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THE SET OF ELEMENTARY TENSORS IS WEAKLY CLOSED IN PROJECTIVE TENSOR PRODUCTS

Published online by Cambridge University Press:  13 May 2024

COLIN PETITJEAN*
Affiliation:
Université Gustave Eiffel, Université Paris Est Creteil, CNRS, LAMA UMR8050, Marne-la-Vallée F-77447, France
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Abstract

We prove that the set of elementary tensors is weakly closed in the projective tensor product of two Banach spaces. As a result, we answer a question of Rodríguez and Rueda Zoca [‘Weak precompactness in projective tensor products’, Indag. Math. (N.S.) 35(1) (2024), 60–75], proving that if $(x_n) \subset X$ and $(y_n) \subset Y$ are two weakly null sequences such that $(x_n \otimes y_n)$ converges weakly in $X \widehat {\otimes }_\pi Y$, then $(x_n \otimes y_n)$ is also weakly null.

Type
Research Article
Copyright
© The Author(s), 2024. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc.

1 Weak convergence in projective tensor product

Let X, Y and Z be real Banach spaces. We denote by $\mathcal {B}(X \times Y , Z)$ the space of continuous bilinear operators from $X \times Y$ into Z. If $Z = \mathbb {R}$ , we simply write $\mathcal {B}(X \times Y)$ . For $x \in X$ and $y \in Y$ , define the elementary tensor $x\otimes y \in \mathcal {B}(X \times Y )^*$ by

$$ \begin{align*} \forall B \in \mathcal{B}(X \times Y), \quad \langle x\otimes y , B \rangle = B(x,y).\end{align*} $$

We then introduce $X \otimes Y := \mathrm {span} \{ x\otimes y : x \in X, \, y \in Y\}$ . Recall that the norm on $\mathcal {B}(X~\times ~Y)$ is defined by $\| B \|_{\mathcal {B}(X \times Y)} = \sup _{x \in B_X, y \in B_Y} |B(x,y)|$ . Let $\|\cdot \|_{\pi }$ be the dual norm of $\| \cdot \|_{\mathcal {B}(X \times Y)}$ . It is well known (see, for example, [Reference Diestel and Uhl1, Proposition VIII.9.a]) that if $u \in X \otimes Y$ , then

$$ \begin{align*} \| u \|_{\pi} = \inf \bigg \{ \sum_{i=1}^n \|x_i\| \|y_i \| : u = \sum_{i=1}^n x_i \otimes y_i \bigg \}.\end{align*} $$

The projective tensor product of X and Y is defined by

$$ \begin{align*}X \widehat{\otimes}_\pi Y = \overline{\mathrm{span}}^{\|\cdot\|_{\pi}} \{ x\otimes y : x \in X, \, y \in Y\} \subseteq \mathcal{B}(X \times Y)^* .\end{align*} $$

As a consequence of the fundamental linearisation property of tensor products, one can easily deduce the isometric identification $(X \widehat {\otimes }_\pi Y)^* \equiv \mathcal {B}(X \times Y)$ . Since $\mathcal {B}(X \times Y) \equiv \mathcal {L}(X, Y^*)$ , where $\mathcal {L}(X, Y^*)$ stands for the space of bounded linear operators from X to $Y^*$ , one also has $\mathcal {L}(X, Y^*) \equiv (X \widehat {\otimes }_\pi Y)^*$ .

The aim of this short note is to answer a question of Rodríguez and Rueda Zoca.

Question 1.1 [Reference Rodríguez and Rueda Zoca5, Question 3.9].

Let X and Y be Banach spaces. Let $(x_n)_{n\in \mathbb {N}}$ and $(y_n)_{n\in \mathbb {N}}$ be weakly null sequences in X and Y, respectively, such that $(x_n \otimes y_n)_{n \in \mathbb {N}}$ is weakly convergent in $X \widehat {\otimes }_\pi Y$ . Is $(x_n \otimes y_n)_{n \in \mathbb {N}}$ weakly null in $X\widehat {\otimes }_\pi Y$ ?

