# Duality of projective and injective tensor product

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.

• I guess you mean $X^* \otimes_\epsilon Y^*$ and not $(X\otimes_\epsilon Y)^*$ in the 3rd line. Ryan adds a hat for the completed tensor products, a notation I've copied below in my answer. Mar 23 at 12:29

• 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.
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.