UPDATE: Started to rewrite Exercise 2.3. Rewrote the lemma, as a beginning.

Author: gitcordier

FA_DM.pdf

@@ -8,7 +8,7 @@
 \usepackage{amsmath}
 \usepackage{
 physics, 
-amssymb, 
+amssymb, amsfonts, 
 mathrsfs, 
 amsthm, 
 mathspec, 
@@ -70,7 +70,7 @@
 \setmathfont(Latin)[Uppercase=Regular,Lowercase=Regular]{\mainfont}
 \setmathfont(Greek)[Uppercase=Regular,Lowercase=Regular]{\mainfont}
 \setmathrm{\mainfont}
-\setmathbb{\mainfont}
+%\setmathbb{\mainfont}
 %\setmathit{\mainfont}
 \setmathtt{CMU Typewriter Text Light}
 %\setmathbf{CMU Serif}
@@ -154,12 +154,15 @@
 
 % Arithmetics
 \newcommand{\ceil}[1]{\lceil #1 \rceilf}
+
 % Analysis
+\newcommand\openinterval[2]{\left]#1, #2\right[}
 %\newcommand\magnitude[1]{\left\lvert\, #1 \,\right\rvert}
 \newcommand\magnitude[1]{\abs{#1}}
 \newcommand{\norma}[2]{\norm{#2}_{#1}}
 %\renewcommand{\norm}[2]{\norm{#2}_{#1}}
 \def\weakstar{\text{weak}^\ast\text{-}}
+
 % Topology
 \newcommand{\localbase}[1]{\mathscr #1}
 
@@ -179,14 +182,21 @@
 \def\iif{{\bf iff} }
 \def\IIF{{\bf iff}}
 \def\wlg{{\bf wlg }} % Without loss og generality
+
+% TODO: Make choices, be consistent
 \def\then{\Rightarrow}
 \def\THEN{\Rightarrow}
 \def\THEREFORE{\Rightarrow}
 \def\therefore{\Rightarrow}
 \def\since{\Leftarrow}
+\def\because{\Leftarrow}
 \def\IF{\Leftarrow}
 \renewcommand{\iff}{\Leftrightarrow}
 
+% Misc
+% phi is varphi:
+\renewcommand{\phi}{\varphi}
+
 % Citations
 \newcommand{\citehere}[2]{\overset{(#1)}{#2}}
 \newcommand{\citeq}[1]{\citehere{#1}{=}}
@@ -221,6 +231,10 @@
 \def\AUTHOR{gitcordier}
 
 \begin{document}
+% Changes the "proof" in proof environment. 
+% source: https://tex.stackexchange.com/questions/8089/changing-style-of-proof
+\let\oldproofname=\proofname
+\renewcommand{\proofname}{{\rm \small PROOF}}
 %\begin{abstract}
 % \input{\ROOT/abstract.tex}
 %\end{abstract}

FA_DM.tex

@@ -1,10 +1,10 @@
 \textit{Suppose %
 %
- $d_1(x,y) = |x-y|, d_2(x,y) = |\varphi(x) - \varphi(y)|$, %
+ $d_1(x,y) = |x-y|, d_2(x,y) = |\phi(x) - \phi(y)|$, %
 %
 where %
 %
- $\varphi(x)={x}/{(1+\lvert x \rvert )}$. %
+ $\phi(x)={x}/{(1+\lvert x \rvert )}$. %
 %
 Prove that $d_1$ and $d_2$ are metrics on $\R$ which induce the same %
 topology, although $d_1$ is complete and $d_2$ is not.
@@ -20,25 +20,25 @@
 \end{align}
 %
 Next, remark that the monotonically increasing mapping %
-$\varphi: \R \to ]\minus 1, 1[$ is odd and that %
+$\phi: \R \to ]\minus 1, 1[$ is odd and that %
 %
 \begin{align}\label{1_12_1}
- \varphi(x) \tendsto{x}{\infty} 1.
+ \phi(x) \tendsto{x}{\infty} 1.
 \end{align}
 %
-$\varphi$ is therefore a $\tau_1$-homeomorphsim of $\R$ onto $]\minus 1, 1[$. %
+$\phi$ is therefore a $\tau_1$-homeomorphsim of $\R$ onto $]\minus 1, 1[$. %
 %
 A first consequence is that, at fixed $a \in \R$, given any positive scalar %
-$\epsilon$, the $\tau_1$-continuousness of $\varphi$ supplies an open ball %
-$B_1(a,\eta)$ on which $|\varphi(a)-\varphi|<\epsilon$. %
+$\epsilon$, the $\tau_1$-continuousness of $\phi$ supplies an open ball %
+$B_1(a,\eta)$ on which $|\phi(a)-\phi|<\epsilon$. %
 In terms of balls $B_i$, this reads as follows, %
 %
 \begin{align}
  B_1(a,\eta) \subset B_2(a,\epsilon).
 \end{align}
 %
 The second consequence is that the $\tau_1$-continuousness of %
-$\varphi^{\, \minus 1}$ yields similar inclusions %
+$\phi^{\, \minus 1}$ yields similar inclusions %
 %
 \begin{align} \label{1_12_6}
 \quad B_2(a, \epsilon') \subset B_1 (a, \eta')
@@ -56,7 +56,7 @@
 Finally, all inequalities $n < i < j$ over $\N$ together yield %
 %
 \begin{align}
- d_2(i, j)=|\varphi(i)-\varphi(j)| \tendsto{n}{\infty} 0.
+ d_2(i, j)=|\phi(i)-\phi(j)| \tendsto{n}{\infty} 0.
 \quad \end{align}
 %
 The sequence $n=0, 1, 2, \dots$ is therefore $\tau_2$-Cauchy. We will %
@@ -67,7 +67,7 @@
 \begin{align}
  d_2(0, \lambda) \geq 
  d_2(0, n) - d_2(\lambda, n) = 
- \varphi(n) -d_2(\lambda, n) \tendsto{n}{\infty} 1.
+ \phi(n) -d_2(\lambda, n) \tendsto{n}{\infty} 1.
 \end{align}
 %
 We then conclude that $d_2$ fails to be complete. 

