UPDATE: Rewrote 1.7, added a proof that uses the given hint. One small change in 2_15.tex.

Author: gitcordier

chapter_1/1_7.tex

@@ -1,183 +1,194 @@
-\textit{
-Let be $X$ the vector space of all complex functions on the unit interval %
-$[0, 1]$, topologized by the family of seminorms %
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+% FunctionalAnalysis 
+% 1_7.tex
+% 
+% encoding: UTF-8 
+% EOL: LF
+%
+% format: LaTeX
+% indent: spaces (2)
+% width: 12
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+\textit{%
+Let be $X$ the vector space of all complex functions on the unit interval $[0, 1]$, topologized by the family of seminorms %
 %
 \begin{align}
- p_x(f)= |f(x)\ \quad (0\leq x\leq 1).\nonumber
+ p_x(f)= \magnitude{f(x)} \quad (0\leq x\leq 1).\nonumber
 \end{align}
 %
-This topology is called the topology of pointwise convergence. %
-Justify this terminology. %
-%
-Show that there is a sequence $\{f_n\}$ in X such that (a) %
-%
- $\{f_n\}$ converges to $0$ as $n \to\infty$, %
-%
-but (b) if $\{\gamma_n\}$ is any sequence of scalars such that %
-%
- $\gamma_n\to\infty$ %
-%
-then $\{\gamma_n f_n\}$ does not converge to $0$. %
-(Use the fact that the collection of all complex sequences converging to $0$ %
-has the same cardinality as $[0, 1]$.) %
-%
-This shows that metrizability cannot be omited in (b) of Theorem 1.28.
+This topology is called the topology of pointwise convergence. Justify this terminology. Show that there is a sequence % 
+$\singleton{f_n}$ in X such that (a) $\singleton{f_n}$ converges to $0$ as $n \to \infty$, but (b) if $\{\gamma_n\}$ is any %
+sequence of scalars such that $\gamma_n \to \infty$ then $\{\gamma_n f_n\}$ does not converge to $0$. (Use the fact that %
+the collection of all complex sequences converging to $0$ has the same cardinality as $[0, 1]$.) This shows that %
+metrizability cannot be omited in (b) of Theorem 1.28.
 }
 %
+
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+% First part: Justification of the terminology
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+\subsection{Justification of the terminology}
 \begin{proof}
-The family of the seminorms $p_x$ is separating: %
-The collection $\mathscr{B}$ of all finite intersections of the sets %
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+% A. tau-convergence => pointwise convergence. 
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+The family of the seminorms $p_x$ is separating: The collection $\mathscr{B}$ of all finite intersections of the sets %
 %
 \begin{align}
  V(x,k) \triangleq \{p_x < 2^{\minus k}\} 
  \quad (x \in [0, 1], k=1, 2, 3, \dots)
 \end{align}
 %
-is therefore a local base for a topology $\tau$ on $X$; %
-see Section 1.37 of \cite{FA}. %
-So, 
+is therefore a local base for a topology $\tau$ on $X$; see Section 1.37 of \cite{FA}. So, 
 %
 \begin{align}
  %
  \label{Inequality boolean series}
  %
- \sum_{n=1}^\infty [f_n \notin \cap_{i=1}^m U_i] \leq 
- \sum_{n=1}^\infty \sum_{i=1}^m [f_n \notin U_i] = 
- \sum_{i=1}^m \sum_{n=1}^\infty [f_n \notin U_i]
- \quad (f_n \in X, U_i \in \tau).
+ \sum_{n=1}^\infty \bigl[f_n \notin \cap_{i=1}^m U_i\bigr] 
+ \leq 
+ \sum_{n=1}^\infty \sum_{i=1}^m \bigl[f_n \notin U_i\bigr] 
+ = 
+ \sum_{i=1}^m \sum_{n=1}^\infty \bigl[f_n \notin U_i\bigr] \quad (f_n \in X, U_i \in \tau).
 \end{align}
 %
 Now assume that $\{f_n\}$ $\tau$-converges to some $f$, \ie 
 %
 \begin{align}
  %
- \sum_{n=1}^\infty [f_n \notin f + W] < \infty \quad (W \in \mathscr{B}).
