Simple constructions of $\mathrm{FBL}(A)$ and $\mathrm{FBL}[E]$

TL;DR

This paper presents new constructions for free Banach lattices, showing they can be obtained as completions of free vector lattices under a maximal lattice seminorm, applicable to both sets and Banach spaces.

Contribution

It introduces explicit completion methods for free Banach lattices derived from free vector lattices and Banach spaces, providing a unified construction approach.

Findings

Construction of $ ext{FBL}(A)$ as a completion of $ ext{FVL}(A)$

Extension of the construction to $ ext{FBL}[E]$ for Banach spaces

Characterization of the maximal lattice seminorm $ u$

Abstract

We show that the free Banach lattice $\mathrm{FBL}(A)$ may be constructed as the completion of $\mathrm{FVL}(A)$ with respect to the maximal lattice seminorm $\nu$ on $\mathrm{FVL}(A)$ with $\nu(a)\le 1$ for all $a\in A$. We present a similar construction for the free Banach lattice $\mathrm{FBL}[E]$ generated by a Banach space $E$.

Equations

$$\lvert f\rvert\leqslant\lambda\bigl{(}\lvert\delta_{a_{1}}\rvert+\dots+\lvert\delta_{a_{n}}\rvert\bigr{)}.$$

$$\Vert f+g\Vert -\epsilon <\nu (f+g)⩽\nu (f)+\nu (g)⩽\Vert f\Vert +\Vert g\Vert .$$

$$f(x^{*})=F\bigl{(}\hat{x}_{1}(x^{*}),\dots,\hat{x}_{n}(x^{*})\bigr{)}=F\bigl{(}x^{*}(x_{1}),\dots,x^{*}(x_{n})\bigr{)}$$

Full text

The author was supported by an NSERC grant.

1. Preliminaries

The free vector lattice over a set $A$, denoted by $\operatorname{FVL}(A)$, goes back to [Bir42]. More recently, a free Banach lattice $\operatorname{FBL}(A)$ has been introduced and investigated; see [dPW15, ART18]. It has been folklore knowledge (and was implicitly mentioned in [dPW15, ART18]) that the norm of $\operatorname{FBL}(A)$ is, in some sense, the greatest lattice norm one can put on $\operatorname{FVL}(A)$. In this note, we make this idea into a formal statement and provide a direct proof. This yields an alternative way of constructing $\operatorname{FBL}(A)$ and $\operatorname{FBL}[E]$.

Let $A$ be a subset of a vector lattice $X$. We say that $X$ is a free vector lattice over $A$ if every function $\varphi\colon A\to Y$, where $Y$ is an arbitrary vector lattice, extends uniquely to a lattice homomorphism $\tilde{\varphi}\colon X\to Y$. For every set $A$ there is a vector lattice $X$ which contains $A$ and is free over $A$. It is easy to see that if $X_{1}$ and $X_{2}$ are both free over $A$ then there exists a lattice isomorphism between $X_{1}$ and $X_{2}$ which fixes $A$. So a free vector lattice over $X$ is determined uniquely up to a lattice isomorphism; we denote it by $\operatorname{FVL}(A)$.

We outline below a construction of $\operatorname{FVL}(A)$ and some of its basic properties; we refer the reader to [Ble73, dPW15] for further details on free vector lattices. Given a set $A$. For every $a\in A$, we write $\delta_{a}$ for the “evaluation functional” of $a$ in the following sense: $\delta_{a}\colon\mathbb{R}^{A}\to\mathbb{R}$ with $\delta_{a}(x)=x(a)$ for $x\colon A\to\mathbb{R}$. Then $\operatorname{FVL}(A)$ may be identified with the sublattice of $\mathbb{R}^{{\mathbb{R}}^{A}}$ generated by $\{\delta_{a}\>:\>a\in A\}$. Identifying $a$ with $\delta_{a}$, one may view $A$ as a subset of $\operatorname{FVL}(A)$. Since $\operatorname{FVL}(A)$ is a sublattice $\mathbb{R}^{{\mathbb{R}}^{A}}$, and the latter is Archimedean, $\operatorname{FVL}(A)$ is also Archimedean.