chapter_1/1_12.tex

@@ -106,22 +106,22 @@ \subsection{Proof (with the given hint)}
 Indeed, the hint suggests that there exists a bijection % 
 %
 \begin{align}
- \varphi: \bigl\{(\theta_n): \theta_n \xrightarrow{n\infty} 0\bigr\} & \to [0, 1] \\
+ \phi: \bigl\{(\theta_n): \theta_n \xrightarrow{n\infty} 0\bigr\} & \to [0, 1] \\
  (\theta_1, \dots, \theta_n, \dots) & \mapsto x. \nonumber
 \end{align}
 %
 \begin{proof}
 We set %
 %
 \begin{align}
- f_n(x) & \Def \theta_n \quad \bigl(x = \varphi(\theta_1, \dots, \theta_n, \dots)\bigr)
+ f_n(x) & \Def \theta_n \quad \bigl(x = \phi(\theta_1, \dots, \theta_n, \dots)\bigr)
 \end{align}
 %
 so that $\singleton{f_n}$ tends pointwise to $0$. Note that, with this construction, the special case %
 %
 \begin{align}
  x_\gamma %
- \Def \varphi\Bigl(1 / \sqrt{1 + \magnitude{\gamma_1}}, \dots, 1 / \sqrt{1 + \magnitude{\gamma_n}}, \dots\Bigr) %
+ \Def \phi\Bigl(1 / \sqrt{1 + \magnitude{\gamma_1}}, \dots, 1 / \sqrt{1 + \magnitude{\gamma_n}}, \dots\Bigr) %
 \end{align}
 %
 outputs %

chapter_1/1_7.tex

@@ -30,7 +30,7 @@
 \newline\newline\noindent
 We will also consider the case $p=0$. Since all supports of %
 %
- $\phi, \varphi', \varphi'', \dots, $ are in $K$, %
+ $\phi, \phi', \phi'', \dots, $ are in $K$, %
 %
 we make a specialization of the mean value theorem: %
 %