Let

$$ \begin{align*} \mathcal T= \{ x \otimes y : x \in X, \, y \in Y \} \end{align*} $$

be the set of elementary tensors in $X \widehat {\otimes }_\pi Y$ . We shall start with a simple but key observation. Recall that a Banach space X has the approximation property (AP in short) if for every $\varepsilon>0$ , for every compact subset $K \subset X$ , there exists a finite rank operator $T\in \mathcal L(X,X)$ such that $\|Tx -x\| \leq \varepsilon $ for every $x \in K$ .

Lemma 1.2. Let $X,Y$ be Banach spaces such that X or Y has the AP. Let $T \in X \widehat {\otimes }_\pi Y$ . Then $T \in \mathcal T$ if and only if every pair of linearly independent families $\{x_1^*,x_2^* \} \subset X^*$ and $\{y_1^{*},y_2^{*} \} \subset Y^{*}$ satisfies

(⋆) $$ \begin{align} \begin{vmatrix} \langle T , x_1^* \otimes y_1^{*} \rangle & \langle T , x_1^* \otimes y_2^{*} \rangle \\ \langle T , x_2^* \otimes y_1^{*} \rangle & \langle T , x_2^* \otimes y_2^{*} \rangle \end{vmatrix} = 0. \end{align} $$

Proof. Thanks to [Reference Ryan6, Proposition 2.8], every $T\in X \widehat {\otimes }_\pi Y$ can be written as

$$ \begin{align*}T = \sum_{n=1}^{\infty} x_n \otimes y_n\end{align*} $$

with $\sum _{n=1}^{\infty } \| x_n \| \| y_n \| \leq 2 \|T\|$ . Moreover, the linear map $\Phi : X \widehat {\otimes }_\pi Y \to \mathcal L(X^*,Y)$ obtained by

$$ \begin{align*}\forall x^* \in X^*, \quad \Phi\bigg( \sum_{n=1}^{\infty} x_n \otimes y_n\bigg)(x^*) = \sum_{n=1}^{\infty} x^*(x_n) y_n\end{align*} $$

defines a bounded operator. Since X or Y has the AP, $\Phi $ is injective (see [Reference Ryan6, Proposition 4.6]).

If $T = x \otimes y \in \mathcal T$ , then it is straightforward to check that condition $(\star )$ is verified:

$$ \begin{align*} \begin{vmatrix} \langle T , x_1^* \otimes y_1^{*} \rangle & \langle T , x_1^* \otimes y_2^{*} \rangle \\ \langle T , x_2^* \otimes y_1^{*} \rangle & \langle T , x_2^* \otimes y_2^{*} \rangle \end{vmatrix} = \begin{vmatrix} x_1^*(x) y_1^{*}(y) & x_1^*(x) y_2^{*}(y) \\ x_2^*(x) y_1^{*}(y) & x_2^*(x) y_2^{*}(y) \end{vmatrix} = 0.\end{align*} $$

Assume now that $T \not \in \mathcal T$ . Then $\Phi (T)$ is an operator of rank greater than 2 in $\mathcal L(X^*,Y)$ . Thus, there exists a linearly independent family $\{x_1^*,x_2^* \} \subset X^*$ such that $\Phi (T)(x_1^*) \ne 0$ , $\Phi (T)(x_2^*) \ne 0$ and $\{\Phi (T)(x_1^*),\Phi (T)(x_2^*) \} \subset Y$ is a linearly independent family. To finish the proof, simply pick a linearly independent family $\{y_1^{*},y_2^{*} \} \subset Y^{*}$ satisfying

$$ \begin{align*} \langle \Phi(T)(x_1^*) , y_1^* \rangle \ne 0 & \quad \langle \Phi(T)(x_1^*) , y_2^* \rangle = 0 \\ \langle \Phi(T)(x_2^*) , y_1^* \rangle = 0 & \quad \langle \Phi(T)(x_2^*) , y_2^* \rangle \ne 0.\\[-2.8pc] \end{align*} $$

Proposition 1.3. Let $X,Y$ be two Banach spaces such that X or Y has the AP. Then the set of elementary tensors $\mathcal T$ is weakly closed in $X \widehat {\otimes }_\pi Y$ .