+ \sum_{n=1}^\infty \bigl[f_n \notin f + W\bigr] < \infty \quad (W \in \mathscr{B}).
 \end{align}
-The special case %
-%
-$W = V(x, k)$ %
-%
-means that, given $k$, %
-%
-$|f_n(x) - f(x)| < 2^{\minus k}$ %
 %
-for almost all $n$. %
+The special case $W = V(x, k)$ means that, given $k$, $\magnitude{f_n(x) - f(x)} < 2^{\minus k}$ for almost all $n$. %
 In other words, $\{f_n(x)\}$ converges to $f(x)$. %
-%
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+% B. tau-divergence => pointwise divergence. 
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
 Conversely, assume that $\{f_n\}$ does not $\tau$-converges in $X$, \ie %
 %
 \begin{align}
  %
  \label{Divergence}
  %
- \forall f \in X, \exists W \in \localbase{B}: 
- \sum_{n=1}^\infty [f_n \notin f + W] = \infty. 
+ \forall f \in X, \exists W \in \localbase{B}: \sum_{n=1}^\infty \bigl[f_n \notin f + W\bigr] = \infty. 
  %
 \end{align}
 %
-$W$ is now the (nonempty) intersection of finitely many $V(x, k)$, say %
-%
- $V(x_1, k_1), \dots, V(x_m, k_m)$. %
-%
-Thus, %
+$W$ is then the intersection of some $V(x_1, k_1), \dots, V(x_m, k_m)$. Hence %
 %
 \begin{align}
- \sum_{i=1}^m \sum_{n=1}^\infty [f_n \notin f + V(x_i, k_i)]
- %
- \overset{(\ref{Inequality boolean series})}{\geq}
- %
- \sum_{n=1}^\infty [f_n \notin f + W]
- % 
- \overset{(\ref{Divergence})}{=}
- %
- \infty .
+ \infty 
+ \citeq{\ref{Divergence}} \sum_{n=1}^\infty \bigl[f_n \notin f + W\bigr] 
+ \citeleq{\ref{Inequality boolean series}} \sum_{i=1}^m \sum_{n=1}^\infty \bigl[f_n \notin f + V(x_i, k_i)\bigr].
 \end{align}
 %
-We can now conclude that, for some index $i$, 
+It is now clear that % 
 %
 \begin{align}
- \sum_{n=1}^\infty [f_n \notin f + V(x_i, k_i)] = \infty .
+ \sum_{n=1}^\infty \bigl[f_n \notin f + V(x_i, k_i)\bigr] = \infty .
 \end{align}
 %
-In other words, $\{f_n(x_i)\}$ fails to converge to $f(x_i)$. We have so %
-proved that $\tau$-convergence is a rewording of pointwise convergence. %
+for some $i$. In other words, $\{f_n(x_i)\}$ fails to converge to $f(x_i)$. The overall conclusion is that %
+$\tau$-convergence is the $X$'s version of pointwise convergence. %
+\end{proof}
 %
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
 % SECOND PART
-We now establish the second part by constructing a specific sequence %
-$\{f_n\}$ that satisfies both (a) and (b). \\
-\\
-The proof will be based on the following well-known result: %
-Each irrational number $\alpha$ has a \textit{unique} binary expansion. %
-More precisely, there exists a bijection %
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
 %
-\begin{align}
- b: [0, 1] \setminus \Q &\to \set{\beta \in \{0, 1\}^{\N_+}}{\beta \text{ is not eventually periodic}}\\
- \alpha &\mapsto (\beta_1, \beta_2, \dots)\nonumber
-\end{align}
-%%
-where $(\beta_1, \beta_2, \dots)$ is the only bit stream such that %
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+% First proof: Use the given hint
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+\subsection{Proof (with the given hint)}
+We now prove the second part by constructing a specific sequence $\{f_n\}$ that simultaneously satisfies (a) and (b). %
+Indeed, the hint says that there exists a one-to-one and \textit{onto} mapping %
 %
 \begin{align}
- \label{definition of alpha}
- \alpha = \sum_{k=1}^\infty \beta_k 2^{\minus k}.
