \right| 70 | \leq \frac{3+2\sqrt[3]{n^2}}{2(n+1)} \to 0\text{ as } n\to\infty, 71 | $$ 72 | by the Squeeze Theorem, we obtain $\lim_{n\to\infty} \sigma_n = 0$ as promised. 73 | \end{proof} 74 | \end{enumerate} 75 | \end{Exercise} -------------------------------------------------------------------------------- /chapter3/exercises/ex_3.2.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 3.2 === 2 | \begin{Exercise} 3 | \begin{solution} 4 | Compute 5 | \begin{align*} 6 | \lim_{n\to\infty} \left( \sqrt{n^2+n}-n \right) 7 | &= \lim_{n\to\infty} \left( \left( \sqrt{n^2+n}-n \right) \left( \frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n} \right) \right) \\ 8 | &= \lim_{n\to\infty} \left( \frac{n}{\sqrt{n^2+n}+n} \right) \\ 9 | &= \lim_{n\to\infty} \left( \frac{1}{\sqrt{1+\frac{1}{n}}+1} \right) \\ 10 | &= \frac{1}{2}. 11 | \end{align*} 12 | \end{solution} 13 | \end{Exercise} -------------------------------------------------------------------------------- /chapter3/exercises/ex_3.3.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 3.3 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | It suffices to prove $\{s_n\}$ is increasing and bounded above by $2$. By Theorem 3.14, it follows that $\{s_n\}$ is convergent. 5 | 6 | \begin{itemize} 7 | \item \textbf{$\{s_n\}$ is increasing.} 8 | \end{itemize} 9 | It is clear that $0 \sqrt{2} = s_1. 13 | $$ 14 | Suppose for $n\leq k$, $s_{k+1}>s_k$ holds. Then for $n = k+1$, we obtain 15 | $$ 16 | s_{k+2} = \sqrt{2+\sqrt{s_{k+1}}} > \sqrt{2+\sqrt{s_k}} = s_{k+1} 17 | $$ 18 | also holds. By induction, $s_{n+1}>s_n$ holds for all $n\in\mathbb{N}$; hence $\{s_n\}$ is increasing. 19 | 20 | \begin{itemize} 21 | \item \textbf{$\{s_n\}$ is bounded above by $2$.} 22 | \end{itemize} 23 | It is clear that $s_1 = \sqrt{2} < 2$. Then suppose for $n=k$, $s_k < 2$ holds. For $n=k+1$, we consider 24 | \begin{alignat*}{7} 25 | \quad&& 0 &0$, we know for $x\in\mathbb{R}^1$, 21 | $$ 22 | 0<|x-0|<\delta \implies |f(x)-f(0)| = 1 \geq \epsilon. 23 | $$ 24 | So we have given a counter-example. 25 | \end{proof} 26 | \end{Exercise} -------------------------------------------------------------------------------- /chapter4/exercises/ex_4.11.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 4.11 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Let $\{x_n\}$ be an arbitrary Cauchy sequence in $X$, then given $\epsilon>0$, there exists $N,\delta>0$ such that 5 | $$ 6 | n,m\geq N \implies d_X(x_n,x_m)<\delta. 7 | $$ 8 | Since $f$ is uniformly continuous, by definition, we know 9 | $$ 10 | d_X(x_n,x_m) <\delta \implies d_Y(f(x_n),f(x_m)) < \epsilon. 11 | $$ 12 | Hence 13 | $$ 14 | n,m\geq N \implies d_Y(f(x_n),f(x_m))<\epsilon. 15 | $$ 16 | This means $\{f(x_n)\}$ is a Cauchy sequence in $Y$. Since $\{x_n\}$ was arbitrary, we complete the proof. 17 | \end{proof} 18 | \end{Exercise} -------------------------------------------------------------------------------- /chapter4/exercises/ex_4.14.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 4.14 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Let $g(x) := f(x) - x$. Since $f(x)$ is continuous on $I$ by hypothesis, and $x$ is also continuous on $I$ trivially. So $g$ is continuous on $I$. 5 | 6 | If $g(0) = 0$ or $g(1) = 0$, these mean $f(x) = x$ for $x=0$ or $x=1$. 7 | 8 | Otherwise, W.L.O.G., we assume $g(0) > 0 > g(1)$, by Theorem 4.23 (Intermediate Value Theorem), there exists $x\in (0,1)$ such that $g(x) = 0$. This means $f(x) = x$ for at least one $x\in I$. A similar argument proves the same result when $g(0) < 0 < g(1)$. 9 | \end{proof} 10 | \end{Exercise} -------------------------------------------------------------------------------- /chapter4/exercises/ex_4.2.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 4.2 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | If $f(\overline{E})$ is empty, then it holds trivially. Suppose $f(\overline{E})$ is nonempty. Let $y\in f(\overline{E})$. Put $y=f(e)$ for some $e\in E\cup E'$. 5 | 6 | If $e\in E$, then $y\in f(E)$. This implies $y\in\overline{f(E)}$ immediately. 7 | 8 | Otherwise, if $e\in E'$, then $N_{\delta}(e)$ contains infinitely many points in $E$ with $\delta>0$. Since $f$ is continuous, then given $\epsilon>0$, we know for all $\delta>0$, 9 | $$ 10 | x\in N_{\delta}(e) \implies f(x)\in N_{\epsilon}(y). 11 | $$ 12 | It follows that $N_{\epsilon}(y)$ contains infinitely many points in $f(E)$. Hence $y$ is a limit point of $f(E)$. 13 | 14 | In either of two cases, we observe $y\in f(E)\cup f(E)' = \overline{f(E)}$. By the arbitrary choice of $y$, we conclude $f(\overline{E})\subset \overline{f(E)}$. 15 | \end{proof} 16 | 17 | \begin{itemize} 18 | \item \textbf{$f(\overline{E})$ can be a proper subset of $\overline{f(E)}$.} 19 | \end{itemize} 20 | \begin{solution} 21 | We define $f:\mathbb{Z}\to\mathbb{R}$ by $f(x) = \frac{1}{x}$. Pick $\delta$ such that $0 < \delta < 1$. Given $\epsilon>0$, for all $x,y\in \mathbb{Z}$, we have 22 | $$ 23 | d(x,y)< \delta \implies x = y. 24 | $$ 25 | Hence $d(f(x),f(y)) = 0 < \epsilon$. This means $f$ is uniformly continuous and so is continuous. 26 | 27 | Now we put $E = \mathbb{Z}$. We see that 28 | $$ 29 | f(\overline{\mathbb{Z}}) 30 | = f(\mathbb{Z}) 31 | = \left\{ \frac{1}{x} : x\in\mathbb{Z} \right\} 32 | \subset \left\{ \frac{1}{x} : x\in\mathbb{Z} \right\} \cup \{0\} 33 | = \overline{f(\mathbb{Z})}. 34 | $$ 35 | \end{solution} 36 | \end{Exercise} -------------------------------------------------------------------------------- /chapter4/exercises/ex_4.20.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 4.20 === 2 | \begin{Exercise} 3 | \begin{enumerate}[(a)] 4 | \item 5 | \begin{proof} 6 | Notice that $d(x,z)\geq 0$ for all $x,z\in E$. 7 | 8 | $(\Longrightarrow)$ 9 | Since $\rho_E(x) = 0$, then 10 | $$ 11 | \inf_{z\in E}d(x,z) = 0. 12 | $$ 13 | Suppose to contrary that $x\notin \overline{E}$, this implies $x\in \overline{E}^c$. Since $\overline{E}$ is closed, we have $\overline{E}^c$ is open. So there exists $r>0$ such that $d(x,y) 0 16 | $$ 17 | which contradict $\rho_E(x) = 0$. So $x\in \overline{E}$. 18 | 19 | \vspace{2ex} 20 | 21 | $(\Longleftarrow)$ 22 | Since $x\in \overline{E}$, then $x\in E$ or $x\in E'$. If $x\in E$, then we pick $z=x\in E$ to complete the proof. Otherwise, if $x\in E'$, given $\epsilon>0$, we can pick $z\in E$ such that $d(x,z)<\epsilon$. This means 23 | $$ 24 | \inf_{z\in E}d(x,z) = 0. 25 | $$ 26 | \end{proof} 27 | 28 | \item 29 | \begin{proof} 30 | We consider three situations as follows. 31 | 32 | \begin{enumerate} 33 | \item[Case 1.] $x,y\in E$ 34 | 35 | From part (a), we know $x,y\in \overline{E}$. Hence 36 | $$ 37 | \left| \rho_E(x) - \rho_E(y) \right| = 0 \leq d(x,y). 38 | $$ 39 | 40 | \item[Case 2.] $x\in E,y\in X\setminus E$ or $x\in X\setminus E,x\in E$ 41 | 42 | W.L.O.G., we suppose $x\in E,y\in X\setminus E$. From part (a), we know $x\in \overline{E}$. Hence 43 | $$ 44 | \left| \rho_E(x) - \rho_E(y) \right| 45 | = |\rho_E(y)| 46 | = \inf_{z\in E}d(y,z) 47 | \leq d(y,x) 48 | = d(x,y). 49 | $$ 50 | By changing the roles of $x$ and $y$, we obtain the same result. 51 | 52 | \item[Case 3.] $x,y\in X\setminus E$ 53 | 54 | Since $\rho_E(x) \leq d(x,z) \leq d(x,y) + d(y,z)$, then we pick $z$ so close to $y$ that given $\epsilon>0$, 55 | $$ 56 | \rho_E(x) \leq d(x,y) + \rho_E(y) + \epsilon. 57 | $$ 58 | This implies 59 | $$ 60 | \rho_E(x) - \rho_E(y) \leq d(x,y) + \epsilon. 61 | $$ 62 | By the arbitrary choice of $z$, we have 63 | 64 | $$ 65 | \rho_E(x) - \rho_E(y) \leq d(x,y). 66 | $$ 67 | By changing the roles of $x$ and $y$, we obtain 68 | $$ 69 | \rho_E(y) - \rho_E(x) \leq d(x,y). 70 | $$ 71 | They follow that 72 | $$ 73 | \left| \rho_E(x) - \rho_E(y) \right| \leq d(x,y). 74 | $$ 75 | \end{enumerate} 76 | Hence we conclude $\rho_E$ is a uniformly continuous function on $X$. 77 | \end{proof} 78 | \end{enumerate} 79 | \end{Exercise} -------------------------------------------------------------------------------- /chapter4/exercises/ex_4.21.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 4.21 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | By Exercise 4.20, we know 5 | $$ 6 | \rho_F(x) = 0 \iff x\in \overline{F}. 7 | $$ 8 | Since $F$ is closed, then $x\in \overline{F} = F$. Notice that $K$ and $F$ are disjoint, this means for all $k\in K, t\in F$, we have $d(k,t) > 0$ and hence $\rho_F(k) > 0$. Because $\rho_F$ is uniformly continuous on $K$ and $K$ is compact, by Theorem 4.16, there exists $k'\in K$ such that $\rho_F(k') = \inf_{x\in K} \rho_F(x)$. 9 | If $p\in K, q\in F$, we consider 10 | $$ 11 | 0 < \frac{\inf_{x\in K} \rho_F(x)}{2} 12 | = \frac{\rho_F(k')}{2} 13 | < \rho_F(p) 14 | \leq d(p,q). 15 | $$ 16 | Hence we pick $\delta = \frac{\rho_F(k')}{2}$ to complete the proof. 17 | \end{proof} 18 | 19 | \begin{itemize} 20 | \item \textbf{The conclusion may fail for two disjoint closed sets if neither is compact.} 21 | \end{itemize} 22 | \begin{solution} 23 | Let 24 | \begin{align*} 25 | K &:= \mathbb{N};\\ 26 | F &:= \{n+\frac{1}{2n}:n\in\mathbb{N} \}. 27 | \end{align*} 28 | Both are closed and not compact; also $K\cap F = \emptyset$. Given $\epsilon>0$, we can pick $n>\frac{1}{2\epsilon}$ and $p\in K,q\in F$ such that 29 | $$ 30 | d(p,q) = \left| n-\left(n+\frac{1}{2n}\right) \right| 31 | = \frac{1}{2n} 32 | < \epsilon. 33 | $$ 34 | This means $d(p,q)$ might equal to $0$. 35 | \end{solution} 36 | \end{Exercise} -------------------------------------------------------------------------------- /chapter4/exercises/ex_4.3.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 4.3 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Notice that $f(Z(f)) = \{0\}$ has only single element. 5 | Then we know $\{0\}$ is closed trivially. 6 | Since $f$ is continuous, by the Corollary of Theorem 4.8, thus we observe $Z(f) = f^{-1}(\{0\})$ is also closed. 7 | \end{proof} 8 | \end{Exercise} -------------------------------------------------------------------------------- /chapter4/exercises/ex_4.4.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 4.4 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{Prove that $f(E)$ is dense in $f(X)$.} 5 | \end{itemize} 6 | \begin{proof} 7 | Let $y := f(x)$ arbitrarily, then we know $x\in E$ or $x\in E'$ by hypothesis that $E$ is dense. It suffices to prove $y\in f(E)$ or $y\in f(E)'$ so that $f(E)$ is dense in $f(X)$. 8 | 9 | if $x\in E$, then $y=f(x)\in f(E)$. Otherwise, if $x\in E'$, then there exists a sequence $\{x_n\}$ in $E$ such that $\lim_{n\to\infty} x_n = x$ and $x_n \neq x$. Since $f$ is continuous, by Theorem 4.2, we observe $\lim_{n\to\infty} f(x_n) = f(x)$. This means there exists a sequence $\{f(x_n)\}$ in $f(E)$ such that $\lim_{n\to\infty} f(x_n) = f(x) = y$ and hence we know $y\in f(E)'$. 10 | 11 | In either of two cases, we have $y\in f(E)$ or $y\in f(E)'$. Because $y$ was arbitrary, we conclude $f(E)$ is dense in $f(X)$. 12 | \end{proof} 13 | 14 | \begin{itemize} 15 | \item \textbf{Prove that $g(p) = f(p)$ for all $p\in X$.} 16 | \end{itemize} 17 | \begin{proof} 18 | By hypothesis, we suppose $g(q) = f(q)$ for all $q\in E$. Given $\epsilon>0$. Let $p\in X$ arbitrarily. Since $f$ is continuous on $X$ and $E\subseteq X$, then $f$ is continuous at $q\in E$. There exists $\delta_1>0$ such that 19 | $$ 20 | d(p,q)<\delta_1 \implies d(f(p),f(q))<\frac{\epsilon}{2}. 21 | $$ 22 | 23 | A similar argument establishes $g$ is also continuous at $q$, then there exists $\delta_2>0$ such that 24 | $$ 25 | d(p,q)<\delta_2 \implies d(g(p),g(q))<\frac{\epsilon}{2}. 26 | $$ 27 | 28 | They follow that 29 | \begin{align*} 30 | d(f(p),g(p)) 31 | &\leq d(f(p), f(q)) + d(f(q), g(q)) + d(g(q), g(p)) \\ 32 | &< \frac{\epsilon}{2} + 0 + \frac{\epsilon}{2} \\ 33 | &= \epsilon. 34 | \end{align*} 35 | 36 | This holds for any $\epsilon>0$, then we know $f(p)=g(p)$. Since $p$ was arbitrary, we conclude $f(p)=q(p)$ for all $p\in X$. 37 | \end{proof} 38 | \end{Exercise} -------------------------------------------------------------------------------- /chapter4/exercises/ex_4.5.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 4.5 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Since $E$ is closed, then $E^c$ is open. 5 | By Exercise 2.29, $E^c$ is the union of an at most countable collection of disjoint segments. 6 | Hence $E^c = \bigcup_{i=1}^{n}(a_i,b_i)$ where $b_i < a_{i+1}$ for each $i$. 7 | For each $i$, we define a function $g$ by 8 | $$ 9 | g(x) = f(a_i) + (x-a_i)\frac{f(b_i)-f(a_i)}{b_i-a_i} 10 | $$ 11 | for $x\in(a_i,b_i)$. 12 | 13 | Let $g(x) = f(x)$ for $x\in E$. 14 | Since it is a linear function on $E^c$ and $f$ is continuous on $E^c$, then $g$ is continuous on interior of $E$. 