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Let $X,Y$ denote (real) vector spaces. The vector space of $n$-linear maps $X^n \to Y$ will be denoted by $L^n(X,Y)$. Unless I'm much mistaken

$$L(X,L(X,Y)) \ \ \ L^2(X,Y) \ \ \ L(X \otimes X,Y)$$

are pairwise isomorphic. One could even view the three as pairwise naturally isomorphic functors from vector spaces to vector spaces (contravariant in $X$ and covariant in $Y$).

But what if we want to work with Banach spaces and bounded linear maps instead?Operator norms allow us to make $B(X,B(X,Y))$ into a Banach space without any headache. A 2-linear map $f:X^2 \to Y$ is continuous in the product topology if and only if the norm $\|f\| = \sup_{\|x\| = \|x'\| = 1} \|f(x,x')\|$ is finite and this makes $B^2(X,Y)$ the set of continuous $2$-linear maps into a Banach space which is isometrically isomorphic to $B(X,B(X,Y))$. But what about $B(X \otimes X, Y)$? According to wikipedia there are various ways of putting a norm on $X \otimes X$. Which is the right one if we want the same sort of equivalence in this setting?

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up vote 10 down vote accepted

You're looking for the projective tensor norm. Explicitly, for $\omega$ in the algebraic tensor product $X \otimes_{\mathbb{R}} Y$ it is given by \[ \Vert \omega\Vert_{\pi} = \inf\left\{\sum \|x_{i}\|\,\|y_{i}\|\,:\,\omega = \sum_{i=1}^{n} x_{i} \otimes y_{i}\right\} \] and the associated tensor product $X \hat{\otimes}_{\pi} Y$ is the completion of the algebraic tensor product with respect to that norm (the above expression obviously defines a semi-norm, in order to show that it is a norm, you have to think a little bit).

Since $X \otimes_{\mathbb{R}} Y$ sits inside $X \hat{\otimes}_{\pi} Y$ you get a bilinear map $m:X \times Y \to X \hat{\otimes}_{\pi} Y$ by sending $(x,y)$ to $x \otimes y$ and by definition $\Vert x \otimes y\Vert_{\pi} \leq \|x\|\,\|y\|$ so that $\|m\| \leq 1$.

The basic thing you have to check is that the above inequality is in fact an equality $\Vert x \otimes y\Vert_{\pi} = \|x\|\,\|y\|$ and using this it is easy to prove that a continuous bilinear map $\Phi: X \times Y \to Z$ yields a unique linear map $\varphi: X \hat{\otimes}_{\pi} Y \to Z$ of the same norm (by the property of the algebraic tensor product $\Phi$ gives a linear map on $X \otimes_{\mathbb{R}} Y$ and if you've proven that this linear map is continuous, it is continuous on a dense subspace, hence extends uniquely to $X \hat{\otimes}_{\pi} Y$).

In other words the map $m: X \times Y \to X \hat{\otimes}_{\pi} Y$ yields an isometric isomorphism $$\displaystyle B(X \hat{\otimes}_{\pi} Y, Z) \to B^2(X,Y;Z) $$ by pre-composition $\varphi \mapsto \varphi \circ m = \Phi$. As you already observed $B^2(X,Y;Z) \cong B(X,B(Y,Z))$ which yields the isometric isomorphisms \[ B(X \hat{\otimes}_{\pi} Y, Z) \cong B(X,B(Y,Z)) \cong B^2(X,Y;Z) \] as in the case of vector spaces. In fact, you can convince yourself that you have to define the projective tensor norm as above if you want the first isomorphism to hold, already for $Z = \mathbb{R}$.

You can find all this and much more in great detail in R.A. Ryan's book Introduction to tensor products of Banach spaces, Springer 2001.

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Thanks for prompt reply and the reference! Accepted. – Mike F Mar 8 '11 at 5:52

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