# Dual Banach space of $B(X,Y)$ when $X$ is finite dimensional

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?
• I would start from the linear isometry $(X\otimes Y^*)^* \sim B(X,Y)^{**}$ (true for any $X$ and $Y$). Dualizing we can compose with $X\otimes Y^*\to (X\otimes Y^*)^{**}\sim B(X,Y^{**})^*\to B(X,Y)^*$: the task is to show that the composition is an isometry for finite dimensional $X$. To start with, is it the case that for finite dimensional $X$ the last map is a left inverse? If $X$ is one dimensional, it is true, since $X^*$ splits in $X^{***}$. – Pietro Majer Oct 21 '13 at 11:15
• (by $X\otimes Y$ above I mean the Banach space tensor product obtained by metric completion from the linear-algebraic tensor product etc. Also I meant "$Y^*$ splits in $Y^{***}$" ). – Pietro Majer Oct 21 '13 at 12:31
• More generally, even when $X$ is infinite-dimensional, you get a tensor product as the dual of the compact operators from $X$ to $Y$. (I forget whether this requires the approximation property.) Of course the tensor product you want is the completion of the algebraic tensor product. – Gerald Edgar Oct 21 '13 at 17:28
• @PietroMajer: What you explain is the (reversed) way I came to my question: I wanted to show that the bidual of $B(X,Y)$ is $B(X,Y^{**})$ isometrically. – Mikael de la Salle Oct 21 '13 at 20:59
• @GeraldEdgar: I think what you claim also requires the Radon-Nykodym property on one of the spaces. – Mikael de la Salle Oct 21 '13 at 21:01

1. If $X=\ell_1^N$, then $B(X,Y)\equiv \ell_\infty^N(Y)$ and the result is easy.

2. If the unit ball of $X$ has $N$ extreme points, then $X$ is a quotient of $\ell_1^N$ and you can deduce what you want from (1).

3. The general case follows from (2) by approximation.

Notice that this is basically Dean's proof of local reflexivity (minus the steps needed to deduce local reflexivity from what you want), and is, to my way of thinking, the "right" approach.

• Great, this is exactly what I was looking for! Thanks. – Mikael de la Salle Oct 21 '13 at 15:06

This isn't a fully answer, but some comments on "conservation of difficulty". It's easy to show that

• $(X\hat\otimes Y^*)^* = B(X,Y^{**})$.

Suppose I know the following, which is a weak version of principle of local reflexivity:

Claim: If $Y$ is Banach and $M\subseteq Y^{**}$ and $N\subseteq Y^*$ are finite-dimensional subspaces and $\epsilon>0$, then there is $T:M\rightarrow Y$ with $\|T\|\leq 1+\epsilon$, and with $\langle m,f \rangle = \langle f,T(m) \rangle$ for $m\in M, f\in N$.

Let $\alpha$ denote the norm on $X\otimes Y^*$ induced by $B(X,Y)^*$. By the triangle-inequality, $\alpha \leq \pi$ the projective tensor norm. If $u\in X\otimes Y^*$ with $\pi(u)>1$ there is $S\in B(X,Y^{**})$ a contraction with $\langle S,u\rangle > 1$. Then $M=S(X)$ is finite-dimensional, there is a finite-dimensional $N$ with $u\in X\otimes N$, and choose $T$ as above, and consider $R=T\circ S\in B(X,Y)$. Then $\langle u,R \rangle = \langle S,u \rangle>1$, so choosing $\epsilon>0$ suitably small gives that $\alpha(u)>1$. So $\alpha\geq \pi$, hence $\alpha=\pi$ as required.

Conversely, suppose we know that $B(X,Y)^* = X\hat\otimes Y^*$. Then $B(X,Y)^{**} = B(X,Y^{**})$ and you can check that the canonical inclusion $B(X,Y) \rightarrow B(X,Y)^{**} = B(X,Y^{**})$ is just what you expect (composition with inclusion of $Y$ into $Y^{**}$). So given $M,N$ as in the claim, letting $X=M$ and $\iota:M\rightarrow Y^{**}$ be inclusion, by a version of Hahn-Banach, there is a net $(T_i)$ of contractions $X=M\rightarrow Y$ with $\lim_i \langle u,T_i \rangle = \langle \iota,u \rangle$ for each $u\in X\otimes Y^*$. By choosing $u$ suitably, this allows you to pick $T$ as in the claim.

So, in some sense, these results are equivalent. You can find elementary proofs of the claim: http://www.ams.org/journals/proc/1999-127-05/S0002-9939-99-04687-0/S0002-9939-99-04687-0.pdf For more on using this approach to principle of local reflexivity see http://www.ams.org/journals/proc/1973-040-01/S0002-9939-1973-0324383-X/S0002-9939-1973-0324383-X.pdf

• In fact the principle of local reflexivity was the motivation for my question; I am therefore looking fora proof that avoids this (this question was part of a problem I gave to my students on local representability, and I did not realize that the question was not that easy) . – Mikael de la Salle Oct 21 '13 at 12:28

I believe the book by Defant and Floret, Tensor Norms and Operator Ideals, contains the proof of this duality when one of the involved spaces is finite-dimensional (see in particular sections 6.4 and 16.1).

• Thanks, I will have a look when I get back to my ofiice. – Mikael de la Salle Oct 21 '13 at 12:34