Extremal Structure of Projective Tensor Products

Abstract

We prove that, given two Banach spaces X and Y and bounded, closed convex sets $C\subseteq X$ and $D\subseteq Y$, if a nonzero element $z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)\subseteq X\widehat{\otimes }_\pi Y$ is a preserved extreme point then $z=x_0\otimes y_0$ for some preserved extreme points $x_0\in C$ and $y_0\in D$, whenever $K(X,Y^*)$ separates points of $X \widehat{\otimes }_\pi Y$ (in particular, whenever X or Y has the compact approximation property). Moreover, we prove that if $x_0\in C$ and $y_0\in D$ are weak-strongly exposed points then $x_0\otimes y_0$ is weak-strongly exposed in ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$ whenever $x_0\otimes y_0$ has a neighbourhood system for the weak topology defined by compact operators. Furthermore, we find a Banach space X isomorphic to $\ell _2$ with a weak-strongly exposed point $x_0\in B_X$ such that $x_0\otimes x_0$ is not a weak-strongly exposed point of the unit ball of $X\widehat{\otimes }_\pi X$.

1 Introduction

One of the most celebrated and earlier results in Functional Analysis is Krein–Milman theorem. This result establishes that if K is a compact convex subset of a locally convex space then $K={\overline{{{\,\textrm{co}\,}}}}({\text {ext}}\left( K\right) )$, where ${\text {ext}}\left( K\right)$ denotes the set of extreme points of K (see e.g. [16, Theorem 3.22]). An example of application is to the unit ball of a dual Banach space $X^*$, where it yields that $B_{X^*}={\overline{{{\,\textrm{co}\,}}}}^{w^*}({\text {ext}}\left( B_{X^*}\right) )$. This result is of capital importance because, thanks to Hahn–Banach theorem, the structure of the geometry of a Banach space X is determined by the dual unit ball $B_{X^*}$. Thus, Krein–Milman theorem tells us that the set ${\text {ext}}\left( B_{X^*}\right)$ codifies all the geometric information of the space.

The identification of the extreme points (and related notions as exposed, denting, or strongly exposed points) on particular classes of Banach spaces has attracted the attention of many researchers in functional analysis, especially in spaces where the definition of the norm is of high complexity, see e.g. [5, 17, 18] for duals of spaces of compact operators, [14] for Orlicz–Lorentz spaces, [11] for Kothe–Bochner spaces, or, more recently, [1,2,3, 8, 9] for Lipschitz-free spaces. In this note, we will focus on projective tensor products. Denoted by $X \widehat{\otimes }_\pi Y$, the projective tensor product is the completion of the algebraic tensor product $X \otimes Y$ endowed with the norm

$$\begin{aligned} \Vert z\Vert _{\pi }:= \inf \left\{ \sum _{i=1}^n \Vert x_i\Vert \Vert y_i\Vert : z = \sum _{i=1}^n x_i \otimes y_i \right\} , \end{aligned}$$

where the infimum is taken over all such representations of z. Recall also that $B_{X\widehat{\otimes }_\pi Y}={\overline{{{\,\textrm{co}\,}}}}(B_X\otimes B_Y)$.

In the analysis of the extremal structure of the projective tensor product we distinguish two lines. The first one is the exhaustive analysis of the extreme points in duals of operators spaces done by Collins and Ruess [5] and by Ruess and Stegall [18]. They established that, given two Banach spaces X and Y, the extreme points of the dual unit ball of the $w^*$-to-w continuous compact operators $K_{w^*}(X^*,Y)$ are the elements of the form $x^*\otimes y^*$, for $x^*\in {\text {ext}}\left( B_{X^*}\right)$ and $y^*\in {\text {ext}}\left( B_{Y^*}\right)$. As a consequence of a classical result of tensor product theory [19, Theorem 5.33], if $X^*$ or $Y^*$ has the Radon–Nikodym property and $X^*$ or $Y^*$ has the approximation property then

$$\begin{aligned} {\text {ext}}\left( B_{X^*\widehat{\otimes }_\pi Y^*}\right) ={\text {ext}}\left( B_{X^*}\right) \otimes {\text {ext}}\left( B_{Y^*}\right) . \end{aligned}$$

Little is known without the duality assumptions. Indeed, up to our knowledge, it is an open question whether every extreme point of $B_{X\widehat{\otimes }_\pi Y}$ must be of the form $x\otimes y$ for $x\in B_X$ and $y\in B_Y$. The situation clarifies for the stronger notions of denting points and strongly exposed points. Ruess and Stegall proved in [17] that

$$\begin{aligned} {\text {strexp}}\left( B_{X\widehat{\otimes }_\pi Y}\right) ={\text {strexp}}\left( B_X\right) \otimes {\text {strexp}}\left( B_Y\right) . \end{aligned}$$

Furthermore, if $x^*$ strongly exposes x in $B_X$ and $y^*$ strongly exposes y in $B_Y$, then $x^*\otimes y^*$ strongly exposes $x\otimes y$ in $B_{X\widehat{\otimes }_\pi Y}$. For denting points, D. Werner proved in [20] an analogous result in a more general framework:

$$\begin{aligned} {\text {dent}}\left( {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)\right) ={\text {dent}}\left( C\right) \otimes {\text {dent}}\left( D\right) \end{aligned}$$

whenever $C\subset X$ and $D\subset Y$ are closed bounded and absolutely convex subsets.

Motivated by the above results, in this note we study the notions of preserved extreme point and weak-strongly exposed point in projective tensor products. Recall that a point $x\in C$ is a preserved extreme point of C (also called weak*-extreme point) if it is an extreme point of ${\overline{C}}^{w^*}\subset X^{**}$; this is a stronger notion than being extreme but weaker than being denting (see e.g. [10]).

With this notation in mind, one of the main results of the present paper is the following one.

Theorem 1.1

Let X and Y be Banach spaces such that $K(X,Y^*)$ is separating for $X\widehat{\otimes }_\pi Y$ (in particular, if the pair $(X,Y^*)$ has the CAP). Let $C\subseteq X$, $D\subseteq Y$ be bounded closed convex subsets. If z is a preserved extreme point of ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)\subset X\widehat{\otimes }_\pi Y$ then $z=x\otimes y$ for some $x\in C$ and $y\in D$. Moreover, if $z\ne 0$ then x and y are preserved extreme points of C and D respectively.