+ \varphi: [0, 1] & \to \set{\theta}{\theta \in \R^{\N_+}, \lim_{\infty} \theta = 0}.\\
+  x & \mapsto (\theta_1, \dots, \theta_n, \dots) \nonumber
 \end{align}
 %
-First, remark that $\beta_1 + \beta_2 + \dots = \infty$ then define %
+\begin{proof}
+We set (under the same notation)%
 %
 \begin{align}
- \label{definition of f_n(alpha)}
- %f_n(\Q) &\Def \singleton{0}, \\
- f_n(\alpha) &\Def \frac{1}{1 + \beta_1 + \dots + \beta_n}, \,\, f_n = 0 \text{ on } \Q, 
+ f_n(x) & \Def \theta_n \tendsto{n}{\infty} 0 \quad \bigl(x = \varphi^{\minus 1}(\theta_1, \dots, \theta_n, \dots)\bigr)
 \end{align}
 %
-so that $f_n$ tends to $0$ (pointwise). %
-%
-Independently, take an arbitrary % 
+so that %
 %
-$\gamma_n \longrightarrow \infty$: % 
+ $x_\gamma = \varphi^{\minus 1}\Bigl( 1 / \sqrt{1 + \magnitude{\gamma_1}}, \dots,1 / \sqrt{1 + \magnitude{\gamma_n}}, \dots\Bigr)$ %
 %
-Given $\counting{p}$, $\gamma_n$ is greater than $2^p$ %
-for all but finitely many $n$. %
-%
-Next, we choose $n_p$ among those \textit{almost all} $n$ that are 
-large enough to additionally satisfy %
+implies %
 %
 \begin{align}
- \label{definition of n_p}
- n - n_{p-1} > p, 
+ \gamma_n f_n(x_\gamma) = \gamma_n / \sqrt{1 + \magnitude{\gamma_n}} %
+ \underset{\infty}{\sim} \sqrt{\gamma_n} %
+ \tendsto{n}{\infty} \infty, 
 \end{align}
 %
-provided $n_0=0$. This way, the distribution of %
+provided $\gamma_n \to \infty$. This proves (b), since 
+%$\bigl(\gamma_n f_n(x_\gamma)\bigr)_n $ $\singleton{\gamma_nf_n(x_\gamma)}_{n=1}^{\infty}$ 
+$\set{\gamma_n f_n(x_\gamma)}{\counting{n}}$ diverges.
+\end{proof}
 %
- $n_1, n_2, \dots$, %
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+% Second proof: No hint, the hard way.
+% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+\subsection{Proving it the hard way (no hint)}
+We will use the following simple proposition about binary expansions: each irrational has an eventually aperiodic binary %
+expansion. %
 %
-\textit{displays no periodic pattern}: %
-Setting $\beta_{\gamma, k} = [k \in \{n_1, n_2, \dots \}]$ yields %
+\begin{proof}
+Formally, there exists a one-to-one and \textit{onto} mapping %
+%
+\def\bin{\text{\fw bin}}
+\begin{align}
+ \bin: [0, 1] \setminus \Q &\to \bigl\{\beta \in \{0, 1\}^{\N_+}: \beta \text{ is eventually aperiodic}\bigr\}\\
+ v &\mapsto (\beta_1, \dots, \beta_n, \dots)\nonumber
+\end{align}
+%%
+where $(\beta_1, \dots, \beta_n, \dots)$ satisfies %
 %
 \begin{align}
- \alpha_\gamma & \Def \sum_{k=1}^\infty \beta_{\gamma, k} 2^{\, \minus k} \notin \Q.
+ \label{definition of alpha}
+ v &= \beta_1 / 2 + \cdots + \beta_n / 2^n + \cdots .
 \end{align}
 %
-Moreover, %
+First, note that $\beta_1 + \beta_2 + \beta_3 + \cdots = \infty$ then define (under the same notation)%
 %
 \begin{align}
- p &= \beta_{\gamma, 1} + \dots + \beta_{\gamma, n_1} + \dots + \beta_{\gamma, n_p}.
