This is problem 3.14.10 from Conway's functional anaylsis (abridged): If $\mathcal{X}$ is a real Banach space with dual $\mathcal{X}^*$, and $(S,d)$ is a metric space, then show $f: S \to \mathcal{X}$ is Lipshitz if $L\circ f: S \to \mathbb{R}$ is Lipshitz for all $L\in \mathcal{X}^*$.
I've hit a road block in the proof I'm working on, inspired by a comment in this post: Lipschitz map between metric and normed spaces. Define $\phi: \mathcal{X}^* \to Lip(S)$ by $\phi(L) = L\circ f$, which is well defined by assumption, and $Lip(S)$ is the Banach space endowed with the norm $\|g\|_{Lip} = \sup_{x\ne y} |g(x)-g(y)|/d(x,y)$. It is fairly routine to show using the closed graph theorem that $\phi$ is bounded (take $L_n \to 0$ and assume $\phi(L_n ) \to y$. It follows from the triangle inequality that $y=0$.)
This is where I'm lost-- now we are supposed to conclude using Hahn-Banach that $f$ is indeed lipshitz. I cannot see how Hahn Banach can be applied here... What is the functional defined on a subspace that we are extending? Can anyone fill in the details?
Boundedness of $\phi$ says that $|L(f(x))-L(f(y))|\leq \|\phi\| d(x,y)$ for all $L \in \mathcal X^{*}$. This implies $\|f(x)-f(y)\|\leq \|\phi\| d(x,y)$.