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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?

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  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Jun 11 '19 at 7:38
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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 $H^*$.

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    $\begingroup$ Thank you so much, I think I got it now! $\endgroup$ – mathnoob777 Jun 11 '19 at 18:55
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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.

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