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 $k＞2$.
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3$\begingroup$ It is rather confusing to have this question be repeatedly editted, which has now invalidated the (previously correct) answer below. I am also curious about the "type $III_1$ factor" hypothesis: can you comment as to why you wish to make this assumption? $\endgroup$– Matthew DawsCommented Sep 1, 2021 at 18:48
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1 Answer
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Since $\rho'$ is faithful, and $p$ is positive (as it's a projection), $\rho'(p)=0$ implies that $p=0$. Thus you can take $\rho=\rho'$
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3$\begingroup$ But the whole point of this answer is that it's impossible to have $\rho'(p)=0$ and $\rho'$ be faithful, unless $p=0$... $\endgroup$ Commented Aug 13, 2021 at 19:17
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$\begingroup$ @mathbeginner: As Matthew says, the answer to your edited question is therefor "No". $\endgroup$ Commented Aug 13, 2021 at 22:14