Timeline for A question on Grothendieck space
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 15, 2020 at 16:39 | comment | added | Bill Johnson | @DirkWerner: So your answer and remark show that $X$ has the property in Q2 if and only if $X^*$ is weakly sequentially complete. | |
| Aug 15, 2020 at 15:22 | vote | accept | Dongyang Chen | ||
| Aug 15, 2020 at 12:56 | comment | added | Dirk Werner | @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. | |
| Aug 15, 2020 at 9:04 | comment | added | Giorgio Metafune | @ Dirk Werner Nice argument! Q2 seems to be false since $B_{X^*}$ is $w$-dense in $B_{X^{***}}$ but we need an $X$ where this density is realized through sequences. Maybe some separabilty of $X^{**}$ suffices and the space should not have the Grothendieck property. | |
| Aug 15, 2020 at 8:46 | comment | added | Dirk Werner | 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 $x^{***}=0$. | |
| Aug 15, 2020 at 7:12 | comment | added | Dirk Werner | @Dongyang: You are right! | |
| Aug 15, 2020 at 0:40 | comment | added | Dongyang Chen | Using the criterion you mentioned, it seems that we can only prove the necessary part of Q1, but can not prove the sufficient part. | |
| Aug 15, 2020 at 0:33 | comment | added | Dongyang Chen | 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 $x^{***}=0$. | |
| Aug 14, 2020 at 17:04 | history | answered | Dirk Werner | CC BY-SA 4.0 |