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The exercises shows that the Riesz Representation does not hold on infinite-dimensional inner product spaces. I need help.

Suppose $C_{\mathbb{R}}([-1, 1])$ is the vector space of continuous real-valued functions on the interval $[-1, 1]$ with inner product given by \begin{equation} \langle \ f\ , \ g \ \rangle = \int_{-1}^{1} f(x) g(x) dx \end{equation} for $f, g \in C_{\mathbb{R}}([-1, 1])$. Let $\varphi$ be the linear functional on $C_{\mathbb{R}}([-1, 1])$ defined by $\varphi (f ) = f (0)$. Show that there does not exist $g \in C_{\mathbb{R}}([-1, 1])$ such that \begin{equation} \varphi(f) = \langle \ f\ , \ g \ \rangle \end{equation} for every $f \in C_{\mathbb{R}}([-1, 1])$.

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  • $\begingroup$ Huh? The Riesz representation theorem for $C(K)'$ says in this case the continuous dual of $C_\mathbb{R}([-1,1])$ is the finite signed Borel measures on $[-1,1]$. And your "counterexample" isn't a counterexample at all because the Dirac delta is a regular Borel probability measure. Are you misquoting? $\endgroup$ – user10354138 Sep 20 '18 at 3:14
  • $\begingroup$ I need to show that the Riesz represetization theorem is not fulfilled for spaces of infinite dimension and the counterexample I must demonstrate the exercise above. $\endgroup$ – Eduardo Maza Sep 20 '18 at 3:29
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    $\begingroup$ I suspect that, by "Riesz representation theorem", you mean this (and @user10354138 means this). The first of these is perfectly valid for infinite dimensions, provided that you are dealing with Hilbert spaces. The latter is exactly the problem in your "counterexample" - it lacks completeness. $\endgroup$ – metamorphy Sep 20 '18 at 6:07
  • $\begingroup$ Do you have any idea on how to approach it? $\endgroup$ – Uskebasi Sep 20 '18 at 8:35
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If there is such a $g$, then $\varphi$ is a continuous linear functional on the given space, and $\varphi(f) = 0$ if $f(0) = 0$. As the set of all such $f$ is (everywhere) dense in the space, this would imply $\varphi \equiv 0$.

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  • $\begingroup$ I have another idea of ​​how to prove it but thank you very much. $\endgroup$ – Eduardo Maza Sep 20 '18 at 14:18

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