Proof. We let I be the set of all vectors $(x_1^*,x_2^* ,y_1^{*},y_2^{*})$ such that $\{x_1^*,x_2^* \} \subset X^*$ and $\{y_1^{*},y_2^{*} \} \subset Y^{*}$ are both linearly independent families. Next, for every $T \in X \widehat {\otimes }_\pi Y$ and $S=(x_1^*,x_2^* ,y_1^{*},y_2^{*}) \in I$ , we define

$$ \begin{align*}D_S(T) = \begin{vmatrix} \langle T , x_1^* \otimes y_1^{*} \rangle & \langle T , x_1^* \otimes y_2^{*} \rangle \\ \langle T , x_2^* \otimes y_1^{*} \rangle & \langle T , x_2^* \otimes y_2^{*} \rangle \end{vmatrix}.\end{align*} $$

The result now directly follows from Lemma 1.2 together with the fact that $D_S$ is continuous with respect to the weak topology. Indeed, one can write $\mathcal T$ as an intersection of weakly closed sets:

$$ \begin{align*} \mathcal{T} =\bigcap_{S \in I} D_S^{-1}(\{0\}).\\[-40pt] \end{align*} $$

The next corollary answers Question 1.1 positively under rather general assumptions.

Corollary 1.4. Let X and Y be Banach spaces such that X or Y has the AP. If $(x_n)_{n\in \mathbb {N}} \subset X$ converges weakly to x, $(y_n)_{n\in \mathbb {N}} \subset Y$ converges weakly to y and $(x_n \otimes y_n)_{n \in \mathbb {N}}$ is weakly convergent in $X \widehat {\otimes }_\pi Y$ , then $(x_n \otimes y_n)_{n \in \mathbb {N}}$ converges weakly to $x \otimes y$ .

Before proving this corollary, let us point out that the canonical basis $(e_n)_{n \in \mathbb {N}}$ of $\ell _2$ shows that if $(x_n)_{n\in \mathbb {N}}\subset X$ and $(y_n)_{n \in \mathbb {N}} \subset Y$ are weakly null sequences, the sequence $(x_n \otimes y_n)_{n \in \mathbb {N}}$ may fail to be weakly null in $X \widehat {\otimes }_\pi Y$ . Indeed, $(e_n \otimes e_n)_{n \in \mathbb {N}}$ is isometric to the $\ell _1$ -canonical basis (see [Reference Ryan6, Example 2.10]).

Proof. Assume first that $(x_n)_{n\in \mathbb {N}} \subset X$ and $(y_n)_{n\in \mathbb {N}} \subset Y$ are weakly null sequences such that $(x_n \otimes y_n)_{n \in \mathbb {N}}$ is weakly convergent in $X \widehat {\otimes }_\pi Y$ . Since $\mathcal T$ is weakly closed, there exists $x \in X$ and $y\in Y$ such that $x_n \otimes y_n \to x \otimes y$ in the weak topology. Arguing by contradiction, suppose that $x \otimes y \neq 0$ . Pick $x^* \in X^*$ and $y^* \in Y^*$ such that $x^*(x) = \|x\| \neq ~0$ and $y^*(y) = \|y\| \neq ~0$ . On the one hand, $x_n \otimes y_n \to x \otimes y$ weakly, so that

$$ \begin{align*}\langle x^* \otimes y^* , x_n \otimes y_n \rangle \to \langle x^* \otimes y^* , x \otimes y \rangle = x^*(x)y^*(y) = \|x\|\|y\| \neq 0.\end{align*} $$

On the other hand, since $(x_n)_{n\in \mathbb {N}}$ and $(y_n)_{n\in \mathbb {N}} $ are weakly null, one readily obtains a contradiction:

$$ \begin{align*}\langle x^* \otimes y^* , x_n \otimes y_n \rangle = x^*(x_n) y^*(y_n) \to 0.\end{align*} $$

Similarly, if $(x_n)_{n\in \mathbb {N}} \subset X$ converges weakly to x, $(y_n)_{n\in \mathbb {N}} \subset Y$ converges weakly to y and $(x_n \otimes y_n)_{n \in \mathbb {N}}$ is weakly convergent in $X \widehat {\otimes }_\pi Y$ , then

$$ \begin{align*}(x-x_n)\otimes (y-y_n) = x \otimes y - x \otimes y_n - x_n \otimes y + x_n \otimes y_n.\end{align*} $$