As a particular case we get:

Corollary 1.2

Let X and Y be Banach spaces such that $K(X,Y^*)$ is separating for $X\widehat{\otimes }_\pi Y$ (in particular, if the pair $(X,Y^*)$ has the CAP). If z is a preserved extreme point of $B_{X\widehat{\otimes }_\pi Y}$, then $z=x\otimes y$ where x and y are preserved extreme points of $B_X$ and $B_Y$ respectively.

Theorem 1.1 points out that, in order to look for preserved extreme points in projective tensor products, we only have to pay attention to basic tensors. We do not know whether the converse holds. However, we will prove a kind of converse for w-strongly exposed points.

A point $x\in C$ is said to be exposed if there exists $x^*\in X^*$ such that $x^*(x)>x^*(y)$ for all $y\in C{\setminus }\{x\}$. We also say that $x^*$ exposes x in C. A point $x\in C$ is said strongly exposed (resp. w-strongly exposed) if there exists $x^*\in X^*$ exposing x and such that for all sequences $(x_n)_n\subset C$ such that $x^*(x_n)\xrightarrow [n]{}x^*(x)$, it follows that $x_n\xrightarrow [n]{}x$ (resp. $x_n\xrightarrow [n]{w}x$). Equivalently, the slices of C produced by $x^*$ are a neighbourhood basis of x for the norm (resp. weak) topology in C.¹ In this case, we write $x\in {\text {strexp}}\left( C\right)$ (resp. $x\in w{\text {-strexp}}\left( C\right)$).

Theorem 1.3

Let X and Y be Banach spaces such that $K(X,Y^*)$ is separating for $X\widehat{\otimes }_\pi Y$ (in particular, if the pair $(X,Y^*)$ has the CAP). Let $C\subseteq X$ and $D\subseteq Y$ be symmetric, bounded closed convex subsets. Assume that $x\otimes y\ne 0$ has a compact neighbourhood system for the weak topology in ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)\subset X\widehat{\otimes }_\pi Y$. Then the following are equivalent:

• (i) $x\otimes y$ is w-strongly exposed in ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$.

• (ii) x and y are w-strongly exposed in C and D, respectively.

In particular, if $C\otimes D$ is weakly compact, then

$$\begin{aligned} w{\text {-strexp}}\left( {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)\right) =w{\text {-strexp}}\left( C\right) \otimes w{\text {-strexp}}\left( D\right) . \end{aligned}$$

The assumption that $x\otimes y$ has a compact neighbourhood system in the above result might seem to be artificial but, surprisingly or not, it cannot be removed. Indeed, in Example 3.8 we find a Banach space X which is isomorphic to $\ell _2$ satisfying that there exists a w-strongly exposed point $x_0\in B_X$ and such that $x_0\otimes x_0$ is not a w-strongly exposed point of $B_{X\widehat{\otimes }_\pi X}$.

2 Notation and Preliminary Results

Throughout the paper we will only deal with real Banach spaces. Let C be a bounded subset of a Banach space X. Given $x^*\in X^*$ and $\alpha >0$, we denote

$$\begin{aligned} S(C, x^*, \alpha )=\{x\in C: x^*(x)>\sup x^*(C)-\alpha \} \end{aligned}$$

the (open) slice of C produced by $x^*$.

We say that $x\in C$ is extreme if the condition $x=\frac{y+z}{2}$ with $y,z\in C$ implies $y=z$. We write $x\in {\text {ext}}\left( C\right)$.

A point $x\in C$ is a preserved extreme point (or a $w^*$-extreme point) if x is an extreme point of ${\overline{C}}^{w^*}$. It can be proved that x is a preserved extreme point if and only if the open slices containing x form a basis for x in the weak topology induced on C (see [15]). This characterization will be used twice in the proof of Theorem 1.1 without further mention. Notice that, in particular, every w-strongly exposed point is a preserved extreme point.

Let us write here the following lemma, which we will use systematically throughout the text. This is a well-known result (see e.g. [13, Lemma 7.21], a preprint version of [12]) but we include a proof for completeness.

Lemma 2.1

Let X be a Banach space. Let A be a bounded subset of X and write $C={\overline{{{\,\textrm{co}\,}}}}(A)$. Let $R:=\sup _{x\in A}\Vert x\Vert$. Then, given $x^*\in X^*$, we have:

• (1) $\sup _{x\in A} x^*(x)=\sup _{x\in C} x^*(x)$.

• (2) Given $0<\varepsilon <\frac{1}{2}$ we have that

  $$\begin{aligned} S(C,x^*,\varepsilon ^2)\subseteq {{\,\textrm{co}\,}}(S(A,x^*,\varepsilon ))+4R\varepsilon B_X. \end{aligned}$$

Proof

(1) is clear. Let’s prove (2). First, take an element of $S({{\,\textrm{co}\,}}(A),x^*, \varepsilon ^2)$, which is a (finite) convex combination of the form $\sum _{n \in {\mathbb {N}}}{\lambda _{n}a_{n}}$ where $a_n\in A$ for every n and

$$\begin{aligned} \sup x^*(A) - \varepsilon ^2 < \sum _{n}\lambda _{n} x^*(a_{n}). \end{aligned}$$

Put $J:= \{ n \in {\mathbb {N}}: \sup x^*(A) - \varepsilon < x^*(a_{n}) \}$. Then

$$\begin{aligned} \sup x^*(A) - \varepsilon ^2&< \sum _{n \in J}\lambda _{n} x^*(a_{n}) + \sum _{n \notin J} \lambda _{n} x^*(a_{n})\\&\leqslant \left( \sum _{n \in J}\lambda _{n}\right) \sup x^*(A) + \left( \sup x^*(A) - \varepsilon \right) \sum _{n \notin J}\lambda _{n}\\&= \sup x^*(A)- \varepsilon \sum _{n \notin J}\lambda _{n} \end{aligned}$$

which allows to deduce that $\sum _{n \notin J}\lambda _{n} < \varepsilon$. Since $a_{n} \in S(A, x^*, \varepsilon )$ for each $n \in J$, the result follows from the following estimation:

$$\begin{aligned} \left\| \sum _{n \in J}\left( \frac{\lambda _{n}}{\sum _{n \in J}\lambda _{n}}\right) a_{n} - \sum _{n \in {\mathbb {N}}}\lambda _{n} a_{n}\right\|&\leqslant \left\| \sum _{n \in J}\left( \frac{\lambda _{n}}{\sum _{n \in J}\lambda _{n}}- \lambda _n \right) a_{n} \right\| + \left\| \sum _{n \notin J}\lambda _{n} a_{n} \right\| \\&\leqslant \left| \frac{1}{\sum _{n \in J}\lambda _{n}} - 1 \right| \cdot \left\| \sum _{n \in J}\lambda _{n} a_{n} \right\| + \left\| \sum _{n \notin J}\lambda _{n} a_{n} \right\| \\&\leqslant \frac{\varepsilon }{1 - \varepsilon } R + \varepsilon R = \frac{2 \varepsilon - \varepsilon ^{2}}{1 - \varepsilon } R. \end{aligned}$$

