Non-continuous linear functional over polynomials with inner product $\langle f,g \rangle = \int_{0}^{1} f(x)g(x)dx$

Question

Let $P(X)$ denote the set of polynomials $p:\mathbb{R} \to \mathbb{R}$, I found that the functional $\phi:P(X) \to \mathbb{R}$ such that $p \mapsto p(0)$ is linear but does not correspond to an inner product $\langle \cdot, v\rangle$. All these is done with the inner product $\langle f,g \rangle = \int_{0}^{1} f(x)g(x)dx$.

According to Riesz-Fréchet theorem, it should be the case except if $\phi$ is not continuous or $P(X)$ is not a Hilbert space with the given norm. What is the problem in this case?

Answer 1

Of course $P(X)$ is not a Hilbert space, so the theorem does not apply.

And regardless, it's clear that $\phi$ is not continuous: Let $p_n(t)=(1-t)^n$. Then you can check that $||p_n||^2=\langle p_n,p_n\rangle\to0$ although $\phi(p_n)=1$ (so in fact $\phi$ is not given by the inner product with any vector in the completion of $P(X)$; not being continuous, there's no reason it should be.)

Answer 2

$P(X)$ is not an Hilbert space since it is not complete by Stone Weirstrass, every continuous function is a limit of a sequence of polynomials.

Answer 3

$P(X)$ is for the set of $L^2$ functions, a bit like what $\mathbb{Q}$ is for $\mathbb{R}$...