Pre-Hilbert spaces and the Riesz Representation Theorem.

Question

I'm looking for an example of a Hilbert space $(H,\langle \cdot,\cdot\rangle)$ that satisfies the following:

In $H$ there exists an element $a$ such that $(H \backslash \{a\},\langle \cdot,\cdot\rangle)$ is a pre-Hilbert space such that the Riesz Representation theorem fails.

I've considered $L^2(\Omega)$ with $\Omega \subset \mathbb{R}$ and the Lebesgue measure but haven't gotten too far. The Riesz Representation theorem states that for $f \in H^*$ there exists a unique $x_o \in H$ such that $f(x)=\langle x, x_o \rangle$. If we were to remove $x_o$ from $H$, then $H \backslash \{x_0\}$ fails to Reiez theorem, but this may fail to be a normed vector space depending on what $x_o$ is chosen.

Answer 1

As pointed out by Berci, removing any one point from a vector space doesn't give a vector subspace (if you remove $x$, and $v\neq x$, then $x=v+(x-v)$, so the new set is not closed under addition), so as stated your question doesn't make much sense. What does make sense to ask for is a non-closed subspace of a Hilbert space (i.e, a pre-Hilbert space) which fails to satisfy the Riesz representation theorem.

For such an example, consider the subspace $C([0,1])$ of the Hilbert space $L^2([0,1])$, and consider the continuous linear functional $\lambda$ on $C([0,1])$ (continuous w.r.t the $2$-norm), given by $$\lambda(f)=\int_0^{1/2}f=\int_0^1f\chi_{[0,1/2]}.$$ Since $\chi_{[0,1/2]}$ is not a.e. equal to a continuous function, it follows that $\lambda$ is not of the form $\lambda(f)=\langle f,g\rangle$ for some $g\in C([0,1])$.