Why every dual element of injective tensor product is a linearization of a bilinear form $X, Y$ are Banach spaces.
I already know that every element of $(X\otimes_\pi Y)^*$ ($^*$ means dual) corresponds to an element of $B(X\times Y)$, a set of bilinear forms on $X\times Y$, via an isometry.
Also, the injective norm of $X\times Y$ is smaller than the projective norm of $X\times Y$(hence the dual norms have the reverse inequality).
However, why this implies every element of $(X\otimes_\epsilon Y)^*$ is a linearization of an element of  $B(X\times Y)$?
Also, could you give me an example that $B(X\times Y)$ is strictly bigger than $(X\otimes_\epsilon Y)^*$? Thanks.
 A: Let $f\in (X\otimes_\epsilon Y)^*$.
Since the injective norm $\epsilon$ is smaller than the projective norm $\pi$, the corresponding operator norms have the reverse inequality 
$\Vert f\Vert_\epsilon\ge\Vert f\Vert_\pi$, so $f\in (X\otimes_\pi Y)^*$.
Since every element of $(X\otimes_\pi Y)^*$ is a linearization of a bounded bilinear form, I answered the first question.
(The second answer haven't finished yet)
For the second question, note every element of $(X\otimes_\epsilon Y)^*$ corresponds to an element of the set of integral bilinear form of $X\times Y$.
Let $K_1, K_2$ be compact sets. Since $C(K_1)\otimes_\epsilon C(K_2)=C(K_1\times K_2)$, $f\in (C(K_1)\otimes_\epsilon C(K_2))^*$ if and only if there is a unique regular Borel measure $\mu$ on $K_1\times K_2$ such that the corresponding bilinear form satisfies $B(f,g)=\int_{K_1\times K_2}f(s)g(t)d\mu(s,t) \forall f\in C(K_1), g\in C(K_2)$.
I believe there is a bilinear form in $B(C(K_1)\times C(K_2)$ that is not of the form of an integral, but I can't found the proper example here.
