# Positiveness of Inverse Of Positive Operator Implies Lattice Isomorphism?

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

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 $$T(x\vee y)=Tx\vee Ty$$.