Timeline for A formally weaker form of the extendable local reflexivity for Banach spaces
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 28, 2014 at 4:18 | vote | accept | Dongyang Chen | ||
| Jul 23, 2014 at 17:14 | comment | added | Dongyang Chen | You are right. Sorry! | |
| Jul 23, 2014 at 16:33 | comment | added | Dongyang Chen | I am not sure of your proof yesterday that we can enlarge $F$ and $E$ such that each is total over the other. Let $Z$ and $\{e_{n+1},...,e_{m}\}$ be as your proof yesterday. Is there a sequence $\{f_{n+1},...,f_{m}\}$ in $Z$ such that $\{e_{n+1},...,e_{m}\}$ and $\{f_{n+1},...,f_{m}\}$ are biorthogonal? | |
| Jul 23, 2014 at 15:40 | comment | added | Dongyang Chen | If $Z$ is a subspace of $X^{*}$ and $\{e_{n+1},...,e_{m}\}$ is a basic sequence in $X^{**}$ such that $Z$ is total over $span\{e_{n+1},...,e_{m}\}$, then is there a sequence $\{f_{n+1},...,f_{m}\}$ in $Z$ such that $\{f_{n+1},...,f_{m}\}$ are biorthogonal to $\{e_{n+1},...,e_{m}\}$? I am not sure of that. | |
| Jul 22, 2014 at 1:03 | comment | added | Bill Johnson | You are right. I was trying to do it so that you can get estimates depending on the dimension of $E$. Forget that and just enlarge $F$ and $E$ so that each is total over the other and choose $u_k$ Auerbach for $E$ and the unique biorthogonal $x_k^*$ in $F$, which must span $F$. Now you have no good estimate on the norms of the $x_k^*$. That does not matter because $\epsilon $ can be chosen small enough to compensate. | |
| Jul 21, 2014 at 21:07 | comment | added | Dongyang Chen | The equality $<(T+S)u_{k},x^{*}_{j}>=<u_{k},x^{*}_{j}>$ only implies that $<(T+S)x^{**},x^{*}>=<x^{**},x^{*}>(x^{**}\in E, x^{*}\in span\{x^{*}_{k}:k=1,2,...n\})$. Is $(x^{*}_{k})_{k=1}^{n}$ a basis for $F$ ? | |
| Jul 20, 2014 at 20:28 | comment | added | Bill Johnson | I gave an incorrect formula for $S$ and so have made an edit to my answer. | |
| Jul 20, 2014 at 20:27 | comment | added | Bill Johnson | The norm condition on $x_k^*$ follows from the fact that $F$ almost norms $E$, which means that the natural mapping from $F$ to $E^*$ is almost (i.e., up to $1+\epsilon$) a quotient map. | |
| Jul 20, 2014 at 20:25 | history | edited | Bill Johnson | CC BY-SA 3.0 |
added 829 characters in body
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| Jul 20, 2014 at 16:47 | comment | added | Dongyang Chen | I am not sure that the operator $T+S$ satisfies $<(T+S)x^{**},x^{*}>=<x^{**},x^{*}>(x^{**}\in E, x^{*}\in F)$ and the norm of $x^{*}_{k}$ is less than and equal to 1+ϵ . | |
| Jul 18, 2014 at 15:45 | history | answered | Bill Johnson | CC BY-SA 3.0 |