Exercise 1.7

Question

\def\since{\Leftarrow}
\def	herefore{\Rightarrow}
\def	hen{\Rightarrow}
\def\eg{	extit{e.g.} }
\def\ie{	extit{i.e.} }
\def\Def{	riangleq}
\def\N{\usualSet{N}}
\def\Q{\usualSet{Q}}
\def\R{\usualSet{R}}
\def\C{\usualSet{C}}
\def\minus{\insmall{-}}
Let be X the vector space of all complex functions on the unit interval 
$[0, 1]$, topologized by the family of seminorms 
%
  \begin{align}
    \mathit{p_{x}(f\,)=|f(x)| \quad\quad (0\leq x\leq 1).
onumber}
  \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 $n 	o\infty$, but (b) if $\{\gamma_n\}$ is any sequence of scalars such 
that $\gamma_n	o\infty$ 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.

Cordier · Accepted Answer

\def\eg{ extit{e.g.} } \def\ie{ extit{i.e.} } % Usual sets ewcommand\usualSet[1]{{\mathbf #1}} \def\N{\usualSet{N}} \def\Q{\usualSet{Q}} \def\R{\usualSet{R}} \def\C{\usualSet{C}} \def\minus{\insmall{-}} % Limits ewcommand{ endsto}[2]{\underset{#1 o #2}{\longrightarrow}} %Local base ewcommand{\localbase}[1]{\mathscr #1} \begin{proof} 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) riangleq \{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 $ au$ on $X$; % see Section 1.37 of extit{Functional Analysis}. % So, % \begin{align} % \label{Inequality boolean series} % \sum_{n=1}^\infty [f_n otin \cap_{i=1}^m U_i] \leq \sum_{n=1}^\infty \sum_{i=1}^m [f_n otin U_i] = \sum_{i=1}^m \sum_{n=1}^\infty [f_n otin U_i] \quad (f_n \in X, U_i \in au). \end{align} % Now assume that $\{f_n\}$ $ au$-converges to some $f$, \ie % \begin{align} % \sum_{n=1}^\infty [f_n otin f + W] < \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 $n$. % In other words, $\{f_n(x)\}$ converges to $f(x)$. % % Conversely, assume that $\{f_n\}$ does not $ au$-converges in $X$, \ie % % \begin{align} % \label{Divergence} % \forall f \in X, \exists W \in \localbase{B}: \sum_{n=1}^\infty [f_n otin f + W] = \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, % % \begin{align} \sum_{i=1}^m \sum_{n=1}^\infty [f_n otin f + V(x_i, k_i)] % \overset{( ef{Inequality boolean series})}{\geq} % \sum_{n=1}^\infty [f_n otin f + W] % \overset{( ef{Divergence})}{=} % \infty . \end{align} % We can now conclude that, for some index $i$, % \begin{align} \sum_{n=1}^\infty [f_n otin f + V(x_i, k_i)] = \infty . \end{align} % In other words, $\{f_n(x_i)\}$ fails to converge to $f(x_i)$. We have so % proved that $ au$-convergence is a rewording of pointwise convergence. % % % 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 extit{unique} binary expansion. % More precisely, there exists a bijection % % \begin{align} b: [0, 1] \setminus \Q & o \{ \beta \in \{0, 1\}^{\N_+}: \beta ext{ is not eventually periodic} \} \end{align} %% where % % $b(\alpha) = (\beta_1, \beta_2, \dots)$ % % is the only bit stream such that % % \begin{align} \label{definition of alpha} \alpha = \sum_{k=1}^\infty \beta_k 2^{\minus k}. \end{align} % First, remark that % % $b(\alpha)_1 + \dots + b(\alpha)_n endsto{n}{\infty}\infty$, % % since % % $b(\alpha)$ % has infinite support. Next, fix % % \begin{equation} \label{definition of f_n(alpha)} f_n(\alpha) riangleq % \frac{1}{b(\alpha)_1 + \cdots + b(\alpha)_n} endsto{n}{\infty} 0 \end{equation} % wherever $b(\alpha)_1+ \cdots + b(\alpha)_n > 0$. % All other values $f_n(x)$ are of no interest. % For instance, put $f_n(x) = 0$. % % Now take an arbitrary % % $\gamma_n \longrightarrow \infty$: % % Given $p=1, 2, 3, \dots$, $\gamma_n$ is greater than $p$ % for all but finitely many $n$. % % Next, we choose $n_p$ among those extit{almost all} $n$ that are large enough to additionally satisfy % % \begin{align} \label{definition of n_p} n_p - n_{p-1} > p \longrightarrow \infty, \end{align} % provided $n_0=0$. This way, the distribution of % % $n_1, n_2, \dots$, % % extit{displays no periodic pattern}. % In other words, the extit{characteristic function} % % $\chi: k\mapsto [k \in \{n_1, n_2, \dots\}]$ % % is not eventually periodic. % Combined with ( ef{definition of alpha}), this establishes that % \def\xgamma{\alpha_{\gamma}} % \begin{align} \xgamma riangleq \sum_{k=1}^\infty \chi_k 2^{\, \minus k} % otin \Q \end{align} % is irrational. Conversely, still with ( ef{definition of alpha}), % % \begin{align} b(\xgamma)_{k} = \chi_{k}. \end{align} % Moreover, it follows from the very definition of $\chi$ that % % \begin{align} \chi_1 + \cdots + \chi_{n_{1}} & + \cdots + \chi_{n_{p}} = p. \end{align} % Hence % % \begin{align} \gamma_{n_p} f_{n_{p}}(\xgamma) = \frac{\gamma_{n_p}}{p} > 1. \end{align} % There so exists a subsequence $\{\gamma_{n_p}\}$ such that % % $\{\gamma_{n_p} f_{\gamma_{n_p}}\}$ % % fails to converge pointwise to $0$. % Since $\{\gamma_{n}\}$ was arbitrary, this proves (b). \end{proof} % END

Exercise 1.7

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Exercise 1.7

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