Duality of projective and injective tensor product

Question

I want a reference of the following statement which I think is true. Let $X$ and $Y$ be Banach spaces with $X$ finite dimensional. Then $(X\otimes_\epsilon Y)^*$ is isometrically isomorphic to $(X^*\otimes_\pi Y^*)$ and $(X\otimes_\pi Y)^*$ is isometrically isomorphic to $(X\otimes_\epsilon Y)^*$ where $\otimes_\epsilon$ and $\otimes_\pi $ denote the injective and projective tensor product of Banach spaces respectively. For tensor product of Banach spaces see the book https://www.amazon.in/Raymond-A-Ryan/e/B001K88BH0/ref=dp_byline_cont_pop_ebooks_1.

Answer 1

Since Ryan's book is mentioned in the original question, here's some pointers for an answer based on this source:

• See Section 3.4 for what the dual space of $X\otimes_\epsilon Y$ is. If you combine Proposition 3.14 (and the comments after) with Proposition 3.22 (see the comments after it) we get $(X\otimes_\epsilon Y)^* = I(X,Y^*) = I(Y,X^*)$ the space of integral operators.
• Now look at Theorem 5.33 which shows that when $X^*$ has the Radon-Nikodym Property and the approximation property (which is true when $X$ is finite-dimensional) then $I(Y,X^*) = X^* \hat\otimes_\pi Y^*$. See also the proof of Proposition 5.52.

For the dual of $X\otimes_\pi Y$ see Section 2.2 where it's shown that $(X\otimes_\pi Y)^* = B(X,Y^*)$ the space of all bounded linear maps. We always have that $X^* \hat\otimes_\epsilon Y^*$ is a subspace of $B(X,Y^*)$, see Section 3.1, and it's easy to see that when $X$ is finite-dimensional, then we have equality here.

Answer 2

The isometry of $(X\otimes_\pi Y)^*$ and $X^*\otimes_\varepsilon Y^*$ for finite dimensional $Y$ is in 6.1 of the book Tensor Norms and Operator Ideals by Defant and Floret and the isometry of $(X\otimes_\varepsilon Y)^*$ and $X^*\otimes_\pi Y^*$ is statement (2) in theorem 6.4.