Timeline for Dual Banach space of $B(X,Y)$ when $X$ is finite dimensional
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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Oct 21, 2013 at 21:01 | vote | accept | Mikael de la Salle | ||
Oct 21, 2013 at 21:01 | comment | added | Mikael de la Salle | @GeraldEdgar: I think what you claim also requires the Radon-Nykodym property on one of the spaces. | |
Oct 21, 2013 at 20:59 | comment | added | Mikael de la Salle | @PietroMajer: What you explain is the (reversed) way I came to my question: I wanted to show that the bidual of $B(X,Y)$ is $B(X,Y^{**})$ isometrically. | |
Oct 21, 2013 at 17:28 | comment | added | Gerald Edgar | More generally, even when $X$ is infinite-dimensional, you get a tensor product as the dual of the compact operators from $X$ to $Y$. (I forget whether this requires the approximation property.) Of course the tensor product you want is the completion of the algebraic tensor product. | |
Oct 21, 2013 at 14:41 | answer | added | Bill Johnson | timeline score: 7 | |
Oct 21, 2013 at 12:31 | comment | added | Pietro Majer | (by $X\otimes Y$ above I mean the Banach space tensor product obtained by metric completion from the linear-algebraic tensor product etc. Also I meant "$Y^*$ splits in $Y^{***}$" ). | |
Oct 21, 2013 at 11:15 | comment | added | Pietro Majer | I would start from the linear isometry $(X\otimes Y^*)^* \sim B(X,Y)^{**}$ (true for any $X$ and $Y$). Dualizing we can compose with $X\otimes Y^*\to (X\otimes Y^*)^{**}\sim B(X,Y^{**})^*\to B(X,Y)^*$: the task is to show that the composition is an isometry for finite dimensional $X$. To start with, is it the case that for finite dimensional $X$ the last map is a left inverse? If $X$ is one dimensional, it is true, since $X^*$ splits in $X^{***}$. | |
Oct 21, 2013 at 11:03 | answer | added | Christian Clason | timeline score: 2 | |
Oct 21, 2013 at 9:50 | answer | added | Matthew Daws | timeline score: 5 | |
Oct 21, 2013 at 7:01 | history | asked | Mikael de la Salle | CC BY-SA 3.0 |