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?