This shows that

$$\begin{aligned} S({{\,\textrm{co}\,}}A,x^*,\varepsilon ^2)\subseteq S({{\,\textrm{co}\,}}A,x^*,\varepsilon )+\frac{2\varepsilon -\varepsilon ^2}{1-\varepsilon }R B_X. \end{aligned}$$

Finally,

$$\begin{aligned} S(C,x^*,\varepsilon ^2)\subset \overline{S({{\,\textrm{co}\,}}(A),x^*,\varepsilon ^2)} \subset S({{\,\textrm{co}\,}}A,x^*,\varepsilon )+4\varepsilon R B_X \end{aligned}$$

since $\frac{2\varepsilon -\varepsilon ^2}{1-\varepsilon }<4\varepsilon$ for $\varepsilon <1/2$. $\square$

Given two Banach spaces X, Y, we denote L(X, Y), K(X, Y) and F(X, Y) the spaces of linear, compact, and finite-rank operators, respectively. Recall that $(X\widehat{\otimes }_\pi Y)^*=L(X, Y^*)$ isometrically. We refer the reader to [19] for basic properties of tensor products. We denote by $\tau _c$ the topology of compact convergence in L(X, Y), i.e. the topology of uniform convergence on compact subsets of X. It is well known that X has the approximation property if $\overline{F(X,X)}^{\tau _c}=L(X,X),$ whereas it has the compact approximation property if $\overline{K(X,X)}^{\tau _c}=L(X,X).$ The definition of the approximation property was extended to pairs of Banach spaces by E. Blonde in [4] as follows: The pair (X, Y) is said to have the approximation property if $\overline{F(X,Y)}^{\tau _c}=L(X,Y)$. In a similar fashion we say that the pair (X, Y) has the compact approximation property (CAP for short) if $\overline{K(X,Y)}^{\tau _c}=L(X,Y)$ (see, for instance, [6]). Notice that for any set $S \subset (X\widehat{\otimes }_\pi Y)^*=L(X, Y^*)$, we have ${\overline{S}}^{\tau _c} \subset {\overline{S}}^{w^*}$. Since for a subspace $Z\subseteq X^*$ to separate points of X is equivalent to the equality ${\overline{Z}}^{w^*}=X^*$, we have the following lemma:

Lemma 2.2

Let X, Y be two Banach spaces. If the pair $(X,Y^*)$ has the CAP, then $K(X,Y^*)$ separates points of $X\widehat{\otimes }_\pi Y$.

We will make use of the previous lemma throughout the text without further mention. We finish this section by recalling that the pair $(X,Y^*)$ has the CAP if and only if the pair $(Y,X^*)$ has the CAP. It is immediate that if X or Y has the compact approximation property or the approximation property then the pair $(X,Y^*)$ has the CAP. As a consequence of [4, Example 4.2], for every $1\leqslant p< 2< q < \infty$ and every subspaces $X \subset \ell _q$ and $Y\subset \ell _p$, the pair (X, Y) has the CAP. Nevertheless, there are such subspaces X and Y failing the compact approximation property.

In order to prove our results about extremal structure, we need the following topological result which is of independent interest.

Theorem 2.3

Let X and Y be two Banach spaces such that $K(X,Y^*)$ is separating for $X\widehat{\otimes }_\pi Y$ (in particular, if the pair $(X,Y^*)$ has the CAP). Let $C\subseteq X$ and $D\subseteq Y$ be two bounded subsets. Then the weak-closure of $C\otimes D$ in $X\widehat{\otimes }_\pi Y$ is equal to ${\overline{C}}^w \otimes {\overline{D}}^w$, that is $\overline{C\otimes D}^w={\overline{C}}^w \otimes {\overline{D}}^w$.

Proof

First, given $x\in {\overline{C}}^w$, we have that the operator $i:Y\rightarrow X\otimes Y$ given by $y\mapsto x\otimes y$ is continuous. Thus, it is also weak-to-weak continuous, so

$$\begin{aligned} \{x\}\otimes {\overline{D}}^w =i({\overline{D}}^w)\subseteq \overline{i(D)}^w=\overline{\{x\}\otimes D}^w. \end{aligned}$$

This shows that ${\overline{C}}^w\otimes {\overline{D}}^w\subseteq \overline{{\overline{C}}^w\otimes D}^w$. Analogously, we get ${\overline{C}}^w\otimes D\subseteq \overline{C\otimes D}^w$ and so

$$\begin{aligned} {\overline{C}}^w\otimes {\overline{D}}^w \subseteq \overline{\overline{C\otimes D}^w}^w=\overline{C\otimes D}^w. \end{aligned}$$

Now, given $z\in \overline{C\otimes D}^w$, take a net $(x_s\otimes y_s)$ in $C\otimes D$ such that $x_s\otimes y_s\rightarrow z$ weakly, and let us prove that $z=x\otimes y$ for certain $x\in {\overline{C}}^w$ and $y\in {\overline{D}}^w$. We denote by ${\overline{C}}^{w^*}$ and ${\overline{D}}^{w^*}$ respectively the closure of C and D in the $w^*$ topology of $X^{**}$ and $Y^{**}$ respectively, which are $w^*$-compact because they are bounded.

