application of Riesz representation theorem into dense subset

Question

Let $\mathcal{H}$ be a Hilbert space with an inner product $\langle\cdot,\cdot\rangle$ and $V\subset\mathcal{H}$ be a dense subspace. We already know that $$\mathcal{H}^*=\{\langle v,\cdot\rangle|v\in\mathcal{H}\}$$ from the Riesz representation theorem where upper star is a continuos dual. Is there any way to represent the element $V^*$ as such form? More precisely, I want to know that if we define $$\tilde{V}:=\{\langle v,\cdot\rangle|v\in V\},$$ ,then $\tilde{V}$ could be a dense subset of $V^*$. Since $V$ might not be a Hilbert space, the Riesz-representation theorem doesn't work for $V$.

Answer 1

Any continuous linear functional on $V$ extends uniquely to a continuous linear functional on $H$ and it can be expressed as $v \to \langle v, x \rangle$ for some $x \in H$. Of course, $x$ need not be in $V$.

Answer 2

As far as I understand your question, the answer is Yes.

Let me recall your definitions, just to be sure. We have a Hilbert space $\mathcal{H}$ and a dense, linear subspace $V \subseteq \mathcal{H}$. Their dual spaces are denote $\mathcal{H}^\ast$ respectively $V^\ast$. The set $\tilde V = \left\{ \langle v , \cdot \rangle \middle| v \in V \right\}$ is a subset of $\mathcal{H}^\ast$.

With this notation, the dual spaces $V^\ast$ and $\mathcal{H}^\ast$ are isomorphic, and $\tilde V$ is a dense subset of them.

Proof of $\mathcal{H}^\ast \cong V^\ast$
The map $L: \mathcal{H}^\ast \ni f \mapsto f \in V^\ast$ that sends $f$ to its restriction to $V$ is a countinuous, linear map. It is injective, since if $f$ and $g$ agree on the dense subset $V$ of $\mathcal{H}$, then they agree everywhere on $\mathcal{H}$. It is surjective, since if $f$ is any functional on $V$, then it extends (uniquely) to a functional on $\mathcal{H}$. So $L$ has a set-theoretic inverse. Lastly, $L$ is norm-preserving. That is, $\lVert f \rVert_{\mathcal{H}^\ast} = \lVert f \rVert_{V^\ast}$ where the $f$ on RHS means the restriction to $V$.

Proof that $\tilde V$ is dense in $\mathcal{H}^\ast$
You already stated in your question that $\mathcal{H}^\ast = \left\{ \langle v , \cdot \rangle \middle| v \in \mathcal{H} \right\}$. But there is even an isomorphism of normed vector spaces:

$$ \mathcal{H} \to \mathcal{H}^\ast, v \mapsto \langle v , \cdot \rangle $$

Now if $\ell \in \mathcal{H}^\ast$ is a functional, then $\ell = \langle x , \cdot \rangle$ for some $x \in \mathcal{H}$. Since $V$ is dense in $\mathcal{H}$, there is a sequence $(v_k) \subset V$ converging to $x$. By the isomorphism $\mathcal{H} \to \mathcal{H}^\ast$, the sequence $(\langle v_k,\cdot \rangle) \subset \tilde V$ converges to $\langle x,\cdot \rangle = \ell$.

I hope that this addresses your actual question; if not, please tell me.