15 | Since $\lim_{x\to a_i} g(x) = f(a_i) = g(a_i)$ and $a_i\in E$, this means $g$ is continuous at $a_i$. 16 | Similarly, $g$ is continuous at $b_i$. 17 | 18 | Hence we know $g$ is continuous on $[a_i,b_i]$ for each $i$. 19 | It follows that $g$ is a continuous extension of $f$. 20 | \end{proof} 21 | 22 | \begin{itemize} 23 | \item \textbf{Show that the result becomes false if the word "closed" is omitted.} 24 | \end{itemize} 25 | \begin{proof} 26 | Consider $f(x) = \frac{1}{x}$ on $(0,1)$. 27 | However, $\lim_{x\to 0}f(x) = \infty \notin \mathbb{R}^1$. 28 | That is, we can not find $g(0)\in \mathbb{R}^1$. 29 | \end{proof} 30 | 31 | \begin{itemize} 32 | \item \textbf{Extend the result to vector-valued functions.} 33 | \end{itemize} 34 | \begin{proof} 35 | Let $f:E\to\mathbb{R}^k$ be a continuous vector-valued function defined by $f = \big( f_1, f_2, \cdots, f_k)$. 36 | Then $f_i$ is continuous on $E$ for each $i$. 37 | We can find $g_i$ is a continuous extension $f_i$ for each $i$. 38 | Hence $g = (g_1,g_2,\cdots,g_k)$ is a continuous extension of $f$. 39 | \end{proof} 40 | \end{Exercise} -------------------------------------------------------------------------------- /chapter4/exercises/ex_4.6.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 4.6 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Let $G$ be the graph of $f$. 5 | Define $g:E\to G$ by $g(x) = \big( x,f(x) \big)$. 6 | Put $d_G\big( g(x), g(y) \big) = d_E\Big( \big(x, f(x) \big), \big( y, f(y) \big)\Big) = \sqrt{(x-y)^2 + \big(f(x)-f(y)\big)^2}$. That is, we consider $G$ is a subset of $\mathbb{R}^2$. 7 | 8 | $(\Longrightarrow)$ 9 | Since $x$, $f(x)$ are continuous on $E$, then $g$ is continuous on $E$. 10 | Notice that $E$ is compact, by Theorem 4.14, it follows that $G$ is compact. 11 | 12 | $(\Longleftarrow)$ 13 | Suppose $f$ is not continuous on $E$, then for some $x\in E$, there is a convergent sequence $\{x_n\}\to x$ such that $f(x_n)$ does not converge to $f(x)$. We consider two cases as follows. 14 | \begin{enumerate}[i)] 15 | \item If no subsequence of $f(x_n)$ converges, $\left\{\big(x_n,f(x_n)\big) \right\}$ has no convergent subsequence. 16 | Hence $G$ is not compact. 17 | 18 | \item If some subsequences of $f(x_n)$ converge, say $f(x_{n_k})\to y$ as $n_k\to \infty$ with $f(x) \neq y$, then the limit point $(x,y)$ does not belong to $G$. 19 | It follows that $G$ is not compact. 20 | \end{enumerate} 21 | As a result, we have proven the contrapositive. 22 | \end{proof} 23 | \end{Exercise} -------------------------------------------------------------------------------- /chapter4/exercises/ex_4.7.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 4.7 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{Prove that $f$ is bounded on $\mathbb{R}^2$.} 5 | \end{itemize} 6 | \begin{proof} 7 | For $(x,y) \neq (0,0)$, we consider $(x-y^2)^2 \geq 0$. Then 8 | \begin{alignat*}{7} 9 | \quad&& 0 &\leq \left| (x-y^2)^2 \right| \\ 10 | \implies&& 0 &\leq \left| x^2 - 2x y^2+ y^4 \right| \leq |x^2+y^4|+|2x y^2| \\ 11 | \implies&& \left| \frac{2 x y^2}{x^2+y^4} \right| &\leq 1 \\ 12 | \implies&& |f(x,y)| &\leq \frac{1}{2} 13 | \end{alignat*} 14 | Otherwise, for $(x,y)=(0,0)$, we have $|f(0,0)| = 0 \leq \frac{1}{2}$ trivially. 15 | 16 | Hence $f(x,y)$ is bounded by $\frac{1}{2}$. 17 | \end{proof} 18 | 19 | \begin{itemize} 20 | \item \textbf{Prove that $g$ is unbounded in every neighborhood of $(0,0)$.} 21 | \end{itemize} 22 | \begin{proof} 23 | Let $x := t^3$ and $y := t$, we observe $\lim_{t\to 0} x = 0$ and $\lim_{t\to 0} y = 0$. Then for $t\neq 0$, we know $(x,y) \neq (0,0)$. Consider 24 | $$ 25 | \lim_{t\to 0} g(x,y) = \lim_{t\to 0} g(t^3,t) = \lim_{t\to 0}\frac{1}{t} = \infty. 26 | $$ 27 | Hence $g$ is unbounded in every neighborhood of $(0,0)$. 28 | \end{proof} 29 | 30 | \begin{itemize} 31 | \item \textbf{Prove that $f$ is not continuous at $(0,0)$.} 32 | \end{itemize} 33 | \begin{proof} 34 | Given $\epsilon = \frac{1}{2}$, for $(x,y)\neq (0,0)$, there exists $\delta > 0$ such that 35 | $$ 36 | d( (0,0),(x,y) ) <\delta \implies d(f(0,0),f(x,y))\geq \epsilon. 37 | $$ 38 | Regardless of $\delta$, we just pick $x=y^2$, then 39 | $$ 40 | d(f(0,0),f(x,y)) = \frac{y^4}{2y^4} = \frac{1}{2} \geq \epsilon. 41 | $$ 42 | Hence $f$ is not continuous at $(0,0)$. 43 | \end{proof} 44 | 45 | \begin{itemize} 46 | \item \textbf{The restrictions of both $f$ and $g$ to every straight line in $\mathbb{R}^2$ are continuous.} 47 | \end{itemize} 48 | \begin{proof} 49 | Since a straight line which doesn't pass through $(0,0)$ is always continuous trivially, then here we only prove the lines passing through $(0,0)$ are continuous. 50 | 51 | Consider the line $y = c x$ where $c$ is constant. Given $\epsilon>0$, we observe 52 | $$ 53 | d((0,0), (x,y)) 54 | = \sqrt{x^2+y^2} 55 | = \sqrt{x^2 + c^2 x^2} 56 | = |x| \sqrt{c^2+1} 57 | < \delta. 58 | $$ 59 | Put 60 | $$ 61 | \delta = \frac{\sqrt{c^2+1}}{c^2}\epsilon. 62 | $$ 63 | For $x\neq 0$, this implies 64 | $$ 65 | d(f(0,0), f(x,y)) 66 | = \left| \frac{x y^2}{x^2+y^4} \right| 67 | = \left| \frac{c^2 x^3}{x^2+c^4 x^4} \right| 68 | = \left| \frac{c^2 x}{1+c^4 x^2} \right| 69 | < |c^2 x| 70 | < c^2 \frac{\delta}{\sqrt{c^2+1}} 71 | = \epsilon. 72 | $$ 73 | So $f$ is continuous on this line. 74 | 75 | Then we consider the line $x=0$. Given $\epsilon>0$, for $y\neq 0$, we observe 76 | $$ 77 | d(f(0,0),f(x,y)) 78 | = \left| \frac{x y^2}{x^2+y^4} \right| 79 | = \frac{0}{y^4} 80 | = 0 81 | < \epsilon 82 | $$ 83 | regardless of the choice of $\delta$ such that $d((0,0),(x,y))<\delta$. Hence $f$ is continuous on this line. 84 | 85 | Finally, we conclude the restriction of $f$ to every straight line in $\mathbb{R}^2$ is continuous. A similar argument proves the restriction of $g$ to every straight line in $\mathbb{R}^2$ is also continuous. 86 | \end{proof} 87 | \end{Exercise} -------------------------------------------------------------------------------- /chapter4/exercises/ex_4.8.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 4.8 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Since $f$ is uniformly continuous on $E$, then there exists $\delta>0$ such that 5 | $$ 6 | |x-y|<\delta,\ x\in E,\ y\in E \implies |f(x)-f(y)|< 1. 7 | $$ 8 | We can pick $M$ so big that $E\subset [-M,M]$ and $N$ such that $N>\frac{2M}{\delta}$ by Archimedean Principle. 9 | This means we can cover $E$ with $N$ closed intervals $I_i$ which intersects $E$ and $|I_i|<\delta$. 10 | For each $i$, pick $x_i\in I_i\cap E$. 11 | 12 | Let $M = 1 + \max_{1\leq i \leq N}\{|f(x_i)\}$. 13 | For every $y\in E$, we know $y\in I_j$ for some $j$. 14 | Since $x_j\in I_j$ and $|I_j|<\delta$, then $|f(y)-f(x_j)|<1$. 15 | It follows that $|f(y)|0$, for $x,y\in \mathbb{R}$, there exists $\delta = \epsilon$ such that 24 | $$ 25 | |x-y|<\delta \implies |f(x)-f(y)| = |x-y| < \delta = \epsilon. 26 | $$ 27 | This means $f$ is uniformly continuous on $\mathbb{R}$. 28 | However, $f$ is not bounded. 29 | \end{proof} 30 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/chapter5.tex: -------------------------------------------------------------------------------- 1 | % === Chapter 5 === 2 | \setcounter{chapter}{4} 3 | \chapter{Differentiation} 4 | 5 | \input{exercises/ex_5.1} 6 | 7 | \vspace{12pt} 8 | 9 | \input{exercises/ex_5.2} 10 | 11 | \vspace{12pt} 12 | 13 | \input{exercises/ex_5.3} 14 | 15 | \vspace{12pt} 16 | 17 | \input{exercises/ex_5.4} 18 | 19 | \vspace{12pt} 20 | 21 | \input{exercises/ex_5.5} 22 | 23 | \vspace{12pt} 24 | \input{exercises/ex_5.6} 25 | 26 | \vspace{12pt} 27 | 28 | \input{exercises/ex_5.7} 29 | 30 | \vspace{12pt} 31 | 32 | \input{exercises/ex_5.8} 33 | 34 | \vspace{12pt} 35 | 36 | \input{exercises/ex_5.9} 37 | 38 | \vspace{12pt} 39 | 40 | \input{exercises/ex_5.10} 41 | 42 | \vspace{12pt} 43 | 44 | \input{exercises/ex_5.11} 45 | 46 | \vspace{12pt} 47 | 48 | \setcounter{Exercise}{16} 49 | \input{exercises/ex_5.17} 50 | 51 | \vspace{12pt} 52 | 53 | \setcounter{Exercise}{21} 54 | \input{exercises/ex_5.22} 55 | 56 | \vspace{12pt} 57 | 58 | \setcounter{Exercise}{24} 59 | \input{exercises/ex_5.25} 60 | -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.1.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.1 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Consider 5 | $$ 6 | |f(x)-f(y)| \leq (x-y)^2 = |x-y|^2. 7 | $$ 8 | For $x\neq y$, we have 9 | $$ 10 | \left| \frac{f(x)-f(y)}{x-y} \right| \leq |x-y|. 11 | $$ 12 | Taking the limit on both sides as $x\to y$, we obtain 13 | $$ 14 | |f'(y)| \leq 0. 15 | $$ 16 | This means $f'(y) = 0$. Since $y$ was arbitrary, then $f'(y) = 0$ for all $y\in\mathbb{R}$. Hence by Theorem 5.11(b), $f$ is constant. 17 | \end{proof} 18 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.10.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.10 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Let $f_1$ and $f_2$ be the real and imaginary parts of $f$. 5 | Let $g_1$ and $g_2$ be the real and imaginary parts of $g$. 6 | Notice that $f_1$, $f_2$, $g_1$, $g_2$ are real functions. 7 | 8 | By Remarks 5.16, we know $f'(x) = f'_1(x) + \ii f'_2(x)$ and $g'(x) = g'_1(x) + \ii g'_2(x)$. 9 | Follow the hint, then we obtain 10 | \begin{align*} 11 | \lim_{x\to 0}\frac{f(x)}{g(x)} 12 | &= \lim_{x\to 0}\left[ \left( \frac{f(x)}{x}-A\right)\cdot\frac{x}{g(x)} + A \cdot \frac{x}{g(x)}\right] \\ 13 | &= \lim_{x\to 0}\left[ \left( \frac{f_1(x)+\ii f_2(x)}{x}-A\right)\cdot\frac{x}{g_1(x)+\ii g_2(x)} + A \cdot \frac{x}{g_1(x)+\ii g_2(x)}\right] \\ 14 | &= \lim_{x\to 0}\left[ \left( \frac{f'_1(x)+\ii f'_2(x)}{1}-A\right)\cdot\frac{1}{g'_1(x)+\ii g'_2(x)} + A \cdot \frac{1}{g'_1(x)+\ii g'_2(x)}\right] \mbox{ 15 | ($\because$~~\textit{L'Hospital's Rule})}\\ 16 | &= \lim_{x\to 0}\left[ \left( \frac{f'(x)}{1}-A\right)\cdot\frac{1}{g'(x)} + A \cdot \frac{1}{g'(x)}\right] \\ 17 | &= (A-A) \cdot \frac{1}{B} + A \cdot \frac{1}{B} \\ 18 | &= \frac{A}{B}. 19 | \end{align*} 20 | Hence we get the desired result. 21 | \end{proof} 22 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.11.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.11 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Consider 5 | $$ 6 | \lim_{h\to 0}\frac{f(x+h) + f(x-h) -2f(x)}{h^2}. 7 | $$ 8 | We observe 9 | \begin{align*} 10 | &\lim_{h\to 0}f(x+h)+f(x-h)-2f(x) = 0; \\ 11 | &\lim_{h\to 0}h^2 =0. 12 | \end{align*} 13 | By L'Hospital's Rule, we differentiate them over $h$, then obtain 14 | \begin{align*} 15 | \lim_{h\to 0}\frac{f(x+h) + f(x-h) -2f(x)}{h^2} 16 | &= \lim_{h\to 0}\frac{f'(x+h) - f'(x-h)}{2h} \\ 17 | &= \frac{1}{2} \lim_{h\to 0}\left( \frac{f'(x+h)-f'(x)}{h} + \frac{f'(x) - f'(x-h)}{h} \right)\\ 18 | &= \frac{1}{2} \big(f''(x) + f''(x) \big) \\ 19 | &= f''(x) 20 | \end{align*} 21 | which is a desired result. 22 | \end{proof} 23 | \begin{itemize} 24 | \item \textbf{An example that the limit may exist even if $f''(x)$ does not.} 25 | \end{itemize} 26 | \begin{answer} 27 | $f(x) = \begin{cases} 28 | x^2 & \mbox{for } x > 0 \\ 29 | -x^2 & \mbox{for } x < 0 \\ 30 | 0 & \mbox{otherwise} 31 | \end{cases}$. 32 | \end{answer} 33 | \begin{proof} 34 | Since $f''(x) = \begin{cases} 35 | 2 & \mbox{for } x> 0 \\ 36 | -2 & \mbox{for } x<0 \\ 37 | \end{cases}$, then $f''(x)$ does not exist at $x=0$. 38 | \end{proof} 39 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.17.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.17 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Follow the hint. 5 | By Theorem 5.15 (Taylor's Theorem), if we put $\alpha = 0, \beta = 1$, then 6 | $$ 7 | f(1) = f(0) + f'(0) + \frac{f''(0)}{2}+\frac{f^{(3)}(s)}{6} 8 | $$ 9 | where $s\in(0,1)$. 10 | If we put $\alpha = 0, \beta = -1$, then 11 | $$ 12 | f(-1) = f(0) - f'(0) + \frac{f''(0)}{2}-\frac{f^{(3)}(t)}{6} 13 | $$ 14 | where $t\in(-1,0)$. 15 | Now we subtract the second equation from the first one so that 16 | $$ 17 | f(1) + f(-1) = 2f'(0) + \frac{f^{(3)}(s)}{6} + \frac{f^{(3)}(t)}{6}. 18 | $$ 19 | Since $f(1) = 1, f(-1) = 0, f'(0) = 0$, we have 20 | $$ 21 | \frac{f^{(3)}(s)}{6} + \frac{f^{(3)}(t)}{6} = 1 22 | $$ 23 | where $s\in(0,1), t\in(-1,0)$. 24 | We deduce 25 | $$ 26 | f^{(3)}(s) + f^{(3)}(t) = 6 27 | $$ 28 | where $s,t\in(-1,1)$. 29 | This is the desired result to conclude 30 | $$ 31 | f^{(3)}(x) \geq 3 32 | $$ 33 | for some $x\in(-1,1)$. 34 | \end{proof} 35 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.2.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.2 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{Prove $f$ is strictly increasing in $(a,b)$.