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Let $A=L^2(X)$ be the space of square integrable functions on a compact Euclidean space $X$. If we equip $A$ with the usual 2-norm, is $A$ compact?

Edit: And if we restrict AA by adding the assumption that the functions are totally bounded, i.e. the supremum norms of all the functions are bounded by a constant?

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    $\begingroup$ No, for two different reasons: 1) it's a vector space 2) it's not even locally compact. $\endgroup$
    – user98602
    Apr 1, 2016 at 17:10
  • $\begingroup$ And if we restrict $A$ by adding the assumption that the functions are totally bounded, i.e. the supremum norm of all the functions are bounded by a constant? $\endgroup$
    – Nadori
    Apr 1, 2016 at 17:22
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    $\begingroup$ Look at my answer. All the functions have sup norm 1. $\endgroup$ Apr 1, 2016 at 17:26
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    $\begingroup$ I remember something along the lines "A subset is compact iff it is closed and almost finite dimensional.". $\endgroup$ Apr 1, 2016 at 17:48

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No. If $X=[0,1]$, then the sequence $\{e^{2\pi inx}\}_{n\in\mathbb{Z}}$ has no convergent subsequence.

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    $\begingroup$ By the way, this shows that even the closed unit ball isn't compact $\endgroup$ Apr 1, 2016 at 17:11
  • $\begingroup$ Do you see what assumption I need to add to make A compact? $\endgroup$
    – Nadori
    Apr 1, 2016 at 17:49
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    $\begingroup$ You could try the Arzela-Ascoli theorem, though that's for continuous functions. $\endgroup$ Apr 1, 2016 at 17:58
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$d(x,y)=\|x-y\|$ is an unbounded metric. Any metric on a compact space is bounded.

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