关键词: $\varepsilon$-phase isometry, phase isometry, Banach space

Abstract: Let $X$ and $Y$ be two normed spaces. Let $\mathcal{U}$ be a non-principal ultrafilter on $\mathbb{N}$. Let $g: X\rightarrow Y$ be a standard $\varepsilon$-phase isometry for some $\varepsilon\geq 0$, i.e., $g(0)=0$, and for all $u,v\in X$,
$$|\; |\|g(u)+g(v)\|\pm \|g(u)-g(v)\||-|\|u+v\|\pm\|u-v\||\;|\leq\varepsilon.$$
The mapping $g$ is said to be a phase isometry provided that $\varepsilon=0$.
In this paper, we show the following universal inequality of $g$: for each $u^*\in w^*$-exp $\|u^*\|B_{X^*}$, there exist a phase function $\sigma_{u^*}: X\rightarrow \{-1,1\}$ and $\varphi$ $\in$ $Y^*$ with $\|\varphi\|= \|u^*\|\equiv \alpha$ satisfying that
$$\;\;\;\;\; |\langle u^*,u\rangle-\sigma_{u^*} (u)\langle \varphi, g(u)\rangle |\leq\frac{5}{2}\varepsilon\alpha ,\;\;{\rm for\;all\;}u\in X.$$ In particular, let $X$ be a smooth Banach space. Then we show the following:
(1) the universal inequality holds for all $u^*\in X^*$;
(2) the constant $\frac{5}{2}$ can be reduced to $\frac{3}{2}$ provided that $Y^\ast$ is strictly convex;
(3) the existence of such a $g$ implies the existence of a phase isometry $\Theta:X\rightarrow Y$ such that $\Theta(u)=\lim\limits_{n,\mathcal{U}}\frac{g(nu)}{n}$ provided that $Y^{**}$ has the $w^*$-Kadec-Klee property (for example, $Y$ is both reflexive and locally uniformly convex).

Key words: $\varepsilon$-phase isometry, phase isometry, Banach space