It is easy to see that if $a_{1},\dots,a_{n}\in A$ then \operatorname{FVL}\bigl{(}\{a_{1},\dots,a_{n}\}\bigr{)} may be identified with the sublattice of $\operatorname{FVL}(A)$ generated by $a_{1},\dots,a_{n}$. For every $f\in\operatorname{FVL}(A)$ there exists a finite subset $\{a_{1},\dots,a_{n}\}$ of $A$ such that $f$ belongs to the sublattice of $\operatorname{FVL}(A)$ generated by $a_{1},\dots,a_{n}$. Furthermore, if $A$ itself is finite, say, $A=\{a_{1},\dots,a_{n}\}$, then $\bigvee_{k=1}^{n}\lvert a_{k}\rvert$ is a strong unit in $\operatorname{FVL}(A)$.

By a lattice-linear expression, we mean an expression formed by finitely many variables and linear and lattice operations. For example, $F(t_{1},t_{2},t_{3})=t_{1}\wedge t_{2}+t_{1}\vee(2t_{3})$ is a lattice-linear expression. Clearly, a lattice-linear expression $F(t_{1},\dots,t_{n})$ induces a positively homogeneous function from $\mathbb{R}^{n}$ to $\mathbb{R}$. On the other hand, if $X$ is an Archimedean vector lattice and $x_{1},\dots,x_{n}\in X$, plugging $x_{1},\dots,x_{n}$ into $F$ instead of $t_{1},\dots,t_{n}$, we can define $F(x_{1},\dots,x_{n})$ as an element of $X$ in a natural way. We say that $F(x_{1},\dots,x_{n})$ is a lattice-linear combination of $x_{1},\dots,x_{n}$. If two lattice-linear expressions $F$ and $G$ agree as functions from $\mathbb{R}^{n}$ to $\mathbb{R}$ then $F(x_{1},\dots,x_{n})=G(x_{1},\dots,x_{n})$. Actually, the calculus of lattice-linear expressions in $X$ is a restriction of Krivine’s function calculus; see, e.g., [BdPvR91, Proposition 3.6]. Observe that the sublattice of $X$ generated by $x_{1},\dots,x_{n}$ is exactly the set of all lattice-linear combinations of $x_{1},\dots,x_{n}$. $\operatorname{FVL}(A)$ may be interpreted as the set of all formal lattice-linear expressions of elements of $A$, where we identify two expressions if they agree as functions from $\mathbb{R}^{n}$ to $\mathbb{R}$. For example, we identify $a_{1}+(a_{2}\vee a_{3})$ and $(a_{1}+a_{2})\vee(a_{1}+a_{3})$. Formally speaking, $\operatorname{FVL}(A)$ consists of equivalence classes of lattice-linear expressions.

In [dPW15], the concept of a free Banach lattice was introduced. Let $A$ be a subset of a Banach lattice $X$. We say that $X$ is a free Banach lattice over a set $A$ if every function $\varphi\colon A\to Y$, where $Y$ is an arbitrary Banach lattice, satisfying \sup_{a\in A}\bigl{\lVert}\varphi(a)\bigr{\rVert}\leqslant 1 extends uniquely to a lattice homomorphism $\tilde{\varphi}\colon X\to Y$ with $\lVert\tilde{\varphi}\rVert\leqslant 1$. It was shown in [dPW15] that for every set $A$ there is a Banach lattice $X$ which contains $A$ and is free over $A$. Again, it is easy to see that such a Banach lattice is unique up to a lattice isometry which fixes $A$; we denote it by $\operatorname{FBL}(A)$. It is easy to see that $A$ is a subset of the unit sphere of $\operatorname{FBL}(A)$.

An alternative way of constructing $\operatorname{FBL}(A)$ was recently obtained in [ART18]. In [ART18], the authors also prove that for every Banach space $E$ there exists a Banach lattice $\operatorname{FBL}[E]$ such that $E$ is a closed subspace of $\operatorname{FBL}[E]$ and every bounded operator $T\colon E\to Y$, where $Y$ is an arbitrary Banach lattice, extends uniquely to a lattice homomorphism $\widetilde{T}\colon\operatorname{FBL}[E]\to Y$ with $\lVert\widetilde{T}\rVert=\lVert T\rVert$. It is easy to see that $\operatorname{FBL}[E]$ is unique up to a lattice isometry preserving $E$. Furthermore, it can be easily verified that \operatorname{FBL}(A)=\operatorname{FBL}\bigl{[}\ell_{1}(A)]\bigr{]} for any set $A$.

In this note, we present constructions of $\operatorname{FBL}(A)$ and $\operatorname{FBL}[E]$ that are somewhat easier than those in [dPW15, ART18].

