Ultraweak topology on B(X): Is the map X\otimes X* -> B(X)* isometric?

Question

Let $X$ be a Banach space. Consider the map $$ \alpha\colon X\hat{\otimes} X^* \to B(X)^*, $$ defined one simple tensors as $$ \alpha(\xi\otimes\eta)(a) = \eta(a(\xi)).\quad (\xi\in X, \eta\in X^*, a\in B(X)) $$ Put differently, we consider the pairing between $X\hat{\otimes} X^*$ (the projective tensor product of $X$ and its dual space $X^*$) and $B(X)$ (the algebra of bounded linear operators on $X$) induced by $$ \langle \xi\otimes\eta, a\rangle = \langle a(\xi), \eta \rangle. $$

Question: Is $\alpha$ isometric?

This question arose when I tried to make sense of the ultraweak topology on $B(X)$ for a (nonreflexive) Banach space $X$. The question asks if $B(X)$ always has a distinguished part of its dual, which could be thought of as a generalization of a predual. However, even if $\alpha$ is isometric, I don't think it necessarily implies that $X\hat{\otimes} X^*$ is a predual of $B(X)$.

One reformulation of the question is as follows: Consider the map $$ \beta\colon B(X) \to B(X^*) $$ which sends an operator $a\in B(X)$ to its transpose $a^*\in B(X^*)$. This map is isometric and we can therefore think of $B(X)$ as a subspace of $B(X^*)$. Then $\alpha$ is isometric if and only if $B(X)$ is $1$-norming in $B(X^*)$ - meaning that the unit ball of $B(X)$ is weak*-dense in the unit ball of $B(X^*)$, for the weak*-topology on $B(X^*)$ induced by the isometric isomorphism $B(X^*)\cong (X\hat{\otimes} X^*)^*$.

Some more remarks:

1. Originally I had written: "We know that $B(X)$ is always weak*-dense in $B(X^*)$. (In general, this is weaker then being $1$-norming.) Translating back to $\alpha$, this means that $\alpha$ is always injective." However, as pointed out by Bill Johnson below, not even that is the case. So the answer to my question is a big "NO".

2. If $X$ is reflexive, then $\beta$ is surjective, and so $\alpha$ is isometric. (This can also be seen more directly.)

3. If $X$ has the metric approximation property, then already the unit ball of $F(X)$, the finite rank operators on $X$, is weak*-dense in the unit ball of $B(X^*)$. Thus, also in this case, $\alpha$ is isometric.

Answer 1

The answer is no.

Let $X$ be a separable Pisier counterexample [P] to Grothendieck’s problem. That is, both $X$ and $X^*$ have cotype 2 and every operator from $X$ or $X^*$ into a Hilbert space is 2-absolutely summing. As Pisier points out, there is a constant $C$ so that if $T$ is a finite rank operator from $X$ (or $X^*$) into a cotype 2 space, then $\pi_2(T) \le C\|T\|$, where $\pi_2(\cdot)$ is the 2-summing norm. This implies that if a finite rank operator $T$ from $X$ into a cotype two space satisfies $\|Tx\| \ge \delta \|x\|$ for all $x$ in some n-dimensional subspace of $X$, then $\|T\| \ge \delta c n^{1/2}$, where $c>0$ depends only on the cotype 2 constant of the range of $T$. Pisier spaces are weird, to be sure, but Pisier proved that every separable cotype 2 space isometrically embeds into a separable Pisier space.

Let $(E_n)$ be an increasing sequence of subspaces of $X$ with $E_n$ having dimension $n$ so that $\cup_{n} E_n$ is dense in $X$, and let $Y$ be the $\ell_1$ sum of $(E_n)$. There is a natural quotient mapping $Q$ from $Y$ onto $X$: given $e_n \in E_n$ with $\sum_n \|e_n\| < \infty$, set $Q(e_n)_n : = \sum _n e_n$. This quotient mapping has local liftings, which implies that $Q^*$ is an isometric embedding of $X^*$ onto a norm one complemented subspace of $Y^*$. Consequently, $X$ embeds isometrically into $Y^{**}$. (Details of this argument can be found in [J]). The nice thing about $Y$ is that it has cotype 2 since $X$ does, and $Y$ obviously also has the metric approximation property (even a monotone finite dimensional decomposition).

Claim: The weak$^*$ operator closure in $B(X,Y^{**})$ of the unit ball of $B(X,Y)$ has empty interior.