} 5 | \end{itemize} 6 | \begin{proof} 7 | Since $f'(x) > 0$ in $(a,b)$, then $f$ is differentiable on $(a,b)$. 8 | By the Mean Value Theorem, we consider 9 | $$ 10 | f(x_1) - f(x_2) = (x_1 - x_2) f'(x) 11 | $$ 12 | for $x_1,x_2\in (a,b)$. This implies 13 | $$ 14 | f'(x) = \frac{f(x_1)-f(x_2)}{x_1-x_2} > 0. 15 | $$ 16 | Then $x_1 > x_2$ if and only if $f(x_1) > f(x_2)$. This means $f$ is strictly increasing in $(a,b)$. 17 | \end{proof} 18 | \begin{itemize} 19 | \item \textbf{Prove that $g$ is differentiable and the formula holds.} 20 | \end{itemize} 21 | \begin{proof} 22 | We have already known $f$ is strictly increasing in $(a,b)$, then $f$ is one-to-one and continuous on $(a,b)$. 23 | Hence $g = f^{-1}$ exists and $g\big( f(x) \big) = x$ for all $x\in(a,b)$; also $f(x)\to f(c)$ as $x\to c$ where $c\in(a,b)$. 24 | 25 | Now we should prove $g$ is differentiable on $f(a,b)$ so that we can differentiate it over $x$ to get the formula. 26 | Consider 27 | $$ 28 | f'(x) = \lim_{x\to c}\frac{f(x)-f(c)}{x-c} 29 | = \lim_{f(x)\to f(c)}\frac{f(x)-f(c)}{g\big( f(x) \big) - g\big( f(c) \big)} 30 | = \frac{1}{g'(f(x))} 31 | $$ 32 | where $c\in(a,b)$. This conclude 33 | $$ 34 | g'\big( f(x) \big) = \frac{1}{f'(x)} 35 | $$ 36 | for $x\in(a,b)$. 37 | \end{proof} 38 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.22.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.22 === 2 | \begin{Exercise} 3 | \begin{enumerate}[(a)] 4 | \item 5 | \begin{proof} 6 | Suppose to contrary that $f$ has more than one fixed point, says $x_1$, $x_2$. 7 | Notice that $x_1\neq x_2$. 8 | Then by Mean Value Theorem, we consider 9 | $$ 10 | f'(t) = \frac{f(x_1)-f(x_2)}{x_1-x_2} 11 | = \frac{x_1-x_2}{x_1-x_2} 12 | =1 13 | $$ 14 | where $t\in (x_1,x_2)$. 15 | This contradicts $f'(t) \neq 1$ for all $t$. 16 | Hence $f$ has at most one fixed point. 17 | \end{proof} 18 | 19 | \item 20 | \begin{proof} 21 | We observe 22 | $$ 23 | f'(t) = 1-\frac{e^t}{(1+e^t)^2}. 24 | $$ 25 | This means $0 0$, we consider 47 | \begin{align*} 48 | | x_{n+k} - x_n | 49 | &\leq |x_{n+k}-x_{n+k-1}| + |x_{n+k-1}-x_{n+k-2}| + \cdots + | x_2 - x_1 | \\ 50 | &= A^{n+k-2} |x_2-x_1| + A^{n+k-3} |x_2-x_1| + \cdots + A^{n-1} |x_2 - x_1| \\ 51 | &\leq k A^{n-1} | x_2-x_1 |. 52 | \end{align*} 53 | Taking the limit on both sides as $n\to\infty$, we have 54 | $$ 55 | \lim_{n\to\infty} |x_{n+k}-x_n| \leq 0. 56 | $$ 57 | Hence $\lim_{n\to\infty}|x_{n+k}-x_n| = 0$, it follows that the sequence $\{x_k\}$ is Cauchy. 58 | Since $\mathbb{R}$ is compact, then by Theorem 3.11(b), we know there is $x\in\mathbb{R}$ such that 59 | $$ 60 | \lim_{n\to\infty} x_n = x. 61 | $$ 62 | Moreover, by definition of $f(x_n)$, we also have 63 | $$ 64 | \lim_{n\to\infty} f(x_n) = x. 65 | $$ 66 | Since $f'(t)$ exists for all $t\in\mathbb{R}$ by hypothesis, then $f$ is continuous on $\mathbb{R}$, we conclude 67 | $$ 68 | f(x) = x 69 | $$ 70 | which means $x$ is a fixed point of $f$. 71 | \end{proof} 72 | 73 | \item 74 | \begin{solution} 75 | Consider the points $\big(x_n, f(x_n)\big)$. 76 | We know when $n\to\infty$, the point must be a fixed point $(x, x)$. 77 | If we start with $\big(x_1, f(x_1)\big)$, then we want to arrive at $\big(x_2, f(x_2)\big)$. We just pick the path 78 | $$ 79 | \big(x_1, f(x_1)\big) \to \big(x_2, f(x_1)\big) \to \big(x_2, f(x_2)\big) 80 | $$ 81 | which means 82 | $$ 83 | \big(x_1, x_2\big) \to \big(x_2, x_2\big) \to \big(x_2, x_3\big). 84 | $$ 85 | Then we arrive at $\big(x_3, f(x_3)\big)$, which means 86 | 87 | $$ 88 | \big(x_2, x_3\big) \to \big(x_3, x_3\big) \to \big(x_3, x_4\big). 89 | $$ 90 | This process continue until we arrive at the fixed point. 91 | By induction, we obtain the zig-zag path. 92 | \end{solution} 93 | \end{enumerate} 94 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.25.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.25 === 2 | \begin{Exercise} 3 | \begin{enumerate}[(a)] 4 | \item 5 | \begin{solution} 6 | Consider the tangent line to a point $(x_n, y_n)$ is 7 | $$ 8 | y-y_n = f'(x_n) (x-x_n). 9 | $$ 10 | Pick $x = x_{n+1}$ and $y = 0$, also put $y_n = f(x_n)$, then we have 11 | \begin{equation}\label{eq:ex_5.25a} 12 | x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}. 13 | \end{equation} 14 | Hence in the geometrical opinion, $x_{n+1}$ is chosen to be the intersection of the tangent line to $\big(x_n, f(x_n)\big)$ and $x$-axis. 15 | \end{solution} 16 | 17 | \item 18 | \begin{proof} 19 | By hypothesis and Intermediate Value Theorem, we know $f(x) > 0$ when $x > \xi$. This implies 20 | $$ 21 | \frac{f(x_n)}{f'(x_n)} > 0. 22 | $$ 23 | We observe $x_{n+1} < x_n$ by \eqref{eq:ex_5.25a}. 24 | 25 | Since $f''(x) \geq 0$, then $f'(x)$ is increasing. This means 26 | $$ 27 | c < x_n \implies f'(c) \leq f'(x_n). 28 | $$ 29 | Pick $c$ such that $\xi < c < x_n$, then by the Mean Value Theorem, we have 30 | $$ 31 | \frac{f(x_n)-f(\xi)}{x_n-\xi} = f'(c) \leq f'(x_n). 32 | $$ 33 | Notice that $f(\xi) = 0$, then rearrange the inequality to obtain 34 | $$ 35 | x_n - \frac{f(x_n)}{f'(x_n)} = x_{n+1}\geq \xi. 36 | $$ 37 | We have $\xi < x_{n+1} < x_n$ so far. 38 | 39 | Construct the sequence $\{x_n\}$. 40 | Since $x_n$ is decreasing and bounded below by $\xi$, by Theorem 3.14, $\{x_n\}$ converges to some $\eta$. 41 | Consider \eqref{eq:ex_5.25a}. 42 | Taking the limit on both sides as $n\to\infty$, we have 43 | $$ 44 | \eta = \eta - \lim_{n\to\infty}\frac{f(x_n)}{f'(x_n)}. 45 | $$ 46 | This implies 47 | $$ 48 | \lim_{n\to\infty}\frac{f(x_n)}{f'(x_n)} = 0. 49 | $$ 50 | Since $f''(x)$ is bounded and so is $f'(x)$, we obtain 51 | $$ 52 | \lim_{n\to\infty}f(x_n) = 0 = f(\xi). 53 | $$ 54 | Notice that $f'(x)$ exists on $[a,b]$, and hence $f$ is continuous on $[a,b]$. 55 | By Theorem 4.2, we conclude that 56 | $$ 57 | \lim_{n\to\infty}x_n = \xi 58 | $$ 59 | as promised eventually. 60 | \end{proof} 61 | 62 | \item 63 | \begin{proof} 64 | Using the Taylor's Theorem, we have 65 | $$ 66 | f(\xi) = f(x_n) + f'(x_n)(\xi-x_n)+\frac{f''(t_n)}{2}(\xi-x_n)^2 67 | $$ 68 | where $t_n\in(\xi, x_n)$. 69 | Since $f'(x_n)> 0$, divides $f'(x_n)$, then we have 70 | $$ 71 | \frac{f(\xi)}{f'(x_n)} = \frac{f(x_n)}{f'(x_n)} + (\xi-x_n) + \frac{f''(t_n)}{2f'(x_n)}(\xi-x_n)^2. 72 | $$ 73 | Since $f(\xi) = 0$, we rearrange the equation and obtain 74 | $$ 75 | x_n - \frac{f(x_n)}{f'(x_n)} - \xi = \frac{f''(t_n)}{2f'(x_n)}(x_n-\xi)^2. 76 | $$ 77 | By definition of $x_{n+1}$, we conclude 78 | \begin{equation}\label{eq:ex_5.25d} 79 | x_{n+1} - \xi = \frac{f''(t_n)}{2f'(x_n)}(x_n-\xi)^2. 80 | \end{equation} 81 | \end{proof} 82 | 83 | \item 84 | \begin{proof} 85 | We have known $0 \leq x_{n+1} - \xi$ from part(b) trivially. 86 | Consider \eqref{eq:ex_5.25d}. 87 | Since $f''(x) \leq M$ and $f'(x) \leq \delta$ by hypothesis, we have 88 | $$ 89 | x_{n+1}-\xi \leq \frac{M}{2\delta}(x_n-\xi)^2 = A(x_n-\xi)^2. 90 | $$ 91 | This inequality holds for $n\in\mathbb{N}$. By induction, we know 92 | \begin{align*} 93 | x_{n+1}-\xi 94 | &\leq A(x_n-\xi)^2 \\ 95 | &\leq A\cdot A^2 (x_{n-1}-\xi)^4 \\ 96 | &\leq \cdots \\ 97 | &\leq A\cdot A^2 \cdots \cdot A^{2^{n-1}} (x_1-\xi)^{2^n} \\ 98 | &= A^{2^n-1} (x_1-\xi)^{2^n} \\ 99 | &= \frac{1}{A} \big[A(x_1-\xi)\big]^{2^n}. 100 | \end{align*} 101 | Hence we deduce 102 | $$ 103 | 0 \leq x_{n+1}-\xi \leq \frac{1}{A} \big[A(x_1-\xi)\big]^{2^n}. 104 | $$ 105 | \end{proof} 106 | 107 | \item 108 | \begin{itemize} 109 | \item \textbf{Show that Newton's method amounts to finding a fixed point of the function $g$.} 110 | \end{itemize} 111 | \begin{solution} 112 | Suppose we have known $\eta$ is a fixed point of the function $g$ such that $g(\eta) = \eta$. Then 113 | $$ 114 | g(\eta) - \eta = \frac{f(\eta)}{f'(\eta)} = 0. 115 | $$ 116 | Since $f''$ is bounded and so is $f'$, then we have $f(\eta) = 0$. 117 | This means $\eta$ is the unique point such that $f(\eta) = 0$. 118 | \end{solution} 119 | \begin{itemize} 120 | \item How does $g'(x)$ behave for $x$ near $\xi$? 121 | \end{itemize} 122 | \begin{solution} 123 | Notice that $g'$ exists on $[a,b]$. Compute 124 | $$ 125 | g'(x) = 1 - \frac{\big[ f'(x) \big]^2 - f''(x) f(x)}{\big[ f'(x) \big]^2} = \frac{f''(x) f(x)}{\big[ f'(x) \big]^2}. 126 | $$ 127 | Taking the limit on both sides as $x\to xi$, we have 128 | $$ 129 | \lim_{x\to\xi}g'(x) = \lim_{x\to\xi}f(x) \lim_{x\to\xi} \frac{f''(x)}{[f'(x)]^2}. 130 | $$ 131 | Since $f$ is continuous near $\xi$, then we know 132 | $$ 133 | \lim_{x\to\xi} f(x) = f(\xi) = 0. 134 | $$ 135 | This means 136 | $$ 137 | \lim_{x\to\xi}g'(x) = 0. 138 | $$ 139 | We deduce $g'(x)$ tends to $0$ as $x$ tends $\xi$. 140 | \end{solution} 141 | 142 | \item 143 | \begin{solution} 144 | Compute 145 | $$ 146 | f'(x) = \frac{1}{3} x^{-\frac{2}{3}}. 147 | $$ 148 | We observe $0$ is a fixed point such that $f(0) = 0$. 149 | However, $f'(0)$ does not exist. 150 | Hence the sequence $\{x_n\}$ diverges. 151 | \end{solution} 152 | \end{enumerate} 153 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.3.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.3 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Pick $\epsilon$ so small that $|\epsilon| < \frac{1}{M}$. 5 | Since $x$ is differentiable and hence $f(x) = x + \epsilon g(x)$ is differentiable. 6 | It follows that 7 | $$ 8 | f'(x) = 1 + \epsilon g'(x). 9 | $$ 10 | Then 11 | $$ 12 | |\epsilon g'(x)| \leq |\epsilon| M < \frac{1}{M} M = 1. 13 | $$ 14 | Hence $0 < f'(x) < 2$. 15 | We know $f$ is strictly increasing and hence is one-to-one. 16 | \end{proof} 17 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.4.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.4 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Put 5 | $$ 6 | f(x) = C_0 x + \frac{C_1}{2} x^2 + \cdots + \frac{C_n}{n+1} x^{n+1}. 7 | $$ 8 | This implies that 9 | $$ 10 | f'(x) = C_0 + C_1 x + \cdots + C_n x^n. 11 | $$ 12 | Observe that 13 | $$ 14 | f(0) = f(1) = 0. 15 | $$ 16 | By the Mean Value Theorem, we know there exists $c\in(0,1)$ such that 17 | $$ 18 | f(1) - f(0) = f'(c) (1-0). 19 | $$ 20 | Hence 21 | $$ 22 | C_0 + C_1 x + \cdots + C_n x^n 23 | $$ 24 | has at least one real root between $0$ and $1$. 25 | \end{proof} 26 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.5.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.5 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Since $f'(x)\to 0$ as $x\to\infty$, then given $\epsilon>0$, there exists $N>0$ such that 5 | $$ 6 | x\geq N \implies |f'(x)| < \epsilon. 7 | $$ 8 | By the Mean Value Theorem, we consider 9 | $$ 10 | g(x) = f(x+1)-f(x) 11 | = \frac{f(x+1)-f(x)}{(x+1)-x} 12 | = f'(c) 13 | $$ 14 | where $c\in(x,x+1)$. 15 | We pick $x \geq N$, then this means $c>x\geq N$. So 16 | $$ 17 | |g(x)| = |f'(c)| < \epsilon. 18 | $$ 19 | It follows that $g(x)\to 0$ as $x\to\infty$. 20 | \end{proof} 21 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.6.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.6 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | It suffice to prove $g'(x) \geq 0$ for all $x>0$. 5 | Since $f$ is continuous on $[0,\infty)$ and $f$ is differentiable on $(0,\infty)$, by the Mean Value Theorem, we have 6 | $$ 7 | f(x) - f(0) = f'(c)(x-0) 8 | $$ 9 | where $c\in(0,x)$. 10 | Using the hypothesis that $f(0) = 0$, we obtain 11 | $$ 12 | f'(c) = \frac{f(x)}{x}. 13 | $$ 14 | Since $f'$ is monotonically increasing, we know 15 | $$ 16 | x\geq c \implies f'(x) \geq f'(c). 17 | $$ 18 | It follows that 19 | $$ 20 | g'(x) 21 | = \frac{xf'(x) - f(x)}{x^2} 22 | \geq \frac{xf'(c) - f(x)}{x^2} 23 | = \frac{x \cdot \frac{f(x)}{x} - f(x)}{x^2} 24 | = 0 25 | $$ 26 | for all $x>0$. 27 | Hence $g$ is monotonically increasing. 28 | \end{proof} 29 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.7.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.7 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Notice that $f(x) = g(x) = 0$. 5 | We consider 6 | $$ 7 | \lim_{t\to x}\frac{f(t)}{g(t)} 8 | = \lim_{t\to x}\frac{f(t)-f(x)}{g(t)-g(x)} 9 | = \lim_{t\to x}\left( \frac{f(t)-f(x)}{t-x} \cdot \frac{t-x}{g(t)-g(x)} \right) 10 | = \frac{f'(x)}{g'(x)}. 11 | $$ 12 | \end{proof} 13 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.8.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.8 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Since $f'$ is continuous on $[a,b]$ and $[a,b]$ is compact, then $f'$ is uniformly continuous on $[a,b]$. 5 | Given $\epsilon>0$, there exists $\delta > 0$ such that 6 | $$ 7 | |t-x|<\delta,\ x\in[a,b],\ t\in[a,b] \implies |f(t)-f(x)|<\epsilon. 8 | $$ 9 | By the Mean Value Theorem, there exists $p$ is between $t$ and $x$ such that 10 | $$ 11 | f(t)-f(x) = f'(p) (t-x). 12 | $$ 13 | Moreover, we have $|x-p|<\delta$ and $x,p\in [a,b]$. 