2. A construction of $\operatorname{FBL}(A)$

Theorem 2.1**.**

There exists a maximal lattice seminorm $\nu$ on $\mathrm{FVL}(A)$ with $\nu(a)\leqslant 1$ for all $a\in A$. It is a lattice norm, and the completion of $\operatorname{FVL}(A)$ with respect to it is $\mathrm{FBL}(A)$.

Proof.

As before, we identify $\operatorname{FVL}(A)$ with the sublattice of $\mathbb{R}^{{\mathbb{R}}^{A}}$ generated by $\{\delta_{a}\>:\>a\in A\}$; by identifying $a\in A$ with $\delta_{a}\in\operatorname{FVL}(A)$, we may view $A$ as a subset of $\operatorname{FVL}(A)$. Let $\mathcal{N}$ be the set of all lattice seminorms $\nu$ on $\operatorname{FVL}(A)$ such that $\nu(\delta_{a})\leqslant 1$ for every $a\in A$.

Let $x\in\mathbb{R}^{A}$ such that $\lvert x(a)\rvert\leqslant 1$ for all $a\in A$. For $f\in\operatorname{FVL}(A)$, put \nu_{x}(f)=\bigl{\lvert}f(x)\bigr{\rvert}. It can be easily verified that $\nu_{x}\in\mathcal{N}$.

For $f\in\operatorname{FVL}(A)$, put $\lVert f\rVert=\sup\limits_{\nu\in\mathcal{N}}\nu(f).$ We claim that this is a lattice norm on $\operatorname{FVL}(A)$.

First, observe that $\lVert f\rVert$ is finite. Find $a_{1},\dots,a_{n}\in A$ such that f\in\operatorname{FVL}\bigl{(}\{a_{1},\dots,a_{n}\}\bigr{)}. Since $\lvert\delta_{a_{1}}\rvert+\dots+\lvert\delta_{a_{n}}\rvert$ is a strong unit in \operatorname{FVL}\bigl{(}\{a_{1},\dots,a_{n}\}\bigr{)}, there exists $\lambda\in\mathbb{R}_{+}$ such that

[TABLE]

It follows that $\nu(f)\leqslant\lambda n$ for every $\nu\in\mathcal{N}$, hence $\lVert f\rVert\leqslant\lambda n<\infty$.

It is straightforward that $\lVert\cdot\rVert$ is positively homogeneous. To verify the triangle inequality, let $f,g\in\operatorname{FVL}(A)$ and fix $\varepsilon>0$. There exists $\nu\in\mathcal{N}$ such that

[TABLE]

It follows that $\lVert f+g\rVert\leqslant\lVert f\rVert+\lVert g\rVert$.

Suppose that $f\neq 0$. Find $a_{1},\dots,a_{n}\in A$ such that f\in\operatorname{FVL}\bigl{(}\{a_{1},\dots,a_{n}\}\bigr{)}. It follows that there is $x\in\mathbb{R}^{A}$ such that $f(x)\neq 0$ and $\operatorname{supp}x\subseteq\{a_{1},\dots,a_{n}\}$. Without loss of generality, scaling $x$ if necessary, $\lvert x(a)\rvert\leqslant 1$ for all $a\in A$. Then \nu_{x}(f)=\bigl{\lvert}f(x)\bigr{\rvert}>0, hence $\lVert f\rVert\neq 0$.

If $\lvert f\rvert\leqslant\lvert g\rvert$ in $\operatorname{FVL}(A)$ then $\nu(f)\leqslant\nu(g)$ for every $\nu\in\mathcal{N}$, hence $\lVert f\rVert\leqslant\lVert g\rVert$. Thus, $\lVert\cdot\rVert$ is a lattice norm on $\operatorname{FVL}(A)$.

Let $X$ be the completion of \bigl{(}\operatorname{FVL}(A),\lVert\cdot\rVert\bigr{)}. We claim that $X$ is a free Banach lattice over $A$.

Let $\varphi\colon A\to Y$, where $Y$ is a Banach lattice and $\sup_{a\in A}\lVert\varphi(a)\rVert\leqslant 1$. Then $\varphi$ extends to a lattice homomorphism $\hat{\varphi}\colon\operatorname{FVL}(A)\to Y$. It suffices to show that $\lVert\hat{\varphi}\rVert\leqslant 1$; it would follow that $\hat{\varphi}$ extends to a contractive lattice homomorphism $\tilde{\varphi}$ from $X$ to $Y$. For $f\in\operatorname{FVL}(A)$, put \nu(f)=\bigl{\lVert}\hat{\varphi}(f)\bigr{\rVert}. It is easy to see that $\nu\in\mathcal{N}$, hence \bigl{\lVert}\hat{\varphi}(f)\bigr{\rVert}=\nu(f)\leqslant\lVert f\rVert.

