Let $X$ be a Banach Lattice and denote by $\mathcal{B}(X)$ the banach space of bounded linear endomorphisms. An operator $T \in \mathcal{B}(X)$ is called positive if $Tx \geq 0$ whenever $x \geq 0.$ We say that a positive operator is a lattice homomorphism if $T(x\vee y) = Tx \vee Ty$ for every $x,y \in X$, where $x\vee y = \sup \{x,y\}.$

Now, consider $T$ a positive operator and assume $T$ is an isomorphism. I wonder if the positiveness of $T^{-1}$ is a suficient condition for $T$ to be a lattice homomorphism (and applying the same condition to $T^{-1}$, a lattice isomorphism). I think this is the most intuitive way of thinking about a lattice homomorphism, but I couldn't find such a result in any Banach Lattices book, so my intuition may be wrong.

Could anyone give me a proof or a counterexample?

Thank you very much


1 Answer 1


It is true, and quite easy to prove. Abstractly, if $T$ is a positive isomorphism with positive inverse, then $T$ preserves the order structure, so of course it preserves suprema.

To make it more explicit: Since $T$ is positive and $x\vee y\geq x,y$, we have $T(x\vee y)\geq Tx, Ty$. Thus $T(x\vee y)\geq Tx\vee Ty$. With the same reasoning for $T^{-1}$ instead of $T$ and $Tx,Ty$ instead of $x,y$ we obtain $T^{-1}(Tx\vee Ty)\geq x\vee y$. Now apply $T$ to get $Tx\vee Ty\geq T(x\vee y)$. Altogether, we have proven $T(x\vee y)=Tx\vee Ty$.


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