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Jul 15, 2021 at 19:17 comment added Mikael de la Salle @YCor Sure, this is what I meant.
Jul 15, 2021 at 11:06 vote accept Overflowian
Jul 15, 2021 at 10:51 comment added YCor @MikaeldelaSalle you probably mean Fredholm rather than compact.
Jul 15, 2021 at 10:51 comment added YCor @WarlockofFiretopMountain see the last few lines in the first page (=page 721) of the linked (1990 Proc AMS) paper.
Jul 15, 2021 at 9:22 comment added Mikael de la Salle I guess that, by Bouldin's result, the characterization extends to arbitrary Hilbert spaces as follows (using Bouldin's notation): $T$ belongs to the closure of compact operators iff $\textrm{essnull} T$ and $ \textrm{essnull} T^*$ are equal or both finite.
Jul 15, 2021 at 9:22 comment added Overflowian Thank you, I am convinced by your proof but the assumption does not look straightforward to me. How did you go from Bouldin's theorem, which gives conditions in terms of the spectral projector of $|T|, |T^*|$, to restate it in terms of closedness of the range and dimension of kernel/cokernel?
Jul 15, 2021 at 8:57 history edited YCor CC BY-SA 4.0
fixed typos
Jul 15, 2021 at 8:40 history answered YCor CC BY-SA 4.0