Functional Analysis – Uniqueness of the Extension in Hahn-Banach Theorem

Question

Thinking about when we can talk about uniqueness in Hahn-Banach theorem i came up with the following conjecture: Let $H$ a Hilbert space and $M$ a closed subspace of it. Take $f\in M'$. The there exist an unique $F\in H'$ such as $F_{|M}=f$ and $||F||=||f||$. Here is my reasoning: since $M$ is a closed subspace of a Hilbert space, it's itself Hilbert, so by Riesz's theorem there exist an unique $u\in M$ such as $f(z)=\langle z,u\rangle,$ for every $z\in M$ and $||f||=||u||$. On the other way, by Hahn-Banach theorem, there exist $F\in H'$ such as $||F||=||f||$ and $F_{|M}=f$. Again, by Riesz's theorem, $F$ is uniquely determined by some vector $v\in H$ such as $||F||=||v||$. Since they are equal in $M$ and have the same norm I think that we can conclude that $u=v$ and that the extension is uniquely. Although I think it fails, the intuition says that it seems true. What do you think? Any help will be very appreciated.

Answer 1

You do not need the Hahn-Banach theorem for Hilbert spaces. The Riesz representation theorem suffices. If $f(z)=\langle z,u\rangle$ for $u\in M$ and all $z\in M,$ then $F(z)=\langle z,u\rangle $ for $z\in H$ is an extension such that $\|F\|=\|f\|=\|u\|.$ Any other extension $G$ should be of the form $G(z)=\langle z,v\rangle$ for some $v\in H$ and would satisfy $\langle z,v\rangle =\langle z,u\rangle$ for $z\in M.$ Hence $v-u\perp M,$ i.e. $v=u+u',$ where $u'\perp M.$ Thus $\|G\|=\|u+u'\|=\sqrt{\|u\|^2+\|u'\|^2}.$ If we require $\|G\|=\|f\|=\|u\|$ then $u'=0,$ hence $v=u.$

The uniqueness of the Hahn-Banach extension holds for strictly convex Banach spaces see, like $L^p$ spaces for $1<p<\infty.$