14 | This implies $|f'(x) - f'(p)|<\epsilon$. 15 | Hence we know 16 | $$ 17 | \left| \frac{f(t)-f(x)}{t-x}-f'(x) \right| < \epsilon 18 | $$ 19 | whenever $0<|t-x|<\delta,\ a\leq x \leq b,\ a\leq t\leq b$. 20 | \end{proof} 21 | 22 | \begin{itemize} 23 | \item \textbf{Does this hold for vector-valued functions too?} 24 | \end{itemize} 25 | \begin{answer} 26 | Yes. 27 | \end{answer} 28 | \begin{proof} 29 | Let $f = (f_1, f_2, \cdots, f_k)$. 30 | Since $f'$ is continuous on $[a,b]$, then $f_i$ is continuous on $[a,b]$. 31 | Moreover, $[a,b]$ is compact, we know $f_i$ is uniformly continuous on $[a,b]$. 32 | On the other hand, $f$ is differentiable on $[a,b]$, so is $f_i$ for each $i$. 33 | 34 | Given $\epsilon>0$ with $t,x\in [a,b]$, for each $i$, there exists $\delta_i >0$ such that 35 | $$ 36 | |t-x|<\delta_i \implies \left| \frac{f(t)-f(x)}{t-x}-f'(x) \right| < \frac{\epsilon}{k}. 37 | $$ 38 | Put $\delta = \min_{1\leq i \leq k}\{\delta_i\}$. Then for $|x-t|<\delta$, we have 39 | \begin{align*} 40 | \left| \frac{f(t)-f(x)}{t-x}-f'(x) \right| 41 | &= \left| \frac{\big( f_1(t), f_2(t),\cdots, f_k(t)\big) - \big( f_1(x), f_2(x),\cdots, f_k(x)\big)}{t-x} - \big( f'_1(x), f'_2(x), \cdots, f'_k(x) \big)\right| \\ 42 | &\leq \left| \frac{f_1(t)-f_1(x)}{t-x}-f'_1(x) \right| + \left| \frac{f_2(t)-f_2(x)}{t-x}-f'_2(x) \right| + \cdots + \left| \frac{f_k(t)-f_k(x)}{t-x}-f'_k(x) \right| \\ 43 | &< \underbrace{ \frac{\epsilon}{k} + \frac{\epsilon}{k} + \cdots + \frac{\epsilon}{k} }_k \\ 44 | &= \epsilon 45 | \end{align*} 46 | \end{proof} 47 | \end{Exercise} -------------------------------------------------------------------------------- /chapter5/exercises/ex_5.9.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 5.9 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | We have already known $\lim_{x\to 0}f'(x) = 3$. 5 | Suppose to contrary that $f'(0)\neq 3$, then $f'$ have a simple discontinuity at $x=3$, this contradicts the Corollary of Theorem 5.12. 6 | 7 | Hence $f'(0)$ exists and is equal to $3$. 8 | \end{proof} 9 | \end{Exercise} -------------------------------------------------------------------------------- /chapter6/chapter6.tex: -------------------------------------------------------------------------------- 1 | % === Chapter 6 === 2 | \setcounter{chapter}{5} 3 | \chapter{The Riemann-Stieltjes Integral} 4 | 5 | \input{exercises/ex_6.1} 6 | 7 | \vspace{12pt} 8 | 9 | \input{exercises/ex_6.2} 10 | 11 | \vspace{12pt} 12 | 13 | \input{exercises/ex_6.3} 14 | 15 | \vspace{12pt} 16 | 17 | \input{exercises/ex_6.4} 18 | 19 | \vspace{12pt} 20 | 21 | \input{exercises/ex_6.5} 22 | 23 | \vspace{12pt} 24 | 25 | \setcounter{Exercise}{6} 26 | \input{exercises/ex_6.7} 27 | 28 | \vspace{12pt} 29 | 30 | \input{exercises/ex_6.8} 31 | 32 | \vspace{12pt} 33 | 34 | \setcounter{Exercise}{9} 35 | \input{exercises/ex_6.10} 36 | 37 | \vspace{12pt} 38 | 39 | \input{exercises/ex_6.11} 40 | 41 | \vspace{12pt} 42 | 43 | \setcounter{Exercise}{14} 44 | \input{exercises/ex_6.15} 45 | -------------------------------------------------------------------------------- /chapter6/exercises/ex_6.1.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 6.1 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Since $|f(x)| \leq 1$ for all $x\in\mathbb{R}$, we know $f$ is bounded. 5 | $f$ is discontinuous at $x_0$, however $\alpha$ is continuous at $x_0$. 6 | By Theorem 6.10, $f\in\mathcal{R}$ follows. 7 | 8 | Consider arbitrary partition of $[a,b]$ which contains a point other than $x_0$, we know $m_i=0$ for all intervals. Hence 9 | $$ 10 | \int_{a}^{b} f\,d\alpha 11 | = \lowint_{a}^{b} f\,d\alpha 12 | = 0 13 | $$ 14 | which is the desired result. 15 | \end{proof} 16 | \end{Exercise} -------------------------------------------------------------------------------- /chapter6/exercises/ex_6.10.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 6.10 === 2 | \begin{Exercise} 3 | \begin{enumerate}[(a)] 4 | % === Part(a) === 5 | \item 6 | \begin{proof} 7 | Consider 8 | $$ 9 | uv = (u^p)^{\frac{1}{p}} (v^q)^{\frac{1}{q}} 10 | = e^{\frac{1}{p} \ln(u^p) + \frac{1}{q} \ln(v^q)}. 11 | $$ 12 | We define a real function $f$ by $f(x) = e^x$. 13 | Since $f'(x) = e^x > 0$, $f$ is monotonically increasing, then $f$ is convex and one-to-one. 14 | Notice that the hypothesis guarantees 15 | $$ 16 | \frac{1}{p} + \frac{1}{q} = 1. 17 | $$ 18 | It follows that 19 | \begin{align*} 20 | uv 21 | &= f\left(\frac{1}{p} \ln(u^p) + \frac{1}{q} \ln(v^q)\right) \\ 22 | &\leq \frac{1}{p} f\big(\ln(u^p)\big) + \frac{1}{q} f\big(\ln(v^q)\big) \\ 23 | &= \frac{u^p}{p} + \frac{v^q}{q}. 24 | \end{align*} 25 | The equality holds if and only if $\ln(u^p) = \ln(v^q)$. 26 | Since $\ln$ is one-to-one, we conclude the equality holds if and only if $u^p = v^q$. 27 | \end{proof} 28 | 29 | % === Part(b) === 30 | \item 31 | \begin{proof} 32 | Since $f,g\in\mathfrak{R}(\alpha)$, then $fg\in\mathfrak{R}(\alpha)$ by Theorem 6.13(a). 33 | Since $f,g\geq 0$, from part(a), we obtain 34 | $$ 35 | fg \leq \frac{f^p}{p}+\frac{g^q}{q} 36 | $$ 37 | holds on $[a,b]$. 38 | This implies that 39 | \begin{align*} 40 | \int_{a}^{b} fg\, d\alpha 41 | &\leq \int_{a}^{b} \left( \frac{f^p}{p}+\frac{g^q}{q} \right)\, d\alpha \\ 42 | &= \frac{1}{p} \int_{a}^{b}f^p\, d\alpha + \frac{1}{q} \int_{a}^{b}g^q\, d\alpha. \\ 43 | &= \frac{1}{p} + \frac{1}{q} \\ 44 | &= 1. 45 | \end{align*} 46 | \end{proof} 47 | 48 | % === Part(c) === 49 | \item 50 | \begin{proof} 51 | Let 52 | \begin{align*} 53 | A &:= \int_{a}^{b}|f|^p\, d\alpha \geq 0; \\ 54 | B &:= \int_{a}^{b}|g|^q\, d\alpha \geq 0. 55 | \end{align*} 56 | We define 57 | \begin{align*} 58 | F(x) := \frac{f(x)}{A^{\frac{1}{p}}}; \\ 59 | G(x) := \frac{g(x)}{B^{\frac{1}{q}}}. 60 | \end{align*} 61 | We have 62 | \begin{align*} 63 | \int_{a}^{b} |F|^p\, d\alpha 64 | &= \frac{1}{A} \int_{a}^{b}|f|^p\, d\alpha 65 | = \frac{A}{A} = 1; \\ 66 | \int_{a}^{b} |G|^q\, d\alpha 67 | &= \frac{1}{B} \int_{a}^{b}|g|^q\, d\alpha 68 | = \frac{B}{B} = 1. 69 | \end{align*} 70 | Since $f,g\in\mathfrak{R}(\alpha)$, then $F,G\in\mathfrak{R}(\alpha)$; furthermore, $|F|, |G|\in\mathfrak{R}(\alpha)$. 71 | From part(b), we obtain 72 | $$ 73 | \int_{a}^{b} |FG|\, d\alpha \leq 1. 74 | $$ 75 | This implies 76 | $$ 77 | \frac{1}{A^{\frac{1}{p}}B^{\frac{1}{q}}} \int_{a}^{b} |fg|\, d\alpha \leq 1. 78 | $$ 79 | It follows that 80 | $$ 81 | \int_{a}^{b} |fg|\, d\alpha 82 | \leq A^{\frac{1}{p}} B^{\frac{1}{q}} 83 | = \left\{\int_{a}^{b}|f|^p\, d\alpha \right\}^{\frac{1}{p}} \left\{\int_{a}^{b}|g|^q\, d\alpha \right\}^{\frac{1}{q}}. 84 | $$ 85 | Notice that by Theorem 6.13(b), we have 86 | $$ 87 | \left| \int_{a}^{b} fg\, d\alpha \right| \leq \int_{a}^{b} |fg|\, d\alpha. 88 | $$ 89 | Hence we conclude 90 | $$ 91 | \left| \int_{a}^{b} fg\, d\alpha \right| 92 | \leq \left\{\int_{a}^{b}|f|^p\, d\alpha \right\}^{\frac{1}{p}} \left\{\int_{a}^{b}|g|^q\, d\alpha \right\}^{\frac{1}{q}}. 93 | $$ 94 | \end{proof} 95 | \end{enumerate} 96 | \end{Exercise} -------------------------------------------------------------------------------- /chapter6/exercises/ex_6.11.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 6.11 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Notice that by Exercise 6.10(c), we have 5 | \begin{align*} 6 | \int_{a}^{b}|f-g| |g-h|\,d\alpha 7 | &= \left| \int_{a}^{b}|f-g| |g-h|\,d\alpha \right| \\ 8 | &\leq \left\{ \int_{a}^{b} |f-g|^2\,d\alpha \right\}^{\frac{1}{2}} \left\{ \int_{a}^{b} |g-h|^2\,d\alpha \right\}^{\frac{1}{2}} \\ 9 | &= \| f-g \|_2~\|g-h\|_2. 10 | \end{align*} 11 | So, 12 | \begin{align*} 13 | \|f-h\|^2_2 14 | &= \int_{a}^{b} |f-h|^2\,d\alpha \\ 15 | &= \int_{a}^{b} |(f-g)+(g-h)|^2\,d\alpha \\ 16 | &= \int_{a}^{b} |(f-g)|^2\,d\alpha + 2\int_{a}^{b} |(f-g)| |(g-h)|\,d\alpha+ \int_{a}^{b}|(g-h)|^2\,d\alpha \\ 17 | &\leq \|f-g\|^2_2 + 2~\|f-g\|_2~\|g-h\|_2 + \|g-h\|^2_2 \\ 18 | &= (\|f-g\|_2 + \|g-h\|_2)^2. 19 | \end{align*} 20 | It follows that 21 | $$ 22 | \|f-h\|_2 \leq \|f-g\|_2 + \|g-h\|_2. 23 | $$ 24 | \end{proof} 25 | \end{Exercise} -------------------------------------------------------------------------------- /chapter6/exercises/ex_6.15.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 6.15 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{Prove that $\int_{a}^{b}x f(x) f'(x)\,x = -\frac{1}{2}$.} 5 | \end{itemize} 6 | \begin{proof} 7 | Integrating by part, we obtain 8 | \begin{align*} 9 | \int_{a}^{b} x f(x) f'(x)\,dx 10 | &= \left. \frac{1}{2}x f^2(x) \right|_{a}^{b} - \frac{1}{2} \int_{a}^{b} f^2(x)\,dx \\ 11 | &=\frac{1}{2}\left( b f^2(b) - a f^2(a) \right) - \frac{1}{2}\cdot 1 \\ 12 | &= -\frac{1}{2}. 13 | \end{align*} 14 | \end{proof} 15 | 16 | \begin{itemize} 17 | \item \textbf{Prove that $\int_{a}^{b}[f'(x)]^2\,dx\cdot \int_{a}^{b}x^2 f^2(x)\,dx > \frac{1}{4}$.} 18 | \end{itemize} 19 | 20 | \begin{proof} 21 | By Schwarz inequality, we know 22 | $$ 23 | \left\{ \int_{a}^{b} [f'(x)]^2\,dx \right\}^{\frac{1}{2}} \left\{ \int_{a}^{b} x^2 f^2(x)\,dx \right\}^{\frac{1}{2}} 24 | \geq \left| \int_{a}^{b} x f(x) f'(x)\,dx \right| 25 | = \frac{1}{2}. 26 | $$ 27 | Taking the square on both sides, we conclude 28 | $$ 29 | \int_{a}^{b} [f'(x)]^2\,dx \cdot \int_{a}^{b} x^2 f^2(x)\,dx > \frac{1}{4}. 30 | $$ 31 | \end{proof} 32 | \end{Exercise} -------------------------------------------------------------------------------- /chapter6/exercises/ex_6.2.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 6.2 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Given $\epsilon>0$, we suppose to contrary that $f(x_0) = \epsilon > 0$ for some $x_0\in[a,b]$. 5 | Since $f$ is continuous on $[a,b]$, then $f$ is uniformly continuous on $[a,b]$. There exists $\delta > 0$ such that 6 | $$ 7 | |x-x_0| < \delta \implies |f(x) - f(x_0)| < \frac{\epsilon}{2}. 8 | $$ 9 | Let $P$ be a partition $\{a, x_0-\delta, x_0+\delta, b\}$ of $[a,b]$. Provided $|x - x_0| < \delta$, we have 10 | $$ 11 | |f(x)-f(x_0)| = |\epsilon - f(x) | < \frac{\epsilon}{2}. 12 | $$ 13 | This means $f(x) > \frac{\epsilon}{2}$. 14 | It follows that 15 | $$ 16 | L(P,f) 17 | = \sum_i m_i \Delta x_i 18 | \geq \inf_{x\in(x_0-\delta, x_0+\delta)} f(x) [(x_0+\delta) - (x_0-\delta)] 19 | > \frac{\epsilon}{2}(2\delta) 20 | = \epsilon \delta 21 | > 0. 22 | $$ 23 | Hence $\sup L(P,f) > 0$. 24 | Notice that $f$ is continuous on $[a,b]$, then $f$ is integrable on $[a,b]$. So 25 | $$ 26 | \int_{a}^{b} f(x)\,d x 27 | = \lowint_{a}^{b} f(x)\,d x 28 | = \sup L(P,f) 29 | > 0. 30 | $$ 31 | This contradicts $\int_{a}^{b} f(x)\,dx = 0$. We conclude $ 32 | f(x) = 0$ for all $x\in[a,b]$. 33 | \end{proof} 34 | \end{Exercise} -------------------------------------------------------------------------------- /chapter6/exercises/ex_6.3.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 6.3 === 2 | \begin{Exercise} 3 | \begin{enumerate}[(a)] 4 | % === Part(a) === 5 | \item 6 | \begin{itemize} 7 | \item \textbf{Prove that $f\in \mathfrak{R}(\beta_1)$ if and only if $f(0+) = f(0)$.} 8 | \end{itemize} 9 | \begin{proof} 10 | $(\Longrightarrow)$ 11 | If $f\in \mathfrak{R}(\beta_1)$, by Theorem 6.6, we know given $\epsilon>0$, there exists a partition $P = \{x_0, x_1, \cdots, x_n\}$ where $x_k = 0$ for $0 < k < n$ such that 12 | $$ 13 | U(P,f,\beta_1) - L(P,f,\beta_1) < \epsilon. 14 | $$ 15 | Let $x_{k-1} = a$ and $x_{k+1} = b$. 16 | We also let 17 | $$ 18 | m_a = \inf_{x\in[a, 0]} f(x); 19 | M_a = \sup_{x\in[a, 0]} f(x); 20 | m_b = \inf_{x\in[0, b]} f(x); 21 | M_b = \sup_{x\in[0, b]} f(x). 22 | $$ 23 | By definition of $\beta_1$, we know 24 | \begin{align*} 25 | U(P,f,\beta_1) 26 | &= \sum_{i=1}^{n}M_i \big[ \beta_1(x_i) - \beta_1(x_{i-1}) \big] 27 | = M_a \big[ \beta_1(0) - \beta_1(a) \big] + M_b \big[ \beta_1(b) - \beta_1(0) \big] 28 | = M_b; \\ 29 | L(P,f,\beta_1) 30 | &= \sum_{i=1}^{n}m_i \big[ \beta_1(x_i) - \beta_1(x_{i-1}) \big] 31 | = m_a \big[ \beta_1(0) - \beta_1(a) \big] + m_b \big[ \beta_1(b) - \beta_1(0) \big] 32 | = m_b. 33 | \end{align*} 34 | This means 35 | $$ 36 | U(P,f,\beta_1) - L(P,f,\beta_1) 37 | = M_b - m_b 38 | < \epsilon. 39 | $$ 40 | Pick $\delta = x_{k+1}$, then for $x\in[0, \delta)$, we have 41 | $$ 42 | m_b \leq f(0), f(x) \leq M_b. 43 | $$ 44 | It follows that 45 | $$ 46 | |f(x) - f(0)| \leq M_b - m_b < \epsilon. 47 | $$ 48 | That is, 49 | $$ 50 | f(0+) := \lim_{x\to 0+} f(x) = f(0). 51 | $$ 52 | 53 | \vspace{2ex} 54 | 55 | $(\Longleftarrow)$ 56 | If $f(0+) = f(0)$, then given $\epsilon > 0$, there exists $\delta > 0$ such that 57 | $$ 58 | 0 \leq x < \delta \implies |f(x) - f(0)| < \epsilon. 59 | $$ 60 | We pick a partition $P = \{x_0, x_1, \cdots, x_n\}$ where $x_k = 0$ for $0 < k < n$ with $x_{k+1} < \delta$. 