Timeline for Continuous choice of Hahn-Banach extensions
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 28, 2011 at 22:35 | comment | added | Bill Johnson | @Yemon, In his Memoirs, Lindenstrauss proved that to get 1-injectivity it is enough to consider the cases where $F/E$ is one dimensional. | |
| Nov 28, 2011 at 22:19 | vote | accept | Itaï BEN YAACOV | ||
| Nov 28, 2011 at 21:47 | comment | added | Yemon Choi | (Clarification of my last comment: the 1-injectivity would be equivalent to requiring norm-preserving extensions for every $E$ and $F$, not just those where $F/E$ is finite-dimensional) | |
| Nov 28, 2011 at 21:45 | comment | added | Yemon Choi | And that last statement would be saying that $C(K)$ is 1-injective in classical terminology, which happens only when $K$ is Stonean by a result of Kelley, Nachbin and someone else IIRC | |
| Nov 28, 2011 at 16:34 | comment | added | Matthew Daws | Hmm. The paper Philip refers to is: projecteuclid.org/euclid.ijm/1255988172 Now Proposition 1 is of interest to us. It shows (via an easy argument) that if there is a map $E_1^* \rightarrow F_1^*$ then as we're mapping into the closed unit ball of $F^*$, the constant $\lambda$ appearing in the estimate $\|\overline T\| \leq \lambda \|T\|$ can be chosen to be 1. Indeed, I think you can chase through the proof of Proposition 1 and show that you need: for every compact Hausdorff $K$ and bdd linear map $T:E\rightarrow C(K)$, we can find a norm-preserving extension of $T$ to $F$. | |
| Nov 28, 2011 at 15:31 | comment | added | Itaï BEN YAACOV | @Matthew: Thanks - this look like a presentable version of what I was thinking. | |
| Nov 28, 2011 at 15:17 | comment | added | Matthew Daws | @Itai: Sure-- pick a basis for $F/E$ of size $n$ say, and then the Open Mapping Theorem shows that $F \cong E \oplus \mathbb R^n$ where you give the direct sum the $\max$ norm, and $\mathbb R^n$, well, any norm. So pick your extension $\overline T$ by setting it to be $T$ on $E$ and $0$ on $\mathbb R^n$. You only need $\|\overline T\| \leq \lambda \|T\|$ and that's ensured by $F$ being isomorphic to $E \oplus \mathbb R^n$. | |
| Nov 28, 2011 at 15:08 | comment | added | Itaï BEN YAACOV | Is it true that if $F/E$ is finite dimensional then $E$ is almost complemented in $F$? (Sorry, just trying to understand the definitions...) If so, I may be OK, it's true I went a bit too quickly over the transfinite induction for the general case, but for my purposes all I need is the finite dimensional case. | |
| Nov 28, 2011 at 13:29 | history | answered | Philip Brooker | CC BY-SA 3.0 |