To see the claim, take any isometric embedding $J$ from $X$ into $Y^{**}$. Take any $n$ dimensional subspace of $JX$; for definiteness we use $JE_n$. Norm $JE_n$ up to $1+\epsilon$ by finitely many unit functionals in $Y^*$. If the claim is false, then there is an operator $J_n: X \to Y$ s.t. $\|J_nx\| \ge (1+\epsilon)^{-1} \|x\| $ for all $x\in E_n$, and the norms $\|J_n\|$ are uniformly bounded (the bound depending on the size of the ball that is contained in the weak$^*$ operator closure of $B(X,Y)$ in $B(X,Y^{**})$). Since $Y$ has the metric approximation property, $J_n$ can be taken of finite rank. But since $Y$ has cotype 2, Pisier’s result forces $\|J_n\| \ge c n^{1/2}$ for some constant $c>0$.

So see that the OP’s question has a negative answer, consider the space $Y \oplus X$.

[J] Johnson, William B. A complementary universal conjugate Banach space and its relation to the approximation problem. Proceedings of the International Symposium on Partial Differential Equations and the Geometry of Normed Linear Spaces (Jerusalem, 1972). Israel J. Math. 13 (1972), 301–310 (1973).

[P] Pisier, Gilles Counterexamples to a conjecture of Grothendieck. Acta Math. 151 (1983), no. 3-4, 181–208.

[EDIT 2/28/15] I should have mentioned that the argument above is similar to the proof of Theorem 3.3 in my paper [JO] with Timur Oikhberg.

[JO] Johnson, William B.; Oikhberg, Timur Separable lifting property and extensions of local reflexivity. Illinois J. Math. 45 (2001), no. 1, 123–137.

Answer 2

I think the OP’s point (1) is not correct; i.e., $B(Z)$ need not be weak$^*$ dense in $B(Z, Z^{**}) (= B(Z^*)$. You don’t need a Pisier space to show this. Instead, take a separable reflexive space $X$ that fails the approximation property (AP) (reflexivity is not so important, but it makes the notation and reasoning slightly simpler). As in my first answer, let $(E_n)$ be an increasing sequence of subspaces of $X$ with $E_n$ having dimension $n$ so that $\cup_{n} E_n$ is dense in $X$, let $Y$ be the $\ell_1$ sum of $(E_n)$, and let $Q: Y\to X$ be the natural quotient map, defined by $Q(e_n)_n : = \sum _n e_n$. As was hinted at in my first answer, $Y^* = W^\perp + Q^*X$ is a direct sum decomposition of $Y^*$ with the projection onto $Q^*X$ having norm one and where $W$ is the kernel of $Q$; i.e., $W $ is the subspace of all $(e_n)$ in $Y$ so that in $X$ $\sum_n e_n = 0$.

Since $X$ fails the AP, by Grothendieck’s duality theory there is a tensor $T$ in $X^* \hat{\otimes} X$, $T = \sum_n x_n^* \otimes x_n$ with $x_n^* \in X^*$, $x_n \in X$, $\sum_n \|x_n^*\| \cdot \|x_n\| < \infty$, and for all finite rank operators $S$ on $X$ we have $\langle S, T\rangle = \sum_n \langle x_n^*, Sx_n \rangle =0$ but $\langle I_X, T\rangle = \sum_n \langle x_n^*, x_n \rangle =1$. Consider now the tensor $T_1$ in $Y^* \hat{\otimes} X$ defined by $T_1 := \sum_n Q^* x_n^* \otimes x_n$. If $J$ denotes the natural isometric embedding from $X$ into $Y^{**} $ induced by $Q$, we see that $\langle T_1, J \rangle = \sum_n \langle Q^* x_n^*, Jx_n \rangle = \sum_n \langle x_n^*, x_n \rangle =1$. Now consider any $S$ in $B(X,Y)$. The operator $S$ is compact because $X$ is reflexive and $Y$ has the Schur property. We claim that $\langle T_1, S\rangle $, which is $\sum_n \langle Q^* x_n^*, Sx_n \rangle$, is zero. If not, since $\| P_kS-S\| \to 0$ as $k \to \infty$ by compactness of $S$ (here $P_k$ is the natural norm one projection from $Y= (\sum_{n=1}^\infty E_n)_1$ onto $(\sum_{n=1}^k E_n)_1$) we can choose $k$ s.t. $\langle T_1, P_k S\rangle \not= 0$. But $$ \langle T_1, P_kS\rangle = \sum_n \langle Q^* x_n^*, P_k Sx_n \rangle = \sum_n \langle x_n^*, QP_k Sx_n =0 $$ because $QP_k S$ has finite rank.

To see that (1) is false, consider the space $Z:= Y\oplus X$.