# Pre-Hilbert spaces and the Riesz Representation Theorem.

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.

• If we remove only one element from a vector space, we just don't get a subspace.. – Berci Mar 3 '19 at 17:12

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 $$g\in C([0,1])$$.