Uniqueness of the extension follows from the fact that any lattice homomorphism extension of $\varphi$ to $X$ has to agree with $\hat{\varphi}$ on $\operatorname{FVL}(A)$, hence with $\tilde{\varphi}$ on $X$ as $\operatorname{FVL}(A)$ is dense in $X$. ∎

It follows that the FBL norm is the greatest norm on $\operatorname{FVL}(A)$ such that $\lVert a\rVert\leqslant 1$ for every $a\in A$.

3. A construction of $\operatorname{FBL}[E]$

For a Banach lattice $E$ and a vector $x\in E$, the evaluation functional $\hat{x}\in E^{**}$ is defined by $\hat{x}(x^{*})=x^{*}(x)$ for $x^{*}\in X^{*}$. In particular, $\hat{x}$ is a function from $E^{*}$ to $\mathbb{R}$, i.e., an element of $\mathbb{R}^{E^{*}}$.

Theorem 3.1**.**

Let $E$ be a Banach space; let $L$ be the sublattice of $\mathbb{R}^{E^{*}}$ generated by $\{\hat{x}\>:\>x\in E\}$. There is a maximal lattice seminorm $\nu$ on $L$ satisfying $\nu(\hat{x})\leqslant\lVert x\rVert$ for all $x\in E$. It is a lattice norm; the completion of $L$ with respect to it is $\mathrm{FBL}[E]$.

Proof.

It is easy to see that the map $x\in E\mapsto\hat{x}\in L$ is a linear embedding, so that we may view $E$ as a linear subspace of $L$. Let $\mathcal{M}$ be the set of all lattice seminorms $\nu$ on $L$ such that $\nu(\hat{x})\leqslant\lVert x\rVert$ for all $x\in E$.

For every $x^{*}\in B_{E^{*}}$, define \nu_{x^{*}}(f)=\bigl{\lvert}f(x^{*})\bigr{\rvert} for $f\in L$. It is easy to see that $\nu_{x^{*}}$ is a seminorm on $L$. Note that $\nu_{x^{*}}$ is a lattice seminorm because \lvert f\rvert(x^{*})=\bigl{\lvert}f(x^{*})\bigr{\rvert}. Furthermore, if $x\in E$ then \nu_{x^{*}}(\hat{x})=\bigl{\lvert}\hat{x}(x^{*})\bigr{\rvert}=\bigl{\lvert}x^{*}(x)\bigr{\rvert}\leqslant\lVert x\rVert, so that $\nu_{x^{*}}\in\mathcal{M}$.

For $f\in L$, define $\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|f\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|=\sup_{\nu\in\mathcal{M}}\nu(f)$. We claim that $\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\cdot\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|$ is a lattice norm on $L$. First, we will show that it is finite. Let $f\in L$. Then $f$ is a lattice-linear expression of $\hat{x}_{1},\dots,\hat{x}_{n}$ for some $x_{1},\dots,x_{n}$ in $E$. Since lattice-linear functions are positively homogeneous, we may assume without loss of generality that $x_{1},\dots,x_{n}\in B_{E}$. Clearly, $\lvert\hat{x}_{1}\rvert+\dots+\lvert\hat{x}_{n}\rvert$ is a strong unit in the sublattice of $L$ generated by $\hat{x}_{1},\dots,\hat{x}_{n}$. It follows that \lvert f\rvert\leqslant\lambda\bigl{(}\lvert\hat{x}_{1}\rvert+\dots+\lvert\hat{x}_{n}\rvert\bigr{)} for some $\lambda>0$, so that $\nu(f)\leqslant\lambda n$ for every $\nu\in\mathcal{M}$, hence $\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|f\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\leqslant\lambda n<\infty$.

It is straightforward that $\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\cdot\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|$ is a lattice seminorm on $L$. Suppose that $0\neq f\in L$. Then $f(x^{*})\neq 0$ for some $x^{*}\in E^{*}$. Since $f$ is positively homogeneous, we may assume without loss of generality that $x^{*}\in B_{E^{*}}$. Then \left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|f\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\geqslant\nu_{x^{*}}(f)=\bigl{\lvert}f(x^{*})\bigr{\rvert}>0. Thus, $\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\cdot\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|$ is a lattice norm on $L$.

