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Operator compactness is characterized by maps the send the unit ball to relatively compact sets. Does anyone have a good justification for why we call this property compactness?

The best justification I've been able to come up with is that on Hilbert spaces a compact operator is a limit of finite rank operators. This conforms with the intuition that compactness is the property of being "almost finite." Can we do any better? For example, is it possible to interpret a compact operator as a compact set in some suitable topological space (my guess is no)? Is there anything more general we can say about the "almost finite" property of compact operators?

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  • $\begingroup$ In Banach spaces, not every compact operator is a norm limit of finite rank operators. Google "approximation property". $\endgroup$ – David Mitra May 2 '15 at 13:31
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The idea of compact came from sequences. You can see how one might come up with the name 'compact' to describe a set where you can't have an infinite set of points that are a minimum fixed positive distance from each other; that would not be a 'compact' set. Finding a cluster point gives you way to find a limit, and that's the importance of a 'compact' set. These notions were being formulated in the last part of the 19th century, around the time that a real number was rigorously defined for the first time!

Around the same time, people began studying differential equations by reformulating them in terms of integral equations. Such formulations were much nicer than direct formulations of differential equations. Iterative techniques allowed people to come up solutions. But you had to be able to find some sort of limit. There's where compactness came in handy.

The work of Fredholm was pivotal in this regard. He used a different type of 'compactness' for functions. Integral operators on finite domains often map bounded sequences of functions to ones with uniformly bounded derivatives (equicontinuous.) So one is able to extract a cluster point from such a sequence, and arrive at a limit that will solve some type of ordinary or partial differential equation. The Arzela-Ascoli Theorem concerning equicontinuous families (sequences) of function goes back to that time in the late 1800's, and imrovements of this theorem remain useful today.

The "Fredholm alternative" came out of compactness. Fredholm was able to prove that null spaces of various differential operators would be finite dimensional, and the deficiency in their ranges would also be finite dimensional for common cases on finite domains. This came out of compactness which, in this case, related compactness and dimension: the closed unit ball in a normed space is compact iff the space is finite-dimensional. (Fredholm also approximated integral equations by discrete equations.) Dimension also 'squashes' sets to be compact.

F. Riesz abstracted these techniques to define a 'compact' operator that was modeled after Fredholm's integral operators which mapped uniformly bounded functions to functions with uniformly bounded derivatives. The word 'compact' was a natural term to describe the abstract operator defined by Riesz because of its origin. What other term would you use? The abstraction of Riesz and the proofs he constructed based on Fredholm's work were so well-done that the material taught today is almost unchanged since the time of Riesz' ~1918 work. Compact operators and the Fredholm alternative are perfectly tied together now in operator algebras. The Fredholm index (nullity - deficiency) remains important and now relates to topological notions as well.

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Every operator is in particular a mapping (in the set theoretical sense) and each mapping $X \to Y$ is (also in set theoretical sense) some subset of the product $X \times Y$ (identified naturally with it's graph). If we are in the functional analysis setting $X$ and $Y$ are (at least) topological vector spaces. When we deal with linear operators then the domain of such an operator should be a linear space which is always unbounded and therefore cannot be compact. So in the naive definition: "compact operator=operator with compact graph" there would be no compact operators at all! You can also define compact operators naively as those operators with compact range-but the range of operator is also linear space so the only compact operator in this sense would be zero operator. Well, so why not to define compact operators as those operators which maps compact sets to compact sets: but then, when you are interested in bounded operators, each bounded operator would be compact (since every continuos mapping have this property of sending compact sets to compact sets). So what remains is this classical definition.

I don't know whether you are familiar with the theory of Hilbert $C^*$-modules. If yes, for a hilbert module $H$ one can define the algebra of all adjointable operators $B(H)$ (being the generalization of the algebra of bounded operators on Hilbert space) and also its subalgebra $K(H)$ of "compact operators" being the norm closure of the linear span of operators of the form $\Theta_{x,y}$ where $\Theta_{x,y}(z)=x\langle y, z\rangle$ (this "inner" product takes values in some $C^*-algebra therefore this operators are no longer finite rank). So the compact operator in the sense of Hilbert modules is in general no longer compact in the classical sense. So in some sense, this alternative definition (via closure of finite rank operators), when properly understood, seems to be more flexible.

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  • $\begingroup$ Why compact operators are not defined as operators taking bounded set to compact sets? @ truebaran $\endgroup$ – Anupam Oct 17 '18 at 16:45
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Suppose $X$ and $Y$ are Banach spaces and $U$ is the open unit ball in $X$. A linear map $T: X\to Y$ is said to be compact if the closure of $T(U)$ is compact in $Y$. It is clear that $T$ is then bounded. Thus $T\in \mathcal B (X, Y)$. Since $Y$ is a complete metric space, the subsets of $Y$ whose closure is compact are precisely the totally bounded ones. Thus $T\in \mathcal B (X, Y)$ is compact if and only if $T(U)$ is totally bounded. Also, $T$ is compact if and only if every bounded sequence $\{x_n\}$ in $X$ contains a subsequence $\{x_{n_k} \}$ such that $\{T(\{x_{n_k}) \}\}$ converges to a point of $Y$.

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