Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Recall the definition of compact operators between Hilbert spaces:

An operator $A$ is called compact if the image $A(\mathcal U_H)$ of the unit ball is relatively compact (i.e. its closure is compact) in the norm topology.

However, I seem to be able to "prove" that the image is, in fact, compact, not just relatively compact. What went wrong?

$\square$: We show that $A(\mathcal U_H)$ is sequentially compact. Since the topology is given by a norm / metric, this implies that it is also compact.

Let $y_n = Ax_n$ be a sequence in $A(\mathcal U_H)$. The unit ball in a Hilbert space is weakly compact, so there is a weakly converging subsequence $x_{n_j} \to^w x$ which gives $Ax_{n_j} \to^w Ax$.

On the other hand, by the sequential compactness of the closure of $A(\mathcal U_H)$, we know that a subsequence of $z_k = x_{n_{j_k}}$ has the property $Az_k \to y$ for some $y$ in the Hilbert space, though we do not yet know that $y$ is also in the image of the unit ball.

But we have $z_k \to^w x$ as well, and hence $A z_k \to^w A x$. Since the weak limit must coincide with the norm limit, we have $y = Ax$, and $y \in A(\mathcal U_H)$ already. In other words, the original sequence $y_n$ has a subsequence that converges in the image of the unit ball. $\square$

I feel very stupid for not finding my mistake that must surely be very elementary. What went wrong? A counterexample would probably help my understanding as well.

share|improve this question
    
It's been a while since functional analysis so I'm mainly guessing here, but is the weak topology on a Hilbert space metrizable? If not, does weak compactness imply weak sequential compactness? –  Miha Habič Jan 5 '13 at 11:23
    
According to Wikipedia, the weak topology is metrizable at least for separable Hilbert spaces. But it appears that you have a good point: in the non-separable case, weak sequential compactness need not follow from weak compactness, but I'm not entirely sure whether this is the case or not. –  Greg Graviton Jan 5 '13 at 12:16
3  
In a Banach Space compactness is equivalent to sequential compactness. This is the Eberlein–Šmulian theorem. –  David Mitra Jan 5 '13 at 12:37
1  
In my previous comment, I meant to say "weak compactness is equivalent to weak sequential compactness". –  David Mitra Jan 5 '13 at 16:00

2 Answers 2

up vote 5 down vote accepted

There is nothing wrong with your argument here, assuming that $\cal U_{\cal H}$ is the closed unit ball. (I imagine that "relative" is included in the definition of compactness since it's needed in the more general setting when defining compact operators between Banach spaces, or when defining compactness of $T$ as "$T(M)$ is relatively compact for any bounded set $M$".)

More generally, if $X$ is reflexive, $M$ is a closed, bounded, and convex subset of $X$ , and $Y$ is a Banach space, then for a compact operator $T:X\rightarrow Y$, the set $T(M)$ is norm-compact in $Y$.

share|improve this answer
    
Thanks a lot! So the qualification "relative" is apparently meant for the case of Banach spaces that are not reflexive. My argument would fail there as the (closed) unit ball would no longer be weakly compact. –  Greg Graviton Jan 6 '13 at 9:23

The open unit ball in a Hilbert space is not compact, nor weakly compact. The closed unit ball is weakly compact. Your $x$ might have $||x||=1$. Take any sequence $x_n \in \mathcal{U_H}$ that converges to some $x \in \mathcal{H}$ with $||x||=1$. Then, any subsequence must converge strongly and weakly to $x$ and, running your argument, you get that $y = Ax$ with $x \notin \mathcal{U_H}$, which doesn't tell you that $y$ is in the image of the open unit ball.

share|improve this answer
    
It is common to see the definition with the open unit ball, so I parsed $\mathcal{U_H}$ to be the open unit ball.. An equivalent definition, as David Mitra noted above and your argument shows, is to require that the image of the closed unit ball be norm-compact. –  levap Jan 5 '13 at 13:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.