Is every Hilbert space the completion of some incomplete pre-Hilbert space?

Question

Is every Hilbert space the completion of some incomplete pre-Hilbert space? It is certainly true of the classical Hilbert space. But for a general Hilbert space, I would imagine that one uses some generalized notion of bases, of which I know nothing.

Answer 1

Every finite-dimensional normed vector space is complete, so a finite-dimensional Hilbert space cannot be the completion of an incomplete pre-Hilbert space. But every infinite-dimensional Hilbert space is the competion of an incomplete pre-Hilbert space. Indeed, it is the completion of any dense subspace, so you just have a to find a dense proper subspace. You can do this, for instance, by taking an orthonormal basis $B$ for the Hilbert space and taking the subspace of all finite linear combinations of elements of $B$.

Answer 2

Every infinite-dimensional Hilbert space $H$ is the completion of some pre-Hilbert space, yes. Just take a discontinuous linear form $\alpha\colon H\longrightarrow\mathbb C$ and consider its kernel. It's a dense subspace of $H$, distinct from it.