A bounded linear functional on a Hilbert space that is a Hahn-Banach extension of one on a subspace Let $M$ be a closed linear subspace of a Hilbert space $H$ and $g\in M*$(all bounded linear functional on $M$). Let $\pi$ be the orthogonal projection of H onto M, then $f=g\circ\pi$ is the only Hahn-Banach extension of g to H.
Can I show this by Riesz's Representation Theorem, g is uniquely determined by an element $v\in M$ and all functional should agree with this element?
 A: Write $f=g\circ \pi + h\circ(1-\pi)$. Wlog assume $\|f\|=\|g\|=1$. Let $x \in M^\perp$, then $f(x)=h(x)=:h$. Suppose this is $>0$ and let $\|x\|=1$. Let $\epsilon>0$, there exists a $y \in M$, $\|y\|=1$ so that $f(y)≥1-\epsilon$.
Note $\|\alpha x+ \beta y\|^2=|\alpha|^2 \|x\|^2+|\beta|^2\|y\|^2$, so if $(\alpha,\beta) \in S^2$ you have that $\|\alpha x +\beta y\|=1$. On the other hand
$$f(\alpha x+ \beta y)=\alpha h+ \beta(1-\epsilon)$$
Can this be made larger than $1$ for any $h \in (0,1]$? The question is equivalent to considering the case $\epsilon=0$, as $\epsilon$ can be made as small as you like, in any case smaller than $\beta \Delta$ where $\Delta$ would be the gap between the evaluation for $\epsilon=0$ and $1$.
Now write $\alpha=\sqrt{1-\beta^2}$ and find a maximum of
$$h\sqrt{1-\beta^2}+\beta$$
You find that it is taken for $\beta=\frac1{\sqrt{1+h^2}}$, and then plugging this into the expression gives
$$f(\alpha x +\beta y)=\frac{1}{\sqrt{1+h^2}}+h\sqrt{1-\frac{1}{1+h^2}}=\frac{1+h^2}{\sqrt{1+h^2}}$$
Which is $>1$ for $h>0$.
So any element of the dual that restricts to $g$ on $M$ but is not $g\circ \pi$ must have norm larger than $\|g\|$ and is not a Hahn-Banach extension of $g$.
