Let $H$ be a Hilbert space. We say that a linear operator $T \colon H \to H$ is compact if it maps bounded sets to precompact ones, that is, if for every bounded sequence $(a_n)$ in $H$, $(Ta_n)$ has got a convergent subsequence. We can characterize compactness in terms of weak convergence: $T$ is compact if and only if it maps weakly convergent sequences into norm convergent ones.

Up to now I would have casually said:

$T$ is compact iff $T$ is continuous when its domain is equipped with weak topology and its range with norm topology.

and I would not have been alone: I've heard this from more than one of my professors. Well, this turns out to be false, as I read on Problem VI.34 of Reed & Simon's Methods of modern mathematical physics:

Show that in a Hilbert space $H$, a map $T \colon H \to H$ is continuous when its domain is given the weak topology and its range the norm topology if and only if $T$ has finite rank!

(the exclamation point is part of the original text).

At the moment this problem is beyond my grasp, since I know almost nothing of the necessary topological tools (that is - nets, Moore-Smith convergence, and the like). But I'm curious about this and wanted to share my curiosity with the community.

Has somebody got any idea on how to prove this, or even some hint that could help me to catch this intuitively?


First you'll need to understand that a basis of the weak topology consists of the sets $$ V_{\epsilon, y_1, ..., y_n} = \{x \in H: |(x, y_i)| \lt \epsilon \} $$ for any $\epsilon \gt 0$ and a finite set of vectors $y_i$. To be more precise: These sets form a neighborhood basis at zero, which is what we'll need for the proof.

Now, if a linear transformation $$ T: H_{weak} \to H_{norm} $$ is continuous, then the inverse image of the open unit ball in $H_{norm}$ has to contain such a basic weak neighborhood $V_{\epsilon, y_1, ..., y_n}$. If H is infinite dimensional, we can find an $x$ such that $(x, y_i) = 0$ for $i = 1,..., n$. This implies that $k x \in V$ for any $k$, so that $T x = 0$ needs to hold (which implies that the operator norm of $T$ has to be zero).

We may assume that the $y_i$ are orthonormal (why?). Given any $x \in H$, we can write $$ z = x - \sum_{i = 1}^{n} (x, y_i) \; y_i $$ Now we know that $A z = 0$ by construction, which means that we know $T x$ for any arbitrary $x$, it is $$ T x = \sum_{i = 1}^{n} (x, y_i) \; y_i $$ So much for your concrete question.

Infinite dimensional normed spaces are fundamentally different from finite normed spaces, for example:

  • The unit ball is compact iff the space is finite dimensional.

Therefore the compact operators are of interest because they retain some properties from the finite dimensional case in infinte dimensions, as is the weak operator topology, that makes them interesting.

BTW: I think your professors were talking about the fact that compact operators are those that are continuous from the weak unit ball of $H_{weak}$ to $H_{norm}$. The restriction to the unit ball is essential, and looking at the proof above you can see which step does not work anymore if we add remove this restriction.

