Let $X$, $Y$ be Banach spaces and $L(X, Y)$ be the space of continuous linear mappings between them equipped with the operator norm. Is there a criterion for when a subset $\mathcal{L} \subseteq L(X, Y)$ is relatively compact? In my textbook, I could only find the well-known theorems about relative compactness in $C(K)$ and $L^p(\mathbb{R})$ for $1 \leq p < \infty$.
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3I would be surprised if there were a useful criterion at that level of generality. Already when $X=Y=L^2(0,1)$ this seems pretty ambitious to me. – Giuseppe Negro Feb 11 '24 at 21:16
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At the very least, when $X$ is locally compact (i.e. finite-dimensional), there is the usual criterion given by the Arzelà-Ascoli theorem. – P. P. Tuong Feb 12 '24 at 10:21
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@P.P.Tuong What form of the Arzelà-Ascoli theorem are you referring to? – Smiley1000 Feb 12 '24 at 12:00
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@Smiley1000 A very general statement for various uniformities can be found in Bourbaki's General Topology, chap. X, § 2, no. 5. Note that since $L(X,Y)$ is Hausdorff and complete, relative compactness in $L(X,Y)$ of a subset of $L(X,Y)$ is equivalent to its precompactness for the induced unifomity – P. P. Tuong Feb 12 '24 at 13:25
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It is possible only (as i see it) if you consider the operator $T: L(X,Y) \to L(X,Y)$ (say $L$) Banach spaces. $T$ here is a linear operator of operators. Then you can use: A linear operator is compact if and only if the image of any bounded set is relatively compact.
Ali Mezher
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1But to decide whether $T$ is compact, we need to know what the compact subsets of $L(X, Y)$ look like, which is what this question is about. – Smiley1000 Feb 12 '24 at 06:37
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