Which class of reflexive spaces $X$ having the property: if a sequence $(x_{n})_{n}\subset B_{X}$ converges to $x$ weakly and $\|x_{n}\|\rightarrow 1$, then the norm of $x$ must be 1. Of course, the classical sequence spaces $l_{p}$ do not have this property.It seems that this condition is too strong.
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asked Oct 16, 2015 at 0:55
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$\begingroup$ For starters, I think no separable reflexive Banach space can have this property. It seems to be pretty straightforward to show that for any separable Banach space $Y$, there is a sequence $f_n \in Y^*$ with $\|f_n\| = 1$ and $f_n \to 0$ weakly-*. (Use the Hahn-Banach theorem to make $f_n$ which vanish on more and more elements of a countable dense subset of $X$.) So if $X$ is reflexive we can apply this with $Y = X^*$. Also of course, there are non-reflexive spaces with this property, like $l_1$ or any other space with Schur's property. $\endgroup$– Nate EldredgeCommented Oct 16, 2015 at 1:20
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$\begingroup$ You are right, Nate. $S_{X^{*}}$ is $weak^{*}$ sequentially dense in $B_{X^{*}}$ for every infinite-dimensional space $X$. $\endgroup$– Dongyang ChenCommented Oct 16, 2015 at 1:39
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$\begingroup$ There are of course spaces with this property: this works (trivially) when weak convergence implies norm convergence, as in finite dimensional spaces $X$ and $X=\ell^1$ (which is not reflexive). $\endgroup$– Christian RemlingCommented Oct 16, 2015 at 2:16
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$\begingroup$ @DongyangChen: I did not quite see how to do the non-separable case, but if you do, then it seems your question is resolved. You can post an answer for it yourself. $\endgroup$– Nate EldredgeCommented Oct 16, 2015 at 14:57
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$\begingroup$ @NateEldredge: Since $S_{X^{*}}$ is $weak^{*}$ sequentially dense in $B_{X^{*}}$ for every infinite-dimensional space $X$(see J.Diestel, Sequences and series in Banach spaces. page.223,Exercise 2), there is a sequence $(x^{*}_{n})_{n}\in S_{X^{*}}$ such that $x^{*}_{n}$ converges to zero $weak^{*}$ly. If $X$ is reflexive, then $x^{*}_{n}$ converges to zero weakly. $\endgroup$– Dongyang ChenCommented Oct 16, 2015 at 15:14
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I think the following space which is called Grothendieck space may be partial answer for your question.
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$\begingroup$ Certainly not. Every reflexive space is Grothendieck, yet don't satisfy the assumption. $\endgroup$ Commented Oct 16, 2015 at 11:25
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$\begingroup$ You are right, Denis. Every infinite-dimensional reflexive space is Grothendieck, yet does not satisfy my question. $\endgroup$ Commented Oct 16, 2015 at 13:34
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$\begingroup$ Since $S_{X^{*}}$ is $weak^{*}$ sequentially dense in $B_{X^{*}}$ for every infinite-dimensional space $X$(see J.Diestel, Sequences and series in Banach spaces. page.223,Exercise 2), there is a sequence $(x^{*}_{n})_{n}\in S_{X^{*}}$ such that $x^{*}_{n}$ converges to zero $weak^{*}$ly. If $X$ is reflexive, then $x^{*}_{n}$ converges to zero weakly. Thus, my question is false. Sorry! $\endgroup$ Commented Oct 16, 2015 at 15:17