# A question on Grothendieck space

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!

• 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. – Giorgio Metafune Aug 14 '20 at 8:03
• Thanks, Giorgio. But it seems that you do not answer my questions. – Dongyang Chen Aug 14 '20 at 8:32
• 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? – Giorgio Metafune Aug 14 '20 at 9:24
• 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. – Dongyang Chen Aug 14 '20 at 15:37
• No, everything happens in $X^*$. – 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.
• 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 $x^{***}=0$. – Dongyang Chen Aug 15 '20 at 0:33
• 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 $x^{***}=0$. – Dirk Werner Aug 15 '20 at 8:46