Let $H$ be a Hilbert space and $M\subset H$ a closed subspace. If $f\in M′$, the dual of M, show there is a unique $g\in H′$such that $g|_{M}=f$ and $‖g‖=‖f‖$.
I am using to solve it the Riesz theorem: "Let H a hilbert space. If $f \ in H'$ there is a unique $\eta \in H, f(x) = <x,\eta> $ and $||f||=||\eta||$ ".
My contribution:
I have that $M = \bar{M}$ cause it is closed. And also it's a Hilbert space. By Riesz, we have that if $f \in M$ there is a unique $\eta \in M$ such that $f(x) = <x,\eta>$ for all $x \in M$.
Since $M$ is a closed subspace of a Hilbert space, it is also a Hilbert space. By Riesz there is some $y\in M$ such that $f(x)=\langle x,y\rangle$ for all $x\in M$. It is known that $||f||=||y||$. There is the obvious extension $g\in H'$, $g(x)=\langle x,y\rangle$ for all $x\in H$. It is indeed a norm preserving extension.
So now we just have to show the uniqueness of the extension. Suppose $g\in H'$ is a norm preserving extension of $f$. Again, by Riesz there is $z\in H$ such that $g(x)=\langle x,z\rangle$ for all $x\in H$. Then $||g||=||z||$, and since $||g||=||f||$ we conclude that $||y||=||z||$. Now, for each $x\in M$ we have $f(x)=g(x)$ and hence:
$0=g(x)-f(x)=\langle x, z-y\rangle$
Hence the vector $z-y$ is orthogonal to $M$. Since $y\in M$ we conclude that $z-y, y$ are orthogonal vectors. So $||z||^2=||z-y+y||^2=||z-y||^2+||y||^2=||z-y||^2+||z||^2$. Hence $||z-y||^2=0$, which implies $y=z$. So $g$ is indeed the functional $g(x)=\langle x,y\rangle$, which proves the uniqueness of the extension.