However, $x \otimes y_n \overset {w}{\underset {n\to +\infty }{\longrightarrow }} x\otimes y$ and $x_n \otimes y \overset {w}{\underset {n\to +\infty }{\longrightarrow }} x\otimes y$ . Therefore, $\big ((x-x_n)\otimes (y-y_n)\big )_{n \in \mathbb {N}}$ converges weakly and, moreover, the weak limit must be 0 thanks to the first part of the proof. This implies that $x_n \otimes y_n \overset {w}{\underset {n\to +\infty }{\longrightarrow }} x\otimes y$ .

In connection with Proposition 1.3, we also wish to mention [Reference García-Lirola, Grelier, Martínez-Cervantes and Rueda Zoca2, Theorem 2.3] which we describe now. If C and D are subsets of X and Y, respectively, then let

$$ \begin{align*}C \otimes D :=\{x \otimes y : x \in C, \, y \in D \} \subset \mathcal T.\end{align*} $$

As stated in [Reference García-Lirola, Grelier, Martínez-Cervantes and Rueda Zoca2, Theorem 2.3], if C and D are bounded, then $\overline {C}^w \otimes \overline {D}^w = \overline {C \otimes D}^w$ in $X \widehat {\otimes }_\pi Y$ . The technique which we introduced in the present note permits us to remove the boundedness assumption in the particular case when C and D are subspaces. It also allows us to slightly simplify the original proof of [Reference García-Lirola, Grelier, Martínez-Cervantes and Rueda Zoca2, Theorem 2.3]. The next lemma is the main ingredient.

Lemma 1.5. Let X and Y be Banach spaces such that X or Y has the AP. Let $(x_s)_s \subset X$ and $(y_s)_s \subset Y$ be two nets such that $x_s \to x^{**}$ in the weak $^*$ -topology of $X^{**}$ , $y_s \to y^{**}$ in the weak $^*$ -topology of $Y^{**}$ and $(x_s\otimes y_s)_s$ converges in the weak $^*$ -topology of $(X \widehat {\otimes }_\pi Y)^{**}$ . Then $(x_s\otimes y_s)_s$ converges weakly $^*$ to $x^{**} \otimes y^{**}$ .

The proof is essentially the same as that of Corollary 1.4, so we leave the details to the reader.

Corollary 1.6. Let X and Y be Banach spaces such that X or Y has the AP. If C and D are subsets of X and Y, respectively, then $\overline {C}^w \otimes \overline {D}^w = \overline {C \otimes D}^w$ if one of the following additional assumptions are satisfied:

  1. (i) C and D are subspaces;

  2. (ii) C and D are bounded.

Proof. First of all, it is readily seen that $\overline {C}^w \otimes \overline {D}^w \subset \overline {C \otimes D}^w$ without any additional assumption on C and D (see the first part of the proof of [Reference García-Lirola, Grelier, Martínez-Cervantes and Rueda Zoca2, Theorem 2.3]). Therefore, we only have to prove the reverse inclusion in both cases.

To prove the result under assumption (i), it suffices to apply Proposition 1.3:

$$ \begin{align*}C \otimes D \subset \overline{C} \otimes \overline{D} \implies \overline{C \otimes D}^w \subset \overline{\overline{C} \otimes \overline{D}}^w = \overline{C} \otimes \overline{D}. \end{align*} $$

To prove the result under assumption (ii), let $z \in \overline {C \otimes D}^w$ . We fix a net $(x_s \otimes y_s)_s\subset C \otimes D$ which converges weakly to z. Thanks to Proposition 1.3, there exist $x \in X$ and $y \in Y$ such that $z = x \otimes y$ . Since C and D are bounded, up to taking a suitable subnet, we may assume that both $x_s \to x^{**}$ in the weak $^*$ -topology of $X^{**}$ and ${y_s \to y^{**}}$ in the weak $^*$ -topology of $Y^{**}$ . Thanks to Lemma 1.5, $x_s \otimes y_s \to x^{**} \otimes y^{**}$ in the weak $^*$ -topology of $(X \widehat {\otimes }_\pi Y)^{**}$ . By uniqueness of the limit, $x^{**} \otimes y^{**} = z = x \otimes y$ . We distinguish two cases.

