UPDATE: notations.tex, added a section about TVS.

Author: gitcordier

FA_DM.pdf

@@ -130,11 +130,15 @@
 \newcommand{\function}[1]{\mathtt{#1}}
 \newcommand{\relation}[2]{{#1}_{#2}}
 \newcommand{\f}[2]{#1(#2)}
-
+\newcommand{\id}[1]{\text{id}_{#1}}
+%\newcommand{\DeclareMathOperator}[1]{\text{#1}}
 
 % Sets
 %\renewcommand{\notin}{\tiny{\not\in}}
 %\renewcommand{\ni}{\small \ni}
+\DeclareMathOperator\opcard{card}
+\newcommand{\card}[1]{\opcard(#1)}
+%
 \def\contains{\supseteq }
 \def\cuts{\cap}
 \DeclareMathOperator\cvxhull{co}
@@ -148,8 +152,10 @@
 % Arithmetics
 \newcommand{\ceil}[1]{\lceil #1 \rceilf}
 % Analysis
-\newcommand\magnitude[1]{\left\lvert\, #1 \,\right\rvert}
-\renewcommand{\norm}[2]{\| \,#2 \, \|_{#1}}
+%\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}

FA_DM.tex

@@ -18,8 +18,8 @@
 \begin{proof}
 Discard the trivial case $\Lambda = 0$ and assume that $\dim Y = n$ %
 for some positive $n$. %
-Let $e$ range over a basis of $B$ of $Y$ then pick in $X$ $W$ an arbitrary %
-neighborhood of the origin: There so exists $V$ a balanced neighborhood of %
+Let $e$ range over a basis of $B$ of $Y$ then pick in $X$ an arbitrary %
+neighborhood $W$ of the origin: There so exists $V$ a balanced neighborhood of %
 the origin of $X$ such that 
 %
  \begin{align}
@@ -59,8 +59,10 @@
  y \in \sum_e \Lambda (V) \subseteq \Lambda(W) 
  \end{align}
 %
-for if $|y_e| < 1/s$; which proves (a); %
-see \hyperref[finite dimensional spaces]{[finite dimensional spaces]}.\\\\
+for if $|y_e| < 1/s$; which proves (a) whether %
+$B$ is the standard basis of $Y = \C^n$ equipped with $\norma{1}{\,\cdot\,}$. %
+%
+The general case is now provided for free by [\ref{notations: vector spaces: finite-dimensional vector spaces}].\\\\
 %
 %
 To prove (b), assume that the null space 

chapter_1/1_10.tex

@@ -0,0 +1,35 @@
+The following permutation
+\begin{align}
+\sigma:&\, \C^2 \to \C^2\\
+& (z,z')\mapsto (z',z) 
+\end{align}
+is an automorphism, \ie an homeomorphism [\cf (a) of 1.21], of the vector space $\C^2$. The key ingredient is that $\sigma$ is open. Moreover, the invariant points under $\sigma$ form the diagonal
+\eq{D\Def \{(z,z)\in\C^2 \}\quad. }\endeq
+As the graph of
+\begin{align*}
+I:& \, \C \to \C\nonumber\quad , \\ 
+&\, \, z\mapsto z \nonumber
+\end{align*}
+$D$ is a (closed) proper subspace of $\C^2$. Thus,
+\eq{D°=\emptyset\quad . }\endeq
+Remark that 
+\eq{\label{1_4_1}
+S\Def \{(z_{1},z_{2})_n :\, n=1,2,3,\dots \}\subseteq B \donc\, \limsup_{n\infty}\, \lvert z_{1,n} \rvert \< \liminf_{n\infty}\, \lvert z_{2,n}\rvert \quad . }\endeq
+For if $S$ converges (\ref{1_4_1}) forces $S$ to do so in $B$, which is then closed.
+\eq{\label{eq_1_4_2}
+\C^2 \setminus B= \{(z_{1},z_{2})\in \C^2:\, \lvert z_{1}\rvert > \lvert z_{2}\rvert \}
+}\endeq 
+is therefore open. So is 
+\eq{E\Def \sigma(\C^2 \setminus B)= \{(z_{1},z_{2})\in \C^2:\, \lvert z_{1}\rvert < \lvert z_{2}\rvert \}\quad , }\endeq
+since $\sigma$ is open.
+\eq{B=E \uplus D \quad . }\endeq
+Let $(z_1,z_2$ be in $B°$ If such a point would lie in $D$, there would be $V$ a neighbourhood of $(0,0)$ such that 
+\eq{(z_1,z_2) + V \subseteq B \quad ,}\endeq
+$(z_1,z_2) + V )\cap E$ is open. So is $(z_1,z_2) + V )\cap D$. contradicting 
+\eq{B°\subseteq B\setminus \Delta \C = B_<}\endeq
+Let $(z_1,z_2)$ be any element of $B°$. Since $(\Delta\C)°=\emptyset$, $(z_1,z_2)$ is in $B_<$. 
+\eq{([z] \in \C^\ast / N \cong \R_+ }\endeq
+Since $\C^2$ is separated, there exists a neighbourhood of the origin such that 
+\eq{C(z_1)+V\cup C(z_2)+ V\quad, }\endeq
+so that
+\eq{z_1 + V \cup z_2 + V}\endeq

