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?


1 Answer 1


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 $\beta(F^*,F)$.

  • $\begingroup$ I am sorry for not saying it explicitely, but $^{F^{*}}$ means the closure in the norm topology on $F^{*}$. $\endgroup$
    – erz
    Aug 21, 2017 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). $\endgroup$ Aug 21, 2017 at 8:31
  • $\begingroup$ Yes, you are right, it should be the strong topology. $\endgroup$
    – erz
    Aug 21, 2017 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, 2017 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). $\endgroup$ Aug 21, 2017 at 11:09

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