If $z = 0$ , then $x^{**} = 0$ or $y^{**} = 0$ . Say $x^{**} = 0$ for instance. This means that $0 \in \overline {C}^w$ . Now pick any $y \in C$ and observe that $z = 0 \otimes y$ , which was to be shown.

If $z \neq 0$ , then it is readily seen that $x^{**} \in \mathrm {span}\{x\}$ and $y^{**} \in \mathrm {span}\{y\}$ . Therefore, $x^{**} \in \overline {C}^{w^*} \cap X = \overline {C}^w$ and $y^{**} \in \overline {D}^{w^*} \cap Y = \overline {D}^w$ , which concludes the proof.

2 Applications to vector-valued Lipschitz free spaces

If M is a pointed metric space, with base point $0 \in M$ , and if X is a real Banach space, then $\operatorname {\mathrm {Lip}}_0(M,X)$ stands for the vector space of all Lipschitz maps from M to X which satisfy $f(0)=0$ . Equipped with the Lipschitz norm,

$$ \begin{align*} \forall f \in \operatorname{\mathrm{Lip}}_0(M,X), \quad \|f\|_L = \sup_{x \neq y \in M} \frac{\|f(x)-f(y)\|_X}{d(x,y)},\end{align*} $$

$\operatorname {\mathrm {Lip}}_0(M,X)$ naturally becomes a Banach space. When $X = \mathbb {R}$ , it is customary to omit the reference to X, that is, $\operatorname {\mathrm {Lip}}_0(M):=\operatorname {\mathrm {Lip}}_0(M,\mathbb {R})$ . Next, for $x\in M$ , we let $\delta (x) \in \operatorname {\mathrm {Lip}}_0(M)^*$ be the evaluation functional defined by $\langle \delta (x) , f \rangle = f(x)$ for all $f\in \operatorname {\mathrm {Lip}}_0(M).$ The Lipschitz free space over M is the Banach space

$$ \begin{align*}\mathcal{F}(M) := \overline{ \mbox{span}}^{\| \cdot \|}\left \{ \delta(x) : x \in M \right \} \subset \operatorname{\mathrm{Lip}}_0(M)^*.\end{align*} $$

The universal extension property of Lipschitz free spaces states that for every $f \in \operatorname {\mathrm {Lip}}_{0}(M,X)$ , there exists a unique continuous linear operator $\overline {f} \in \mathcal {L}(\mathcal {F}(M),X)$ such that:

  1. (i) $f=\overline {f} \circ \delta $ ; and

  2. (ii) $\| \overline {f} \|_{\mathcal {L}(\mathcal {F}(M),X)} = \| f \|_L$ .

In particular, we have the isometric identification

$$ \begin{align*}\operatorname{\mathrm{Lip}}_0(M,X) \equiv \mathcal{L}(\mathcal{F}(M),X).\end{align*} $$

A direct application (in the case $X = \mathbb {R}$ ) provides another basic yet important identification, namely

$$ \begin{align*}\operatorname{\mathrm{Lip}}_0(M) \equiv \mathcal{F}(M)^*.\end{align*} $$

It also follows from basic tensor product theory that $\operatorname {\mathrm {Lip}}_0(M,X^*) \equiv (\mathcal {F}(M) \widehat {\otimes }_\pi X)^*$ , which leads to the next definition (see [Reference García-Lirola, Petitjean and Rueda Zoca4] for more details).

Definition 2.1 (Vector-valued Lispschitz free spaces).

Let M be a pointed metric space and let X be a Banach space. We define the X-valued Lipschitz free space over M to be $\mathcal {F}(M,X) := \mathcal {F}(M) \widehat {\otimes }_\pi X$ .