Since $(x_s)_s\subset {\overline{C}}^{w^*}$ and $(y_s)_s\subset {\overline{D}}^{w^*}$ we can assume, up to taking a suitable subnet, that both $x_s\rightarrow x^{**}$ in the $w^*$-topology of $X^{**}$ and $y_s\rightarrow y^{**}$ in the $w^*$-topology of $Y^{**}$. $\square$

Claim 2.4

For any compact operator $K:X \longrightarrow Y^*$, we have that

$$\begin{aligned} K(x_s)(y_s) \rightarrow K^{**}(x^{**})(y^{**}). \end{aligned}$$

Proof of the claim

First, recall that $K^{**}:X^{**} \longrightarrow Y^{***}$ is a compact operator which satisfies $K^{**}(X^{**})\subseteq Y^{*}$. Fix $\varepsilon >0$. We claim that there exists $s_0$ such that $|K(x_s)(y_s)-K^{**}(x^{**})(y^{**})|<\varepsilon$ for every $s \geqslant s_0$. Namely, we know that, since $K^{**}$ is compact, $K^{**}(x^{**})\in Y^*$ and $y_s\rightarrow y^{**}$ in the $w^*$-topology, there exists $s_0$ such that

$$\begin{aligned} \Vert K(x_s)-K^{**}(x^{**})\Vert<\varepsilon /(2R) \quad \text{ and } \quad |K^{**}(x^{**})(y_s)-K^{**}(x^{**})(y^{**})|<\varepsilon /2 \end{aligned}$$

for every $s \geqslant s_0$, where $R>0$ is such that $D \subset RB_{Y}$. Then

$$\begin{aligned} \begin{aligned} |K(x_s)(y_s)-K^{**}(x^{**})(y^{**})|&\leqslant \Vert K(x_s)-K^{**}(x^{**})\Vert \Vert y_s\Vert \\&+|K^{**}(x^{**})(y_s)-K^{**}(x^{**})(y^{**})|<\varepsilon \end{aligned} \end{aligned}$$

for every $s \geqslant s_0$ as desired. $\square$

Now, we claim we can assume $x^{**}\ne 0$ and $y^{**}\ne 0$. Indeed, if $x^{**}=0$ this would imply $0\in {\overline{C}}^{w}$. Moreover, since $z(K)=(0\otimes y^{**})(K)=0$ holds for every $K\in K(X,Y^*)$, which is separating for $X\widehat{\otimes }_\pi Y$, we would get that $z=0$ so, taking any $y\in D$, we have $z=0\otimes y\in {\overline{C}}^w\otimes {\overline{D}}^{w}$ and the proof would be finished. Henceforth, we assume $x^{**}\ne 0$ and $y^{**}\ne 0$ and, clearly, the above mentioned equality $z(K)=(x^{**}\otimes y^{**})(K)$ holding true for every $K\in K(X,Y^*)$ implies $z\ne 0$ too.

Claim 2.5

$x^{**}\in X$ and $y^{**}\in Y$.

Proof of the claim

Let us prove that $x^{**}$ is $w^*$-continuous, the proof for $y^{**}$ being completely analogous. Take $y^*\in S_{Y^*}$ such that $y^{**}(y^*)\ne 0$. Now we have that

$$\begin{aligned} x^{**}(x^*)=\frac{(x^{**}\otimes y^{**})(x^*\otimes y^*)}{y^{**}(y^*)}=\frac{(x^*\otimes y^*)(z)}{y^{**}(y^*)} \quad \forall x^*\in X^*. \end{aligned}$$

Thus, to see that $x^{**}$ is weak*-continuous it suffices to show that $(x_s^*\otimes y^*)(z)\rightarrow (x^*\otimes y^*)(z)$ whenever $x^*_s{\mathop {\rightarrow }\limits ^{w^*}} x^*$. This is a consequence of the fact that the operator $X^*\rightarrow L(X,Y^*)$ given by $x^*\mapsto x^*\otimes y^*$ is $w^*$-to-$w^*$-continuous as being the adjoint of the operator $X\widehat{\otimes }_\pi Y\rightarrow X$ given by $x\otimes y\mapsto y^*(y)x$. $\square$

At this point we will save notation calling $x:=x^{**}\in X$ and $y:=y^{**}\in Y$. Now we have that $K(z)=K(x\otimes y)$ holds for every $K\in K(X,Y^*)$. Since $K(X,Y^*)$ is separating for $X\widehat{\otimes }_\pi Y$, we deduce that $z=x\otimes y$. Moreover, observe that $x_s\rightarrow x$ in the weak topology of X. Since $x_s\in C$ for every s we conclude that $x\in {\overline{C}}^w$. Analogously, $y\in D$, so $z=x\otimes y\in {\overline{C}}^w\otimes {\overline{D}}^w$, which finishes the proof. $\square$

In spite of the fact that, under the approximation property, the tensor product of weakly closed sets is weakly closed (see e.g. Theorem 2.3), it is interesting to notice that if C and D are weakly compact, it does not follow that $C\otimes D$ is weakly compact in $X\widehat{\otimes }_\pi Y$ (for instance, if we take $C=D=B_{\ell _2}$, then the sequence $(e_n\otimes e_n)_n$ is equivalent to the $\ell _1$-basis, c.f. e.g. [19, Example 2.10]).

3 Main Results

The aim of this section is to present the proof of Theorems 1.1 and 1.3. We start with the proof of Theorem 1.1.

Proof of Theorem 1.1

Since z is a preserved extreme point, there is a neighbourhood basis $\{S_\alpha \}$ of z for the weak-topology of ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$ so that $S_\alpha$ is a slice for every $\alpha$. Now, since $S_\alpha$ is a slice of ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$ we can find $x_\alpha \otimes y_\alpha \in S_\alpha \cap (C\otimes D)$ for every $\alpha$. Since $S_\alpha$ is a weak basis for the weak topology at z we get that $z\in \overline{\{x_\alpha \otimes y_\alpha \}}^w$, and now Theorem 2.3 and the fact that C and D are weakly closed imply that $z=x\otimes y$ for certain $x\in C$ and $y\in D$.

If $z\ne 0$ it is not difficult to prove that x and y are preserved extreme points of C and D. Indeed, if $S({\overline{{{\,\textrm{co}\,}}}}(C\otimes D),T_\alpha , \beta _\alpha )$ is a neighbourhood system of $x\otimes y$ for the weak topology in ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$, then the family of slices $S_\alpha '$ defined as

$$\begin{aligned} S_\alpha ':=\{x'\in C: T_\alpha (x')(y)>1-\beta _\alpha \} \end{aligned}$$

is a neighbourhood system of x for the weak topology of X. $\square$

An immediate consequence of Theorem 1.1 is the following corollary.

Corollary 3.1

Let X and Y be Banach spaces such that $K(X,Y^*)$ is separating for $X\widehat{\otimes }_\pi Y$ (in particular, if the pair $(X,Y^*)$ has the CAP). Let $C\subseteq X$ and $D\subseteq Y$ be convex bounded subsets. If z is a w-strongly exposed point of ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$ then $z=x\otimes y$ for some $x\in C$ and $y\in D$. Moreover, if $z\ne 0$ then x and y are w-strongly exposed points of C and D respectively.

