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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