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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!

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    $\begingroup$ 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
    Commented Aug 14, 2020 at 8:03
  • $\begingroup$ Thanks, Giorgio. But it seems that you do not answer my questions. $\endgroup$
    – Dongyang Chen
    Commented Aug 14, 2020 at 8:32
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    $\begingroup$ 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
    Commented Aug 14, 2020 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
    Commented Aug 14, 2020 at 15:37
  • $\begingroup$ No, everything happens in $X^*$. $\endgroup$
    – Giorgio Metafune
    Commented Aug 14, 2020 at 15:42

2 Answers 2

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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.

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  • $\begingroup$ 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
    Commented Aug 15, 2020 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
    Commented Aug 15, 2020 at 0:40
  • $\begingroup$ @Dongyang: You are right! $\endgroup$
    – Dirk Werner
    Commented Aug 15, 2020 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
    Commented Aug 15, 2020 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
    Commented Aug 15, 2020 at 12:56
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Q1 is already answered by Prof. Dirk Werner above. I simply list a number of equivalent conditions that seems to be related, but not the same as the Grothendieck property.

The following are indeed equivalent:

  1. every weak*-null sequence in $X^{\ast}$ has a weakly Cauchy subsequence.
  2. every bounded weak*-sequentially compact subset of $X^{\ast}$ is weakly precompact (every sequence has a weakly Cauchy subsequence).
  3. for every bounded $T:X\to c_0$, the adjoint $T^{\ast}:\ell^1\to X^{\ast}$ is weakly precompact (i.e., $T^{\ast}$ maps bounded sets onto weakly precompact sets).
  4. for every bounded $T:X\to Y$, where $Y$ is another Banach space with weak* sequentially compact dual ball, the adjoint $T^{\ast}$ is weakly precompact.
  5. no weak*-null sequence in the unit ball of $X^{\ast}$ contains an $\ell^1$-subsequence.
  6. there is no surjective bounded $T:X\to c_0$.

It is well known that $X$ is a Grothendieck space iff $X^{\ast}$ is weakly sequentially complete and (6)

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    $\begingroup$ While your post might contain useful information (although no proof or reference is given), I don't think that it answers the question. It looks more like a (long) comment. $\endgroup$
    – Alex M.
    Commented Apr 25, 2021 at 10:29
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    $\begingroup$ This is essentially Theorem 3.1.1 in arxiv.org/pdf/2102.03838.pdf $\endgroup$
    – Tomasz Kania
    Commented Apr 25, 2021 at 13:21
  • $\begingroup$ @Tomasz Kania: Thank you for the reference. I was not aware that there were so many open problems pertaining to Grothendieck spaces. It will be a pleasure to read through the paper, and it sure will take some time for me. $\endgroup$
    – Onur Oktay
    Commented Apr 25, 2021 at 13:56
  • $\begingroup$ @AlexM. I think there is a place on MO for answers in the "too long to be a comment" category which provide useful information and context for a question. I agree it would be nice to have some references. $\endgroup$
    – Tim Campion
    Commented Apr 25, 2021 at 14:42
  • $\begingroup$ AlexM @TimCampion: Sorry folks, I didn't include a reference since I do not have a good reference for the whole equivalence except my notes. If I included a proof myself, the post would be longer than the question and the answer. However, as Tomasz Kania graciously pointed out, the proof is almost verbatim similar to Theorem 3.1.1 in the reference he's given. $\endgroup$
    – Onur Oktay
    Commented Apr 25, 2021 at 15:10

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