Denote $B(X,Y)$ the Banach space of bounded operators between Banach spaces $X$ and $Y$.


When $X$ and $Y$ are both finite dimensional, it follows from the formula
$$\|u\|\_{B(X,Y)} = \sup\_{\|x\|\_X < 1,\|\xi\|\_{Y^\*}< 1} \xi(u(x))$$
and the Hahn-Banach separation theorem that that the dual Banach space of $B(X,Y)$ is the projective tensor product $X \hat \otimes Y^\*$ (= $X \otimes Y^\*$ as a vector space with unit ball the convex hull of the $x \otimes \xi$ for $\|x\|\_X < 1,\|\xi\|\_{Y^\*}< 1$).


When only $X$ is finite dimensional, this identification of $B(X,Y)^\*$ with $X \hat \otimes Y^\*$ still holds isometrically. For a long time I thought that this was as elementary as the case when both spaces are finite dimensional, but I recently realized that I could not find an elementary proof. Can anybody help me, or provide me with a reference? (I think the only reference I know is Grothendieck's memoire).


Some elementary facts~:


* The dual of $B(X,Y)$ is $X \otimes Y^\*$ as a vector space.
* The formula above and Hahn-Banach tell me that the closed unit ball of $B(X,Y)^\*$ corresponds to the **weak-$\*$** closure of the unit ball of $X \hat \otimes Y^\*$. So the question is why is the closed unit ball of $X \hat \otimes Y^\*$ weak-$\*$ closed?



Dual Banach space of $B(X,Y)$ when $X$ is finite dimensional
1. If $X=\ell\_1^N$, then $B(X,Y)\equiv \ell\_\infty^N(Y)$ and the result is easy.
2. If the unit ball of $X$ has $N$ extreme points, then $X$ is a quotient of $\ell\_1^N$ and you can deduce what you want from (1).
3. The general case follows from (2) by approximation.


Notice that this is basically Dean's proof of local reflexivity (minus the steps needed to deduce local reflexivity from what you want), and is, to my way of thinking, the "right" approach.