chapter_1/1_4_comments.txt

@@ -1 +1 @@
-%!TEX root = /Volumes/HD_2/Rudin/Rudin_DM.tex
+%

chapter_1/1_8.tex

@@ -2,7 +2,7 @@
 Let X be the normed space of all real polynomials in one variable, with %
 %
  \begin{align*}
- \norm{}{f}=\int_0^1\lvert f(t)\rvert\ dt.
+ \norm{f}=\int_0^1\lvert f(t)\rvert\ dt.
 \end{align*}
 %
 Put %
@@ -16,7 +16,7 @@
 %
  \begin{align}\label{magnitude of B(f, g)}
  \magnitude{B(f, g)} 
- & < \norm{}{f} \cdot \max_{[0,1]} \magnitude{g} ; 
+ & < \norm{f} \cdot \max_{[0,1]} \magnitude{g} ; 
  \end{align}
 %
 which is sufficient (\citeresultFA{1.18}) to assert that any %
@@ -32,7 +32,7 @@
 There so exists a positive $M$ such that, 
 %
  \begin{align}\label{magnitude of B(f, g) H_c}
- \magnitude{B(f, g)} < M \norm{}{f}\norm{}{g}.
+ \magnitude{B(f, g)} < M \norm{f}\norm{g}.
  \end{align}
 %
 Put %
@@ -44,7 +44,7 @@
 so that 
 %
  \begin{align}\label{norm of f_n}
- \norm{}{f_n} = \frac{2 \sqrt{n}}{n+1} \tendsto{n}{\infty}0.
+ \norm{f_n} = \frac{2 \sqrt{n}}{n+1} \tendsto{n}{\infty}0.
  \end{align}
 %
 On the other hand,
@@ -61,7 +61,7 @@
 %
 and so obtain
 \begin{align}
- 1 < B(f_n, f_n) < M \norm{}{f_n}^2 \tendsto{n}{\infty} 0.
+ 1 < B(f_n, f_n) < M \norm{f_n}^2 \tendsto{n}{\infty} 0.
 \end{align}
 Our continuousness assumption is then contradicted. So ends the proof.
 \end{proof}