61 | Define $m_b$, $M_b$ be the same as previous mentions. 62 | We know for $x\in [0, \delta)$, 63 | $$ 64 | U(P,f,\beta_1) - L(P,f,\beta_1) 65 | = M_b - m_b 66 | \leq |f(x) - f(0)| 67 | < \epsilon. 68 | $$ 69 | Since $\epsilon$ was arbitrary, we obtain $f\in\mathfrak{R}(\beta_1)$ by Theorem 6.6. 70 | \end{proof} 71 | 72 | \begin{itemize} 73 | \item \textbf{$\int f\,d\beta_1 = f(0)$ under the hypothesis.} 74 | \end{itemize} 75 | \begin{proof} 76 | By definition, we know 77 | $$ 78 | L(P,f,\beta_1) \leq \int f\,d\beta_1 \leq U(P,f,\beta_1). 79 | $$ 80 | We follow the notations from previous arguments. 81 | Thus we have 82 | \begin{align*} 83 | U(P,f,\beta_1) &= M_b \geq f(0); \\ 84 | L(P,f,\beta_1) &= m_b \leq f(0). 85 | \end{align*} 86 | Given $\epsilon > 0$, it follows that 87 | \begin{align*} 88 | f(0)-\int f\,d\beta_1 \leq U(P,f,\beta_1) - L(P,f,\beta_1) < \epsilon; \\ 89 | \int f\,d\beta_1-f(0) \leq U(P,f,\beta_1) - L(P,f,\beta_1) < \epsilon. 90 | \end{align*} 91 | Hence 92 | $$ 93 | \left| \int f\,d\beta_1-f(0) \right| < \epsilon. 94 | $$ 95 | Since $\epsilon$ was arbitrary, we conclude 96 | $$ 97 | \int f\,d\beta_1 = f(0) 98 | $$ 99 | as desired. 100 | \end{proof} 101 | 102 | % === Part(b) === 103 | \item 104 | \begin{itemize} 105 | \item \textbf{Prove that $f\in \mathfrak{R}(\beta_2)$ if and only if $f(0-) = f(0)$.} 106 | \end{itemize} 107 | \begin{proof} 108 | This is similar to part(a). Here we just state the difference between this and part(a). 109 | We have 110 | \begin{align*} 111 | U(P,f,\beta_2) 112 | &= \sum_{i=1}^{n}M_i \big[ \beta_2(x_i) - \beta_2(x_{i-1}) \big] 113 | = M_a \big[ \beta_2(0) - \beta_2(a) \big] + M_b \big[ \beta_2(b) - \beta_2(0) \big] 114 | = M_a; \\ 115 | L(P,f,\beta_2) 116 | &= \sum_{i=1}^{n}m_i \big[ \beta_2(x_i) - \beta_2(x_{i-1}) \big] 117 | = m_a \big[ \beta_2(0) - \beta_2(a) \big] + m_b \big[ \beta_2(b) - \beta_2(0) \big] 118 | = m_a. 119 | \end{align*} 120 | And then replace $m_b$, $M_b$ by $m_a, M_a$, respectively. 121 | Besides, we pick $\delta = x_{k-1}$ and consider $x\in (\delta, 0]$. 122 | \end{proof} 123 | 124 | \begin{itemize} 125 | \item \textbf{$\int f\,d\beta_2 = f(0)$ under the hypothesis.} 126 | \end{itemize} 127 | \begin{proof} 128 | By definition, we know 129 | $$ 130 | L(P,f,\beta_2) \leq \int f\,d\beta_2 \leq U(P,f,\beta_2). 131 | $$ 132 | We follow the notations from previous arguments. 133 | Thus we have 134 | \begin{align*} 135 | U(P,f,\beta_2) &= M_a \geq f(0); \\ 136 | L(P,f,\beta_2) &= m_a \leq f(0). 137 | \end{align*} 138 | Given $\epsilon > 0$, it follows that 139 | \begin{align*} 140 | f(0)-\int f\,d\beta_2 \leq U(P,f,\beta_2) - L(P,f,\beta_2) < \epsilon; \\ 141 | \int f\,d\beta_2-f(0) \leq U(P,f,\beta_2) - L(P,f,\beta_2) < \epsilon. 142 | \end{align*} 143 | Hence 144 | $$ 145 | \left| \int f\,d\beta_2-f(0) \right| < \epsilon. 146 | $$ 147 | Since $\epsilon$ was arbitrary, we conclude 148 | $$ 149 | \int f\,d\beta_2 = f(0) 150 | $$ 151 | as desired. 152 | \end{proof} 153 | 154 | % === Part(c) === 155 | \item 156 | \begin{proof} 157 | We follow the notations from previous arguments. 158 | 159 | Consider 160 | \begin{align*} 161 | U(P,f,\beta_3) 162 | &= M_a \big[ \beta_3(0) - \beta_3(a) \big] + M_b \big[ \beta_3(b) - \beta_3(0) \big] 163 | = \frac{1}{2}(M_a + M_b); \\ 164 | L(P,f,\beta_3) 165 | &= m_a \big[ \beta_3(0) - \beta_3(a) \big] + m_b \big[ \beta_3(b) - \beta_3(0) \big] 166 | = \frac{1}{2}(m_a + m_b). 167 | \end{align*} 168 | Given $\epsilon > 0$, as from part(a), we know 169 | \begin{equation} 170 | M_a - m_a < \epsilon \iff f(0+) = f(0);\label{eq:ex_6.3c1} 171 | \end{equation} 172 | as from part(b), we know 173 | \begin{equation} 174 | M_b - m_b < \epsilon \iff f(0-) = f(0).\label{eq:ex_6.3c2} 175 | \end{equation} 176 | Hence 177 | $$ 178 | U(P,f,\beta_3) - L(P,f,\beta_3) < \epsilon \iff f(0+) = f(0) = f(0-). 179 | $$ 180 | Since $\epsilon$ was arbitrary, we obtain $f\in\mathfrak{R}$ by Theorem 6.6. 181 | Notice that $f(0+) = f(0-) = f(0)$ if and only if $f$ is continuous at $0$. 182 | We conclude $f\in\mathfrak{R}(\beta_3)$ if and only if $f$ is continuous at $0$. 183 | \end{proof} 184 | 185 | % === Part(d) === 186 | \item 187 | \begin{proof} 188 | Since $f$ is continuous at $0$, then $f(0+) = f(0) = f(0-)$. 189 | We have 190 | $$ 191 | \int f\,d\beta_1 = f(0) 192 | $$ 193 | from part(a); also 194 | $$ 195 | \int f\,d\beta_2 = f(0) 196 | $$ 197 | from part(b). 198 | Notice that \eqref{eq:ex_6.3c1} and \eqref{eq:ex_6.3c2} hold, and 199 | \begin{align*} 200 | f(0) 201 | &\leq U(P,f,\beta_3) 202 | = \frac{1}{2}(M_a + M_b) 203 | \leq \int f\,d\beta_3; \\ 204 | f(0) 205 | &\geq L(P,f,\beta_3) 206 | = \frac{1}{2}(m_a + m_b) 207 | \geq \int f\,d\beta_3. 208 | \end{align*} 209 | Hence 210 | $$ 211 | \left| \int f\,d\beta_3 - f(0) \right| < \epsilon. 212 | $$ 213 | Since $\epsilon$ was arbitrary, 214 | $$ 215 | \int f\,d\beta_3 = f(0) 216 | $$ 217 | follows. 218 | \end{proof} 219 | \end{enumerate} 220 | \end{Exercise} -------------------------------------------------------------------------------- /chapter6/exercises/ex_6.4.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 6.4 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Consider any partition of $P = \{x_0, x_1, \cdots, x_n\}$ of $[a,b]$. 5 | Since rational and irrational numbers are dense in $\mathbb{R}$, we know there exists a rational number and an irrational one in the interval $[x_{i-1}, x_i]$ for each $i = 1,2,\cdots, n$. 6 | This means 7 | $$ 8 | M_i = 1; m_i = 0 9 | $$ 10 | for each $i$. 11 | It follows that 12 | \begin{align*} 13 | U(P,f) &= \sum_{i=1}^{n} M_i\Delta x_i = b-a; \\ 14 | L(P,f) &= \sum_{i=1}^{n} m_i\Delta x_i = 0. 15 | \end{align*} 16 | Given $\epsilon$ with $0 < \epsilon \leq b-a$, we observe 17 | $$ 18 | U(P,f) - L(P,f) = b-a \geq \epsilon > 0. 19 | $$ 20 | By Theorem 6.6, we conclude $f\notin \mathfrak{R}$ on $[a,b]$. 21 | Since $a$, $b$ were arbitrary with $a0$. 8 | Consider 9 | $$ 10 | \int_{0}^{1} f(x)\,dx = \int_{0}^{c}f(x)\, dx + \int_{c}^{1}f(x)\,dx. 11 | $$ 12 | This implies that 13 | $$ 14 | \left| \int_{0}^{1} f(x)\,dx - \int_{c}^{1}f(x)\,dx \right| 15 | = \left| \int_{0}^{c}f(x)\, dx \right| 16 | \leq \int_{0}^{c} |f(x)|\, dx 17 | \leq cM. 18 | $$ 19 | Since $cM\to 0$ as $c \to 0$, we conclude 20 | $$ 21 | \int_{0}^{1} f(x)\, dx = \lim_{c\to 0}\int_{c}^{1}f(x)\, dx. 22 | $$ 23 | \end{proof} 24 | 25 | % ============================ 26 | % !!! Is f integrable on [c,1]? 27 | % ============================ 28 | % === Part(b) === 29 | \item 30 | \begin{answer} 31 | \begin{equation} 32 | f(x) = n(-1)^{n+1} \quad\mbox{ for } \frac{1}{n+1} < x \leq \frac{1}{n}. \label{eq:ex_6.7b} 33 | \end{equation} 34 | \end{answer} 35 | \begin{proof} 36 | Consider \eqref{eq:ex_6.7b}. 37 | Since $c\in(0,1]$, there exists an index $N$ such that 38 | $$ 39 | \frac{1}{N+1} < c \leq \frac{1}{N}. 40 | $$ 41 | Since 42 | $$ 43 | 0 44 | \leq \frac{1}{N} - c 45 | \leq \frac{1}{N} - \frac{1}{N+1} 46 | = \frac{1}{N(N+1)}, 47 | $$ 48 | we observe 49 | \begin{align*} 50 | \int_{c}^{1} f(x)\, dx 51 | &= N(-1)^{N+1}\left( \frac{1}{N} - c \right) + \sum_{i=1}^{N-1} \frac{(-1)^{i+1}}{i} \\ 52 | &\leq \sum_{i=1}^{N} \frac{(-1)^{i+1}}{i}. 53 | \end{align*} 54 | This is the famous alternating harmonic series, we know 55 | $$ 56 | \lim_{N\to\infty} \int_{c}^{1} f(x)\, dx 57 | \leq \lim_{N\to\infty} \sum_{i=1}^{N} \frac{(-1)^{i+1}}{i} 58 | = \ln 2 59 | $$ 60 | Notice that $c\to 0$ as $N\to\infty$ by the Squeeze Theorem. 61 | Hence we have 62 | $$ 63 | \lim_{c\to 0} \int_{c}^{1} f(x)\, dx = \ln 2. 64 | $$ 65 | 66 | On the other hand, we observe 67 | 68 | % ??? This is incredible. 69 | \begin{align*} 70 | \int_{c}^{1} |f(x)|\, dx 71 | &= N \left( \frac{1}{N} - c \right) + \sum_{i=1}^{N-1} \frac{1}{i} \to \infty \text{ as } N\to\infty 72 | \end{align*} 73 | By the Comparison Test, we know 74 | $$ 75 | \lim_{c\to 0} \int_{c}^{1} |f(x)|\, dx 76 | $$ 77 | diverges. 78 | \end{proof} 79 | \end{enumerate} 80 | \end{Exercise} -------------------------------------------------------------------------------- /chapter6/exercises/ex_6.8.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 6.8 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Assume $c>1$. 5 | Let $n$ be the greatest integer such that $c \leq n < c+1$. 6 | Let $P_n$ be a partition $\{x_0=1, 2, \cdots, n = x_{n-1}\}$ of $[1,n]$. 7 | We observe 8 | \begin{align*} 9 | \int_{1}^{c} f(x)\,dx 10 | &\geq \int_{1}^{n} f(x)\,dx & (\because f(x)\geq 0) \\ 11 | &\geq L(P_n, f) \\ 12 | &= \sum_{i=1}^{n-1}m_i \Delta x_i \\ 13 | &= \sum_{i=1}^{n-1} f(i+1). & (\because f \text{ decrease monotonically}) 14 | \end{align*} 15 | Taking the limit on both sides as $c\to\infty$, provided $n\to\infty$ simultaneously, we obtain 16 | $$ 17 | \lim_{c\to\infty} \int_{1}^{c} f(x)\,dx 18 | \geq \lim_{n\to\infty} \sum_{i=1}^{n-1} f(i+1). 19 | $$ 20 | By the Comparison Test, if $\int_{1}^{\infty} f(x)\,dx$ converges, then $\sum_{i=1}^{\infty} f(i)$ follows. 21 | 22 | Conversely, let $P_{n+1}$ be a partition $\{x_0=1, 2, \cdots, n+1 = x_{n}\}$ of $[1,n+1]$. 23 | We observe 24 | \begin{align*} 25 | \int_{1}^{c} f(x)\,dx 26 | &\leq \int_{1}^{n+1} f(x)\,dx & (\because f(x)\geq 0) \\ 27 | &\leq U(P_{n+1}, f) \\ 28 | &= \sum_{i=1}^{n}M_i \Delta x_i \\ 29 | &= \sum_{i=1}^{n} f(i). & (\because f \text{ decrease monotonically}) 30 | \end{align*} 31 | Taking the limit on both sides as $c\to\infty$, provided $n\to\infty$ simultaneously, we obtain 32 | $$ 33 | \lim_{c\to\infty} \int_{1}^{c} f(x)\,dx 34 | \leq \lim_{n\to\infty} \sum_{i=1}^{n} f(i). 35 | $$ 36 | By the Comparison Test, if $\sum_{i=1}^{\infty} f(i)$ converges, then $\int_{1}^{\infty} f(x)\,dx$ follows. 37 | \end{proof} 38 | \end{Exercise} -------------------------------------------------------------------------------- /chapter7/chapter7.tex: -------------------------------------------------------------------------------- 1 | % === Chapter 7 === 2 | \setcounter{chapter}{6} 3 | \chapter{Sequence and Series of Functions} 4 | 5 | \input{exercises/ex_7.1} 6 | 7 | \vspace{12pt} 8 | 9 | \input{exercises/ex_7.2} 10 | 11 | \vspace{12pt} 12 | 13 | \setcounter{Exercise}{3} 14 | \input{exercises/ex_7.4} 15 | 16 | \vspace{12pt} 17 | 18 | \input{exercises/ex_7.5} 19 | 20 | \vspace{12pt} 21 | 22 | \input{exercises/ex_7.6} 23 | 24 | \vspace{12pt} 25 | 26 | \input{exercises/ex_7.7} 27 | 28 | \vspace{12pt} 29 | 30 | \setcounter{Exercise}{8} 31 | \input{exercises/ex_7.9} 32 | 33 | \vspace{12pt} 34 | 35 | \setcounter{Exercise}{15} 36 | \input{exercises/ex_7.16} 37 | 38 | \vspace{12pt} 39 | 40 | \setcounter{Exercise}{19} 41 | \input{exercises/ex_7.20} 42 | -------------------------------------------------------------------------------- /chapter7/exercises/ex_7.1.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 7.1 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Let $f_n\to f$ uniformly on a set $E$ and $f_n$ is bounded for each $n$. 5 | That is, for each $n$, there exists $M_n > 0$ such that 6 | $$ 7 | |f_n(x)| \leq M_n 8 | $$ 9 | for all $x\in E$; besides, given $\epsilon = 1$, there exists $N>0$ such that 10 | $$ 11 | m,n \geq N \implies |f_n(x) - f_m(x)| < 1 12 | $$ 13 | for all $x\in E$. 14 | Hence for $n\geq N$, we have 15 | $$ 16 | |f_n(x)| 17 | \leq |f_n(x) - f_N(x)| + |f_N(x)| + |f_N(x)| 18 | \leq 1 + M_N. 19 | $$ 20 | Pick $M' := \max\{M_1,M_2,\cdots,M_{N-1}\}$, then we know 21 | $$ 22 | |f_n(x)| \leq M_n \leq M' \mbox{ , } n=1,2,\cdots,N-1 23 | $$ 24 | for all $x\in E$. 25 | Finally, we pick $M := \max\{1 + M_N, M'\}$ to get 26 | $$ 27 | |f_n(x)| \leq M \mbox{ , } \forall x\in E,\ n=1,2,\cdots. 28 | $$ 29 | Hence we conclude $\{f_n\}$ is uniformly bounded. 30 | \end{proof} 31 | \end{Exercise} -------------------------------------------------------------------------------- /chapter7/exercises/ex_7.16.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 7.16 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Let $\epsilon > 0$ be given. 5 | Since $\{f_n\}$ is an equicontinuous sequence of functions on a compact set $K$, then for all $n$, there exists $\delta>0$ such that 6 | $$ 7 | d(x,y)<\delta,\ x\in K,\ y\in K \implies |f_n(x) - f_n(y)| < \epsilon. 8 | $$ 9 | Let $V(x,\delta)$ be the set of all $y\in K$ with $d(x,y)<\delta$. 10 | Since $K$ is compact, then there are finitely many points $x_1,x_2,\cdots,x_m\in K$ such that 11 | $$ 12 | K\subset V(x_1,\delta)\cup V(x_2,\delta)\cup \cdots \cup V(x_m,\delta). 13 | $$ 14 | Since $\{f_n\}$ converges pointwise on $K$, then there exists $N>0$ such that 15 | $$ 16 | n\geq N \implies |f_n(x_i) - f(x_i)|<\frac{\epsilon}{4} 17 | $$ 18 | for $i=1,2,\cdots,m$. 