Is the point-spectrum of bounded linear integral operators always countable?

Question

I would like to know whether bounded linear integral operators (defined on a separable Hilbert space of functions) always have a countable point-spectrum. Or if not, what would be a practical counterexample.

The following situations for instance already lead to a countable point spectrum:

• self-adjoint (or symmetrizable) operator
• kernel integral operator with bounded kernel

But what about bounded integral operators in general? I couldn't find any helpful theorem.

Answer 1

I don't think that there is a simple answer, since the question is very general. The approach that I am familiar with usually works along the lines of:

• Higher regularity implies compact embeddings, such as $H_0^1\hookrightarrow\hookrightarrow L^2$ (and there are many more, just google "compact embeddings sobolev spaces").
• Integral operators often are bounded maps from $L^2\to H_0^1$ (or from one Sobolev space into a space with higher regularity). Due to the compactness of the embeddings, this would imply the compactness of the integral operator, implying an at most countable point spectrum.

There is a plethora of compact embeddings of Sobolev spaces, and it can be a challenge of its own to find suitable embeddings. The above argument hinges on the common property of integral operators having some kind of regularization property, but I would expect there to be counter-examples in some spaces. Note that the properties of the spectrum are tied to the space the operator is considered in, so it is an even harder question to decide whether an integral operator does not have a countable spectrum in any space.

Often, the approach mentioned above is used in reverse for differential operators, for which the inverse is a compact operator, and similar statements for the spectrum follow.