Surjective nonsymmetric norm preserving maps between ${\rm{Lip}}(X,d,\tau)$-algebras

Document Type : Original Article

Authors

Esmat Ahmadloo ¹

Davood Alimohammadi ²

¹ Department of Mathematics‎, ‎Faculty of Science‎, ‎Arak University‎‎, ‎Arak‎, ‎Iran

² Department of Mathematics, Faculty of Science, Arak University, Arak, Iran

10.22034/kjm.2024.426471.3059

Abstract

Let $A$ be a real subalgebra of $C(X,\tau)$ containing ${\rm{Lip}}(X,d,\tau)$ and $B$ be a real subalgebra of $C(Y,\eta)$ containing ${\rm{Lip}}(Y,\rho,\eta)$‎, ‎where $(X,d)$ and $(Y,\rho)$ are compact metric spaces and $\tau$ and $\eta$ are Lipschitz involutions on $(X,d)$ and $(Y,\rho)$‎, ‎respectively‎. ‎In this paper‎, ‎we first study surjective multiplicatively uniform norm preserving maps $T$ from $A$ to $B$ and determine their structures‎. ‎Next‎, ‎we give a description of surjective maps $T:A\rightarrow B$ satisfying the nonsymmetric uniform norm condition $\|T(f)T(g)-1_Y\|_Y=\|fg-1_X\|_X$ for all $f,g\in A$‎. ‎For such a map $T$‎, ‎we show that $(T(1_X))^2=1_Y$ and $T$ is multiplicatively uniform norm preserving‎. ‎In the case that $A={\rm{Lip}}(X,d,\tau)$ and $B={\rm{Lip}}(Y,\rho,\eta)$‎, ‎we prove that‎ $T(A_\mathbb{R})$ is a subset of $B_\mathbb{R}$ and there exists a bijective map $\varphi\colon Y\rightarrow X$ with $\varphi\circ\eta=\tau\circ\varphi$ on $Y$ such that $\varphi$ is continuous at each $y\in Y$ with $\eta(y)=y$‎, ‎${\varphi}^{-1}$ is continuous at each $x\in X$ with $\tau(x)=x$ and $T(f)=T(1_X)\cdot(f\circ\varphi)$ on $Y$ for all $f\in A_\mathbb{R}$‎, ‎where $ A_\mathbb{R}$ is the set of all $f\in A$ for which $f$ is real-valued on $X$.

Keywords

Multiplicatively norm preserving map‎

Nonsymmetric norm preserving map

‎Real Lipschitz algebra‎

‎Lipschitz involution‎

‎Uniform norm

Subjects

• 47 Operator theory

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