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).



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




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


