How to verify the completeness of a given Hilbert space?

Question

Hilbert space is defined as a complete inner product space. It is also said that a finite dimensional vector space with inner product is trivially complete. I have two questions. How can I verify whether the $L^2$ (space of square integrable functions is complete in terms of Cauchy sequence.). I also want to see in an explicit example that a finite dimensional vector space with inner product is complete. For example, how can I be sure that the 2-dimensional vector space associated with spin $s=1/2$ (for example) particle is complete?

I have limited mathematical knowledge about Hilbert spaces and therefore an answer without technical terms will be helpful. I know about Cauchy sequence.

Answer 1

One way to introduce the $L_p(\mathcal{D})$ spaces is by defining them as the completion of the corresponding underlying $C_p^{\infty}(\mathcal{D})$ with respect to the $p$-norm, that is $$ ||f||_p = \left(\int_{\mathcal{D}}\textrm{d}\mu\,|f|^p\right)^{1/p} $$ where $\textrm{d}\mu$ is the Lebesgue measure. There are some Cauchy sequences which do not converge within the space whose set of limit points can be added as completion to the underlying space to define a set that will then by definition be complete.

In general, proof of completeness follow by applying by brute force methods the statement that every Cauchy sequence must converge within the space. When you spot out that there are some points limit of Cauchy sequence that do not belong to the space, you can define a further space adding those culprit points, which will by definition be complete (the standard example is the completion of the rational numbers). Along the same lines it trivially follows that every finite dimensional vector space provided with an inner product must be complete: every Cauchy sequence does converge within the space (this is a standard exercise that you can find in almost every textbook on Calculus more or less advanced).

For all the other more complicated cases it is not straightforward at all and more or less complicated proofs may be found case by case.

EDIT: Also see comments below, I edited the notation accordingly.

Answer 2

Suppose $\mathcal H$ is the inner product space you want to test for completeness. There are basically two general strategies:

1. Show that every Cauchy sequence in $\mathcal H$ converges to a point in $\mathcal H$;
2. Show that $\mathcal H$ is closed in a greater Hilbert space $\mathcal K$ (in this case, the inner product on $\mathcal H$ must be the restriction of the Hilbert inner product of $\mathcal K$).

If, for some reason, you start with a Banach space $\mathcal H$ with the norm $\| \|$, then $\mathcal H$ is Hilbert iff $\| \|$ satisfies the parallelogram rule (Jordan-Fréchet-von Neumann theorem).

Example. Suppose we know that $L^p$ spaces are Banach. It turns out that only $\| \|_2$ (norm of $L^2$) satisfies the parallelogram identity, hence only $L^2$ spaces are Hilbert into the $L^p$ realm ($\ell^2$ are here seen as a special case of $L^2$).

In general, given a metric space $(X,d)$, this can be completed, i.e., there exists a unique (up to isometries) map $i: X \to Y$, $(Y, d_Y)$ complete metric space (technically, $i$ is a dense range isometry, linear if $X$ is normed). $Y$ is a set of equivalence classes of $X$: precisely, $\{x_n\}$ and $\{x_n'\}$ Cauchy sequences in $X$ are equivalent in $X$ iff $d(x_n, x_n') \to 0$ as $n\to \infty$.

So, it does really matter if an inner product space is complete or not? One could argue that would not a big deal, since the completion could be always considered. This is true but in principle some trouble can happen. For example, the equivalence relation could not be well suited to produce the right slices, so states which we would like consider as different could collapse in the same equivalence class, and also the contrary. I have never seen this problem in practise, mostly because one usually selects a priori the space of states as a complete one invoking some physical request.

In QM it assumed that $\mathcal H$ is Hilbert and that it is separable. Separable Hilbert spaces are all isometrically isomorphic to $\ell^2(\mathbb N)$ (Riesz-Fischer theorem), so the problem is reduced to show that the last one is complete. This can be done by hand following strategy 1, and completeness of finite dimensional inner product spaces can be regarded as a corollary of this general fact (so applying strategy 2, if you want).

Remark. For the last conclusion it is essential that the isomorphism is also isometric. Completeness is not a topological property, so not preserved under homeomorphism nor isomorphism.