How can we use Hahn-Banach Theorem to prove a functional is linear and bounded?

Question

Let $V$ and $W$ be Banach spaces, bounded linear operators $T: V \to W$ and $S: W' \to V'$ such that \begin{align*} f(T(v)) = S(f)(v), \qquad \forall v \in V, \quad \forall f \in W' \end{align*}

1. Prove $f$ is linear, by using Hahn-Banach Theorem.
2. Prove $f$ is bounded.

To prove $f$ is linear, we usually need to show $f(aw_1 + bw_2) = af(w_1) + bf(w_2)$ for $w_1, w_2 \in W$ and $a,b \in \mathbb{C}$. But how can we apply Hahn-Banach Theorem for this? Also, for the boundedness.

Answer 1

As stated, the question makes little sense. If you say that $S$ is bounded, you are requiring that the elements in its domain belong to a normed space. Compounded with the fact that you state the relation between $S$ and $T$ using all $f$, but then as about properties of $f$.

It makes more sense that the problem is about showing that the adjoint is bounded, so the goal is to show that $S$ is linear and bounded.

The linearity of $S$ is direct, since $$ S(af+g)(v)=(af+g)(Tv)=a\,f(Tv)+g(Tv)=a\,S(f)v+S(g)v=(aSf+Sg)v. $$ for boundedness, $$ |S(f)v|=|f(Tv)|\leq\|f\|\,\|Tv\|\leq\|f\|\,\|T\|\,\|v\|. $$ The above inequality show that $\|Sf\|\leq\|\,\|T\|\,\|f\|$, so $\|S\|\leq\|T\|$.