# Using Riesz Representation Theorem to show a set of functionals is dense in H*

I am currently struggling with the following question:

Let $$H$$ be a Hilbert space with basis $$\mathcal{E}$$ and $$φ_e (h)=〈h,e⟩$$. Show that the set of functionals, span⁡{$$φ_e:e\in \mathcal{E}$$}, is dense in $$H^*$$ (which is identified with $$H$$ by the Riesz representation theorem).

Here is an outline of my sketch:

1. Suppose $$\mathcal{E}=$${$$e_1,e_2,...,e_n$$} is a basis in $$H$$. Let $$y_f \in \mathcal{E}$$ where $$y_f= e_1e_1+...+e_ne_n$$.
2. Let $$f\in H^*$$ with $$f(x)=〈x,y_f⟩$$ and $$x_0$$=$$\frac{y_f}{‖y_f‖}$$. Then use $$‖x_0‖=1$$ to show that $$‖f‖≥|f(x_0 )|=|〈x_0,y_f ⟩|=‖y_f‖$$.
3. Show that $$‖f‖\le‖y_f‖$$. Hence $$‖f‖=‖y_f‖$$.
4. Verify the uniqueness property. That is, if $$f∈x^*$$, there exist $$y_f$$, $$z_f∈x$$ such that $$f(x)=〈x,y_f⟩=〈x,z_f⟩$$, then we have $$y_f=z_f$$.

Although I have completed all the steps, I can't help but feel that there is something wrong. If possible, could you please tell me where I made mistakes in?

• It is not given the Hilbert space is finite dimensional. A general Hilbert space has an orthonormal basis but the basis need not even be countable. – Kavi Rama Murthy Jun 11 '19 at 7:38

As stated in the comments, you cannot assume that the basis is finite; not even that it is countable. You should think about this in a "topological way" (and a piece of advice in general, try starting from the most relevant notions; if this is not enough, then try to decide what information you have in hand will help you bypass your problem): You should probably understand that if $$X,Y$$ are topological spaces and $$f:X\to Y$$ is a topological isomorphism (that is, $$f$$ is a continuous bijection with continuous inverse), then $$f$$ carries dense subsets of $$X$$ to dense subsets of $$Y$$.
So take $$X=H, Y=H^*$$ and $$f=x\mapsto\langle\cdot,x\rangle$$. The Riesz representation theorem states that $$f$$ is a bijection. You can easily see using the Cauchy Schwarz inequality that this $$f$$ is continuous with continuous inverse. Since $$\mathcal{E}$$ is an orthonormal basis, the set $$span\{e: e\in\mathcal{E}\}$$ is dense in $$H$$. This is carried through $$f$$ to $$span\{\varphi_e: e\in\mathcal{E}\}$$, therefore your set is dense in $$H^*$$.
The assignement $$H\ni x \mapsto \langle \cdot,x\rangle\in H^*$$ is an isometric isomorphism. In particular it carries dense sets to dense sets. Since span$$\{\varepsilon\}$$ is dense in $$H$$, it follows that your set is dense.