2.1 Weak closure of $\delta (M,X)$

From [Reference García-Lirola, Petitjean, Procházka and Rueda Zoca3, Proposition 2.9], $\delta (M) = \{\delta (x) : x \in M \}$ is weakly closed in $\mathcal {F}(M)$ provided that M is complete. Our first aim is to prove the vector-valued counterpart. For this purpose, we need to identify a set that corresponds to $\delta (M)$ in the vector-valued case. A legitimate set to look at is

$$ \begin{align*}\delta(M,X) := \{ \delta(y) \otimes x : y \in M, \, x \in X \} \subset \mathcal{F}(M,X).\end{align*} $$

Notice that this does not exactly correspond to $\delta (M)$ in the case $X = \mathbb {R}$ since we have $\delta (M,\mathbb {R}) = \mathbb {R} \cdot \delta (M)$ . This discrepancy is not a major issue since $\mathbb {R} \cdot \delta (M)$ is also a weakly closed set when M is complete. The next result is thus a natural extension to the vector valued setting of [Reference García-Lirola, Petitjean, Procházka and Rueda Zoca3, Proposition 2.9].

Proposition 2.2. Let M be a complete pointed metric space and X be a Banach space such that $\mathcal {F}(M)$ or X have the approximation property. Then $\delta (M,X)$ is weakly closed in $\mathcal {F}(M,X)$ .

Proof. In what follows, $\mathcal T$ denotes the elementary tensors in $\mathcal {F}(M) \widehat {\otimes }_\pi X$ . Consider a net $(\delta (m_{\alpha })\otimes x_\alpha )_\alpha \subset \delta (M,X)$ which is weakly convergent. Since $\delta (M,X) \subset \mathcal T$ and $\mathcal T$ is weakly closed (Proposition 1.3), there exist $\gamma \in \mathcal {F}(M)$ and $x\in X$ such that the net goes to $\gamma \otimes x$ in the weak topology. We may assume that $x \neq 0$ , otherwise there is nothing to do. Pick $x^* \in X^*$ such that $x^*(x)\neq 0$ . Then, for every $f \in \operatorname {\mathrm {Lip}}_0(M)$ , we have $f(m_\alpha )x^*(x_\alpha ) \to f(\gamma ) x^*(x)$ . So the net $\big (({x^*(x_\alpha )}/{x^*(x)}) \delta (m_\alpha )\big )_\alpha \subset \mathbb {R} \cdot \delta (M)$ weakly converges to $\gamma $ . Since $\mathbb {R} \cdot \delta (M)$ is weakly closed, there are $\lambda \in \mathbb {R}$ and $m \in M$ such that $\gamma = \lambda \delta (m)$ . Consequently, $\gamma \otimes x = \delta (m) \otimes \lambda x \in \delta (M,X)$ .

2.2 Natural preduals

Next, following [Reference García-Lirola, Petitjean, Procházka and Rueda Zoca3, Section 3], $S \subset \operatorname {\mathrm {Lip}}_0(M)$ is a natural predual of $\mathcal {F}(M)$ if $S^* \equiv \mathcal {F}(M)$ and $\delta (B(0,r))$ is $\sigma (\mathcal {F}(M), S)$ -closed for every $r\geq 0$ . A reasonable extension of this notion in the vector-valued setting is as follows.

Definition 2.3. Let M be a pointed metric space and X be a Banach space with $\dim (X)\geq 2$ . We say that a Banach space S is a natural predual of $\mathcal {F}(M,X^*)$ if $Y^*\equiv \mathcal {F}(M,X^*)$ and

$$ \begin{align*}\delta(B(0,r),X^*) = \{\delta(m) \otimes x^* : m \in B(0,r), \, x^* \in X^* \} \subset \mathcal{F}(M,X^*)\end{align*} $$

is $\sigma (\mathcal {F}(M,X^*),S)$ -closed for every $r \geq 0$ .

Notice again that $\delta (B(0,r),\mathbb {R}) = \mathbb {R} \cdot \delta (B(0,r))$ . In the next statement, $\operatorname {\mathrm {lip}}_0(M)$ denotes the subspace of $\operatorname {\mathrm {Lip}}_0(M)$ of all uniformly locally flat functions. Recall that $f \in \operatorname {\mathrm {Lip}}_0(M)$ is uniformly locally flat if

$$ \begin{align*} \lim\limits_{d(x,y) \to 0} \frac{|f(x) - f(y)|}{d(x,y)} = 0.\end{align*} $$

Lemma 2.4. Let M be a separable pointed metric space. Suppose that $S\subset \operatorname {\mathrm {lip}}_0(M)$ is a natural predual of $\mathcal {F}(M)$ . Then, for every $r \geq 0$ , $\mathbb {R} \cdot \delta (B(0,r))$ is weak $^*$ closed in $\mathcal {F}(M)$ .

