# Conditions for compactness of operator

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 $$\langle e_n, A e_m \rangle$$?

• 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) Sep 5, 2020 at 4:54
• @KaviRamaMurthy, thanks for your comment and pointing this out. I edited to make it more sensible incase someone comes up with a clever argument.
– PPR
Sep 5, 2020 at 5:09
• 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 $c_n$.
– Ruy
Sep 5, 2020 at 17:24

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.

• 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 $$\lim_{m,n\to\infty} m \max_{i,j\in{1,\dots,m}}|\langle e_i, A e_j\rangle| = 0\,?$$
– PPR
Sep 5, 2020 at 21:11
• 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.
– Ruy
Sep 6, 2020 at 2:40