19 | If $x\in K$, we have $x\in V(x_i,\delta)$ for some $i$. 20 | It follows that 21 | $$ 22 | |f_n(x)-f_n(x_i)|<\frac{\epsilon}{4} 23 | $$ 24 | for all $n$. 25 | For $j\geq N$, $k\geq N$, we consider 26 | \begin{align*} 27 | |f_j(x)-f_k(x)| 28 | &\leq |f_j(x)-f_j(x_i)| + |f_j(x_i) - f(x_i)| + |f(x_i) - f_k(x_i)| + |f_k(x_i) - f_k(x)| \\ 29 | &< \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{4} \\ 30 | &= \epsilon. 31 | \end{align*} 32 | By Theorem 7.8, we conclude $\{f_n\}$ converges uniformly on $K$. 33 | \end{proof} 34 | \end{Exercise} -------------------------------------------------------------------------------- /chapter7/exercises/ex_7.2.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 7.2 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{Prove $\{f_n+g_n\}$ converges uniformly on $E$.} 5 | \end{itemize} 6 | \begin{proof} 7 | Since $\{f_n\}$, $\{g_n\}$ converge uniformly on a set $E$, we know given $\epsilon>0$, there exists $N_1>0$ such that 8 | $$ 9 | n,m\geq N_1,\ x\in E \implies |f_n(x) - f_m(x)| < \frac{\epsilon}{2}. 10 | $$ 11 | And there exists $N_2>0$ such that 12 | $$ 13 | n,m\geq N_2,\ x\in E \implies |g_n(x) - g_m(x)| < \frac{\epsilon}{2}. 14 | $$ 15 | Then we pick $N = \max\{N_1,N_2\}$. For $n,m\geq N$ and $x\in E$, we have 16 | \begin{align*} 17 | \Big| \big( f_n(x) + g_n(x) \big) - \big( f_m(x) + g_m(x) \big) \Big| 18 | &\leq | f_n(x) - f_m(x) | + | g_n(x) - g_m(x) | \\ 19 | &< \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ 20 | &= \epsilon. 21 | \end{align*} 22 | This means $\{f_n + g_n\}$ converges uniformly on $E$. 23 | \end{proof} 24 | 25 | \begin{itemize} 26 | \item \textbf{Prove $\{f_n g_n\}$ converges uniformly on $E$.} 27 | \end{itemize} 28 | \begin{proof} 29 | Since $\{f_n\}$, $\{g_n\}$ converge uniformly on $E$ and bounded, by Exercise 7.1, we know $\{f_n\}$, $\{g_n\}$ are uniformly bounded. 30 | Hence there exists $M_f > 0$, $M_g > 0$ such that $|f_n(x)| \leq M_f$, $|g_n(x)| \leq M_g$ for each $n$ and for all $x\in E$. 31 | Since $\{f_n\}$ converges uniformly on $E$, then there exists $N_1 > 0$ such that 32 | $$ 33 | n\geq N_1,\ x\in E \implies |f_n(x) - f(x)| < 1. 34 | $$ 35 | Then 36 | $$ 37 | |f(x)| 38 | \leq |f(x) - f_{N_1}(x)| + |f_{N_1}(x)| 39 | < 1 + M_f. 40 | $$ 41 | Moreover, given $\epsilon>0$, there exists $N_2 > 0$ such that 42 | $$ 43 | n\geq N_2,\ x\in E \implies |f_n(x) - f(x)| < \frac{\epsilon}{2M_g}. 44 | $$ 45 | Since $\{g_n\}$ converges uniformly on $E$, then there exists $N_3 > 0$ such that 46 | $$ 47 | n\geq N_3,\ x\in E \implies |g_n(x) - g(x)| < \frac{\epsilon}{2(1+M_f)} 48 | $$ 49 | Pick $N := \max\{N_1,N_2,N_3\}$. 50 | For $n\geq N$ and $x\in E$, we consider 51 | \begin{align*} 52 | |f_n(x) g_n(x) - f(x) g(x)| 53 | &\leq |f_n(x) g_n(x) - f(x) g_n(x)| + |f(x) g_n(x) - f(x) g(x)| \\ 54 | &= |g_n(x)| \cdot |f_n(x) - f(x)| + |f(x)| \cdot |g_n(x) - g(x)| \\ 55 | &< M_g \cdot \frac{\epsilon}{2M_g} + (1+M_f) \cdot \frac{\epsilon}{2(1+M_f)} \\ 56 | &= \epsilon 57 | \end{align*} 58 | Hence $\{f_n g_n\}$ converges uniformly on $E$. 59 | \end{proof} 60 | \end{Exercise} -------------------------------------------------------------------------------- /chapter7/exercises/ex_7.20.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 7.20 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Let $P(x)$ be an arbitrary polynomial such that 5 | $$ 6 | P(x) = a_n x^n + \cdots + a_1 x + a_0. 7 | $$ 8 | We observe 9 | $$ 10 | \int_{0}^{1} f(x) P(x)\,dx 11 | = a_n \int_{0}^{1} f(x)x^n\,dx + \cdots + a_1 \int_{0}^{1} f(x)x\,dx + a_0 \int_{0}^{1} f(x)\,dx 12 | = 0. 13 | $$ 14 | By the Stone-Weierstrass Theorem, we can pick a sequence of polynomials $P_n(x)$ such that 15 | $$ 16 | \lim_{n\to\infty} P_n(x) = f(x) 17 | $$ 18 | uniformly on $[0,1]$. 19 | By Theorem 7.16, we know 20 | $$ 21 | 0 22 | = \lim_{n\to\infty}\int_{0}^{1}f(x) P_n(x)\,dx 23 | % ↓ Here is maybe incredible. 24 | = \int_{0}^{1} \lim_{n\to\infty} f(x) P_n(x)\,dx 25 | = \int_{0}^{1} f(x) \lim_{n\to\infty} P_n(x)\,dx 26 | = \int_{0}^{1} f^2(x)\,dx 27 | $$ 28 | Since $f$ is continuous on $[0,1]$, and so is $f^2$. 29 | Moreover, $f^2(x) \geq 0$ for all $x\in[0,1]$. 30 | By Exercise 6.2, we know $f^2(x) = 0$ on $[0,1]$. 31 | It follows that $f(x) = 0$ on $[0,1]$ immediately. 32 | \end{proof} 33 | \end{Exercise} -------------------------------------------------------------------------------- /chapter7/exercises/ex_7.4.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 7.4 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{For what values of $x$ does the series converges absolutely?} 5 | \end{itemize} 6 | \begin{answer} 7 | For all $x\in \mathbb{R}$ except $x=0$ or $x=-\frac{1}{n^2}$ where $n\in\mathbb{N}$. 8 | \end{answer} 9 | \begin{solution} 10 | We split it into four cases to discuss. 11 | \begin{enumerate}[i)] 12 | \item For $x=0$. 13 | 14 | Consider 15 | $$ 16 | \sum_{n=1}^{\infty} \left| \frac{1}{1 + n^2 \cdot 0} \right| 17 | = \sum_{n=1}^{\infty} \left| 1 \right| = \infty. 18 | $$ 19 | This means the series diverges. 20 | 21 | \item For $x>0$. 22 | 23 | Consider 24 | $$ 25 | \sum_{n=1}^{\infty} \left| \frac{1}{1 + n^2 x} \right| 26 | = \left| \frac{1}{x} \right| \sum_{n=1}^{\infty} \left| \frac{1}{\frac{1}{x} + n^2} \right| 27 | \leq \left| \frac{1}{x} \right| \sum_{n=1}^{\infty} \left| \frac{1}{n^2} \right| 28 | \leq \sum_{n=1}^{\infty} \left| \frac{1}{n^2} \right| 29 | = \sum_{n=1}^{\infty} \frac{1}{n^2} 30 | $$ 31 | By the p-Series Test and the Comparison Test, we know the series converges absolutely. 32 | 33 | \item For $x<0$ with $x = -\frac{1}{n^2}$ for some $n\in\mathbb{N}$. 34 | 35 | The $n$th term is not defined, and hence the series is undefined. 36 | 37 | \item For $x<0$ with $x \neq -\frac{1}{n^2}$ for all $n\in\mathbb{N}$. 38 | 39 | Consider 40 | $$ 41 | \sum_{n=1}^{\infty} \left| \frac{1}{1 + n^2 x} \right| 42 | = \left| \frac{1}{x} \right| \sum_{n=1}^{\infty} \left| \frac{1}{n^2} \right| \left| \frac{1}{\frac{1}{n^2 x} + 1} \right| 43 | \leq \sum_{n=1}^{\infty} \left| \frac{1}{n^2} \right| \left| \frac{1}{\frac{1}{n^2 x} + 1} \right|. 44 | $$ 45 | Notice that $\sum_{n=1}^{\infty} \left| \frac{1}{n^2} \right|$ converges. 46 | Since $\left\{\frac{1}{\frac{1}{n^2 x} + 1} \right\}_{n\in\mathbb{N}}$ is bounded by $1$ and is decreasing, by Exercise 3.8 and the Comparison Test, we know the series converges absolutely. 47 | \end{enumerate} 48 | \end{solution} 49 | 50 | \begin{itemize} 51 | \item \textbf{On what intervals does it converge uniformly? (Omitted)} 52 | \end{itemize} 53 | 54 | \begin{itemize} 55 | \item \textbf{On what intervals does it fail to converge uniformly? (Omitted)} 56 | \end{itemize} 57 | 58 | \begin{itemize} 59 | \item \textbf{Is $f$ continuous wherever the series converges? (Omitted)} 60 | \end{itemize} 61 | 62 | \begin{itemize} 63 | \item \textbf{Is $f$ bounded? (Omitted)} 64 | \end{itemize} 65 | \end{Exercise} -------------------------------------------------------------------------------- /chapter7/exercises/ex_7.5.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 7.5 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{Show that $\{f_n\}$ converges to a continuous function, but not uniformly.} 5 | \end{itemize} 6 | \begin{proof} 7 | Considering for $x\leq 0$ and $x > 1$, we observe $f_n(x) = 0$ for all $n\in\mathbb{N}$. 8 | Then $f(x) = \lim_{n\to\infty}f_n(x) = 0$ trivially. 9 | For any other value $x$, we just choose $N$ such that $N>\frac{1}{x}$ to make 10 | $$ 11 | n\geq N \implies x > \frac{1}{n}. 12 | $$ 13 | Then $f(x) = \lim_{n\to\infty}f_n(x) = 0$, too. 14 | So far, we observe $\{f_n\}$ converges to $f(x) = 0$ which is continuous. 15 | 16 | To show that $f_n\to f$ not uniformly, let $0<\epsilon<1$ be given. 17 | We consider any $n$, then there always exists a point $x = \frac{2}{2n+1}$ such that 18 | $$ 19 | f_n(x) 20 | = \sin^2 \frac{\pi}{x} 21 | = \sin^2 \left(n\pi + \frac{\pi}{2} \right) 22 | = 1. 23 | $$ 24 | It follows that 25 | $$ 26 | |f_n(x) - f(x)| = 1 \geq \epsilon. 27 | $$ 28 | Hence $f_n\to f$ not uniformly. 29 | \end{proof} 30 | 31 | \begin{itemize} 32 | \item \textbf{Show that absolute convergence does not imply uniform convergence.} 33 | \end{itemize} 34 | \begin{proof} 35 | Notice that $|f_n| = f_n$ for all $x$. 36 | Since $\{f_n\}$ converges to $f$, then $\{|f_n|\}$ also converges to $f$. 37 | Hence $\sum f_n$ converges absolutely. 38 | However, $\{f_n\}$ does not converge uniformly from the previous argument. 39 | \end{proof} 40 | \end{Exercise} -------------------------------------------------------------------------------- /chapter7/exercises/ex_7.6.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 7.6 === 2 | \begin{Exercise} 3 | \begin{itemize} 4 | \item \textbf{The series converges uniformly in every bounded interval.} 5 | \end{itemize} 6 | \begin{proof} 7 | Given a bounded interval $[a,b]$. 8 | Define 9 | $$ 10 | f_m = \sum_{n=1}^{m}(-1)^n \frac{x^2+n}{n^2}. 11 | $$ 12 | Put $M = \max\{|a|, |b|\}$. 13 | Since both $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$ and $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ converge, then given $\epsilon > 0$, there exist $N_1,N_2$ such that 14 | \begin{align*} 15 | p>q\geq N_1 &\implies \left| \sum_{n=q}^{p}\frac{(-1)^n}{n}\right| < \frac{\epsilon}{2}; \\ 16 | p>q\geq N_2 &\implies \left| \sum_{n=q}^{p}\frac{(-1)^n}{n^2}\right| < \frac{\epsilon}{2 M^2}. 17 | \end{align*} 18 | Pick $N = \max\{N_1, N_2\}$. 19 | For $p > q \geq N$, we observe 20 | \begin{align*} 21 | \left| f_p(x) - f_q(x) \right| 22 | &= \left| \sum_{n=q}^{p}\frac{(-1)^n}{n} + x^2 \sum_{n=q}^{p}\frac{(-1)^n}{n^2} \right| \\ 23 | &\leq \left| \sum_{n=q}^{p}\frac{(-1)^n}{n} \right| + x^2 \left| \sum_{n=q}^{p}\frac{(-1)^n}{n^2} \right| \\ 24 | &< \frac{\epsilon}{2} + x^2 \frac{\epsilon}{2 M^2} \\ 25 | &\leq \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ 26 | &= \epsilon. 27 | \end{align*} 28 | By Cauchy criterion, we conclude 29 | $$ 30 | \sum_{n=1}^{\infty}(-1)^n \frac{x^2+n}{n^2} 31 | $$ 32 | uniformly converges on $[a,b]$. 33 | 34 | Since $[a,b]$ was arbitrary, then the series uniformly converges in every bounded interval. 35 | \end{proof} 36 | 37 | \begin{itemize} 38 | \item \textbf{The series does not converge absolutely for any value of $x$.} 39 | \end{itemize} 40 | \begin{proof} 41 | For any value of $x$, we consider 42 | $$ 43 | \sum_{n=1}^{\infty} \left| (-1)^n \frac{x^2+n}{n^2} \right| 44 | = \sum_{n=1}^{\infty} \left| \frac{x^2+n}{n^2} \right| 45 | = \sum_{n=1}^{\infty} \left| \frac{x^2}{n^2} + \frac{1}{n} \right| 46 | \geq \sum_{n=1}^{\infty} \left| \frac{1}{n} \right|. 47 | $$ 48 | Since $\sum_{n=1}^{\infty} \left| \frac{1}{n} \right|$ diverges by the p-Series Test, then by the Comparison Test, we know 49 | $$ 50 | \sum_{n=1}^{\infty} \left| (-1)^n \frac{x^2+n}{n^2} \right| 51 | $$ 52 | diverges. 53 | \end{proof} 54 | \end{Exercise} -------------------------------------------------------------------------------- /chapter7/exercises/ex_7.7.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 7.7 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | In order to show that $\{f_n\}$ converges uniformly to a function $f$, we use Theorem 7.9. 5 | Now we need to prove 6 | $$ 7 | \sup_{x\in\mathbb{R}} |f_n(x) - f(x)| \to 0 \text{ as } n\to\infty. 8 | $$ 9 | First we try to find the global maximum of $f_n$, 10 | $$ 11 | f_n'(x) 12 | = \frac{(1+n x^2) - 2n x^2}{(1+n x^2)^2} 13 | = \frac{1-n x^2}{(1+n x^2)^2}. 14 | $$ 15 | $f_n'(x) = 0$ implies $x = \pm\frac{1}{\sqrt{n}}$. 16 | Notice that $f_n(-\frac{1}{\sqrt{n}}) < f_n(\frac{1}{\sqrt{n}})$. 17 | Then we observe 18 | $$ 19 | \sup_{x\in\mathbb{R}}|f_n(x)| 20 | = \left| f_n(\frac{1}{\sqrt{n}}) \right| 21 | = \left| \frac{1}{2\sqrt{n}} \right| 22 | = \frac{1}{2\sqrt{n}}. 23 | $$ 24 | 25 | Since $f(x) = \lim_{n\to\infty}f_n(x) = 0$ for all $x\in\mathbb{R}$, then we know 26 | $$ 27 | \sup_{x\in\mathbb{R}} |f_n(x) - f(x)| 28 | = \sup_{x\in\mathbb{R}} |f_n(x)| 29 | = \frac{1}{2\sqrt{n}} \to 0 30 | $$ 31 | as $n\to\infty$. 32 | Hence $\{f_n\}$ converges uniformly to $f$ which is a zero function. 33 | 34 | Notice that 35 | $$ 36 | \lim_{n\to\infty}f_n'(x) 37 | = \lim_{n\to\infty}\frac{1-n x^2}{1 + 2 n x^2 + n^2 x^4} 38 | = \begin{cases} 39 | 0 & \mbox{ for } x=0 \\ 40 | 1 & \mbox{ otherwise} 41 | \end{cases}. 42 | $$ 43 | Since $f(x) = 0$ for all $x\in\mathbb{R}$ and hence $f'(x) = 0$ trivially. 44 | It follows that $f'(x) = \lim_{n\to\infty}f'_n(x)$ is correct if $x\neq 0$, but false if $x = 0$. 45 | \end{proof} 46 | \end{Exercise} -------------------------------------------------------------------------------- /chapter7/exercises/ex_7.9.