UPDATE: Rewriting of 1.10.

Author: gitcordier

FA_DM.pdf

@@ -1,8 +1,16 @@
-\textit{Suppose that X and Y are topological vector spaces,
-%
- $\dim Y < \infty$,
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+% FunctionalAnalysis 
+% 1_10.tex
+% 
+% encoding: UTF-8 
+% EOL: LF
 %
-$\Lambda : X \to Y$ is linear, and $\Lambda(X) = Y$.
+% format: LaTeX
+% indent: spaces (2)
+% width: 127
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+\textit{Suppose that X and Y are topological vector spaces, $\dim Y < \infty$, $\Lambda : X \to Y$ is linear, and %
+$\Lambda(X) = Y$.
 %
 \begin{enumerate}
  \item{
@@ -16,68 +24,47 @@
 }
 %
 \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 $X$ an arbitrary %
-neighborhood $W$ of the origin: There so exists $V$ a balanced neighborhood of %
-the origin of $X$ such that 
+Discard the trivial case $\Lambda = 0$ and assume that $\dim Y > 0$. From now on, $Y$ has a basis $B= \singleton{e, e', \dots}$ %
+and $W$ is an arbitrary neighborhood of $0$. Since addition is continuous, there exists a balanced open $V\subset Y$ such that %
 %
  \begin{align}
  \label{definition of v}
- \sum_{e} V \subset W, 
+ \sum_{e} V \subset W. 
  \end{align}
 %
-since addition is continuous. Moreover, for each $e$, there exists $x_e$ %
-in $X$ such that 
-%
- $\Lambda(x_e)=e$, 
-% 
-simply because $\Lambda$ is onto: Given $y$ in $Y$, %
-of $e$-component(s) $y_e$, we now obtain
+Moreover, the surjectivity of $\Lambda$ allows us to pick $x_e$ with $\Lambda x_e = e$. This implies that %
+every $y = \sum_e y_e e $ ($y_e \in \C$) can be expressed as %
 %
 \begin{align}\label{1_10_sum}
- y = \sum_{e} y_e \Lambda (x_e).
+ y = \sum_{e} y_e \Lambda x_e. 
 \end{align}
 %
-As a finite set, $\set{x_e}{e\in B}$ is bounded: There so exists a positive %
-scalar $s$ such that 
+Note that $\set{x_e}{e\in B}$ is bounded (as a finite set). Hence %
 %
 \begin{align}
- \forall e\in B, x_e \in s V.
+ \set{x_e}{e \in B} \subset sV 
 \end{align}
 %
-Combining this with (\ref{1_10_sum}) shows that 
+for some $s > 0$. Combining this with (\ref{1_10_sum}) yields % 
 %
  \begin{align}
  \label{y in sum of lambda V}
- y \in \sum_e y_e \,s \Lambda (V).
+ y \in \sum_e s y_e \Lambda (V).
  \end{align}
 %
-We now come back to (\ref{definition of v}) and so conclude that %
+Using (\ref{definition of v}) and the balancedness of $\Lambda(V)$, we conclude that
 %
  \begin{align}
- y \in \sum_e \Lambda (V) \subset \Lambda(W) 
+ y \in \sum_e \Lambda (V) \subset \Lambda(W)  \quad (\norma{\infty}{y} < 1/s). 
  \end{align}
 %
-for if $|y_e| < 1/s$; which proves (a) whether %
-$B$ is the standard basis of $Y = \C^n$ equipped with $\norma{\infty}{\,\cdot\,}$. %
-%
-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 
-%
- $\{\Lambda = 0\}$ %
-% 
-is closed and let $f, \pi$ be as in Exercise 1.9, %
-%
- $\{\Lambda = 0\}$ %
+This proves (a) in the case where $Y = \C^n$ and $B$ is the standard basis. The general case now follows from %
+[\ref{notations: vector spaces: finite-dimensional vector spaces}].\\\\
 %
-playing the role of $N$. Since $\Lambda$ is onto, the first isomorphism %
-theorem (see Exercise 1.9) asserts that $f$ is an isomorphism of $X/N$ %
-onto $Y$. %
-We now conclude with the help of [\ref{notations: vector spaces: finite-dimensional vector spaces}] %
-that $f$ is a homeomorphism of $X/N$ onto $Y$. %
+To prove (b), assume that the null space $\{\Lambda = 0\}$ is closed and let $f, \pi$ be as in Exercise 1.9, %
+$\{\Lambda = 0\}$ playing the role of $N$. Since $\Lambda$ is onto, the first isomorphism theorem (see Exercise 1.9) asserts %
+that $f$ is an isomorphism of $X/N$ onto $Y$. We now conclude with the help of %
+[\ref{notations: vector spaces: finite-dimensional vector spaces}] that $f$ is a homeomorphism of $X/N$ onto $Y$. %
 We have thus established that $f$ is continuous: So is $\Lambda = f\circ \pi$.
 \end{proof}
 % END

