2
$\begingroup$

Let $A$ be an bounded operator on a Hilbert space with ONB $\{e_n\}_n$. I am looking for precise conditions on $\langle e_n, A e_m \rangle$ to guarantee that $A$ is compact (i.e. the limit of finite rank operators). It is known that $A$ is trace-class if $$ \sum_{m} \sum_n|\langle e_n, A e_m \rangle| < \infty $$ and the trace class operators are within the compacts. However, what if $A$ is not trace class? If $A$ were diagonal we know the sufficient and necessary condition is that $$\langle e_n, A e_n \rangle\to 0 \qquad( n\to\infty)\,.$$

However, what if $A$ is not diagonal? Is there a criterion in terms of the decay of $\langle e_n, A e_m \rangle$?

$\endgroup$
3
  • 2
    $\begingroup$ For the identity operator $ \langle e_n, Ae_m \rangle \to 0$ as $n \to \infty$ or $m \to \infty$. There is no simple characterization in terms of these inner products (and that is why you don't find any such theorem in books) $\endgroup$ Sep 5, 2020 at 4:54
  • $\begingroup$ @KaviRamaMurthy, thanks for your comment and pointing this out. I edited to make it more sensible incase someone comes up with a clever argument. $\endgroup$
    – PPR
    Sep 5, 2020 at 5:09
  • 1
    $\begingroup$ To illustrate how hard this problem can be assume that there exists a sequence of scalars $\{c_n\}_{n\geq0}$ such that $⟨e_n, Ae_m⟩ = c_{n+m}$. In this case $A$ is called a Hankel operator. Then Hartman's (resp Nehari's) Theorem says that $A$ is compact (resp. bounded) iff there exists a continuous (resp. bounded measurable) function on the circle whose positive Fourier coefficients are given by the $c_n$. $\endgroup$
    – Ruy
    Sep 5, 2020 at 17:24

1 Answer 1

2
$\begingroup$

For $m\geq n$, let $A^{m, n}$ be the $m\times m$ matrix given by

$$ A^{m, n}_{i, j} = \left\{\matrix{ 0, & \text{if } i, j\leq n, \cr \langle e_i, Ae_j\rangle , & \text{otherwise}. } \right. $$ For example, $$ A^{3,2} = \pmatrix{ 0 & 0 &\langle e_1, Ae_3\rangle \cr 0 & 0 &\langle e_2, Ae_3\rangle \cr \langle e_3, Ae_1\rangle & \langle e_3, Ae_2\rangle & \langle e_3, Ae_3\rangle }. $$

Theorem. The operator $A$ is compact if and only if $\lim_{m,n\to\infty} \|A^{m, n}\| =0$.

Proof. Let $P_n$ be the orthogonal projection onto the span of the first $n$ basis vectors and notice that, for $m\geq n$, the matrix of $$ P_mAP_m - P_nAP_n $$ coincides with $A^{m,n}$ inside the top left $m\times m$ block of entries, and has zero entries everywhere else. It follows that $$ \|A^{m, n}\| = \| P_mAP_m - P_nAP_n\|, $$ so the condition regarding the limit in the statement is equivalent to $\{P_nAP_n\}_n$ being a Cauchy sequence.

As we are working within the space of bounded operators on a Hilbert space, which is complete, we may replace "Cauchy" by "converging" in the sentence above.

Assuming that this condition is true, that is, that $P_nAP_n$ converges, then the limit operator has the same matrix as $A$ (I'm assuming we are using Physicist's inner-product, which is conjugate linear in the first variable), and hence coincides with $A$. In other words $$ \lim_nP_nAP_n = A. $$ Since $P_nAP_n$ is finite rank, hence compact, and since the space of compact operators is closed, it follows that $A$ is compact. This proves the "if" part. As for the "only if" part, suppose that $A$ is compact. Using that $\{P_n\}_n$ is uniformly bounded and converges to the identity operator in the strong (pointwise) topology one shows that $P_nAP_n$ converges to $A$ in norm, so $\{P_nAP_n\}_n$ is a Cauchy sequence which we have already agreed to be equivalent to the condition in the statement.


Remarks:

  1. Since $A^{m, n}$ is a finite matrix, its norm is (in principle) computable in terms of the $\langle e_i, Ae_j\rangle $, as required.

  2. Computing norms of finite matrices is a hard numerical problem, so this criterion might not be as useful as the OP would like.

$\endgroup$
2
  • $\begingroup$ Thanks Ruy, actually interesting claim, I would be interested in a proof. Since we know norms on finite vector spaces are equivalent, this could be translated to a claim on the largest of the matrix elements multiplied by $m$, no? I mean something like if $$ \lim_{m,n\to\infty} m \max_{i,j\in{1,\dots,m}}|\langle e_i, A e_j\rangle| = 0\,? $$ $\endgroup$
    – PPR
    Sep 5, 2020 at 21:11
  • $\begingroup$ OK, I have added a proof. Regarding the equivalence of norms, it is true that all norms are equivalent on a finite dimensional space but the way in which these equivalences behave (the constants involved) vary in an uncontrollable way and I doubt that a cleaner condition may be obtained, such as the one you propose in your comment. $\endgroup$
    – Ruy
    Sep 6, 2020 at 2:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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