Proof

Theorem 1.1 provides points $x\in C$ and $y\in D$ such that $z=x\otimes y$. It remains to prove that x and y are w-strongly exposed points if $z\ne 0$. Let $T\in L(X,Y^*)$ w-strongly exposing $x\otimes y$ in ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$, and define $f\in X^*$ by $f(v):=T(v)(y)$. It is immediate that f w-strongly exposes x in C. The argument for y is analogous. $\square$

Now we will analyse a possible converse for Corollary 3.1. The first result we find is the following.

Proposition 3.2

Let X, Y be Banach spaces. Let $C\subseteq X$ and $D\subseteq Y$ be bounded and symmetric convex subsets. Let $x_0$ be a strongly exposed point of C and $y_0$ be a w-strongly exposed point of D. Then $x_0\otimes y_0$ is a w-strongly exposed point of ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$.

Proof

By homogeneity, we may assume that $C\subseteq B_X$ and $D\subseteq B_Y$, so $R:=\sup _{z\in C\otimes D}\Vert z\Vert \leqslant 1$. Assume that $x^*$ strongly exposes $x_0$ in C and $y^*$ w-strongly exposes $y_0$ in D. We may also assume that $\sup x^*(C)=1=\sup y^*(D)$. Let us prove that $x^*\otimes y^*$ w-strongly exposes $x_0\otimes y_0$ in ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$. To this end, pick $U:=\bigcap \limits _{i=1}^n S({\overline{{{\,\textrm{co}\,}}}}(C\otimes D), T_i,\alpha _i)$ to be a relatively weakly open subset of ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$ containing $x_0\otimes y_0$, with $\left\| T_i\right\| =1$ for each i, and let us prove that $S({\overline{{{\,\textrm{co}\,}}}}(C\otimes D), x^*\otimes y^*,\beta )\subseteq U$ for a suitable $\beta$. Notice that

$$\begin{aligned} (x^*\otimes y^*)(x_0\otimes y_0)=x^*(x_0)y^*(y_0)=1=\sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)} (x^*\otimes y^*)(z) \end{aligned}$$

(here we use Lemma 2.1).

Since $x_0\otimes y_0\in U$, we have $T_i(x_0)(y_0)>\sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i(z)-\alpha _i$ for every $1\leqslant i\leqslant n$. Thus we can find $\varepsilon _0>0$ so that $T_i(x_0)(y_0)>\sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i(z)-\alpha _i+\varepsilon _0$ for every i.

Since $x^*$ strongly exposes $x_0$, there is $\delta '>0$ such that ${{\,\mathrm{diam\,}\,}}(S(C,x^*,\delta '))<\frac{\varepsilon _0}{4}$. Moreover, notice that

$$\begin{aligned} y_0\in \bigcap \limits _{i=1}^n \left\{ y\in D: T_i(x_0)(y)>\sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i(z)-\alpha _i+\varepsilon _0\right\} , \end{aligned}$$

which is a relatively weakly open subset of D containing $y_0$. Since $y_0$ is weakly exposed by $y^*$ we can find $\delta ''>0$ such that

$$\begin{aligned} S(D,y^*,\delta '')\subseteq \bigcap \limits _{i=1}^n \left\{ y\in D: T_i(x_0)(y)>\sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i(z)-\alpha _i+\varepsilon _0\right\} . \end{aligned}$$

Take $\delta :=\min \{\delta ',\delta '', \varepsilon _0/4\}$. Consider finally the slice $S({\overline{{{\,\textrm{co}\,}}}}(C\otimes D),x^*\otimes y^*,\eta ^2)$, where $0<\eta <\delta /4$. Let us prove that the previous slice is contained in U. To this end, notice that

$$\begin{aligned} S({\overline{{{\,\textrm{co}\,}}}}(C\otimes D),x^*\otimes y^*,\eta ^2)&\subseteq {{\,\textrm{co}\,}}(S(C\otimes D,x^*\otimes y^*,\eta ))+4\eta B_{X\widehat{\otimes }_\pi Y}\\&\subseteq {{\,\textrm{co}\,}}(S(C\otimes D,x^*\otimes y^*,\delta ))+\delta B_{X\widehat{\otimes }_\pi Y}=:A \end{aligned}$$

thanks to Lemma 2.1 and the choice of $\eta$. So, it suffices to prove that $A\subseteq U$. To this end, pick $x\otimes y\in S(C\otimes D,x^*\otimes y^*,\delta )$. This means that $x^*(x)y^*(y)>1-\delta$, from where $x^*(x)>1-\delta \geqslant 1-\delta '$ and $y^*(y)>1-\delta \geqslant 1-\delta ''$. By the definition of $\delta '$ and $\delta ''$ we get that $\Vert x-x_0\Vert <\frac{\varepsilon _0}{4}$ and $T_i(x_0)(y)>\sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i(z)-\alpha _i+\varepsilon _0$ for every i. Hence

$$\begin{aligned} \begin{aligned} T_i(x)(y)\geqslant T_i(x_0)(y)-\Vert T_i\Vert \Vert x-x_0\Vert \Vert y\Vert&>\sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i(z)-\alpha _i+\varepsilon _0-\frac{\varepsilon _0}{4}\\&=\sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i(z)-\alpha _i+\frac{3\varepsilon _0}{4}. \end{aligned} \end{aligned}$$

An easy convexity argument implies that

$$\begin{aligned} T_i(u)>\sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i(z)-\alpha _i+\frac{3\varepsilon _0}{4} \quad \forall u\in {{\,\textrm{co}\,}}(S(C\otimes D,x^*\otimes y^*,\delta )). \end{aligned}$$

Now, given $u\in A$, we have $u=v+w$ with $v\in {{\,\textrm{co}\,}}(S(C\otimes D,x^*\otimes y^*,\delta ))$ and $\Vert w\Vert \leqslant \delta \leqslant \varepsilon _0/4$. Then,

$$\begin{aligned} \begin{aligned} T_i(u)=T_i(v)+T_i(w)&\geqslant \sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i(z)-\alpha _i+\frac{3\varepsilon _0}{4}-\Vert w\Vert \\&\geqslant \sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i(z)-\alpha _i+\frac{\varepsilon _0}{2}>\sup _{z\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i(z)-\alpha _i \end{aligned} \end{aligned}$$

for each i. We conclude that $u\in U$, which proves that $A\subseteq U$ and the proof is finished. $\square$

Note that in Proposition 3.2 we obtain a compact operator ($T:X\rightarrow Y^*$ given by $T(x)=x^*(x)y^*$) providing a neighbourhood basis for $x_0\otimes y_0$ for the weak topology in ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$. This motivates to consider the following notion.

