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Let $E, F$ be locally convex $\mathbb{K}$ vector spaces and $T: E\to F$ linear and continuous. Let $T': F'\to E'$ be defined through $T'(g)=g\circ T$

$T': F'\to E'$ is weak-$\ast$-continuous (hence continuous when $F'$ and $E'$ are provided with the weak-$\ast$ topology)

This question is related to $T: E\to F$ weak continuous, $E, F$ locally convex and unfortunatly I do not know where to start with the proof. I struggle with the definition of the weak topology and the weak-$\ast$-topology.

Can you help me, or give me a hint. Thanks in advance.

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Assume $g_{\delta}\rightarrow g$ in $w^{\ast}$ of $F'$, and let $x\in E$, then $T(x)\in F$ and hence $|T'(g_{\delta})(x)-T'(g)(x)|=|g_{\delta}\circ T(x)-g\circ T(x)|=|g_{\delta}(T(x))-g(T(x))|\rightarrow 0$.

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  • $\begingroup$ This might be a dumb question, but why is it enough to show this? And why holds $|g_\delta(T(x))-g(T(x))|\to 0$ $\endgroup$ – Cornman Nov 28 '17 at 18:31
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    $\begingroup$ The point is to show that the seminorm $p_{x}(T'(g_{\delta})-T'(g))\rightarrow 0$ (actually it is sufficient to show at the zero, that is, $g=0$), here $p_{x}(\eta)=|\eta(x)|$, $\eta\in E'$. Why that holds? Because $g_{\delta}$ converges to $g$ $w^{\ast}$ of $F'$, that means, for any $y\in F$, $p_{y}(g_{\delta}-g)=|g_{\delta}(y)-g(y)|\rightarrow 0$. $\endgroup$ – user284331 Nov 28 '17 at 18:33
  • $\begingroup$ Is $p_x$ a special seminorm, or just a seminorm? I could also assume $g_\delta\to 0$ in $w^\ast$, since it is sufficient to show it at zero. Hence $|T'(g_\delta(x)-T'(0)(x)|=|g_\delta(T(x))-0|\to 0$ since $g_\delta\to 0$ $\endgroup$ – Cornman Nov 28 '17 at 18:43
  • $\begingroup$ Yes, as I said it suffices to do for $0$. The weak star topology is generated by the bunch of seminorms $p_{x}(\eta)=|\eta(x)|$, and the convergence must satisfy all these seminorms, that is, for any fixed $x$, one must have $p_{x}(\eta_{\delta})=|\eta_{\delta}(x)|\rightarrow 0$. $\endgroup$ – user284331 Nov 28 '17 at 18:45
  • $\begingroup$ Do you have a hint on the linked question too? I would love to figue that out mostly on my own, but I am stuck. :( $\endgroup$ – Cornman Nov 28 '17 at 18:49

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