Note also that $\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\hat{x}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|=\lVert x\rVert$ for all $x\in E$, so that we may view $\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\cdot\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|$ as an extension of $\lVert\cdot\rVert$ from $E$ to $L$. Indeed, $\nu(\hat{x})\leqslant\lVert x\rVert$ for every $\nu\in\mathcal{M}$, hence $\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\hat{x}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\leqslant\lVert x\rVert$. On the other hand, let $x^{*}\in B_{E^{*}}$ such that $x^{*}(x)=\lVert x\rVert$; then \left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\hat{x}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\geqslant\nu_{x^{*}}(\hat{x})=\bigl{\lvert}\hat{x}(x^{*})\bigr{\rvert}=\lVert x\rVert.

Let $X$ be the completion of \bigl{(}L,\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\cdot\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\bigr{)}. We claim that $X=\operatorname{FBL}[E]$.

Let $T\colon E\to Y$ be a linear operator from $E$ to an arbitrary Banach lattice $Y$ with $\lVert T\rVert=1$. We define $\widehat{T}\colon L\to Y$ as follows. Let $f\in L$. Then $f$ is a lattice-linear combination of $\hat{x}_{1},\dots,\hat{x}_{n}$ for some $x_{1},\dots,x_{n}\in E$. Without loss of generality, $x_{1},\dots,x_{n}$ are linearly independent in $E$. We define $\widehat{T}f$ to be the same lattice-linear combination of $Tx_{1},\dots,Tx_{n}$ in $Y$. That is, suppose that $f=F(\hat{x}_{1},\dots,\hat{x}_{n})$ for some formal lattice-linear expression $F(t_{1},\dots,t_{n})$; we then put $\widehat{T}f=F(Tx_{1},\dots,Tx_{n})$. Note that $\widehat{T}f$ is well-defined, i.e., does not depend on a particular choice of a lattice-linear combination representing $f$. Indeed, suppose that $f=G(\hat{x}_{1},\dots,\hat{x}_{n})$, where $G(t_{1},\dots,t_{n})$ is another formal lattice-linear expressions. Since $L$ is a sublattice of $\mathbb{R}^{E^{*}}$, the lattice operations in $L$ are point-wise, hence

[TABLE]

in $\mathbb{R}$ for every $x^{*}\in X^{*}$. Similarly, f(x^{*})=G\bigl{(}x^{*}(x_{1}),\dots,x^{*}(x_{n})\bigr{)}. Since $x_{1},\dots,x_{n}$ are linearly independent, this means that $F(t_{1},\dots,t_{n})=G(t_{1},\dots,t_{n})$ for all $t_{1},\dots,t_{n}\in\mathbb{R}$. Therefore, $\widehat{T}$ is well-defined.

The definition of $\widehat{T}$ immediately yields that it is a lattice homomorphism. Clearly, $\widehat{T}$ extends $T$ in the sense that $\widehat{T}\hat{x}=Tx$ for every $x\in E$. It follows that $\lVert\widehat{T}\rVert\geqslant 1$. We claim that that $\lVert\widehat{T}\rVert=1$. Indeed, for $f\in L$, define $\nu(f)=\lVert\widehat{T}f\rVert$ in $Y$. It is easy to see that $\nu$ is a lattice seminorm on $L$. For every $x\in X$, one has $\nu(\hat{x})=\lVert\widehat{T}\hat{x}\rVert=\lVert Tx\rVert\leqslant\lVert x\rVert.$ It follows that $\nu\in\mathcal{M}$, so that $\lVert\widehat{T}f\rVert=\nu(f)\leqslant\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\left|f\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|\mathchoice{{}\mkern-4.5mu}{{}\mkern-4.5mu}{{}\mkern-2.5mu}{}\right|,$ so that $\lVert\widehat{T}\rVert\leqslant 1$. It follows that $\widehat{T}$ extends to a contractive lattice homomorphism $\widetilde{T}\colon X\to Y$.

Again, uniqueness of the extension follows from the fact that any contractive lattice homomorphism that extends $T$ to $X$ has to agree with $\widehat{T}$ on $L$ and, therefore, with $\widetilde{T}$ on $X$. ∎