The Riesz representation states that
Let $(H, \langle \cdot, \cdot \rangle)$ be a Hilbert space. Then for any linear and continuous functional $x^{\ast}:H \to \mathbb{K}$ there is a unique element $y \in H$ such that $x^\ast(x) = \langle x, y\rangle\ \forall x \in X$, and conversely, if $y \in H$, then $x^\ast:H \to K, x^\ast(x) = \langle x, y \rangle\ \forall x \in H$, is linear and continuous, with $\|x^\ast\|=\|y\|$.
Now the question is, can we replace "Hilbert space" with "prehilbertian space"?
Well, for the moment I don't think so. My thoughts so far are that I have to find a non complete space $X$, such as $C[a, b]$ with the inner product $\langle f, g \rangle = \int_a^b f(t)g(t)dt$, and in that space, find a functional for which there exists $x \in X$ such that $x^\ast(x) \neq \langle x, y\rangle\ \forall y \in X$.
Something else that I could do, is find a functional for which there exists $y, y' \in X$, $y \neq y'$ such that $\forall x \in X$, $x^\ast(x) = \langle x, y\rangle = \langle x, y' \rangle$.
For the moment, I didn't manage to find a counter example, so if you could give me directions for where to look, that would be great.
Tanks in advance for any answer.
Take a Hilbert space made from a pre-Hilbert space (such as $L^2[a,b]$ from $C[a,b]$). Since $C[a,b]$ embeds in $L^2[a,b]$, take any functional on $L^2[a,b]$ coming from a function that does not have a continuous representative. It will be a linear functional on $C[a,b]$, but will not be given by inner product with any element of $C[a,b]$. -- Neal