chapter_2/2_12.tex

@@ -18,37 +18,66 @@
  \end{enumerate}
 %
 By (i), the complement $V_n$ of $\overline{E}_n$ is a dense open set. %
-Since $X$ is an F-space, it follows from the Baire's theorem that %
+Now pick $x$ in $X$: $x + V_n$ is dense open as well, 
+since the translation by $x$ is an homeomorphism of $X$. %
 %
-  the intersection $S$ of the $V_n$'s is dense in $X$:  %
+Note that the density is also a special case of \citeresultFA{1.3 (b)}, as follows, %
 %
-So is $x+S$ ($x\in X$). To see that, remark that %
+\begin{align}\label{2_15_2}
+ X = x + X \subseteq \overline{x + V_n}.
+\end{align} 
 %
- \begin{align}\label{2_15_2}
- X = x + \overline{S} \subseteq \overline{x + S}
- \end{align} 
+We now apply the Baire's theorem twice to establish that %
 %
-follows from \citeresultFA{1.3 (b)}. %
-Since $S$ and $x+S$ are both dense open subsets of $X$, %
-the Baire's theorem asserts that %
+\begin{enumerate}
+ \item every intersection $W_n = V_n \cap (x+V_n)$ is dense in $X$;
+ \item so is the intersection $\displaystyle{\bigcap_{n=1}^\infty W_n}$.
+\end{enumerate}
 %
- \begin{align}
- \overline{(x+S)\cap S} = X.
- \end{align} 
+Bear in mind that every dense subset of $X$ is nonempty and remark that %
+\begin{align}\label{2_15_3}
+ \bigcap_{n=1}^\infty W_n \subseteq \bigcap_{n=1}^\infty V_n \subseteq Y
+\end{align} 
+holds, since %
 %
-Thus, 
+ $X\setminus Y \subseteq \bigcup_{n=1}^\infty \overline{E}_n$. %
 %
- \begin{align}\label{2_15_4}
- (x+S)\cap S\neq\emptyset.
- \end{align}
+Now pick $w$ in $\bigcap_{n=1}^\infty W_n$, so that %
+$w$ is an element of $Y$ (by (\ref{2_15_3})) that lies in every $W_n$. %
+The key ingredient is that $W_n$ a subset of $x + V_n$. %
+Hence %
 %
-Moreover, it follows from (ii) that %
+\begin{align}
+ w \in x + V_n. 
+\end{align}
 %
- $X\setminus Y \subseteq \bigcup_n \closure{E}_n$, \ie 
- $Y \contains S$. %
-Combined with (\ref{2_15_4}), this shows that $x+Y$ cuts $Y$. 
-Therefore, our arbitrary $x$ is an element of the subgroup $Y$. %
-We have thus established that $X \subseteq Y$, which achieves the proof.
+Finally, 
+\begin{align}
+ w - x \in \bigcap_{n=1}^\infty V_n \subseteq Y; 
+\end{align}
+%
+again with (\ref{2_15_3}).
+It is now clear that $x + Y$ cuts $Y$ in $x + (w - x ) = w$, which establishes that $x$ lies is $Y - Y$. %
+As a conclusion, 
+\begin{align}
+ X \subseteq Y, 
+\end{align}
+%
+since $x$ was arbitrary and that $Y$ is a subgroup. %
+So ends the proof.
+%
+%
+% \begin{align}\label{2_15_4}
+% (x+S)\cap S\neq\emptyset.
+% \end{align}
+%
+%Moreover, it follows from (ii) that %
+%
+% $X\setminus Y \subseteq \bigcup_n \closure{E}_n$, \ie 
+% $Y \contains S$. %
+%Combined with (\ref{2_15_4}), this shows that $x+Y$ cuts $Y$. 
+%Therefore, our arbitrary $x$ is an element of the subgroup $Y$. %
+%We have thus established that $X \subseteq Y$, which achieves the proof.
 \end{proof}
 \renewcommand{\labelenumi}{(\alph{enumi})} 
 %

chapter_2/2_15.tex

@@ -1,4 +1,4 @@
-%!TEX root = /Volumes/HD_2/Rudin/Rudin_DM.tex
+%
 \textit{
 Put $K=[-1,1]$; define $\D_{K}$ as in section 1.46 
 (with $\R$ in place of $\R^{n}$). 
@@ -16,7 +16,7 @@
  \int_{\minus 1}^1 f_n (t)\phi (t) dt\
  \right\rvert
  \leq 
- M \norm{\infty}{D^{p}} 
+ M \norma{\infty}{D^{p}} 
  \nonumber
  \end{align}
 for all $n$.

chapter_2/2_3/2_3.tex

@@ -4,9 +4,9 @@
 %
 %
  \begin{align}\label{2.3. Mean value inequality.}
- \norm{\infty}{D^\alpha \phi} 
+ \norma{\infty}{D^\alpha \phi} 
  \leq 
- \norm{\infty}{D^p \phi} \left(\frac{\lambda}{2}\right)^{p-\alpha}
+ \norma{\infty}{D^p \phi} \left(\frac{\lambda}{2}\right)^{p-\alpha}
  \quad (\alpha = 0, 1, \dots, p) 
  \end{align}
 %
@@ -59,7 +59,7 @@
  \begin{align}\label{2.3. Mean Value (1).}
  \lvert D^\alpha \phi(x) \rvert
  \leq 
- \norm{\infty}{D^p \phi} 
+ \norma{\infty}{D^p \phi} 
  \left(\frac{\lambda}{2}\right)^{p-\alpha}
  \quad (x \in [a, c])
  \end{align}
@@ -74,7 +74,7 @@
  \begin{align}\label{2.3. Mean Value (2).}
  \lvert D^\alpha \phi(x) \rvert
  \leq 
- \norm{\infty}{D^p \phi} 
+ \norma{\infty}{D^p \phi} 
  \left(\frac{\lambda}{2}\right)^{p-\alpha} 
  % \quad (x \in [c, b]).
  \end{align}
@@ -87,9 +87,9 @@
 and so obtain %
 %
  \begin{align}
- \norm{\infty}{D^\alpha \phi} 
+ \norma{\infty}{D^\alpha \phi} 
  \leq 
- \norm{\infty}{D^p \phi} 
+ \norma{\infty}{D^p \phi} 
  \left(\frac{\lambda}{2}\right)^{p-\alpha}.
  \end{align}
 %