chapter_2/2_3/2_3.tex

@@ -1,96 +1,47 @@
 If 
 %
- $\phi\in\D_{[a, b]}$, then %
-%
+ $\phi\in\D_{[a, b]}$, then, for all $k \leq p$ in $\N$, 
 %
  \begin{align}\label{2.3. Mean value inequality.}
- \norma{\infty}{D^\alpha \phi} 
- \leq 
- \norma{\infty}{D^p \phi} \left(\frac{\lambda}{2}\right)^{p-\alpha}
- \quad (\alpha = 0, 1, \dots, p) 
+ \norma{\infty}{D^k \phi} \leq \norma{\infty}{D^p \phi} \left(\frac{b-a}{2}\right)^{p-k} .
  \end{align}
 %
-at every order $\integers{p}$; %
-where $\lambda$ is the length $\magnitude{b-a}$. %
-%
 \begin{proof}
-Let $x_0$ be in %
-% 
- $(a, b)$ .
-% 
-We first consider the case $x_0 \leq c = (a+b)/2$: %
-The mean value theorem asserts that there exists %
-%
- $x_{1}$ ($a <x_{1} < x_{0} $), 
-%
-such that 
+Choose $a < x_0 \leq (a+b)/2$ first. By the mean value theorem, there exists $x_1 \in \openinterval{a}{x_0}$ such that % 
 %
  \begin{align}
- \phi(x_0) =\phi(x_0) - \phi(a)= D\phi(x_{1})(x_{0} -a).
+ \phi(x_0) = \phi(x_0) - \phi(a)= D\phi(x_{1})(x_{0} - a).
  \end{align}
 %
-Since every %
-%
- $D^p\phi$ lies in $\D_{[a, b]}$, %
+Repeating the same reasoning for $D\phi, D^2 \phi, \dots, D^p \phi \in \D_{[a, b]}$ yields % 
 %
-a straightforward proof by induction shows that there exists a partition % 
+\begin{align}
+ \phi(x_0) & = D^0 \phi(x_0) \\ 
+ & = D^1\phi(x_{1})(x_0 - a) \\
+ & = D^2\phi(x_{2})(x_1 - a)(x_0 - a) \\
+ & \nonumber \mspace{10mu} \vdots \\
+ & = D^p\phi(x_{p})(x_{p-1} - a) \cdots (x_0-a),
+\end{align}
 %
- $a < \cdots < x_p < \cdots < x_0 $ 
+for some $x_1, \dots ,x_p \in \openinterval{a}{x_0}$. Hence %
 %
-such that 
- \begin{align}
- \phi(x_0) & = D^0 \phi(x_0) \\ 
- & = D^1\phi(x_{1})(x_{0} - a ) \\
- % & = D^2\phi(x_{2})(x_{0} - a )(x_{1} - a)\\
- & = \cdots \nonumber\\
- & = D^p\phi(x_{p})(x_{0} - a)\cdots(x_{p-1}-a),
- \end{align}
-%
-for all $p$. %
-More compactly, 
-%
- \begin{align}
- D^\alpha \phi (x_0) = D^p\phi(x_p) \prod_{k= \alpha}^{p-1}(x_k - a);
- %(x_\alpha -a )\cdots (x_{p-1} -a).
- \end{align}
+\begin{align}\label{2.3. Mean Value (1).}
+ \magnitude{\phi(x_0)} \leq \norma{\infty}{D^p \phi} \left(\frac{b-a}{2}\right)^p .
+\end{align}
 %
-which yields, 
-%
- \begin{align}\label{2.3. Mean Value (1).}
- \lvert D^\alpha \phi(x) \rvert
- \leq 
- \norma{\infty}{D^p \phi} 
- \left(\frac{\lambda}{2}\right)^{p-\alpha}
- \quad (x \in [a, c])
- \end{align}
-%
-The case $x_0 \geq c$ outputs a ``reversed'' result, with %
-%
- $b > \cdots > x_p > \cdots > x_0$ %
-%
-and $x_k -b$ playing the role of $x_k-a$: %
+Similarly, if $b > x_0 \geq (a+b)/2$ (interchanging the roles of $a$ and $b$), the same inequality (\ref{2.3. Mean Value (1).}) holds. %
 So, %
 %
- \begin{align}\label{2.3. Mean Value (2).}
- \lvert D^\alpha \phi(x) \rvert
- \leq 
- \norma{\infty}{D^p \phi} 
- \left(\frac{\lambda}{2}\right)^{p-\alpha} 
- % \quad (x \in [c, b]).
- \end{align}
-%
-Finally, we combine %
-%
- (\ref{2.3. Mean Value (1).}) with %
- (\ref{2.3. Mean Value (2).}) %
+\begin{align}\label{2.3. Mean value (2).}
+ \magnitude{\phi(x_0)} \leq \norma{\infty}{D^p \phi} \left(\frac{b-a}{2}\right)^p \quad ( a < x_0 < b ), 
+\end{align}
 %
-and so obtain %
+which establishes the case $k=0$. %
+Finally, applying (\ref{2.3. Mean value (2).}) to $D^k \phi $ in place of $\phi$ gives % 
 %
- \begin{align}
- \norma{\infty}{D^\alpha \phi} 
- \leq 
- \norma{\infty}{D^p \phi} 
- \left(\frac{\lambda}{2}\right)^{p-\alpha}.
- \end{align}
+\begin{align}
+ \norma{\infty}{D^k \phi} \leq \norma{\infty}{D^p \phi} \left(\frac{b-a}{2}\right)^{p-k}
+\end{align}
 %
+for all $0 \leq k \leq p$. %
 \end{proof}