Hilbert spaces and unique extensions of linear functions.

So I’m pondering the statement:

Show that a continuous linear functional $f$ on a subspace $V$ of a Hilbert space $H$ has a unique norm preserving extension $h$ on $H$

Here is my thought.

Consider the closure of $V$, $\overline{V}$. Then since $V$ is dense in $\overline{V}$, there exists a unique extension $\overline{f}$ of $f$ to $\overline{V}$.

Moreover since $\overline{V}$ is a closed subspace of $H$ then it is also a Hilbert space, so by the Frechet-Riesz theorem we have $\overline{f}(v)$ = $\langle v,y_{\space\overline{f}} \rangle$ for some unique element $y_{\space\overline{f}} \in \overline{V}$

Take $h(x) = \langle x,y_{\space\overline{f}} \rangle$.

How do I get uniqueness though?

Solutions Collecting From Web of "Hilbert spaces and unique extensions of linear functions."

Let me simplify things by assuming (as you point out, without loss of generality) that $V$ is closed. Then, indeed, $f(x) = \langle x,y \rangle$ on $V$ for a unique $y \in V$, yielding the immediate extension $h(x) := \langle x,y\rangle$ of $f$ to $H$; one can readily check that $$\|f\|_{V^\ast} = \|y\|_V = \|y\|_H = \|h\|_{H^\ast},$$ so that this extension is norm-preserving.

Now, suppose that $k \in H^\ast$ is another extension of $f$ to $H$, so that $k(x) = \langle x,z\rangle$ for a unique $z \in H$. Then for any $x \in V$, $$0 = f(x) – f(x) = h(x) – k(x) = \langle x,y\rangle – \langle x,z\rangle = \langle x,y-z\rangle,$$ so that $y-z \in V^\perp$. Hence, $y$ and $y-z$ are orthogonal, so that
$$\|k\|_{H^\ast} = \|z\|_H = \|y – (y-z)\|_H = \sqrt{\|y\|^2_H + \|y-z\|^2_H}.$$ Thus, $\|k\|_{H^\ast} = \|f\|_{V^\ast}$ if and only if $\sqrt{\|y\|^2_H + \|y-z\|^2_H} = \|y\|_H$, if and only if $\|y-z\| = 0$, if and only if $y =z$, if and only if $h=k$. Hence, $h$ is indeed the unique norm-preserving extension.