How to show that $\lambda\in V^{**}$ is in the image of $V$ iff it is weak-* continuous?

Question

Let $V$ be a Banach space and define the inclusion $\iota: V\to V^{**}, v\mapsto \left(\iota(v): f\in V^{*}\mapsto f(v)\in\mathbb{R}\right)$. I already showed that $\iota$ is a linear isometry, that it is a homeomorphism onto its image if $V$ has the weak topology and $V^{**}$ has the weak-* topology, and that $\iota(V)$ is weak-* dense in $V^{**}$. Now I have to prove that if $\lambda\in V^{**}$ then $\lambda\in \iota(V)$ if and only if $\lambda:V^*\to\mathbb{R}$ is weak-* continuous. The “only if” part is immediate, however I seem to can’t wrap my head around the “if” part. Has anyone a hint? Please hints only!

Answer 1

If $\lambda$ is weak-* continuous, then the set $\{f \in V^* : |\lambda(f)| < 1\}$ is a weak-* open neighborhood of 0. Now there is a nice basis of open sets for the weak-* topology, which perhaps you know. This will relate $\lambda$ to a finite number of elements $v_1, \dots, v_n \in V$. By looking at how $\ker \lambda$ relates to $\ker \iota(v_i)$, you can show that $\lambda$ is some linear combination of the $\iota(v_i)$.

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