Does a completion exist in general?

Question

Let $\mathcal H_0$ be a space of functions equipped with an inner product. Is it possible in general to define a “completion” $\mathcal H$ such that $\mathcal H_0 \subset \mathcal H$ and cauchy sequences converge in this space? I thought that it was in fact possible, but I am working on a HW problem that asks me to show (for a specific definition of $\mathcal H_0$ and the inner product) that the completion exists and that cauchy sequences do actually converge.

Answer 1

Basically, yes, but there are some details to pay attention to. Importantly, the particular construction of the completion is irrelevant (any one would do, all leading to isometric models). Further, the entire inner product space structure extends uniquely to the completion. This is true by a general principle: any uniformly continuous function on a space extends uniquely to all of the completion. And, when that is done, the completed space's metric agrees with the one induced by the extended norm (obtained from the inner product).

In short: any inner product space (resp. normed space) has an essentially unique completion and that completion is uniquely an inner product space (resp. normed space) in a way that extends the original structure and agreeing with the metric. This is a standard fact of Banach space and Hilbert space theory showing that any inner product space embeds essentially uniquely in a Hilbert space and similarly for normed spaces in Banach spaces. It should be noted though that very often having a particular model of the completed space is very important. Working directly with any of the standard methods that produce a completion of a metric space (or, more generally, a uniform space) is not a good idea. In particular, the completion of the space of continuous functions with respect to the $L_p$ norm is the $L_p$ space --- that space simply exists as a completion but its model in terms of measurable functions is very important.