I relate to this one question about corollary 4.10.2 pag. 198 of "Introduction to Hilbert Spaces - Debnath, Mikusinki" third edition, that states

Let $$A$$ be a compact self-adjoint operator on an infinite-dimensional Hilbert space $$\mathcal{H}$$. Then $$\mathcal{H}$$ has a complete orthonormal system $$\{v_k\}$$ consisting of eigenvectors of $$A$$

So I interpret it like "if an Hilbert space admits a compact self-adjoint operator, it will be separable" but this is not true (also this).

So what are the authors really telling, that I'm not understanding, with that proposition?

Little note:

$$\ker{A}$$ is sequentially closed subspace of $$\mathcal{H}$$, because $$A$$ is linear and compact (hence bounded, hence continuos), so by decomposition theorem $$\mathcal{H} = \ker{A} \oplus \ker{A}^{\perp\mathcal{H}}$$ Now, Hilbert-Schmidt theorem basically says that exists a sequence of eigenvectors $$u_k$$ of $$A$$ associated to non-null eigenvalues, such that $$\text{cl}\,\text{Span}\{u_k\} = \ker{A}^{\perp\mathcal{H}}$$ but nothing is said about $$\ker{A}$$ that may also be non-separable, so that's why I cited $$\ker{A}$$ in the title

• What about the case where $A=0$ on a non-separable space? Commented Jul 27, 2022 at 5:26
• @DisintegratingByParts Exactly! $A=0$ is linear, bounded, compact and self-adjoint with $\ker{A}=\mathcal{H}$, right? So what are the authors telling? Commented Jul 27, 2022 at 10:38
• If the orthonormal system has to be countable then the statement is false, as the case $A=0$ on any non-separable Hilbert space shows.
– daw
Commented Jul 27, 2022 at 14:15

Every element of $$\ker A$$ is an eigenvector for $$A$$ with eigenvalue $$0$$. So you just choose an orthonormal basis of $$\ker A$$, countable or not, and you put it together with an orthonormal basis of $$(\ker A)^\perp$$ made out of eigenvectors, to get a full basis of eigenvectors.
It is certainly true that the orthogonal complement to the kernel decomposes as (the closure of) a countable direct sum of finite-dimensional eigenspaces, since the non-zero eigenvalues' only possible limit point is $$0$$. But (as alluded-to in @MartinArgerami's answer), this says nothing about the separability-or-not of the kernel itself. Yes, this does say that a compact operator on a non-separable Hilbert space must have a very large (=inseparable) kernel. Meanwhile, on separable Hilbert spaces, $$0$$ need not be an eigenvalue at all (though it is inevitably in the spectrum).
• Well, just to repeat, on an infinite-dimensional separable Hilbert space, a compact self-adjoint operator can possibly have only non-zero eigenvalues. Yes, since the spectrum is a closed, and eigenspaces are finite-dimensional, $0$ must be a limit point, so is definitely in the spectrum. There are examples where $0$ is an eigenvalue, and, also, where it is not. That is, there are examples where the kernel is trivial... Commented Jul 27, 2022 at 21:20