Search

Let $\mathrm{Lip}(X)$, $\mathrm{Lip}^b(X)$, $\mathrm{Lip}^{\mathrm{loc}}(X)$ and $\mathrm{Lip}^\mathrm{pt}(X)$ be the vector spaces of Lipschitz, bounded Lipschitz, locally Lipschitz and pointwise Lipschitz (real-valued) functions defined on a metric space $(X, d_X)$, respectively. We show that if a weighted composition operator $Tf=h\cdot f\circ \varphi$ defines a bijection between such vector spaces preserving Lipschitz constants, local Lipschitz constants or pointwise Lipschitz constants, then $h= \pm1/\alpha$ is a constant function for some scalar $\alpha>0$ and $\varphi$ is an $\alpha$-dilation. Let $U$ be open connected and $V$ be open, or both $U,V$ are convex bodies, in normed linear spaces $E, F$, respectively. Let $Tf=h\cdot f\circ\varphi$ be a bijective weighed composition operator between the vector spaces $\mathrm{Lip}(U)$ and $\mathrm{Lip}(V)$, $\mathrm{Lip}^b(U)$ and $\mathrm{Lip}^b(V)$, $\mathrm{Lip}^\mathrm{loc}(U)$ and $\mathrm{Lip}^\mathrm{loc}(V)$, or $\mathrm{Lip}^\mathrm{pt}(U)$ and $\mathrm{Lip}^\mathrm{pt}(V)$, preserving the Lipschitz, locally Lipschitz, or pointwise Lipschitz constants, respectively. We show that there is a linear isometry $A: F\to E$, an $\alpha>0$ and a vector $b\in E$ such that $\varphi(x)=\alpha Ax + b$, and $h$ is a constant function assuming value $\pm 1/\alpha$. More concrete results are obtained for the special cases when $E=F=\mathbb{R}^n$, or when $U,V$ are $n$-dimensional flat manifolds., Comment: to appear in "Annals of Mathematical Sciences and Applications"

Let ${\bf M}_n(\mathbb{F})$ be the algebra of $n\times n$ matrices over an arbitrary field $\mathbb{F}$. We consider linear maps $\Phi: {\bf M}_n(\mathbb{F}) \rightarrow {\bf M}_r(\mathbb{F})$ preserving matrices annihilated by a fixed polynomial $f(x) = (x-a_1)\cdots (x-a_m)$ with $m\ge 2$ distinct zeroes $a_1, a_2, \ldots, a_m \in \mathbb{F}$; namely, $$ f(\Phi(A)) = 0\quad\text{whenever} \quad f(A) = 0. $$ Suppose that $f(0)=0$, and the zero set $Z(f) =\{a_1, \dots, a_m\}$ is not an additive group. Then $\Phi$ assumes the form \begin{align}\label{eq:standard} A \mapsto S\begin{pmatrix} A \otimes D_1 &&\cr & A^{T} \otimes D_2& \cr && 0_s\cr\end{pmatrix}S^{-1}, \tag{$\dagger$} \end{align} for some invertible matrix $S\in {\bf M}_r(\mathbb{F})$, invertible diagonal matrices $D_1\in {\bf M}_p(\mathbb{F})$ and $D_2\in {\bf M}_q(\mathbb{F})$, where $s=r-np-nq\geq 0$. The diagonal entries $\lambda$ in $D_1$ and $D_2$, as well as $0$ in the zero matrix $0_s$, are zero multipliers of $f(x)$ in the sense that $\lambda Z(f) \subseteq Z(f)$. In general, assume that $Z(f) - a_1$ is not an additive group. If $\Phi(I_n)$ commutes with $\Phi(A)$ for all $A\in {\bf M}_n(\mathbb{F})$, or if $f(x)$ has a unique zero multiplier $\lambda=1$, then $\Phi$ assumes the form \eqref{eq:standard}. The above assertions follow from the special case when $f(x) = x(x-1)=x^2-x$, for which the problem reduces to the study of linear idempotent preservers. It is shown that a linear map $\Phi: {\bf M}_n(\mathbb{F}) \rightarrow {\bf M}_r(\mathbb{F})$ sending disjoint rank one idempotents to disjoint idempotents always assume the above form \eqref{eq:standard} with $D_1=I_p$ and $D_2=I_q$, unless ${\bf M}_n(\mathbb{F}) = {\bf M}_2(\mathbb{Z}_2)$.