showing uniqueness of a Hahn Banach extension

Question

I am trying to prove the following:

If $H$ is a Hilbert space and $G\subseteq H$ is a closed linear subspace, then any bounded linear functional on $G$ has a unique Hahn-Banach extension on $H$.

So far I have proven the first half:

Recall by the Riesz Representation Theorem, that if $X$ is a Hilbert space, and $h\in X^*$, then there is a unique $y\in X$ such that $h(x)=(x,y)$ for all $x\in X$. Also, since $G$ is a closed linear subspace of $H$, then $G$ itself is a Hilbert space. Thus, there exists a unique $y\in G$ such that $f_G(x)=(x,y)$ for all $x\in G$, where $f_G\in G^*$. Let $f\in H^*$ be defined by $f(x)=(x,y)$. Then we have $f|_G=f_G$, since $g\subseteq H$, so $f$ is an extension of $f_G$ on $H$. Now suppose there exists another extension $g\in H^*$ of $f_G$ such that $g(x)=(x,z)$ for some unique $z\in G$. Then...??

How can I show uniqueness? It seems kind of difficult. Thank you for your help!!

Answer 1

Hint: If $G$ is a closed linear subspace of $H$, it has an orthogonal complement $K$. A bounded linear functional on $H$ is determined by its restrictions to $G$ and $K$.

Answer 2

Using your notation, let $P$ be the orthogonal projection onto $G$. Then, for $x\in G$, $$ (x,y)=(x,z)=(Px,z)=(x,Pz). $$ So $(x, y-Pz)=0$ for all $x\in G$, implying that $y=Pz$. You also have that $\|y\|=\|h\|=\|g\|=\|z\|$. So $\|Pz\|=\|z\|$, which implies that $z\in G$, and so $z=y$.