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  • $\begingroup$ I made some small corrections, but I didn't want to mess too much with your nice text. One thing I left unchanged since I didn't see a very easy fix that would without doubt be okay with you is that your basis is of course only a neighborhood basis at zero. $\endgroup$ – t.b. May 25 '11 at 18:26
  • $\begingroup$ Thanks! I'll see what I can do :-) $\endgroup$ – Tim van Beek May 26 '11 at 9:48
  • $\begingroup$ I'm sorry for this long wait, I have had to postpone this question because of my exam schedule. Ok, I understand this argument: at a point you ask "why can we suppose $y_1 \ldots y_n$ to be othonormal?", and I would say, because if they are not we could take $z_1 \ldots z_k$, yielded by a Gram-Schmidt on $y_1 \ldots y_n$, and we have $$\max_{j=1 \ldots n} \lvert (x, y_j) \rvert \le C\max_{l=1\ldots k} \lvert (x, z_l)\rvert$$ for some constant $C$. [to be continued...] $\endgroup$ – Giuseppe Negro Jun 3 '11 at 18:39
  • $\begingroup$ To see this, let $P$ denote orthogonal projection over $\mathrm{span}(y_1 \ldots y_n) = \mathrm{span} (z_1 \ldots z_k)$: then $$\max_{j=1 \ldots n} \lvert(x, y_j)\rvert \lesim \lVert Px \rVert = \sqrt{\sum_{l=1}^k\lvert (z, y_l) \rvert^2} \lesim \max_{l=1\ldots k} \lvert (x, z_l)\rvert$$ where $\lesim$ stands for: "lesser or equal up to a multiplicative constant". $\endgroup$ – Giuseppe Negro Jun 3 '11 at 18:45
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    $\begingroup$ Anyway, as for what my professors say: I believe that they simply forget to say "sequentially" in ($T$ is compact iff $T$ is weak-norm sequentially continuous). IMHO this is not a mortal sin for somebody not working in strictly theoretical issues. $\endgroup$ – Giuseppe Negro Jun 3 '11 at 18:47

There is the following characterization of compactness:

An operator $T: H \to H$ is compact if and only if $T|_{B} : B \to H$ is continuous, where $B$ is the closed unit ball of $H$ equipped with the weak topology and $H$ is equipped with its usual norm topology.

You can find a proof of this fact e.g. in Pedersen's Analysis Now as part of Theorem 3.3.3. However, this result relies heavily on net techniques, but you could take the opportunity to learn about that in Chapter 1 of that book.

To see that continuity implies compactness in your sense, simply observe that continuity implies that $T(B)$ is compact, and as $T(B)$ is metrizable, every sequence in $B$ has a subsequence whose image under $T$ converges. The other direction is a bit more complicated but not really difficult.

Now suppose that $T: H \to H$ is weak-norm continuous. Then the seminorm on $H$ given by $x \mapsto \|Tx\|$ is continuous as a map $(H, \text{weak}) \to \mathbb{R}$, so by a standard result on locally convex spaces and the definition of the weak topology (see also Tim's answer to your question), there are points $x_{1},\ldots,x_{n} \in H$ such that $\|Tx\| \leq \max{\{|\langle x,x_{i}\rangle|\,:\,i = 1,\ldots,n\}}$ for all $x \in H$. But this means that $T$ must vanish on the orthogonal complement of the span of $x_1, \ldots, x_n$, so certainly $T$ has finite rank.

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  • $\begingroup$ Great! This is exactly the answer I expected. Now I need to chew it up a little. Thank you for the bibliographical reference also. $\endgroup$ – Giuseppe Negro May 25 '11 at 17:46
  • $\begingroup$ Great! Take your time... I'll think about how I could make the thing a bit easier to digest. $\endgroup$ – t.b. May 25 '11 at 17:48
  • $\begingroup$ @dissonance: the standard result I was referring to is of course exactly what Tim is explaining in his answer. The condition $\|Tx\| \leq \max{\{\ldots\}}$ is equivalent to the statement that $T$ is continuous at zero from $(H,\text{weak}) \to (H,\text{norm})$ and this in turn is equivalent to the statement that $x \mapsto \|Tx\|$ is a continuous semi-norm on $(H,\text{weak})$. This is completely analogous to the statement you're used to for (semi-)normed or Banach spaces. $\endgroup$ – t.b. May 25 '11 at 18:32
  • $\begingroup$ You may also be interested in this answer of mine. $\endgroup$ – t.b. May 26 '11 at 6:00
  • $\begingroup$ Thanks to you and Tim we have solved Reed & Simon's problem for linear $T$. However, the text does talk of a generic map, not necessarily linear: in fact the problem is marked as very difficult! I'm certainly not willing to go into this, since it is far beyond my possibilities at the moment. Just wanted to point it out for the curious (and advanced) people. $\endgroup$ – Giuseppe Negro Jun 4 '11 at 11:09

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