chapter_1/1_10.tex

@@ -192,11 +192,11 @@ \subsubsection{Topology of a finite-dimensional vector space}
  \item $W$ and $Y$ are homeomorphic to each other, they are normable.
 \end{enumerate}
 %
-Furthermore, the norms on $W$ and $Y$ are \textit{equivalent}. That is, for any given norm on $Y$ and any given norm on $W$, 
-there exists a postive constant (termed as ``modulus of continuity'') $C=C_h$ such that %
-%
+Furthermore, the norms on $W$ and $Y$ are \textit{equivalent}. That is, for any given norm $\norm{\, \cdot \, }$ on $Y$ and %
+any given norm $\norma{W}{\,\cdot \,}$ on $W$, there exists a postive constant (termed as ``modulus of continuity'') $C= C_h$ %
+such that %
 \begin{align}
- \norma{}{w} & \leq C \norm{y} \quad\quad \left((y, w) \in h \right), 
+ \norma{W}{w} & \leq C \norm{y} \quad\quad \left((y, w) \in h \right), 
 \end{align}
 %
 as $h$ is continuous. The special case $W=Y$ is that all norms on $Y$ are equivalent, in the sense that %
@@ -205,14 +205,14 @@ \subsubsection{Topology of a finite-dimensional vector space}
  \norma{{\text{copy of }Y}}{h(y)} \leq C \norm{y}.
 \end{align}
 %
-\subsubsection{The standard norms $\norma{1}{\,\cdot\,}$, $\norma{2}{\,\cdot\,}$, $\norma{\infty}{\,\cdot\,}$}
-When $\C^n$ is equipped with standard norms $1, 2, \infty$, the sharp constant\ie the smallest %
-$C_{i, j}$ ($i, j = 1, 2, \infty$) such that %
+\subsubsection{The standard norms $\norma{1}{\,\cdot\,}$, $\norma{2}{\,\cdot\,}$, $\norma{\infty}{\,\cdot\,}$} When $\C^n$ %
+is equipped with standard norms $1, 2, \infty$, the sharp constant\ie the smallest $C_{i, j}$ such that %
+%
 \begin{align}
 \norma{j}{z} \leq C_{i, j} \norma{i}{z} %\quad \left(z \in \C^n \right)
 \end{align}
 %
-is easily derived from definitions - see [1.19] of \cite{FA} - with the noticeable exception of $C_{2, 1}=\sqrt{n}$. %
+is easily derived from definitions - see [1.19] of \cite{FA} - with the noticeable exception of $C_{2, 1} = \sqrt{n}$. %
 Indeed, $\norma{1}{z} \leq \sqrt{n} \norma{2}{z}$ is a special Cauchy-Schwarz inequality, see (1) in [12.2] of \cite{FA}. %
 The steps of this classical trick are left to the reader. %
 %