0
$\begingroup$

Let $F$ be a metrizable locally convex space (you may assume it is a Banach space), and let $E$ be a complete locally convex space (you may assume it is a Frechet space). Let $T$ be a continuous linear map from $F$ into $E$ and let $H\subset E^{*}$ be such a subspace that $\overline{T^*H}^{F^*}=\overline{T^*E^*}^{F^*}$ (the closure in the strong topology; or in the norm-topology, if $F$ is a normed space, or equivalently, in the weak topology of $F^{*}$).

Let $B\subset F$ be an absolutely convex bounded closed set (you may assume that it is the unit ball).

Does $\sigma(E,H)$-compactness of $\overline{TB}^E$ imply that $\overline{TB}^E$ is $\sigma(E,E^{*})$-compact?

$\endgroup$
1
$\begingroup$

If the condition $\overline{T^*H}^{F^*}=\overline{T^*E^*}^{F^*}$ refers to the weak$^*$ topology on $F^*$ a counterexample is provided by any non-reflexive Banach space $X$ with $E=F=X^*$ (with the dual Banach space norm), $T=$ id, $H=X\subseteq X^{\ast\ast}$ and $B$ the unit ball of $X^*$: Since $X$ is $\sigma(X^{\ast\ast},X^*)$-dense you have $\overline{T^*H}^{F^*}=\overline{T^*E^*}^{F^*}=X^{\ast\ast}$, $B$ is $\sigma(E,H)=\sigma(X^*,X)$-compact by Alaoglu but not $\sigma(E,E^\ast)=\sigma(X^\ast,X^{\ast\ast})$-compact because then $X^{\ast\ast}$ and hence $X$ would be reflexive.


If $F$ is normed and $B$ is the unit ball of $F$ then the required condition $T^*E^* \subseteq \overline{T^*H}$ (closure with respect to the dual norm) implies $\sigma(E,H)|_A=\sigma(E,E^*)|_A$ for $A=\overline{T(B)}$:

Since $A$ is absolutely convex and $\sigma(E,E^*)$ is clearly finer than $\sigma(E,H)$ we have to show that for every $0$-neighbourhood $U\in {\scr U}_0(E,\sigma(E,E^*))$ there is $V\in {\scr U}_0(E,\sigma(E,H))$ with $V\cap A\subseteq U$. A typical $U$ is a finite intersection of sets of the form $\{|\varphi|<1\}$ with $\varphi\in E^*$. Since $\frac 12 B^\circ$ is a $0$-neighbourhood in $F^*$ (it is the closed ball around $0$ with radius $\frac 12$) the density of $T^*H$ gives $h\in H$ with $T^*\varphi -T^*h\in \frac 12 B^\circ$ and hence $$ \varphi -h \in \frac 12 \left(T^*\right)^{-1}(B^\circ)= \frac 12 T(B)^\circ = \frac 12 A^\circ.$$ This implies $\{|h| <\frac 12\} \cap A \subseteq \{|\varphi|<1\}$, and taking finite intersections gives a desired $V$.

This works for arbitrary locally convex spaces if you replace the dual norm by the strong topology $\beta(F^*,F)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I am sorry for not saying it explicitely, but $^{F^{*}}$ means the closure in the norm topology on $F^{*}$. $\endgroup$ – erz Aug 21 '17 at 8:16
  • $\begingroup$ But if, as you wrote, $F$ is a metrizable locally convex space there is no norm topology on $F^\ast$ (there are many topologies on $F^*$ among them the weak$^*$ topology is the most prominent). $\endgroup$ – Jochen Wengenroth Aug 21 '17 at 8:31
  • $\begingroup$ Yes, you are right, it should be the strong topology. $\endgroup$ – erz Aug 21 '17 at 9:02
  • $\begingroup$ Thank you! Do you know if the statement is valid if we only know that $T(B)$ is $\sigma(E,H)$-precompact? $\endgroup$ – erz Aug 21 '17 at 10:23
  • $\begingroup$ If two locally convex topologies coincide on an absolutely convex set then the induced uniformities also coincide (I am quite sure that this is true but I don't know where I learned that, it might be in Bourbaki or Köthe). $\endgroup$ – Jochen Wengenroth Aug 21 '17 at 11:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.