Proof Clarification weak convergence

Question

I have been looking at this solution for a bit now and can't for the life of me figure out why $g \circ T \in X'$. I was hoping someone could shed some light on this for me. Thanks!

Answer 1

It is clear (as is noted in the comments) that $g \circ T$ is continuous and linear as a composition of continuous linear maps.

To see that $g \circ T$ is really in $X'$ notice that $T:X \to Y$ and $g: Y \to \mathbb{F}$ implies that $g \circ T: X \to \mathbb{F}$ (and not $g \circ T: Y \to \mathbb{F}$) so that $g \circ T \in X'$ (and not $g \circ T \in Y'$).