A Banach space $X$ is said to be Grothendieck if the weak and the weak* convergence of sequences in $X^{*}$ coincide. I have the following two questions.

Question 1. A Banach space $X$ is Grothendieck if and only if every weak*-Cauchy sequence in $X^{*}$ is weakly Cauchy?

Question 2. If $(x^{*}_{n})_{n}$ is a weak Cauchy sequence and a weak*-null sequence in $X^{*}$, is $(x^{*}_{n})_{n}$ a weak-null sequence?

Thank you!

  • 3
    $\begingroup$ If $(x_n^*)$ is $w^*$ Cauchy, then for every $x \in X$, $x_n^*(x) $ has a limit, call it $f(x)$. Then $f$ is linear. Moreover, $\|x_n^*\| \le C$, by uniform boundedness, hence $f$ is bounded, too. Then $w^*$ Cauchy means $w^*$ convergent. $\endgroup$ – Giorgio Metafune Aug 14 '20 at 8:03
  • $\begingroup$ Thanks, Giorgio. But it seems that you do not answer my questions. $\endgroup$ – Dongyang Chen Aug 14 '20 at 8:32
  • 4
    $\begingroup$ The argument applies both to $w$ and $w^*$ convergence, so Cauchy means convergent in both topologies and Q1 is equivalent to the definition. Q2 follows similarly, if $x_n^* \to 0$ $w^*$ and is $w$-Cauchy, then it converges weak to some $z^*$ and then $z^*=0$ since $w$ imples $w^*$. Did I overlook some point? $\endgroup$ – Giorgio Metafune Aug 14 '20 at 9:24
  • $\begingroup$ If $x^{*}_{n}\rightarrow 0$ $ w^{*}$ and is $w$-Cauchy, then it converges weak to some $x^{**}\in X^{**}$, not $z^{*}\in X^{*}$. I think that you overlook this point. $\endgroup$ – Dongyang Chen Aug 14 '20 at 15:37
  • $\begingroup$ No, everything happens in $X^*$. $\endgroup$ – Giorgio Metafune Aug 14 '20 at 15:42

I find the following criterion useful: A sequence $(x_n)$ is Cauchy iff for all subsequences $(x_{n_{k+1}}-x_{n_k})$ tends to $0$. This works for the norm topology, the weak topology and the weak$^*$ topology. This answers Q1 in the positive.

As for Q2, if $(x_n^*)$ is weakly Cauchy and weak$^*$ null, it has a limit $x^{***}$ for the weak$^*$ topology of $X^{***}$; decompose $x^{***}=x^* + x_s^{***}$, where $x_s^{***}$ is the ``singular part'' in the annihilator of $X$ in $X^{***}$. By, the assumption of Q2, $x^*=0$; i.e., $x^{***}$ is singular. This seems to be as good as it gets in a general Banach space.

  • $\begingroup$ Since $(x^{*}_{n})$ is weak*-null, $x^{***}(x)=0$ for all $x\in X$,i.e.,$x^{***}$ is in the annihilator of $X$ in $X^{***}$. But this does not necessarily imply that $x^{***}=0$. We have to prove that $x^{***}=0$. $\endgroup$ – Dongyang Chen Aug 15 '20 at 0:33
  • $\begingroup$ Using the criterion you mentioned, it seems that we can only prove the necessary part of Q1, but can not prove the sufficient part. $\endgroup$ – Dongyang Chen Aug 15 '20 at 0:40
  • $\begingroup$ @Dongyang: You are right! $\endgroup$ – Dirk Werner Aug 15 '20 at 7:12
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    $\begingroup$ Call the condition in Q1 Cauchy Grothendieck. Let $X$ have this property. If $(x_n^*)$ is w$^*$ null, it has a limit $x^{***}\in X^{***}$. To show that it is $0$, consider the w$^*$ null sequence $(x_1^*, 0, x_2^*, 0, \dots)$ interlacing the given sequence with $0$. It has a limit $y^{***}\in X^{***}$ since the space is Cauchy Grothendieck. Now along the odd integers, the new sequence tends tends to $x^{***}$, along the even integers it tends to $0$. Hence $x^{***}=0$. $\endgroup$ – Dirk Werner Aug 15 '20 at 8:46
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    $\begingroup$ @Giorgio: Yes, separability of the bidual is good enough, e.g., the bidual of the James space is separable. And a separable Grothendieck space is reflexive. -- Q2 is true in a Grothendieck space (or any other space with a wsc dual); but I read the question as about general Banach spaces. $\endgroup$ – Dirk Werner Aug 15 '20 at 12:56

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