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 7.9 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Since $f_n\to f$ uniformly on $E$, then given $\epsilon>0$, there exists $N_1>0$ such that 5 | $$ 6 | n\geq N_1,\ y\in E \implies |f_n(y) - f(y)| < \frac{\epsilon}{2}. 7 | $$ 8 | Since $x_n\in E$, we have 9 | $$ 10 | n\geq N_1 \implies |f_n(x_n) - f(x_n)| < \frac{\epsilon}{2}. 11 | $$ 12 | Notice that $\{f_n\}$ is a sequence of continuous functions on $E$ and $f_n\to f$ uniformly, by Theorem 7.12, we know $f$ is continuous on $E$. 13 | Notice that $x_n, x\in E$. 14 | Hence there exists $N_2 > 0$ such that 15 | $$ 16 | n\geq N_2 \implies |f(x_n) - f(x)| < \frac{\epsilon}{2}. 17 | $$ 18 | Put $N := \max\{N_1, N_2\}$. 19 | Then for $n\geq N$, we have 20 | \begin{align*} 21 | |f_n(x_n) - f(x)| 22 | &\leq |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)| \\ 23 | &< \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ 24 | &= \epsilon. 25 | \end{align*} 26 | Since $\epsilon$ was arbitrary, it follows that 27 | $$ 28 | \lim_{n\to\infty} f_n(x_n) = f(x). 29 | $$ 30 | \end{proof} 31 | 32 | \begin{itemize} 33 | \item \textbf{Is the converse of the statement true?} 34 | \end{itemize} 35 | \begin{answer} 36 | No. 37 | \end{answer} 38 | \begin{proof} 39 | Consider $f_n(x) = \frac{x}{n}$. 40 | Put $x_n = 1$ for all $n\in\mathbb{N}$ and $E = [0,1]$. 41 | Then $\{x_n\}$ converges to $1$ trivially. 42 | Notice that 43 | $$ 44 | f(0) = \lim_{n\to\infty} f_n(0) = 0. 45 | $$ 46 | We observe 47 | $$ 48 | \lim_{n\to\infty}f_n(x_n) 49 | = \lim_{n\to\infty}f_n(1) 50 | = \lim_{n\to\infty} \frac{1}{n} 51 | = 0 52 | = f(0). 53 | $$ 54 | This means $\{x_n\}$ converges to $0$, which is a contradiction. 55 | \end{proof} 56 | \end{Exercise} -------------------------------------------------------------------------------- /chapter8/chapter8.tex: -------------------------------------------------------------------------------- 1 | % === Chapter 8 === 2 | \setcounter{chapter}{7} 3 | \chapter{Some Special Functions} 4 | 5 | \setcounter{Exercise}{6} 6 | \input{exercises/ex_8.7} 7 | -------------------------------------------------------------------------------- /chapter8/exercises/ex_8.7.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 8.7 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Put $f(x) = \dfrac{2x}{\pi} - \sin x$. 5 | Since $f(0) = f(\dfrac{\pi}{2}) = 0$, by Rolle's Theorem, there exists $c\in\left(0, \dfrac{\pi}{2}\right)$ such that $f'(c) = 0$. 6 | Consider $f''(x) = \sin x > 0$ on $\left(0,\dfrac{\pi}{2}\right)$. 7 | This means $f'$ is strictly increasing on $\left(0,\dfrac{\pi}{2}\right)$. 8 | So, such $c$ is the only point, then $f(x) < 0$ for $0 < x < \dfrac{\pi}{2}$. 9 | That is, 10 | \begin{equation*} 11 | \frac{2}{\pi} < \frac{\sin x}{x}.\qquad\mbox{ $(0 < x < \frac{\pi}{2}$)} 12 | \end{equation*} 13 | 14 | On the other hand, put $g(x) = \sin x - x$. 15 | Then $g'(x) = \cos x - 1 < 0$ on $\left( 0, \dfrac{\pi}{2} \right)$. 16 | Therefore $g$ is strictly decreasing on this interval. 17 | Since $g(0) = 0$, we know $g(x) < 0$ on this interval. 18 | 19 | Hence we conclude for $0 < x < \dfrac{\pi}{2}$, 20 | $$ 21 | \frac{2}{\pi} < \frac{\sin x}{x} < 1. 22 | $$ 23 | \end{proof} 24 | \end{Exercise} -------------------------------------------------------------------------------- /chapter9/chapter9.tex: -------------------------------------------------------------------------------- 1 | % === Chapter 9 === 2 | \setcounter{chapter}{8} 3 | \chapter{Functions of Several Variables} 4 | 5 | \setcounter{Exercise}{5} 6 | \input{exercises/ex_9.6} 7 | 8 | \vspace{12pt} 9 | 10 | \setcounter{Exercise}{13} 11 | \input{exercises/ex_9.14} 12 | 13 | \vspace{12pt} 14 | 15 | \input{exercises/ex_9.15} 16 | 17 | \vspace{12pt} 18 | 19 | \setcounter{Exercise}{18} 20 | \input{exercises/ex_9.19} 21 | 22 | \vspace{12pt} 23 | 24 | \setcounter{Exercise}{26} 25 | \input{exercises/ex_9.27} 26 | 27 | 28 | -------------------------------------------------------------------------------- /chapter9/exercises/ex_9.14.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 9.14 === 2 | \begin{Exercise} 3 | \begin{enumerate}[(a)] 4 | \item 5 | \begin{proof} 6 | For $(x,y) = (0,0)$, we observe 7 | \begin{align*} 8 | D_1 f(0,0) 9 | &= \lim_{h\to 0} \frac{f(h,0)-f(0,0)}{h-0} 10 | = \frac{h-0}{h} 11 | = 1; \\ 12 | D_2 f(0,0) 13 | &= \lim_{k\to 0} \frac{f(0,k)-f(0,0)}{k-0} 14 | = \frac{0-0}{k} 15 | = 0. 16 | \end{align*} 17 | Notice that $|2xy| \leq x^2+y^2$. 18 | For $(x,y) \neq (0,0)$, we have 19 | \begin{align*} 20 | |D_1 f(x,y)| 21 | &= \left| \frac{x^4+3x^2 y^2}{(x^2+y^2)^2} \right| 22 | \leq \frac{3(x^4+x^2 y^2)}{(x^2+y^2)^2} 23 | = \frac{3x^2}{x^2+y^2} 24 | \leq 3; \\ 25 | |D_2 f(x,y)| 26 | &= \left| \frac{2x^3 y}{(x^2+y^2)^2} \right| 27 | \leq \frac{x^2(x^2+y^2)}{(x^2+y^2)^2} 28 | = \frac{x^2}{x^2+y^2} 29 | \leq 1. 30 | \end{align*} 31 | Hence $D_1 f$ and $D_2 f$ are bounded functions on $\mathbb{R}^2$. 32 | \end{proof} 33 | 34 | \item 35 | \begin{proof} 36 | Put $u = (\cos \theta, \sin \theta)$. 37 | Consider 38 | $$ 39 | |D_u f(0,0)| 40 | = \left| \lim_{h\to 0}\frac{f(hu) - f(0,0)}{\|hu - (0,0) \|}\right| 41 | = \left| \lim_{h\to 0} \frac{h\cos^3 \theta - 0}{h} \right| 42 | = |\cos^3 \theta|. 43 | $$ 44 | Since $|\cos\theta|\leq 1$ and then $|\cos^3 \theta| \leq 1$ follows, we conclude $|D_u f(0,0)|\leq 1$. 45 | \end{proof} 46 | 47 | \item 48 | \begin{proof} 49 | (Omitted) 50 | \end{proof} 51 | 52 | \item 53 | \begin{proof} 54 | Put $u = \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$. 55 | Consider 56 | $$ 57 | D_u f(0,0) = \left(\frac{\sqrt{2}}{2}\right)^3 58 | = \frac{\sqrt{2}}{4}. 59 | $$ 60 | On the other hand, 61 | $$ 62 | D_1 f(0,0) u_1 + D_2 f(0,0) u_2 63 | = \frac{\sqrt{2}}{2}\cdot 1 + \frac{\sqrt{2}}{2}\cdot 0 64 | = \frac{\sqrt{2}}{2}. 65 | $$ 66 | These mean $D_u f(0,0) \neq D_1 f(0,0) u_1 + D_2 f(0,0) u_2$. 67 | That is, formula (40) fails. 68 | 69 | We conclude $f$ is not differentiable at $(0,0)$. 70 | \end{proof} 71 | \end{enumerate} 72 | \end{Exercise} -------------------------------------------------------------------------------- /chapter9/exercises/ex_9.15.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 9.15 === 2 | \begin{Exercise} 3 | \begin{enumerate}[(a)] 4 | \item 5 | \begin{proof} 6 | Since $(x^2-|y|)^2 \geq 0$, then $2x^2 |y| \leq x^4+y^2$. 7 | It follows that 8 | $$ 9 | 4x^4y^2\leq (x^4+y^2)^2. 10 | $$ 11 | Given $\epsilon > 0$, there exists $\delta = \sqrt{-1+\sqrt{1+\epsilon} } > 0$ such that 12 | \begin{align*} 13 | \| (x,y) - (0,0) \| < \delta 14 | &\implies | f(x,y) - f(0,0) | \\ 15 | &= \left| x^2 + y^2 -2x^2 y - \frac{4x^6 y^2}{(x^4+y^2)^2} \right| \\ 16 | &\leq x^2 + y^2 + 2x^2 |y| + \left| \frac{4x^6 y^2}{(x^4+y^2)^2} \right| \\ 17 | &\leq x^2 + y^2 + 2x^2 |y| + x^2 \\ 18 | &= 2x^2 + y^2 + (x^4+y^2) \\ 19 | &\leq 2(x^2+y^2) + (x^2+y^2)^2 \\ 20 | &< 2\delta^2 + \delta^4 \\ 21 | &= \epsilon. 22 | \end{align*} 23 | This means $f$ is continuous at $(0,0)$. 24 | Besides, $f$ is continuous on $\mathbb{R}^2\setminus (0,0)$ trivially. 25 | Hence $f$ is continuous on $\mathbb{R}^2$. 26 | \end{proof} 27 | 28 | \item 29 | \begin{proof} 30 | Observe 31 | $$ 32 | g_{\theta}(t) = t^2-2t^3\cos^2 \theta \sin \theta - \frac{4t^4\cos^6\theta\sin^2\theta}{(t^2\cos^4\theta+\sin^2\theta)^2}. 33 | $$ 34 | Hence $g_{\theta}(0) = 0$, $g'_{\theta}(0) = 0$ and $g''_{\theta}(0) = 2$. 35 | 36 | Since $g''_{\theta}(0) > 0$ and $g'_{\theta}(0) = 0$, we know $g_{\theta}$ has therefore a strict local minimum at $t=0$. 37 | \end{proof} 38 | 39 | \item 40 | \begin{proof} 41 | We pick $y = x^2$, then $f(x,x^2) = -x^4 < 0$ for $(x,y)$ near $(0,0)$. 42 | Consequently, $f$ has no local minimum at $(0,0)$. 43 | \end{proof} 44 | \end{enumerate} 45 | \end{Exercise} -------------------------------------------------------------------------------- /chapter9/exercises/ex_9.19.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 9.19 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Put 5 | \begin{align*} 6 | f_1(x,y,z,u) &= 3x+y-z+u^2; \\ 7 | f_2(x,y,z,u) &= x-y+2x+u; \\ 8 | f_3(x,y,z,u) &= 2x+2y-3z+2u. 9 | \end{align*} 10 | Then $f_1 = f_2 = f_3 = 0$ implies that $f_1 - f_2 - f_3 = 0$. 11 | Consequently, $3u - u^2 = 0$. 12 | Hence when $u = 0$ or $u = 3$, the system has a solution. 13 | That is, the system can not be solved generally for $x,y,z$ in terms of $u$. 14 | 15 | For the other three statements, we give examples as a proof. 16 | \begin{itemize} 17 | \item \textbf{The system can be solved for $x,y,u$ in terms of $z$.} 18 | \end{itemize} 19 | $$ 20 | (x,y,u) = \begin{cases} 21 | \left( -\dfrac{z}{4}, \dfrac{7}{4}z, 0 \right) &, \mbox{(u = 0)}\\ 22 | \left( -\dfrac{z+9}{4}, \dfrac{7z+3}{4}, 3 \right) &, \mbox{(u = 3)}\\ 23 | \end{cases} 24 | $$ 25 | 26 | \begin{itemize} 27 | \item \textbf{The system can be solved for $x,z,u$ in terms of $y$.} 28 | \end{itemize} 29 | $$ 30 | (x,z,u) = \begin{cases} 31 | \left( -\dfrac{y}{7}, \dfrac{4}{7}y, 0 \right) &, \mbox{(u = 0)}\\ 32 | \left( \dfrac{4y+60}{7}, \dfrac{4y-3}{7}, 3 \right) &, \mbox{(u = 3)}\\ 33 | \end{cases} 34 | $$ 35 | 36 | \begin{itemize} 37 | \item \textbf{The system can be solved for $y,z,u$ in terms of $x$.} 38 | \end{itemize} 39 | $$ 40 | (y,z,u) = \begin{cases} 41 | \left( -7x, -4x, 0 \right) &, \mbox{(u = 0)}\\ 42 | \left( \dfrac{7}{4}x-15, -4x+9, 3 \right) &, \mbox{(u = 3)}\\ 43 | \end{cases} 44 | $$ 45 | \end{proof} 46 | 47 | \end{Exercise} -------------------------------------------------------------------------------- /chapter9/exercises/ex_9.27.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 9.27 === 2 | \begin{Exercise} 3 | \begin{enumerate}[(a)] 4 | \item 5 | We prove three functions are continuous in $\mathbb{R}^2$ one by one. 6 | \begin{itemize} 7 | \item \textbf{$f$ is continuous in $\mathbb{R}^2$.} 8 | \end{itemize} 9 | \begin{proof} 10 | $f$ is continuous on $\mathbb{R}^2\setminus (0,0)$ trivially, so we consider for $(x,y) = (0,0)$. 11 | 12 | Given $\epsilon>0$, pick $\delta = \sqrt{2\epsilon}$ such that 13 | $$ 14 | \| (x,y) - (0,0) \| = \sqrt{x^2+y^2} < \delta 15 | $$ 16 | implies 17 | \begin{align*} 18 | |f(x,y)-f(0,0)| 19 | &= \left| \frac{xy(x^2-y^2)}{x^2+y^2} \right| \\ 20 | &\leq \frac{1}{2}|x^2-y^2| & (\because 2|xy| \leq x^2+y^2) \\ 21 | &\leq \frac{1}{2} |x^2+y^2| \\ 22 | &< \epsilon. 23 | \end{align*} 24 | This means $f$ is continuous at $(0,0)$. 25 | Hence we conclude $f$ is continuous in $\mathbb{R}^2$. 26 | \end{proof} 27 | 28 | \begin{itemize} 29 | \item \textbf{$D_1 f$ is continuous in $\mathbb{R}^2$.} 30 | \end{itemize} 31 | \begin{proof} 32 | Notice that 33 | $$ 34 | D_1 f(0,0) 35 | = \lim_{h\to 0} \frac{f(h,0)-f(0,0)}{h} 36 | = 0. 37 | $$ 38 | Calculate 39 | \begin{align*} 40 | D_1 f(x,y) = \frac{x^4 y + 4x^2 y^3 - y^5}{(x^2+y^2)^2}, && (x,y)\neq(0,0). 41 | \end{align*} 42 | $D_1 f$ is continuous on $\mathbb{R}^2\setminus (0,0)$ trivially, so we consider for $(x,y) = (0,0)$. 43 | 44 | Observe that 45 | \begin{align*} 46 | |D_1 f(x,y)-D_1 f(0,0)| 47 | &= \left| \frac{x^4 y + 4x^2 y^3 - y^5}{(x^2+y^2)^2} \right| \\ 48 | &= \left| \frac{y(x^2-y^2)}{x^2+y^2} + \frac{4x^2y^3}{(x^2+y^2)^2} \right| \\ 49 | &\leq \left| \frac{y(x^2-y^2)}{x^2+y^2} \right| + \left| \frac{4x^2y^3}{(x^2+y^2)^2} \right| \\ 50 | &\leq |y| + \left| \frac{4x^2y^3}{(x^2+y^2)^2} \right| & (\because |x^2-y^2| \leq |x^2+y^2|) \\ 51 | &\leq |y| + |y| & (\because 2|xy|\leq x^2+y^2) \\ 52 | &= 2|y| \to 0 \text{ as } (x,y) \to (0,0). 53 | \end{align*} 54 | Since $D_1 f(0,0) = 0 = \lim_{(x,y)\to(0,0)}$, then $D_1 f$ is continuous at $(0,0)$. 55 | Hence we conclude $D_1 f$ is continuous in $\mathbb{R}^2$. 56 | \end{proof} 57 | 58 | \begin{itemize} 59 | \item $D_2 f$ is continuous in $\mathbb{R}^2$. 60 | \end{itemize} 61 | \begin{proof} 62 | Notice that 63 | $$ 64 | D_2 f(0,0) 65 | = \lim_{k\to 0} \frac{f(0,k) - f(0,0)}{k} 66 | = 0. 67 | $$ 68 | Calculate 69 | \begin{align*} 70 | D_2 f(x,y) = \frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}, && (x,y)\neq(0,0). 71 | \end{align*} 72 | $D_2 f$ is continuous on $\mathbb{R}^2\setminus (0,0)$ trivially, so we consider for $(x,y) = (0,0)$. 73 | 74 | Observe that 75 | \begin{align*} 76 | |D_2 f(x,y)-D_2 f(0,0)| 77 | &= \left| \frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2} \right| \\ 78 | &= \left| \frac{x(x^2-y^2)}{x^2+y^2} - \frac{4x^3y^2}{(x^2+y^2)^2} \right| \\ 79 | &\leq \left| \frac{x(x^2-y^2)}{x^2+y^2} \right| + \left| \frac{4x^3y^2}{(x^2+y^2)^2} \right| \\ 80 | &\leq |x| + \left| \frac{4x^3y^2}{(x^2+y^2)^2} \right| & (\because |x^2-y^2| \leq |x^2+y^2|) \\ 81 | &\leq |x| + |x| & (\because 2|xy|\leq x^2+y^2) \\ 82 | &= 2|x| \to 0 \text{ as } (x,y) \to (0,0). 