I think the question as expressed in the title should be clear. I do not know whether there is a known "characterization" of the weakly compact convex sets in $c_0(\mathbb N_0)$ but testing examples has lead me to conjecture that sequences in $\ell^1(\mathbb N_0)$ converging to zero uniformly on these sets also converge to zero in the norm topology. Is this conjecture correct/know/well-know/is there an explicit reference?
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1$\begingroup$ Check out Schur's property in Banach space theory. $\endgroup$– bathalf15320Commented Mar 22, 2022 at 5:06
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$\begingroup$ @bathalf15320 To use it, one should be able to prove that convergence on convex weakly compact sets in $c_0$ implies convergence at vectors in $\ell^{\,+\infty}$. Do you have any idea for this? $\endgroup$– TaQCommented Mar 22, 2022 at 7:42
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1$\begingroup$ Is there a reason why you consider convex sets only? Since the closed convex hull of a weakly compact set is again weakly compact, it doesn't seem to make a difference if we just drop "convex". $\endgroup$– Jochen GlueckCommented Mar 22, 2022 at 10:44
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1$\begingroup$ Jochen Glueck is right, this is a theorem of Krein (it holds in all quasi-complete locally convex spaces). The topology of uniform convergence on all weakly compact avsolutely convex sets is the Mackey topology on $\ell^1$ for the dual pair $\langle c_0,\ell^1\rangle$. $\endgroup$– Jochen WengenrothCommented Mar 22, 2022 at 11:30
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1$\begingroup$ No, of course not (weak quasi-completeness is equivalent to semi-reflexivity). The theorem of Krein applies to the Banach space $c_0$. $\endgroup$– Jochen WengenrothCommented Mar 22, 2022 at 11:52
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The answer is yes.
Proof. Assume to the contrary that a sequence $(x_n)$ in $\ell^1$ converges, say to $0$, uniformly on convex weakly compact subsets of $c_0$, but is not norm convergent and hence not norm convergent to $0$.
Note that the sequence even converges uniformly on all weakly compact subsets of $c_0$, be they convex or not (since the closed convex hull of a weakly compact set is again weakly compact). So if we construct a sequence $(y_n)$ in $c_0$ that converges weakly to $0$ but such that $\langle y_n, x_n \rangle \not\to 0$, then we have a contradiction.
After replacing $(x_n)$ with a subsequence we may assume that $\|x_n\| \ge \varepsilon$ for some $\varepsilon > 0$ and all $n$. Moreover, since $(x_n)$ converges, in particular, weakly to $0$, it also converges pointswise to $0$. So after replacing $(x_n)$ with yet another subsequence we may assume that, for all $n$, we have $\sum_{k=1}^{n-1} |x_n(k)| < \varepsilon/3$. Hence, there exist numbers $N_n \ge n$ such that $\sum_{k=n}^{N_n} |x_n(k)| \ge \varepsilon / 3$ for each $n$.
Now simply choose $y_n(k)$ to be zero for $k$ outside the set $\{n,\dots, N_n\}$ and to be the complex conjugate of the (complex) sign of $x_n(k)$ for $k$ inside this set. Then we have $\langle y_n, x_n \rangle \ge \varepsilon / 3$ for each $n$. Moreover, the sequence $(y_n)$ in $c_0$ is bounded and converges pointwise to $0$; hence, it also converges weakly to $0$.
answered Mar 22, 2022 at 11:58
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$\begingroup$ Great! The closed convex hull of a weakly compact set is weakly compact was the most important observation that was missing in my head. $\endgroup$– TaQCommented Mar 22, 2022 at 13:02
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Please take this post as an appendix to Jochen Glueck's answer. Let $X$ be a Banach space and $(f_n)$ be a sequence in $X^*$. It is fairly routine to show the equivalence of the following.
- $\sup\{|f_n(x)|:x\in A\}\to 0$ for every weakly compact $A\subseteq X$.
- $f_n(x_n)\to 0$ for every weakly null sequence $(x_n)$ in $X$.
- $f_n(x_n)\to 0$ for every weakly Cauchy sequence $(x_n)$ in $X$.
- $\sup\{|f_n(x)|:x\in A\}\to 0$ for every weakly precompact $A\subseteq X$.
Now suppose, in addition, that $X$ does not contain an isomorphic copy of $\ell^1$. Then, the unit ball $B$ of $X$ is weakly precompact. Thus, (1-4) above is also equivalent to
- $\|f_n\| = \sup\{|f_n(x)|:x\in B\}\to 0$.
Clearly, $c_0$ is a Banach space that doesn't contain an isomorphic copy of $\ell^1$.
answered Mar 22, 2022 at 13:03
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1$\begingroup$ The unit ball of every normed space is weakly precompact -- so what is the point about the containment of $\ell^1$? $\endgroup$ Commented Mar 23, 2022 at 10:14
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$\begingroup$ @JochenWengenroth For example, the unit ball of $\ell^1$ is not weakly precompact. The standart basis $(e_n)$ of $\ell^1$, as a sequence, doesn't have a weakly Cauchy subsequence in $\ell^1$. $\endgroup$ Commented Mar 23, 2022 at 16:07
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$\begingroup$ @OnurOktay In a precompact/totally bounded uniform space every net, in particular every sequence, has a Cauchy subnet but need not have a Cauchy subsequence. The concept of subnet is more involved than that of subsequence since one may have to consider another directed (index)set. Btw., what is the "fairly routine" argument (apart from Jochen Glueck's answer) to show that any of 1,2 or 3 implies 4? $\endgroup$– TaQCommented Mar 23, 2022 at 22:00
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$\begingroup$ I use the term "weakly precompact" as it is customary to use in Banach space terminology: "A subset of a Banach space $A$ is weakly precompact if every sequence in $A$ has a weakly Cauchy subsequence". Perhaps this usage caused ambiguity. The well-known theorem of Rosenthal provides that if $A$ contains no sequence that spans an isomorphic copy of $\ell^1$, then $A$ is weakly precompact. $\endgroup$ Commented Mar 23, 2022 at 23:02
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$\begingroup$ It is also well-known that the unit ball $B^{**}$ of $X^{**}$ is weak$^*$ compact by Banach-Alaoglu theorem, and the unit ball $B$ of $X$ is weak$^*$ dense in $B^{**}$ by Goldstine's theorem. Clearly the restriction of the weak$^*$ topology to $B$ is the same as the weak topology on $B$. Thus, it's true that every net in $B$ has weak$^*$ limit points in $B^{**}$, and every $x^{**}\in B^{**}$ is a weak$^*$ limit of a net in $B$. $\endgroup$ Commented Mar 23, 2022 at 23:08