chapter_2/2_3/2_3_0_lemma.tex

@@ -19,8 +19,8 @@
  \leq 
  \int_\R \left\lvert u\phi \right\rvert 
  \leq
- \norm{L^1}{u}
- \quad(\norm{\infty}{\phi} \leq 1)
+ \norma{L^1}{u}
+ \quad(\norma{\infty}{\phi} \leq 1)
  \end{align} 
 %
 imply that every linear functional 
@@ -55,12 +55,12 @@
  \sup\set{
  \magnitude{\bra{u}\ket{\phi}}
  }{ 
- \norm{\infty}{\phi} = 1
+ \norma{\infty}{\phi} = 1
  } 
- = \norm{L^1}{u}.
+ = \norma{L^1}{u}.
  \end{align} 
 %
-Note that, in the latter equality, $\leq \norm{L^1}{u}$ comes from %
+Note that, in the latter equality, $\leq \norma{L^1}{u}$ comes from %
 %
  (\ref{2.3 g bounded operator (1).}), %
 %
@@ -156,7 +156,7 @@
  }
  \item{
  The sequence $\singleton{\phi_j}$ converges (pointwise) to %
- $1_{[0, 1]} - 1_{[\minus 1, 0]}$, and $\norm{\infty}{\phi_{j}} = 1$. %
+ $1_{[0, 1]} - 1_{[\minus 1, 0]}$, and $\norma{\infty}{\phi_{j}} = 1$. %
  }
  \end{enumerate}
 %
@@ -168,21 +168,21 @@
  2 \int_{0}^1 g_n(t) \phi^{+}_{j}(t) d t
  \tendsto{j}{\infty} 
  2 \int_{0}^1 g_n(t) d t
- = \norm{L^1}{g_n} = n.
+ = \norma{L^1}{g_n} = n.
  \end{align}
 %
 Finally, %
 %
  \begin{align}\label{2.3. Norm in the dual equals norm L1 (2).}
- \norm{L^1}{g_n}
+ \norma{L^1}{g_n}
  \citeleq{\ref{2.3. Norm in the dual equals norm L1 (1).}} 
  \sup\set{
  \magnitude{\bra{g_n}\ket{\phi}}
  }{
- \norm{\infty}{\phi} = 1
+ \norma{\infty}{\phi} = 1
  }
  \citeleq{\ref{2.3 g bounded operator (2).}}
- \norm{L^1}{g_n};
+ \norma{L^1}{g_n};
  \end{align} %
 %
 which is the desired result. %
@@ -195,13 +195,13 @@
  \bra{g_n}\ket{\phi} 
  \right\rvert 
  \leq 
- C_{n} \quad (\norm{\infty}{\phi} = 1); 
+ C_{n} \quad (\norma{\infty}{\phi} = 1); 
  \end{align}
  %
 see (\ref{2.3 g bounded operator (1).}). 
 Furthermore, % 
 %
- $\norm{L^1}{g_n}$ is actually the best, \ie lowest, possible $C_{n}$; see %
+ $\norma{L^1}{g_n}$ is actually the best, \ie lowest, possible $C_{n}$; see %
  %
  (\ref{2.3. Norm in the dual equals norm L1 (2).}). %
  %
@@ -217,7 +217,7 @@
 %
  $\langle{g_n}\lvert{\phi_{\rho(n)}}\rangle$ %
 % 
-is greater than, say, $n -0.01$, as $\norm{\infty}{\phi_{\rho(n)}} = 1$. %
+is greater than, say, $n -0.01$, as $\norma{\infty}{\phi_{\rho(n)}} = 1$. %
 %
 Consequently, there is no bound $M$ such that %%
 %
@@ -226,7 +226,7 @@
  \bra{g_n}\ket{\phi} 
  \right\rvert 
  \leq M
- \quad (\norm{\infty}{\phi} = 1; \counting{n}).
+ \quad (\norma{\infty}{\phi} = 1; \counting{n}).
  \end{align}
 %
 In other words, the $g_n$ have no \textit{uniform bound} in ${L^1}$, %