3
$\begingroup$

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 $1 \leq p < \infty$.

$\endgroup$
4
  • 3
    $\begingroup$ I 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. $\endgroup$ Feb 11 at 21:16
  • $\begingroup$ At the very least, when $X$ is locally compact (i.e. finite-dimensional), there is the usual criterion given by the Arzelà-Ascoli theorem. $\endgroup$ Feb 12 at 10:21
  • $\begingroup$ @P.P.Tuong What form of the Arzelà-Ascoli theorem are you referring to? $\endgroup$
    – Smiley1000
    Feb 12 at 12:00
  • $\begingroup$ @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 $\endgroup$ Feb 12 at 13:25

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ But 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. $\endgroup$
    – Smiley1000
    Feb 12 at 6:37
  • 1
    $\begingroup$ @Smiley1000 I edited my post. $\endgroup$
    – Ali Mezher
    Feb 12 at 16:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .