Let $M$ be a diffuse von Neumann algebra and $\varphi$ be a faithful normal state on $M$.
If $s,t$ are two positive elements in $M$ such that $\varphi(s)=\lambda \varphi(t)$, where $\lambda$ is the point spectrum of $\varphi$. Then we can find $\epsilon>0$ and two non-zero projections $p,q\in M$ such that $\epsilon p\leq s,\epsilon q\leq t$ and $\varphi(p)=\lambda \varphi(q)$.
Since $M$ is diffuse, there is no minimal projections in $M$, there exist projection $p\leq I$, but how to assure that $\epsilon p\leq s I$?
Best Answer
Because $s$ is not zero, there exists $\epsilon'>0$ such that $\sigma(a)\cap[\epsilon',\infty)\ne\varnothing$. Take $\epsilon''$ to be the same thing for $t$, and take any $\epsilon>0$ with $\epsilon≤\min\{\epsilon',\epsilon''\}$. Let $$p'=\ 1_{[\epsilon,\infty)}(s),\qquad\qquad q'=\ 1_{[\epsilon,\infty)}(t).$$ Then $$\epsilon p'\leq s,\qquad\qquad \epsilon q'\leq t.$$ Because $M$ is diffuse and $\varphi$ is faithful and normal, there exist subprojections $p\leq p'$ with $\varphi(p)$ any number in $(0,\varphi(p')]$. Now choose numbers $a\in(0,\varphi(p')]$ and $b\in(0,\varphi(q')]$ with $a=\lambda b$. Then let $p\leq p'$ with $\varphi(p)=a$ and $q\leq q'$ with $\varphi(q)=b$.