Is every projection the support of a normal state? Let $M$ be a von Neumann algebra and $p \in M$ a projection.
Does there exist a normal state $\varphi \in M_*$ such that $s(\varphi)=p$?
Here $s(\varphi)$ is the smallest projection $e$ with $\varphi = e\varphi = \varphi e$.
I find it hard to start proving this, mainly because I don't know how to 'calculate' $s(\varphi)$ for a normal functional $\varphi$. If necessary, I can assume that the projection $p$ is $\sigma$-finite. Any help, hints, comments, etc are highly appreciated!
 A: There is a normal state with support $p$ if and only if $p$ is $\sigma$-finite. First note that there is a $1$-to-$1$-correspondence between normal states with support $p$ and faithful normal states on $pMp$:
If $\phi$ is a state with support $p$, then the restriction $\psi$ to $pMp$ is a normal state. If $q\in pMp$ is a projection, then $q\leq p$. Hence $1=\psi(q)=\phi(q)$ only if $p=q$ since $p$ is the support of $\phi$. Thus $\psi$ has support $p$, that is, $\psi$ is faithful.
Conversely, if $\psi$ is a faithful normal state on $pMp$, let $\phi\colon M\to\mathbb C,\,x\mapsto\psi(pxp)$. This is clearly a normal state. Now let $q\in M$ be a projection with $1=\phi(q)=\psi(pqp)$. Then $0=\psi(p-pqp)$ and $pqp\leq p$ imply $q=pqp\in pMp$. Since $\psi(q)=\phi(pqp)=1$, we have $q=p$ since $\psi$ is faithful.
To conclude, all you need to know is that $pMp$ is $\sigma$-finite if and only if $p$ is and a von Neumann algebra admits a faithful normal state if and only if it is $\sigma$-finite.
