Question on difference between closedness and completeness of Hilbert spaces and subspaces of Hilbert spaces?

Question

in Stein's book, we have the following definitions: 1)A subspace S of a Hilbert space H is closed if whenever $(f_n) ⊂ S$ converges to some f ∈ H, then f also belongs to S. 2) The set L^2 is complete if every Cauchy sequence $(f_n)$ in $L^2(\mathbb R)$ converges to a function $f \in L^2(\mathbb R).$

By assumption a Hilbert space H is complete. Now what is the difference between a subspace S of H being complete and being closed? Isn't any convergent (in the norm) sequence of functions, a cauchy sequence? So checking for closedness S is equivalent to checking for completness of S, ie whether every cauchy sequence in S conveges in S(in the norm)?

I will take any help I can get as this is really confusing me. Thanks.

Answer 1

If you have $S$ a subspace of a Hilbert space $H$, then $S$ being complete is equivalent to $S$ being closed in $H$.

However, completeness is an absolute property, and closedness is a relative property.

For example, let $c_{00}:=\{x \in \mathbb{R}^{\mathbb{N}} \mid x(n) \neq 0 \text{ for finitely many terms}\}$, with inner product $\langle x,y \rangle$ given by $\sum_\limits n x(n)y(n)$.

Consider the sequence $y_n$ in $c_0$ given by $(y_n)(m)=1/m$ for $m \leq n$ and $0$ for $m>n$. It is clear that $y_n$ is Cauchy. However, $y_n$ can't converge to any $x \in c_{00}$, since the distance from $y_n$ to any given $x$ only grows up after $n$ sufficiently large.

Note that the argument was entirely in $c_{00}$. This illustrate that completeness is an intrinsic property.

However, we can see $c_{00}$ as a subspace of the Hilbert space $l^2:=\{x \in \mathbb{R}^{\mathbb{N}} \mid \sum\limits_n (x(n))^2<\infty\}$. Now, my sequence $y_n$ previously defined clearly converges to $x$ given by $x(n)=1/n$. And $x$ is clearly not in $c_{00}$, hence $c_{00}$ is not closed (and therefore not complete by the statement in the yellow bar).

This may seem more straightforward, however it comes at a cost. We must know beforehand some manageable Hilbert space in which our candidate for completeness lives. This is not so easy in general.

Note also that the advantage of completeness relies heavily on the fact that you can assure a sequence will converge by analysing itself, not finding a candidate beforehand. The situation is analogous: the desire for an intrinsic process.

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Let's consider extensively a more elementary example, since you seem to know the concept of metric spaces in the comments. Consider the example discussed above: $\big((0,1),d\big)$, where $d(x,y)=|x-y|$.

Let $x_n=1/n$. Note that $d(x_n,x_m)=|1/n-1/m|< 1/\min\{n,m\}$. Therefore, $x_n$ is Cauchy. Indeed, given $\epsilon>0$, take $N$ such that $1/N < \epsilon$. Therefore, if $n,m>N$, then $\min\{n,m\}>N$ and $d(x_n,x_m)=|1/n-1/m|< 1/\min\{n,m\}<1/N < \epsilon$.

However, $x_n$ does not converge. Indeed, for any $x \in (0,1)$, we have that there exists $N$ such that $x> 1/N$, and therefore, for $n>N$, $d(x_n,x)=|x-1/n|=x-1/n>x-1/N$. That is, for $\epsilon:=x-1/N$, there exists no $n$ etc etc.

It follows, by definition, that $\big((0,1),d\big)$ is not complete.

Note that if $(0,1)$ is a subset of another metric space $(X,d')$ for which $d'|_{(0,1) \times (0,1)}=d$, then given a sequence $x_n \in (0,1)$ and an element $x\in (0,1)$,

$x_n$ is a Cauchy sequence in $\big((0,1),d\big)$ if and only if $x_n$ is a Cauchy sequence in $\big(X,d').$

and

$x_n$ converges to $x$ in $\big((0,1),d\big)$ if and only if $x_n$ converges to $x$ in $\big(X,d').$

It doesn't matter who $X$ is, as long as the induced metric is the metric on $X$. This is precisely why we call completeness an "absolute", or "intrinsic" property: it doesn't depend on where we are, as long as it induces the structure we originally have (obviously, otherwise it would be senseless to compare).

Closedness is very sensitive to the metric/topology of the ambient space. $(0,1)$ is closed on $\big((0,1),d\big)$ (as is any metric space as a subset of itself), but $(0,1)$ is not closed on $\big(\mathbb{R},d_{can}\big)$ for example, even though the metric is the induced one.

To be very explicit, when we talk about completeness, the following phrase is meaningful:

The metric space $(X,d)$ is complete.

When talking about closedness, we need the following phrase in order to have an entire meaningful information:

The subset $A \subset X$ is closed in (X,d).

Answer 2

Yes, for a subspace $X$ of a complete metric space $Y$, $X$ is complete as a metric space with the restricted metric if and only $X$ is closed as a subset of $Y$. Even if $Y$ is not complete, completeness of $X$ with the restricted metric implies that $X$ is closed in $Y$, because, as you said, convergent sequences are Cauchy. (But if $Y$ is not complete, then closedness of $X$ does not imply completeness, as even taking $X=Y$ shows.)

But when we're working with complete spaces, why make the distinction? (Is that part of your question?) First of all, the equivalence only applies once you're already sitting in a complete space, so for our "big" space $Y$, we need to use the concept of completeness. But then for a subspace $X$ of a complete space $Y$, why not check completeness instead of closedness? I would say that part of the reason is that it's easier to work with and show closedness, and then you get completeness for free when it is convenient. To show closedness you start by assuming a sequence converges and then just have to show the limit is in your subspace. Working with completeness for subspaces in general would add another unnecessary step of pointing out that there is a limit of a Cauchy sequence because of completeness of the big space, then you still have to show the limit is also in the subspace.