Definition 3.3

Let X and Y be Banach spaces, and let $C\subseteq X$, $D\subseteq Y$ be two subsets. We say that $x\otimes y\in C\otimes D$ has a compact neighbourhood system for the weak topology in ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$ if, given any weakly open subset U containing $x_0\otimes y_0$, there are slices $S({\overline{{{\,\textrm{co}\,}}}}(C\otimes D), T_i,\alpha _i)$ given by compact operators $T_i\in K(X, Y^*)$ such that

$$\begin{aligned} x_0\otimes y_0 \in \bigcap \limits _{i=1}^n S({\overline{{{\,\textrm{co}\,}}}}(C\otimes D), T_i,\alpha _i)\subseteq U. \end{aligned}$$

Remark 3.4

• (a) The above definition has an easy interpretation in terms of nets: $x\otimes y$ has a compact neighbourhood system for the weak topology in ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$ if, and only if, given a net $(z_\alpha )_\alpha \subset {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$, the condition $T(z_\alpha )\rightarrow T(x\otimes y)$ for every $T\in K(X,Y^*)$ implies $z_\alpha \rightarrow x\otimes y$ in the weak topology on $X\widehat{\otimes }_\pi Y$. Equivalently, $x\otimes y$ is a point of continuity of the formal identity

  $$\begin{aligned} I:({\overline{{{\,\textrm{co}\,}}}}(C\otimes D),w)\longrightarrow ({\overline{{{\,\textrm{co}\,}}}}(C\otimes D),\sigma (X\widehat{\otimes }_\pi Y, K(X,Y^*))). \end{aligned}$$

• (b) In the case that $K(X,Y^*)$ is separating for $X\widehat{\otimes }_\pi Y$ (in particular, if the pair $(X,Y^*)$ has the CAP) and $C\otimes D$ is weakly compact, ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$ is also weakly compact by Krein–Smulyan theorem (see e.g. [7, Theorem II. 2.11]) and $\sigma (X\widehat{\otimes }_\pi Y, K(X,Y^*))$ is Hausdorff because $K(X,Y^*)$ is separating for $L(X, Y^*)$. Thus, the identity map I above is a homeomorphism and so every $x\otimes y\in C\otimes D$ has a compact neighbourhood system.

Now we are ready to present the proof of Theorem 1.3.

Proof of Theorem 1.3

(i)$\Rightarrow$ (ii) follows from Corollary 3.1.

(ii)$\Rightarrow$(i). Write $R:=\sup _{z\in C\otimes D}\Vert z\Vert$. Take $x_0^*\in X^*$ and $y_0^*\in Y^*$ w-strongly exposing $x_0$ and $y_0$ in C and D, respectively, with $x_0^*(x_0)=\sup x^*_0(C)=1$, and $y_0^*(y_0)=\sup y^*_0(D)=1$. Pick U to be a weak neighbourhood of $x_0\otimes y_0$ in ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$. By the assumption, we can assume that $U=\bigcap \limits _{i=1}^n S({\overline{{{\,\textrm{co}\,}}}}(C\otimes D), T_i,\alpha _i)$ for certain compact operators $T_1,\ldots , T_n:X\rightarrow Y^*$. Furthermore, we can assume $\sup _{{\overline{{{\,\textrm{co}\,}}}}(C\otimes D)}T_i=1$ for every i. Let $\eta$ small enough so that $T_i(x_0\otimes y_0)>1-\alpha _i+\eta$ holds for every $1\leqslant i\leqslant n$. Moreover, observe that $x_0\in \bigcap \limits _{i=1}^n \{z\in C: T_i(z)(y_0)>1-\alpha _i+\eta \}$, which is a relatively weakly open subset of C. Since $x_0^*$ w-strongly exposes $x_0$ then there exists $\delta '>0$ so that

$$\begin{aligned} x_0\in S(C, x_0^*,\delta ')\subseteq \bigcap \limits _{i=1}^n \{z\in C: T_i(z)(y_0)>1-\alpha _i+\eta \}. \end{aligned}$$

Now, for every $1\leqslant i\leqslant n$, the set $T_i(S(C, x_0^*,\delta '))$ is a relatively compact subset of $Y^*$. Using the compactness condition on all the $T_i's$ we can find a finite set $x_1,\ldots , x_m\in S(C, x_0^*,\delta ')$ so that $B(T_i(x_j),\frac{\eta }{8}), 1\leqslant j\leqslant m$, is a covering of $T_i(S(C, x_0^*,\delta '))$ for every $1\leqslant i\leqslant n$. Observe that $T_i(x_j)(y_0)>1-\alpha _i+\eta$ holds for every $1\leqslant i\leqslant n$ and $1\leqslant j\leqslant m$. Consequently,

$$\begin{aligned} y_0\in \bigcap \limits _{i=1}^n\bigcap \limits _{j=1}^m\{y\in D: T_i(x_j)(y)>1-\alpha _i+\eta \}. \end{aligned}$$

Since $y_0^*$ w-strongly exposes $y_0$ we can find $\delta ''>0$ so that

$$\begin{aligned} y_0\in S(B_Y,y_0^*,\delta '')\subseteq \bigcap \limits _{i=1}^n\bigcap \limits _{j=1}^m\{y\in D: T_i(x_j)(y)>1-\alpha _i+\eta \}. \end{aligned}$$

We claim now that

$$\begin{aligned} S(C, x^*_0, \delta ')\otimes S(D, y^*_0, \delta '')\subset \bigcap _{i=1}^n S\left( {\overline{{{\,\textrm{co}\,}}}}(C\otimes D), T_i, \alpha _i-\frac{\eta }{2}\right) . \end{aligned}$$

Indeed, let $x\in S(C, x^*_0,\delta ')$ and $y\in S(D, y^*_0, \delta '')$. We have, for every $i\in \{1,\ldots , n\}$, an index $j_i\in \{1,\ldots , m\}$ such that $\Vert T_i(x)-T_i(x_{j_i})\Vert <\frac{\eta }{2}$. On the other hand, since $S(D,y_0^*,\delta '')\subseteq \bigcap \limits _{i=1}^n\bigcap \limits _{j=1}^m\{y\in D: T_i(x_j)(y)>1-\alpha _i+\eta \}$ we have that, for every $1\leqslant i\leqslant n$, $T_i(x_{j_i})(y)>1-\alpha _i+\eta$. Consequently

$$\begin{aligned} T_i(x)(y)\geqslant T_i(x_{j_i})(y)-\Vert T_i(x_{j_i})-T_i(x)\Vert >1-\alpha _i+\eta -\frac{\eta }{2} =1-\alpha _i+\frac{\eta }{2}. \end{aligned}$$