Proof. Let us fix $r \geq 0$ . Let $(\lambda _n \delta (x_n))_{n} \subset \mathbb {R} \cdot \delta (B(0,r))$ be a sequence converging to some $\gamma \in \mathcal {F}(M)$ in the weak $^*$ topology. We assume that $\gamma \neq 0$ , otherwise there is nothing to do. Since a weak $^*$ convergent sequence is bounded, and by weak $^*$ lower semi-continuity of the norm, we may assume that there exists $C>0$ such that for every n:

$$ \begin{align*} 0 < \frac{\|\gamma\|}{2} \leq |\lambda_n| \| \delta(x_n) \| = |\lambda_n| d(x_n,0) \leq C.\end{align*} $$

Thus, $d(x_n,0) \neq 0$ and $\lambda _n \neq 0$ for every n. Up to extracting a further subsequence, we may assume that the sequence $(\lambda _n d(x_n,0))_{n}$ converges to some $\ell \neq 0$ . Since $(x_n)_{n} \subset B(0,r)$ , we also assume that $(d(x_n,0))_{n}$ converges to some d. We will distinguish two cases.

If $d \neq 0$ , then $(\lambda _n)_{n}$ converges to $\lambda :={\ell }/{d}$ and so $(\delta (x_n))_{n}$ weak $^*$ converges to ${\gamma }/{\lambda }$ . Since S is a natural predual of $\mathcal {F}(M)$ , $\delta (B(0,r))$ is weak $^*$ closed in $\mathcal {F}(M)$ . So there exists $x \in M$ such that $\gamma = \lambda \delta (x)$ .

If $d = 0$ , then $(\delta (x_n))_{n}$ converges to $0$ in the norm topology (and $(\lambda _n)_{n}$ tends to infinity). Note that we may write

$$ \begin{align*} \lambda_n \delta(x_n) = \lambda_n d(x_n,0) \frac{\delta(x_n) - \delta(0)}{d(x_n,0)}. \end{align*} $$

Since $S\subset \operatorname {\mathrm {lip}}_0(M)$ , the sequence $({(\delta (x_n) - \delta (0))}/{d(x_n,0)})_{n}$ weak $^*$ converges to 0. Moreover, the sequence $(\lambda _n d(x_n,0))_{n}$ converges to $\ell \neq 0$ . Consequently, $(\lambda _n \delta (x_n))_{n}$ weak $^*$ converges to 0 and so $\gamma = 0$ , which is a contradiction.

Before going further, we need to introduce the injective tensor product of two Banach spaces. Recall that to define the projective tensor product, we introduced $x\otimes y$ as an element of $\mathcal {B}(X \times Y)^{*}$ . For the injective tensor product, we change the point of view since we now consider $x\otimes y$ as an element of $\mathcal {B}(X^* \times Y^*)$ defined as follows:

$$ \begin{align*}\forall (x^*,y^*) \in X^* \times Y^*, \quad \langle x\otimes y , (x^*,y^*) \rangle = x^*(x) y^*(y).\end{align*} $$

In this case, we denote by $\| \cdot \|_{\varepsilon }$ the canonical norm on $\mathcal {B}(X^* \times Y^*)$ . Thus, if $u = \sum _{i=1}^n x_i \otimes y_i \in X \otimes Y$ , then

$$ \begin{align*}\| u \|_{\varepsilon} = \sup \bigg \{ \bigg| \sum_{i=1}^n x^*(x_i) y^*(y_i) \bigg| : x^* \in B_{X^*}, y^* \in B_{Y^*} \bigg \}.\end{align*} $$

The injective tensor product of X and Y is defined by

$$ \begin{align*}X \widehat{\otimes}_\varepsilon Y = \overline{\mathrm{span}}^{\|\cdot\|_{\varepsilon}} \{ x\otimes y : x \in X, \, y \in Y\} \subseteq \mathcal{B}(X^* \times Y^*) .\end{align*} $$

In what follows, we will use a classical result from tensor product theory (see, for example, [Reference Ryan6, Theorem 5.33]): if $X^*$ or $Y^*$ has the Radon–Nikodým property (RNP in short) and $X^*$ or $Y^*$ has the AP, then $(X \widehat {\otimes }_\varepsilon Y)^* \equiv X^* \widehat {\otimes }_\pi Y^*$ . The RNP has many characterisations (see [Reference Diestel and Uhl1, Section VII.6] for a nice overview).

