Exercise 3.11

Question

\def\weakstar{	ext{weak}^\ast	ext{-}}

Let $X$ be an infinite-dimensional Fréchet space. %
Prove that $X^\ast$, with its $\weakstar$topology, %
is of the first category in itself.

Cordier · Accepted Answer

\def\since{\Leftarrow} \def hen{\Rightarrow} \def\eg{ extit{e.g.} } \def\ie{ extit{i.e.} } \def\weakstar{ ext{weak}^\ast ext{-}} \def\Def{ riangleq} \def\N{\usualSet{N}} \def\R{\usualSet{R}} \def\C{\usualSet{C}} \def\minus{\insmall{-}} This is actually a consequence of the below lemma, % which we prove first. % The proof that $X^\ast$ is of the first category in itself comes right after, % as a corollary.% % \paragraph{Lemma.}% If $X$ is an infinite dimensional topological vector space whose dual % % $X^\ast$ % % separates points on $X$, then the polar % \begin{align} K_A\Def \{ \Lambda \in X^\ast:\, \magnitude{\Lambda} \leq 1 ext{ on } A\} \end{align} % of any absorbing subset $A$ is a $\weakstar$closed set that has empty interior. % \begin{proof}% Let $x$ range over $X$. The linear form % % $\Lambda \mapsto \Lambda x$ % % is $\weakstar$continuous; see \citeresultFA{3.14}. % Therefore, % % $P_x= \set{\Lambda \in X^\ast}{\magnitude{\Lambda x} \leq 1}$ % % is $\weakstar$ closed: % %In particular, every $P_a$ ($a \in A$) is $\weakstar$closed: % As the intersection of $\set{P_a}{a\in A}$, % $K_A$ is also a $\weakstar$closed set. % We now prove the second half of the statement. % % % ewline ewline oindent % From now on, $X$ is assumed to be endowed with its weak topology: % $X$ is then locally convex, but its dual space is still % % $X^\ast$ (see \citeresultFA{3.11}). % % Put % % \begin{align} W \Def \bigcap_{x\in F} \set{\Lambda \in X^\ast}{\magnitude{\Lambda x} < r_x}, \end{align} % where $r_x$ runs on $\R_+$, % as $F$ runs through the nonempty finite subsets of $X$. % % Clearly, the collection of all such $W$ is a local base of $X^\ast$. % Pick one of those $W$ and remark that the following subspace % % \begin{align} M \Def ext{span}(F) \end{align} % is finite dimensional. % Assume, to reach a contradiction, that $A\subset M$. % So, every $x$ lies in $t_xM=M$ for some $t_x>0$, since $A$ is absorbing. % As a consequence, $X=M$ is finite dimensional, which is a desired contradiction. % We have just established that $A ot\subset M$: % Now pick $\varit{a}$ in $A\setminus M$ and so conclude that %f % \begin{align} b\Def \frac{a}{t_a}\in A \end{align} % Remark that $b otin M$ (otherwise, $a = t_a b \in t_a M=M$ would hold) % and that $M$ is closed (see \citeresultFA{1.21 (b)}): % By the Hahn-Banach theorem \citeresultFA{3.5}, % there exists $\Lambda_a$ in $X^\ast$ such that % \begin{align}\label{3.11. Polar_4.} \Lambda_a b > 2 \end{align} % and % \begin{align}\label{3.11. Polar_4bis.} \Lambda_a (M) = \singleton{0}. \end{align} % The latter equality implies that $\Lambda_a$ vanishes on $F$; % hence $\Lambda_a$ is an element of $W$. % On the other hand, given an arbitrary $\Lambda \in K_A$, % the following inequalities % % \begin{align} \magnitude{\Lambda_a b + \Lambda b} \geq 2 - \magnitude{\Lambda b} > 1. \end{align} % show that $\Lambda + \Lambda_a$ is not in $K_A$. % % We have thus proved that % \begin{align} \Lambda + W ot\subset K_A. \end{align} % Since $W$ and $\Lambda$ are both arbitrary, this achieves the proof. % \end{proof} % % ewline ewline oindent We now give a proof of the original statement. % % \paragraph{Corollary.}% If $X$ is an infinite-dimensional Fréchet space, % then $X^\ast$ is meager in itself. % \begin{proof}% From now on, $X^\ast$ is only endowed with its $\weakstar$topology. % Let $d$ be an invariant distance that is compatible with the topology of $X$, % so that the following sets % \begin{align} B_n \Def \set{x\in X}{d(0, x) < 1/n}\quad (\counting{n}) \end{align} % form a local base of $X$. % % If $\Lambda$ is in $X^\ast$, then % % \begin{align} \magnitude{\Lambda} \leq m ext{ on } B_n \end{align} % for some $(n, m) \in \singleton{1, 2, 3, \dots}^2$; see \citeresultFA{1.18}. % % Hence, $X^\ast$ is the countable union of all % % \begin{align}\label{3.11. Countable union.} m\cdot K_n \quad (\counting{m,n}), \end{align} % where $K_n$ is the polar of $B_n$. % Clearly, showing that every $m\cdot K_n$ is nowhere dense % is now sufficient. % To do so, we use the fact that $X^\ast$ separates points; % see \citeresultFA{3.4}. % As a consequence, the above lemma implies % % \begin{align} \left({\overline{K}_n} ight)^\circ = \left({{K}_n} ight)^\circ=\emptyset. \end{align} % Since the multiplication by $m$ is an homeomorphism (see \citeresultFA{1.7}), % this is equivalent to % % \begin{align}\label{3.11. Nowhere dense.} \left(\,{\overline{m\cdot K_n}}\, ight)^\circ = %\left(\, m \cdot \overline{{{K}_n}} \, ight)^\circ % = %\left(m \cdot K_n ight)^\circ % = m\cdot \left({{K}_n} ight)^\circ = \emptyset. \end{align} % So ends the proof.% \end{proof}

Exercise 3.11

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Exercise 3.11

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