Take $\delta :=\min \{\delta ',\delta '', \frac{\eta }{8},\frac{\eta }{8R}\}$ and consider $S:=S({\overline{{{\,\textrm{co}\,}}}}(C\otimes D),x_0^*\otimes y_0^*, \delta ^2)$. Observe that $x_0\otimes y_0\in S$. Moreover,

$$\begin{aligned} S\subseteq {{\,\textrm{co}\,}}(S(C\otimes D,x_0^*\otimes y_0^*,\delta ))+4R\delta B_{X\widehat{\otimes }_\pi Y} \end{aligned}$$

in virtue of Lemma 2.1. Now, given $1\leqslant i\leqslant n$, since $1-\delta >\max \{1-\delta ',1-\delta ''\}$ we conclude that every element $x\otimes y$ of $S(C\otimes D,x_0^*\otimes y_0^*,\delta )$ satisfies $x_0^*(x)>1-\delta '$ and $y_0^*(y)>1-\delta ''$, so $T_i(x)(y)>1-\alpha _i+\frac{\eta }{2}$. Since $T_i$ is a linear continuous functional on $X\widehat{\otimes }_\pi Y$ we conclude that $T_i(z)\geqslant 1-\alpha _i+\frac{\eta }{2}$ holds for every $1\leqslant i\leqslant n$ and every $z\in {{\,\textrm{co}\,}}(S(C\otimes D,x_0^*\otimes y_0^*,\delta ))$. Henceforth, given $z\in S$ we can find $u\in {{\,\textrm{co}\,}}(S(C\otimes D,x_0^*\otimes y_0^*,\delta ))$ and $v\in B_{X\widehat{\otimes }_\pi Y}$ so that $z=u+4R\delta v$. Now, given $1\leqslant i\leqslant n$ we get

$$\begin{aligned} T_i(z)=T_i(u)+4\delta R T_i(v)\geqslant 1-\alpha _i+\frac{\eta }{2}-4R\delta >1-\alpha _i, \end{aligned}$$

from where we conclude that $z\in \bigcap \limits _{i=1}^n S({\overline{{{\,\textrm{co}\,}}}}(C\otimes D), T_i,\alpha _i)=U$. This proves that $S\subseteq U$.

Summarising, we have proved that every relatively weakly open subset of ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$ containing $x_0\otimes y_0$ actually contains a slice $S({\overline{{{\,\textrm{co}\,}}}}(C\otimes D), x_0^*\otimes y_0^*,\alpha )$. Moreover,

$$\begin{aligned}(x_0^*\otimes y_0^*)(x_0\otimes y_0)=\sup x_0^*(C)\sup y_0^*(D)=\sup (x_0^*\otimes y_0^*)({\overline{{{\,\textrm{co}\,}}}}(C\otimes D)).\end{aligned}$$

Thus, $x_0^*\otimes y_0^*$ w-strongly exposes $x_0\otimes y_0$. $\square$

Remark 3.5

The hypothesis of symmetry of C and D in Theorem 1.3 and Proposition 3.2 is needed. Indeed, let $C\subseteq X$ be any bounded subset with more than one point such that 0 is a strongly exposed point of C, and let $D\subseteq Y$ be a bounded set with a strongly exposed point $y\in D$ satisfying that $-y\in D$ too. In spite of 0 and y being strongly exposed, the basic tensor $0\otimes y=0\in {\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$ is not an extreme point, since

$$\begin{aligned} 0\otimes y=0=\frac{1}{2}(x\otimes y+x\otimes (-y)), \end{aligned}$$

for any $x\in C\setminus \{0\}$. Similarly, the result in [20, Theorem 1] about the denting points of ${\overline{{{\,\textrm{co}\,}}}}(C\otimes D)$ does not hold when C or D are non-symmetric.

At this point one can wonder whether the assumption of the existence of the compact neighbourhood system can be removed in Theorem 1.3. We will show that the answer is negative. Let us consider first the following example, where the set is not symmetric.

Example 3.6

Consider $X=Y=\ell _2$, let $K:={\overline{{{\,\textrm{co}\,}}}}\{e_n:n\in {\mathbb {N}}\}$ and $f:=\sum _{k=1}^\infty 2^{-k}e_k^*$. It is known that 0 is w-strongly exposed by f in K. Indeed, assume $(x_n)_{n=1}^\infty \subset K$ and $\lim _{n\rightarrow \infty }\langle f,x_n\rangle =0$. Since $K\subset \ell _2$, each $x_n$ can be expressed as $x_n=\sum _{k=1}^\infty a_k ^n e_k$ with $a_k\geqslant 0$ and $\sum _{k=1}^\infty a_k^n \leqslant 1$. Therefore

$$\begin{aligned} 0 = \lim _{n\rightarrow \infty } \langle f, x_n\rangle = \lim _{n\rightarrow \infty } \sum _{k=1}^\infty 2^{-k} a_k^n \geqslant \lim _{n\rightarrow \infty } 2^{-k} a_k^n. \end{aligned}$$

This means that $\lim _{n\rightarrow \infty }\langle e_k, x_n\rangle = 0$ for each $k\in {\mathbb {N}}$ and so $x_n{\mathop {\rightarrow }\limits ^{w}}0$. However, $f\otimes f$ does not weak-strongly exposes 0 in $K\otimes K\subseteq \ell _2\widehat{\otimes }_\pi \ell _2$. Even more, 0 is not weakly strongly exposed in $K\otimes K$, i.e. there is no bilinear form $B:\ell _2\times \ell _2\rightarrow {\mathbb {R}}$ such that $z_k\rightarrow 0$ weakly whenever $z_k\in K\otimes K$ satisfies that $B(z_k)\rightarrow 0$.

In order to prove that, take a bilinear form B. For every $n\in {\mathbb {N}}$, the sequence $\{B(e_n,e_k)\}_{k\in \mathbb N}\rightarrow 0$ since $e_k$ is weakly null in $\ell _2$. Thus there is a subsequence $(e_{k_n})_n$ of $(e_n)_n$ satisfying that $B(e_n, e_{k_n})\rightarrow 0$. Observe that $e_n\otimes e_{k_n}\in K\otimes K$ for every $n\in {\mathbb {N}}$.

