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Let $X$ be a subset of a Banach space $Y$. Please can you give me a definition of what "$X$ is weakly compact" means? I want one which is in terms of sequences and boundedness, as opposed to one with topology and stuff like that. Thank you.

I have searched the internet for days to avail for such a nice definition.

I did receive an answer in this thread, however the answer makes no reference to the set $Y$. Also a citation to a text would be nice.

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    $\begingroup$ "Weakly compact" means "compact in the weak topology". That is usually not good to describe in terms of sequences. $\endgroup$ – Daniel Fischer Mar 10 '14 at 13:57
  • $\begingroup$ How about if $Y$ is a Hilbert space? And if $X$ is a closed subspace? Maybe reflexivity may help $\endgroup$ – maximumtag Mar 10 '14 at 13:59
  • $\begingroup$ But, by the Eberlein–Šmulian Theorem, a subset of a Banach space is weakly compact if and only if it is weakly sequentially compact. $\endgroup$ – David Mitra Mar 10 '14 at 14:02
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As said in comments, "$X$ is weakly compact" means "If we endow $Y$ with the weak topology, and consider its restriction to $X$, then $X$ is compact."

Since the weak topology is not metrizable, we cannot completely describe it using sequences; e.g., we don't get a sequential characterization of the closure of a set. However, we can describe compactness in this topology using sequences. This is what Eberlein–Šmulian theorem theorem says: a set $X\subset Y$ is weakly compact if and only if every sequence of elements of $X$ has a weakly convergent subsequence. This applies to any Banach space $Y$, reflexive or not.

For the weak* topology we do not have an analog of Eberlein–Šmulian theorem: for example, the unit ball of $\ell_\infty = \ell_1^*$ is compact in the weak* topology, but the sequence of "standard basis" vectors $e_n = (0,0,\dots, 0,1,0,\dots)$ in $\ell_\infty$ has no weak* convergent subsequence.

If $Y$ is reflexive, then the weak topology on $Y$ is also the weak* topology, considering $Y= (Y^*)^*$. So you get the best of both worlds: Banach-Alaoglu theorem provides compactness, while Eberlein–Šmulian theorem provides a description of that compactness in terms of sequences.

Recommended text: A Course in Functional Analysis by John B. Conway, Chapter 5.

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  • $\begingroup$ Hi, one question on this to be sure (because english is not my native language) I have understood. The sentence "$X\subset Y$ is weakly compact" means the following: Let $(Y, \tau)$ be a topological space, where $\tau$ is a weak topology on $Y$, then $(X, \tau)$ is a topological space where $X$ is compact. $\endgroup$ – jjepsuomi Mar 18 '18 at 11:09

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