83 | \end{align*} 84 | Since $D_2 f(0,0) = 0 = \lim_{(x,y)\to(0,0)}$, then $D_2 f$ is continuous at $(0,0)$. 85 | Hence we conclude $D_2 f$ is continuous in $\mathbb{R}^2$. 86 | \end{proof} 87 | 88 | \item 89 | \begin{proof} 90 | We know $D_{12} f(0,0) = 1$ and $D_{21} f(0,0) = -1$ from part(c). 91 | Calculate patiently, for $(x,y)\neq (0,0)$, 92 | $$ 93 | D_{12} f(x,y) 94 | = D_{21} f(x,y) 95 | = \frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}. 96 | $$ 97 | Hence $D_{12} f$, $D_{21} f$ exist at every point of $\mathbb{R}^2$ and are continuous on $\mathbb{R}^2\setminus (0,0)$. 98 | 99 | In order to achieve they are not continuous at $(0,0$) indeed, taking the limit along $x=y$, we have 100 | $$ 101 | \lim_{x\to 0}D_{12} f(x,x) 102 | = \lim_{x\to 0} 0 103 | = 0 104 | \neq D_{12} (0,0). 105 | $$ 106 | A similar argument shows that $\lim_{x\to 0}D_{21} f(x,x) \neq D_{21}f(0,0)$. 107 | These mean $D_{12} f$, $D_{21} f$ are not continuous at $(0,0)$, as desired. 108 | \end{proof} 109 | 110 | \item 111 | \begin{proof} 112 | Calculate 113 | \begin{align*} 114 | D_{12} f(0,0) 115 | &= \lim_{h\to 0} \frac{D_2(h,0) - D_2(0,0)}{h} 116 | = \lim_{h\to 0} \frac{h}{h} 117 | = 1; \\ 118 | D_{21} f(0,0) 119 | &= \lim_{k\to 0} \frac{D_1(0,k) - D_1(0,0)}{k} 120 | = \lim_{k\to 0} \frac{-k}{k} 121 | = -1. 122 | \end{align*} 123 | \end{proof} 124 | \end{enumerate} 125 | \end{Exercise} -------------------------------------------------------------------------------- /chapter9/exercises/ex_9.6.tex: -------------------------------------------------------------------------------- 1 | % === Exercise 9.6 === 2 | \begin{Exercise} 3 | \begin{proof} 4 | Calculate 5 | \begin{align*} 6 | D_1 f(x,y) 7 | &= \frac{y(y^2-x^2)}{(x^2+y^2)^2}; \\ 8 | D_2 f(x,y) 9 | &= \frac{x(x^2-y^2)}{(x^2+y^2)^2}. 10 | \end{align*} 11 | For $(x,y)\neq (0,0)$, $D_1 f(x,y)$, $D_2 f(x,y)$ exist. 12 | Then consider $(x,y) = (0,0)$, 13 | \begin{align*} 14 | D_1 f(0,0) 15 | &= \lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h} 16 | = 0; \\ 17 | D_2 f(0,0) 18 | &= \lim_{k\to 0}\frac{f(0,k)-f(0,0)}{k} 19 | = 0. 20 | \end{align*} 21 | So $D_1 f$, $D_2 f$ exist at every point of $\mathbb{R}^2$. 22 | 23 | However, along $x = y$, we have $f(x,x) = \dfrac{1}{2}$ for $x\neq 0$. 24 | Since 25 | $$ 26 | \lim_{x\to 0} f(x,x) = \frac{1}{2} \neq f(0,0), 27 | $$ 28 | we conclude $f$ is not continuous at $(0,0)$. 29 | \end{proof} 30 | \end{Exercise} -------------------------------------------------------------------------------- /config.tex: -------------------------------------------------------------------------------- 1 | % === Config === 2 | \let\oldproofname=\proofname 3 | \renewcommand{\proofname}{\color{red}{\rm\bf{\oldproofname}}} 4 | 5 | \newenvironment{solution} 6 | {\begin{proof}[\color{red}{$\mathbf{Solution}$}]} 7 | {\end{proof}} 8 | 9 | \newenvironment{sketch} 10 | {\begin{proof}[$\mathbf{Sketch}$]} 11 | {\end{proof}} 12 | 13 | \newenvironment{answer} 14 | {\begin{proof}[\color{purple}{$\mathbf{Answer}$}]\renewcommand{\qedsymbol}{}} 15 | {\end{proof}} 16 | 17 | \newtheorem{theorem}{Theorem}[chapter] 18 | \newtheorem{lemma}[theorem]{\color{blue}{Lemma}} 19 | 20 | \pagestyle{fancy} 21 | \fancyhf{} 22 | \rhead{\rightmark} 23 | \rfoot{\thepage} 24 | 25 | \renewcommand{\dateseparator}{/} 26 | \renewcommand{\qedsymbol}{$\blacksquare$} 27 | 28 | \def\upint{\mathchoice% 29 | {\mkern13mu\overline{\vphantom{\intop}\mkern7mu}\mkern-20mu} 30 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu} 31 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu} 32 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu} 33 | \int} 34 | \def\lowint{\mkern3mu\underline{\vphantom{\intop}\mkern7mu}\mkern-10mu\int} 35 | 36 | \DeclareMathOperator{\ii}{\mathrm{i}} -------------------------------------------------------------------------------- /cover.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % University Assignment Title Page 3 | % LaTeX Template 4 | % Version 1.0 (27/12/12) 5 | % 6 | % This template has been downloaded from: 7 | % http://www.LaTeXTemplates.com 8 | % 9 | % Original author: 10 | % WikiBooks (http://en.wikibooks.org/wiki/LaTeX/Title_Creation) 11 | % 12 | % License: 13 | % CC BY-NC-SA 3.0 (http://creativecommons.org/licenses/by-nc-sa/3.0/) 14 | % 15 | % Instructions for using this template: 16 | % This title page is capable of being compiled as is. This is not useful for 17 | % including it in another document. To do this, you have two options: 18 | % 19 | % 1) Copy/paste everything between \begin{document} and \end{document} 20 | % starting at \begin{titlepage} and paste this into another LaTeX file where you 21 | % want your title page. 22 | % OR 23 | % 2) Remove everything outside the \begin{titlepage} and \end{titlepage} and 24 | % move this file to the same directory as the LaTeX file you wish to add it to. 25 | % Then add \input{./title_page_1.tex} to your LaTeX file where you want your 26 | % title page. 27 | % 28 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 29 | 30 | %---------------------------------------------------------------------------------------- 31 | % PACKAGES AND OTHER DOCUMENT CONFIGURATIONS 32 | %---------------------------------------------------------------------------------------- 33 | 34 | \begin{titlepage} 35 | % === Math Font Setup === 36 | $\mathbb{}$ 37 | 38 | \newcommand{\HRule}{\rule{\linewidth}{0.5mm}} % Defines a new command for the horizontal lines, change thickness here 39 | 40 | \center % Center everything on the page 41 | 42 | %---------------------------------------------------------------------------------------- 43 | % HEADING SECTIONS 44 | %---------------------------------------------------------------------------------------- 45 | 46 | 47 | %---------------------------------------------------------------------------------------- 48 | % TITLE SECTION 49 | %---------------------------------------------------------------------------------------- 50 | 51 | ~\\[3cm] 52 | \HRule \\[0.4cm] 53 | { \huge \bfseries Solutions to \\ 54 | ~\\ 55 | Principles of Mathematical Analysis\\ 56 | (Rudin 3/e) }\\[0.4cm] % Title of your document 57 | \HRule \\[1.2cm] 58 | 59 | %---------------------------------------------------------------------------------------- 60 | % AUTHOR SECTION 61 | %---------------------------------------------------------------------------------------- 62 | 63 | ~\\[2cm] 64 | \begin{minipage}{0.4\textwidth} 65 | \begin{flushleft} \large 66 | \emph{Author:}\\ 67 | Cheng-Mao Lee 68 | \end{flushleft} 69 | \end{minipage} 70 | ~ 71 | \begin{minipage}{0.4\textwidth} 72 | \begin{flushleft} \large 73 | \emph{Email:} \\ 74 | chengmao.lee@gmail.com 75 | \end{flushleft} 76 | \end{minipage}\\[6cm] 77 | 78 | 79 | %---------------------------------------------------------------------------------------- 80 | % DATE SECTION 81 | %---------------------------------------------------------------------------------------- 82 | 83 | {\large \today\ \ (Version 1.0) }\\[3cm] % Date, change the \today to a set date if you want to be precise 84 | 85 | %---------------------------------------------------------------------------------------- 86 | % LOGO SECTION 87 | %---------------------------------------------------------------------------------------- 88 | 89 | %\includegraphics{Logo}\\[1cm] % Include a department/unisity logo - this will require the graphicx package 90 | 91 | %---------------------------------------------------------------------------------------- 92 | 93 | \vfill % Fill the rest of the page with whitespace 94 | 95 | \end{titlepage} -------------------------------------------------------------------------------- /license/CC_BY-NC-SA.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/howeverforever/Principles-of-Mathematical-Analysis/2da7700e37d855f18ddbf98004a2cd861919220a/license/CC_BY-NC-SA.jpg -------------------------------------------------------------------------------- /main.tex: -------------------------------------------------------------------------------- 1 | \documentclass{report} 2 | 3 | \usepackage[utf8]{inputenc} 4 | \usepackage[a4paper, total={6in, 8in}]{geometry} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{amssymb} 8 | \usepackage{fancyhdr} 9 | \usepackage{natbib} 10 | \usepackage{mathpazo} 11 | \usepackage{pgf,tikz} 12 | \usepackage{mathrsfs} 13 | 14 | \usepackage[titletoc]{appendix} 15 | \usetikzlibrary{arrows} 16 | 17 | \usepackage[bookmarks=true]{hyperref} 18 | \usepackage{bookmark} 19 | 20 | \usepackage{graphicx} 21 | \usepackage{epstopdf} 22 | \usepackage{caption} 23 | 24 | \usepackage{titletoc} 25 | \usepackage{import} 26 | \usepackage[yyyymmdd]{datetime} 27 | 28 | 29 | \usepackage{enumerate} 30 | \usepackage{type1cm} 31 | \usepackage{exercise,chngcntr} 32 | 33 | % === Environment Settings === 34 | \include{style} 35 | \include{config} 36 | % === -------------------- === 37 | 38 | \begin{document} 39 | \counterwithin{Exercise}{chapter} 40 | 41 | % === Cover Page === 42 | \include{cover} 43 | 44 | % === Contents Pages === 45 | \pagenumbering{roman} 46 | 47 | % === Preface === 48 | \include{abstract} 49 | 50 | % === Notation Table === 51 | \include{notation} 52 | 53 | % === Table === 54 | \include{table} 55 | 56 | % === Main Pages === 57 | \newpage 58 | \pagenumbering{arabic} 59 | \subimport*{chapter2/}{chapter2} 60 | \subimport*{chapter3/}{chapter3} 61 | \subimport*{chapter4/}{chapter4} 62 | \subimport*{chapter5/}{chapter5} 63 | \subimport*{chapter6/}{chapter6} 64 | \subimport*{chapter7/}{chapter7} 65 | \subimport*{chapter8/}{chapter8} 66 | \subimport*{chapter9/}{chapter9} 67 | 68 | \end{document} -------------------------------------------------------------------------------- /notation.tex: -------------------------------------------------------------------------------- 1 | % === Notation Table === 2 | \setcounter{page}{2} 3 | 4 | \phantomsection 5 | \addcontentsline{toc}{chapter}{\textbf{Notation Table}} 6 | \chapter*{Notation Table} 7 | 8 | \begin{table}[h] 9 | \centering 10 | \resizebox{\textwidth}{0.20\textheight}{\begin{minipage}{\textwidth} 11 | \begin{tabular}{|c|l|c|} 12 | \hline 13 | Symbol & \multicolumn{1}{c|}{Explanation} & Used where \\ \hline \hline 14 | $\emptyset$ & The empty set. & Chapter 2 \\ \hline 15 | $\mathbb{N}$ & The set of all positive integers excluding zero. & \\ \hline 16 | $\mathbb{Z}$ & The set of all integers. & \\ \hline 17 | $\mathbb{Q}$ & The set of all rational numbers. & \\ \hline 18 | $\mathbb{R}$ & The set of all real numbers. & \\ \hline 19 | $E$' & The set of all limit points of $E$. & Chapter 2 \\ \hline 20 | $\overline{E}$ & The union set of $E$ and all limit points of $E$. & Chapter 2 \\ \hline 21 | $d_E(x,y)$ & The distance from $x$ to $y$ under metric $E$. & Chapter 2 \\ \hline 22 | $N_r(p)$ & The neighborhood of a point $p$ where the radius is $r$. & Chapter 2 \\ \hline 23 | $[a,b]$ & $\{ x\in\mathbb{R} : a \leq x \leq b \}$ & \\ \hline 24 | $(a,b)$ & $\{ x\in\mathbb{R} : a < x < b \}$ & \\ \hline 25 | $|x|$ & The absolute value of $x$. & \\ \hline 26 | $\mathcal{R}$ & The set of all Riemann-integrable functions. & Chapter 6 \\ \hline 27 | $f_n\to f$ & The sequence of functions $f_n$ converge to the function $f$. & Chapter 7 \\ \hline 28 | $D_uf(x,y)$ & The directional derivatives of $f$ at $(x, y)$ in the direction of unit vector $u$. & Chapter 9 \\ \hline 29 | $\|x \|$ & The norm of $x$. & Chapter 9 \\ \hline 30 | \end{tabular} 31 | \end{minipage} } 32 | \end{table} 33 | 34 | \thispagestyle{plain} 35 | 36 | -------------------------------------------------------------------------------- /style.tex: -------------------------------------------------------------------------------- 1 | % === Custom Chapter Style ===$ 2 | \usepackage{fncychap} 3 | 4 | \pagestyle{fancy} 5 | \fancyhf{} 6 | \rhead{\rightmark} 7 | \rfoot{\thepage} 8 | \fancypagestyle{plain}{ 9 | \fancyhf[]{} 10 | \fancyhead[LO]{\slshape \leftmark} 11 | \fancyfoot[RO]{\thepage} 12 | } 13 | 14 | \ChNumVar{\fontsize{76}{80}\usefont{OT1}{pzc}{m}{n}\selectfont} 15 | \ChTitleVar{\raggedleft\Large\sffamily\bfseries} -------------------------------------------------------------------------------- /table.tex: -------------------------------------------------------------------------------- 1 | % === Table === 2 | \setcounter{page}{3} 3 | 4 | \titlecontents{chapter} 5 | [0pt] 6 | {\addvspace{1em}} 7 | {\bfseries\chaptername\ \thecontentslabel\quad} 8 | {} 9 | {\bfseries\hfill\contentspage} 10 | 11 | \tableofcontents --------------------------------------------------------------------------------