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I would like to understand why compact operators are considered so special to consider them as an extra class of operators.

Over Hilbert spaces these (as far as I know) these are the ones with separable range - limits of finite rank operators. However in general Banach spaces this is not what makes a bounded operator compact. So I'm still wondering what the crucial point is to consider compact operators (something like "bounded operators are precisely the continuous ones and moreover turns the class of bounded operators into a banach space itself").

Does somebody have a good solid reasoning (some theorem characterizing compact operator)? Everything welcome of course =)

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    $\begingroup$ Well, compact operators can be approximated by finite rank operators. $\endgroup$
    – user40276
    May 31, 2014 at 20:24
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    $\begingroup$ @user40276: Not in general (only in special situations)!!! $\endgroup$
    – C-star-W-star
    May 31, 2014 at 20:27
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    $\begingroup$ Well, as far as I know, compact operators started to be studied by considering those Hilbert-Schimidit integral operators. $\endgroup$
    – user40276
    May 31, 2014 at 20:33
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    $\begingroup$ In response to your "only in special situations": compact operators can be approximated by finite rank operators on any Banach space with the approximation property. This includes all Banach spaces with a Schauder basis and thus the earliest and most influential examples of Banach spaces. Until Enflo came up with a counterexample in 1973, it was conceivable that every Banach space has the approximation property and thus might have seemed plausible that the compact operators were exactly the closure of the finite-rank operators. $\endgroup$
    – Tom Cooney
    May 31, 2014 at 20:43
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    $\begingroup$ Thats what I mean: There is Banach spaces which do not possess the approximation property. But your right: The idea that compact operators are precisely the limits of finite rank operators seems to be central for considering compact operators, or? What do u think? $\endgroup$
    – C-star-W-star
    May 31, 2014 at 20:50

1 Answer 1

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Differential operators are badly discontinuous in general, and not defined for all functions. This was recognized as a problem early in the study of PDEs of classical Math/Physics. However, it was found that the inverse problems written as Fredholm integral equations gave rise to operators that are very continuous, and, in modern terms, often compact. This has to do with the Arzela-Ascoli Theorem for equicontinuous families of functions, which goes back to around 1880. Integral operators, such as the Poisson integral, etc., and resolvents for $PDE's$, often map bounded sequences to equicontinuous sequences (which have convergent subsequences.)

A lot of results of such analysis, including existence of solutions, the Fredholm alternative, discrete spectrum, etc., were successfully summarized in F. Riesz's abstract setting of compact operators. The Riesz abstraction was so successful and so clean that the subject is still taught in virtual the same way it was originally presented in 1918.

After the subject of operator algebras began to develop, it was realized that the compact operators are a left and right ideal, and that Fredholm operators could be viewed as invertible modulo the ideal of compact operators. So the ideas coming out of Fredholm's late 19th century work on integral equations continued to bear fruit in more modern settings. Atiyah-Singer index theory then connected Fredholm indices with topological ones. I doubt that this is the end of the story connecting Compact Operators, Physics, differential/integral equations, Functional Analysis, Operator Algebras, Pseudo-Differential operators, Fredholm Indices, Topological Indices, and Geometry.

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