Skip to main content
deleted 9 characters in body
Source Link

Let $M$ be a type III$_1$ factor and $\rho$ be a faithful normal state on $M$. If $p$ is a projection in $M$, can we find another normal state $\rho'$ on $M$ such that $\rho'(p)=0$ and $\|\rho-\rho'\|=k\rho(p)$ for some $k>2$.

Let $M$ be a type III$_1$ factor and $\rho$ be a faithful normal state on $M$. If $p$ is a projection in $M$, can we find another normal state $\rho'$ on $M$ such that $\rho'(p)=0$ and $\|\rho-\rho'\|=k\rho(p)$ for some $k>2$.

Let $M$ be a type III$_1$ factor and $\rho$ be a normal state on $M$. If $p$ is a projection in $M$, can we find another normal state $\rho'$ on $M$ such that $\rho'(p)=0$ and $\|\rho-\rho'\|=k\rho(p)$ for some $k>2$.

edited title
Link

The norm of the difference of two faithful normal states

deleted 9 characters in body
Source Link

Let $M$ be a type III$_1$ factor and $\rho$ be a faithful normal state on $M$. If $p$ is a projection in $M$, can we find another faithful normal normal state $\rho'$ on $M$ such that $\rho'(p)=0$ and $\|\rho-\rho'\|=k\rho(p)$ for some $k>2$.

Let $M$ be a type III$_1$ factor and $\rho$ be a faithful normal state on $M$. If $p$ is a projection in $M$, can we find another faithful normal state $\rho'$ on $M$ such that $\rho'(p)=0$ and $\|\rho-\rho'\|=k\rho(p)$ for some $k>2$.

Let $M$ be a type III$_1$ factor and $\rho$ be a faithful normal state on $M$. If $p$ is a projection in $M$, can we find another normal state $\rho'$ on $M$ such that $\rho'(p)=0$ and $\|\rho-\rho'\|=k\rho(p)$ for some $k>2$.

added 7 characters in body
Source Link
Loading
Source Link
Loading