+ \label{definition of f_n(alpha)}
+ f_n(v) &\Def 1/2^{\beta_1 + \cdots + \beta_n} \tendsto{n}{\infty} 0.
 \end{align}
 %
-Hence %
+Now pick an arbitrary $\gamma_n \to \infty$: Given a positive integer $k$, $\gamma_n > 4^k$ for almost all $n$, %
+say $\tilde{n}$. Next, choose $n=n_k$ from $\singleton{\tilde{n}}$ so large that %
 %
 \begin{align}
- \gamma_{n_p} f_{n_{p}}(\alpha_\gamma) 
- = \frac{\gamma_{n_p}}{p+1} 
- > \frac{2^p}{p+1}
- \tendsto{p}{\infty} \infty.
+ \label{definition of n_k}
+ n_{k+1} - n_k > k + 1 \to \infty .
 \end{align}
 %
-There so exists a subsequence $\{\gamma_{n_p}\}$ that prevents %
+This way, $\chi = 1_{\{ n_1, n_2, \dots \}}$ \textit{is not eventually periodic on }$\N_+$! Moreover, one easily checks that %
+the specialization $\beta = \chi = \bin (v_\gamma)$ yields %
+%
+\begin{align} \label{p sum of bits}
+ \beta_1 + \dots + \beta_{n_1} + \dots + \beta_{n_k} = k.
+\end{align}
 %
-$\{\gamma_{n_p} f_{\gamma_{n_p}}\}$ %
+Finally, combining (\ref{definition of f_n(alpha)}) with (\ref{p sum of bits}) shows that %
 %
-to converge pointwise to $0$. %
-Since $\{\gamma_{n}\}$ was arbitrary, this proves (b).
+\begin{align}
+ \gamma_{n_k} f_{n_{k}}(v_\gamma) 
+ = {\gamma_{n_k}}/{2^k} 
+ > 4^k / 2 ^k 
+ > 2^k 
+ \tendsto{k}{\infty}\infty.
+\end{align}
+%
+As a conclusion, every $\gamma_n \to\infty$ contains a subsequence $\{\gamma_{n_k}\}$ that keeps %
+$\gamma_{n_k} f_{n_k}(v)$ away from $0$ (for some $v=v_\gamma$) when $k \to \infty$; which is (b). 
 \end{proof}
+%
 % END

chapter_2/2_15.tex

@@ -18,20 +18,19 @@
 % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
 Assume that $X = V_n \sqcup \overline{E}_n$ ($n=1, 2, 3, \dots$) such that %
 %
- \renewcommand{\labelenumi}{(\roman{enumi})} 
- \begin{enumerate}
- \item $X = \overline{V}_n$; 
- \item $ X\setminus Y = \bigcup_{n=1}^\infty E_n$.
- \end{enumerate}
- \renewcommand{\labelenumi}{(\alph{enumi})} 
-%
-First, let $x$ range over $X$, so that $x + V_n$ is open and dense as well %
-(because%
+\renewcommand{\labelenumi}{(\roman{enumi})} 
+\begin{enumerate}
+ \item $X = \overline{V}_n$; 
+ \item $ X\setminus Y = \bigcup_{n=1}^\infty E_n$.
+\end{enumerate}
+\renewcommand{\labelenumi}{(\alph{enumi})} 
+%
+First, let $x$ range over $X$, so that $x + V_n$ is open and dense as well (because%
 \footnote{
  This is also a special case of \citeresultFA{1.3 (b)}, where: %
  $X = x + X \subset \overline{x + V_n}$.
 }
-the translation by $x$ is an homeomorphism of $X$ onto $X$).
+the translation by $x$ is a homeomorphism of $X$ onto $X$).
 %
 % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
 % B. Where Baire's theorem is involved.
@@ -69,6 +68,7 @@
 \end{align}
 %
 since $Y$ is a subspace (subgroup) of $X$. We have so established that %
+%
 \begin{align}
  X \subset Y, 
 \end{align}