Let $\mathcal{H}$ be a Hilbert space. Let $T:\mathcal{H} \to \mathcal{H}'$ be bounded and linear with the following unique continuation property:
$\langle Tx,x \rangle_{\mathcal{H}',\mathcal{H}} =0 \implies x=0$
Then this quadratic form defines a norm on $\mathcal{H}$. Let $F$ be the completion of $\mathcal{H}$ with respect to this new norm. Then $T$ is an isomorphism from $F \to F'$.
This is clear to me until the final sentence. How do we even know that $T$ is defined on $F$. I can see that $T$ should be one-to-one with dense range, but that is it. Any help would be appreciated.
Suppose $H$ is a real Hilbert space. Suppose $T$ self-adjoint. Suppose $\langle Tx,x\rangle \ge0$ for all $x$. Then $$ \|x\|_T:= \sqrt{\langle Tx,x\rangle} $$ defines the new norm, which is induced by the new inner product $$ ( x,y)_T:=\langle Tx,x\rangle. $$
Let us show that $T$ is continuous from $(H,\|\cdot\|_T)$ to its dual space $(H,\|\cdot\|_T)'$. Let me denote the norm in the dual space by $\|\cdot\|_{T'}$. Take $x\in H$, then $$ \|Tx\|_{T'}=\sup_{y: \|y\|_T\le 1} (Tx)(y) = \sup_{y: \|y\|_T\le 1} \langle Tx,y\rangle = \sup_{y: \|y\|_T\le 1} (x,y)_T =\|x\|_T. $$ Hence $T$ is bounded with respect to the new norm. As $H$ is dense in $F$ with respect to the new norm by construction, we can uniquely extend $T$ to a linear and continuous operator from $F$ to $(H,\|\cdot\|_T)'$. The extension has the same norm, thus $T:F\to (H,\|\cdot\|_T)'$ is an isometry.
Since $F$ is the completion of $(H,\|\cdot\|_T)$ it follows that $F'$ and $(H,\|\cdot\|_T)'$ are isometric isomorph: The restriction of a functional $f\in F'$ is in $(H,\|\cdot\|_T)'$, whereas the unique extension of $h\in (H,\|\cdot\|_T)'$ is in $F'$. The restriction and extension operations preserve norms.
Maybe the answer is overly complicated and verbose. The idea behind everything is that $T$ can be regarded as isometric isomorphism between $F$ and $F'$. In fact, $T$ is the Riesz isomorphism of the (Hilbert!) space $F$.
The operator $T$ is not an isometry if $T$ is not supposed to be self-adjoint. Take $H=\mathbb R^2$ and $$ T=\pmatrix{1 & 10 \\ -10 & 1}. $$ Then $x^TTx=\|x\|_2^2$ and $\|x\|_T=\|x\|_2$. Then the operator norm of $T$ with respect to $\|\cdot\|_2$ and $\|\cdot\|_T$ coincide. But $\|T\|_2\ne 1$ obviously.