Let $E, F$ be locally convex $\mathbb{K}$-vector spaces and let $T: E\to F$ be a continuous, linear function. Let $T': F'\to E'$ be defined by $T'(g)=g\circ T$. Then holds:
$T:E\to F$ is a topological isomorphism regarding the weak topology, iff $T'\colon F'\to E'$ is a topological isomorphism regarding the weak topology.
I already know, that $T$ and $T'$ are continuous regarding the weak topology. Here: weak-$\ast$-continuous, topological isomorphism, functional analysis
"$\Rightarrow$"
Since $T$ is a topological isomorphism we have a continuous $T^{-1}$. I have to show, that $T'$ is a topological isomorphism. Hence I have to give an inverse $(T')^{-1}$ which is continuous regarding the weak-$\ast$ topology.
I define $(T')^{-1}: E'\to F'$ by $(T')^{-1}(f)=f\circ T^{-1}$
Then $((T')^{-1})(T'(g))=((T')^{-1})(T(g))=T^{-1}(T(g))=g$ and
$T'((T')^{-1}(f))=T'(T^{-1}(f))=T(T^{-1}(f))=f$
Therefore my defined $(T')^{-1}$ is indeed inverse to $T'$. It is continuous, since it is a composition of continuous functions.
I write my attempts to "$\Leftarrow$" once we checked this. Thanks in advance.