However, $(e_n\otimes e_{k_n})_n$ does not converge weakly to 0 since it is isometrically equivalent to the $\ell _1$ basis; this follows by the same argument as for the diagonal $e_n\otimes e_n$, see e.g. [19, Example 2.10].

Remark 3.7

The above argument also proves that 0 is not weakly strongly exposed in ${\overline{{{\,\textrm{co}\,}}}}(K\otimes K)$. It also follows that $f\otimes f$ exposes 0 in ${\overline{{{\,\textrm{co}\,}}}}(K\otimes K)$. Indeed, since f exposes 0 in K we have that, given any $z\in K$, $f(z)=0$ if and only if $z=0$. Consequently, given $x\otimes y\in K\otimes K$ we have that $(f\otimes f)(x\otimes y)=f(x)f(y)=0$ implies that either x or y equals 0 and so $x\otimes y=0$. From this, and the fact that $\{e_n\otimes e_m\}_{n,m}$ is a Schauder basis for $\ell _2\widehat{\otimes }_\pi \ell _2$, it follows that $(f\otimes f)(z)=0$ if and only if $z=0$ for every $z\in {\overline{{{\,\textrm{co}\,}}}}(K\otimes K)$.

In order to find an example showing that Theorem 1.3 does not hold without the assumption of the existence of a compact neighbourhood system, we will use the set K from Example 3.6 to construct an equivalent renorming $|\cdot |$ on $\ell _2$ satisfying that the new unit ball $B_{(\ell _2, |\cdot |)}$ has a w-strongly exposed point $x_0$ such that $x_0\otimes x_0$ is not w-strongly exposed in $B_{(\ell _2,|\cdot |)\widehat{\otimes }_\pi (\ell _2,|\cdot |)}$.

Example 3.8

Set an equivalent norm $|\cdot |$ on $\ell _2$ so that the new unit ball is ${\overline{{{\,\textrm{co}\,}}}}((K-e_1)\cup (-K+e_1)\cup \frac{1}{8} B_{\ell _2})$, where K is the set described in Example 3.6. We claim that $-e_1\in B_{(\ell _2,|\cdot |)}$ is w-strongly exposed by $f:=\sum _{k=1}^\infty 2^{-k}e_k^*$. Observe that $f(-e_1)=-\frac{1}{2}$. Call $A:=K-e_1$ for simplicity. Since f is linear, it is clear that

$$\begin{aligned} \sup _{z\in B_{(\ell _2,|\cdot |)}} \vert f(z)\vert =\sup \limits _{z\in A\cup -A\cup \frac{1}{8} B_{\ell _2}} \vert f(z)\vert . \end{aligned}$$

Observe that the above supremum equals $\sup _{z\in A}\vert f(z)\vert$ since $\vert f(z)\vert \leqslant \frac{1}{8}$ on $\frac{1}{8}B_{\ell _2}$ and by a symmetry argument.

On the other hand, given $z\in A$ we have $z=v-e_1$ for $v\in K$. Now $f(v-e_1)=-\frac{1}{2}+f(v)\geqslant -\frac{1}{2}$ since $f\geqslant 0$ on K. This proves that $\Vert f\Vert =1/2=\vert f(-e_1)\vert$. Observe, moreover, that $f(v-e_1)\leqslant 0$ holds for every $v\in K$. In order to see that f w-strongly exposes $-e_1$ it remains to prove that if $f(z_n)\rightarrow -\frac{1}{2}$ with $(z_n)_n\subset B_{(\ell _2,|\cdot |)}$ then $z_n\rightarrow -e_1$ weakly. In order to do so, by a density argument, we can assume with no loss of generality that $z_n\in {{\,\textrm{co}\,}}(A\cup -A\cup \frac{1}{8} B_{\ell _2})$. For every n, we can write

$$\begin{aligned} z_n=\alpha _n a_n+\beta _n(-a_n')+\gamma _n x_n \end{aligned}$$

for $a_n,a_n'\in A, x_n\in \frac{1}{8} B_{\ell _2}$ and $\alpha _n,\beta _n,\gamma _n\in [0,1]$ with $\alpha _n+\beta _n+\gamma _n=1$ for every n.

Observe that

$$\begin{aligned} f(z_n)=\alpha _n f(a_n)+\beta _n f(-a_n')+\gamma _n f(x_n)\geqslant \alpha _n f(a_n)+\gamma _n (-1/8) \end{aligned}$$

since $f(-a_n')\geqslant 0$ for every $n\in {\mathbb {N}}$ and since $\vert f(x_n)\vert \leqslant 1/8$. Since $f(z_n)\rightarrow -1/2$ the unique possibility is that $\alpha _n\rightarrow 1$ (which implies $\beta _n\rightarrow 0$ and $\gamma _n\rightarrow 0$). Moreover, it is immediate that $f(a_n)\rightarrow -1/2$. Since f w-strongly exposes $-e_1$ in A we conclude that $a_n\rightarrow -e_1$ weakly, so $z_n\rightarrow -e_1$ weakly, as desired.

Finally, if we consider $e_1\otimes e_1$, we get that it is not w-strongly exposed in $B_{(\ell _2,|\cdot |)\widehat{\otimes }_\pi ( \ell _2,|\cdot |)}$. As in Example 3.6, given any bilinear and continuous form B, we can find a strictly increasing sequence $(k_n)_n$ such that $B(e_n,e_{k_n})\rightarrow 0$, so $B(-e_1+e_n,-e_1+e_{k_n})\rightarrow B(e_1,e_1)$. However, if $k_1<k_2<\cdots$ we have that $\{e_n\otimes e_{k_n}\}$ is equivalent to the $\ell _1$ basis since $(\ell _2,|\cdot |)$ and $\ell _2$ are isomorphic (it follows for instance from [19, Proposition 2.3]) and therefore $((-e_1+e_n)\otimes (-e_1+e_{k_n}))_n$ is not weakly convergent to $e_1 \otimes e_1$.

We end the paper with some open questions.

Question 3.9

Is $x\otimes y$ a preserved extreme point of $B_{X\widehat{\otimes }_\pi Y}$ whenever x and y are preserved extreme points of $B_X$ and $B_Y$?

Question 3.10

Is every (preserved) extreme point of $B_{X\widehat{\otimes }_\pi Y}$ a basic tensor?