Assume now that there exists a subspace S of $\operatorname {\mathrm {Lip}}_0(M)$ such that $S^* \equiv \mathcal {F}(M)$ . Then

$$ \begin{align*} \mathcal F(M,X^*) = \mathcal F(M) \widehat\otimes_\pi X^* \equiv (S \widehat{\otimes}_\varepsilon X)^{*}\end{align*} $$

whenever either $\mathcal F(M)$ or $X^*$ has the AP and either $\mathcal F(M)$ or $X^*$ has the RNP. It is quite natural to wonder whether there are conditions which ensure that $S \widehat {\otimes }_\varepsilon X$ is a natural predual of $\mathcal {F}(M,X^*)$ . The next result asserts that this sometimes relies on the scalar case.

Proposition 2.5. Let M be a separable pointed metric space, $S\subset \operatorname {\mathrm {lip}}_0(M)$ be a natural predual of $\mathcal {F}(M)$ and X be a Banach space (with $\dim (X) \geq 2$ ). Assume moreover that either $\mathcal F(M)$ or $X^*$ has the AP and either $\mathcal F(M)$ or $X^*$ has the RNP. Then $S \widehat {\otimes }_\varepsilon X$ is a natural predual of $\mathcal {F}(M,X^*)$ .

Proof. To show that $S \widehat {\otimes }_\varepsilon X$ is a natural predual, we essentially follow the proof of Proposition 2.2. First of all, we show that $\mathcal T:= \{ \gamma \otimes x^* : \gamma \in \mathcal {F}(M), \, x \in X^* \}$ is weak $^*$ closed in $\mathcal {F}(M,X^*)$ . Indeed, it is not hard to show that if $T \in \mathcal {F}(M,X^*)$ , then $T \in \mathcal T$ if and only if for every pair of linearly independent families $\{f_1,f_2 \} \subset S$ and $\{x_1,x_2\} \subset X$ ,

$$ \begin{align*}\begin{vmatrix} \langle T , f_1 \otimes x_1 \rangle & \langle T , f_1 \otimes x_2 \rangle \\ \langle T , f_2 \otimes x_1 \rangle & \langle T , f_2 \otimes x_2 \rangle \end{vmatrix} = 0 .\end{align*} $$

Accordingly, $\mathcal T$ is weak $^*$ closed. Fix $r{\kern-1pt}>{\kern-1pt}0$ and consider a net $(\delta (m_{\alpha })\otimes x_\alpha ^*)_\alpha {\kern-1pt}\subset{\kern-1pt} \delta (B(0,r),X^*)$ which weak $^*$ converges to some $\gamma \otimes x^* \in \mathcal T$ . We may assume that $x^* \neq 0$ otherwise there is nothing to do. Consider $x \in X$ such that $x^*(x)\neq 0$ . Then, for every $f \in S$ , we have $f(m_\alpha )x^*(x_\alpha ) \to f(\gamma ) x^*(x)$ . So the net $\big (({x^*(x_\alpha )}/{x^*(x)}) \delta (m_\alpha )\big )_\alpha \subset \mathbb {R} \cdot \delta (M)$ weak $^*$ converges to $\gamma $ . Since $\mathbb {R} \cdot \delta (M)$ is weak $^*$ closed (Lemma 2.4), there are $\lambda \in \mathbb {R}$ and $m \in M$ such that $\gamma = \lambda \delta (m)$ .

Acknowledgements

The author thanks Christian Le Merdy and Abraham Rueda Zoca for useful discussions.

Footnotes

The author was partially supported by the French ANR project No. ANR-20-CE40-0006.

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