The Hahn-Banach Theorem for Normed Space:
Let $X$ be a real or complex normed space and let $W$ be a linear subspace of $X$. If $f_W \in W'$ (the dual of $W$), then there exists an extension $f \in X'$ such that $\|f\|=\|f_w\|$.
How if I extend to a Hilbert Space?
Best Answer
Alright, I think that the following proof works but I am by no means an expert in linear functional analysis so please consider this carefully and decide if it is convincing to you.
Without loss of generality, we can take $W \subset H$ to be a closed linear subspace of $H$ since any continuous map on $W$ can be uniquely extended to $\overline W$ by density.
Since $W$ is closed, it is itself a Hilbert space and so by the Riesz representation theorem, there is $x_f \in W$ such that $$f(w) = \langle w, x_f \rangle, \,\,\,\, \forall w \in W.$$ Using this, we can define and extension of$f$ to $H$. Indeed, define $$F(h) := \langle h, x_f \rangle, \,\,\,\, \forall h \in H$$ and certainly $F$ is a linear extension of $f$ and we will have $$\|F \|_{H'} = \|x_f \|_H = \|f \|_{W'}.$$
Now let $G \in H'$ be another linear extension of $f$ to $H$ guaranteed by the Hanh-Banach theorem. Applying the Riesz Representation Theorem to $G$, we find $x_G \in H$ such that $$G(h) = \langle h, x_G \rangle, \,\,\,\,\, \forall h \in H$$ and since both theorems preserve norms, we have $$\|f\|_{W'} = \|G \|_{H'} = \|x_G\|_H.$$
Now since $F,G$ both extend $f$, they agree on $W$: $$F(w) = G(w) \,\,\, \Longleftrightarrow \langle w, x_f - x_G \rangle =0, \,\,\,\,\, \forall w \in W.$$ In particular, since $x_f \in W$, this shows that $x_f$ and $x_f - x_G$ are orthogonal. Then using our norm equalities and the Pythagorean theorem, we have $$\|x_f\|_H = \|f\|_{W'} = \|G\|_{H'} = \|x_G\|_{H} = \|x_f +(x_G - x_f)\|_H = \sqrt{\|x_f\|^2_H + \|x_f - x_G\|_H^2}.$$ This is possible iff $\|x_f - x_G \|_H = 0$ and so we conclude that $x_f = x_G$ which implies that $F = G$. Thus any extension of $f$ is equal to $F$.