# application of Riesz representation theorem into dense subset

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 $$V$$.

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 $$V$$.

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 $$\langle x,\cdot \rangle = \ell$$.