Continuously embedded subspace (with own topolog. structure) of a separable real Hilbert space is itself separable

Question

Let $H$ be a separable real Hilbert space and let $V$ be another real Hilbert space. Additionally, $V$ is a dense subspace of $H$ and is continuously embedded, i. e. $$ \exists C>0: \forall v\in V: \left\| v \right\|_H \leq C \cdot \left\| v \right\|_V. $$ Note, that $V$ has its own scalar product and has therefore a different topological structure.

Is V a separable space w. r. t. $\left\| . \right\|_V$? If so, how can we prove it?

Answer 1

Let $H$ be a separable Hilbert space, let $V$ be a nonseparable Hilbert space, and let $T : V \to H$ be linear, injective, and $\|T(v)\|_H \le C\|v\|_V$. That is, $T$ is a bounded linear operator. We must show this is impossible.

Let $T^* : H \to V$ be the adjoint operator. Since $T$ is injective, we know that $T^*$ has dense range.

Suppose $T^*(H)$ is not dense in $V$. There is $v \in V$ with $v \ne 0$ and $v \perp T^*(H)$. For every $h \in H$,$$0 = \langle v,T^*h\rangle_V = \langle Tv,h\rangle_H$$ Thus $Tv=0$. So $T$ is not injective.

Now $T^*(H)$ is a continuous image of a separable set, so $T^*(H)$ is itself separable. But it is dense in $V$, so $V$ is separable.