Suppose that $M$ is a von Neumann algebra with no minimal projections. Let $p$ be a nonzero projection in $M$ and $\rho$ be a normal state on $M$.


For any $\epsilon>0$, can we find a projection $e$ in $M$ such that $0\leq e\leq p$ and $\rho(e)=\epsilon \rho(p)$?



Projections in a diffuse von Neumann algebra
I think the proof goes something like this. We need to assume that $\epsilon\le 1$ as pointed out.


Let $M$ be a von Neumann algebra with no minimal projections. Let $\rho$ be a normal state on $M$.


Claim-1: Given $\epsilon>0$, there exists a non-zero projection $p\in \text{Proj}(M)$ such that $\rho(p) <\epsilon$.


Proof. Towards a contradiction, let us assume otherwise. Then, $\inf\{\rho(p) : p \in \text{Proj}(M), p\ne 0\}>0$. Let us denote the infimum by $l$.
Given $\epsilon> 0$, we can find a non-zero projection $p\in M$ such that $l\le \rho(p) < l +\epsilon$. Since $M$ contains no minimal projections, we can find a non-zero projection $q<p$. Let us now observe that
$l \le \rho(q)$ and $0\le\rho(p-q)<l+\epsilon-l=\epsilon.$ Now, since $p-q$ is a projection, choosing $\epsilon<l$ gives us a contradiction. Note that since $q\le p$, $pq=qp=q$.


Claim-2: The map $\rho: \text{Proj}(M)\to [0,1]$ is onto.


Proof: Let $t\in [0,1]$. We want to show that there exists $q\in \text{Proj}(M)$ such that $\rho(q)=t$. We let $T=\{p\in \text{Proj}(M): \rho(p)\le t\}$. We give it the natural order, i.e., $p\le p'\iff p'-p\ge 0$. We just need to show that there is a maximal element in the set. Let $\{p\_i\}$ be an increasing net of projections (bounded above in norm by 1). Then $p\_i\to p$ in the strong operator topology. Moreover, $p$ is the supremum of $\{p\_i\}$ under the natural order. Since $\rho$ is a normal state, $\rho$ is strong operator continuous on the unit ball of $M$. Therefore, $\rho(p)=\lim\_i\rho(p\_i)\le t$. By Zorn's lemma, there exists a maximal element which we call $q$. We want to show that $\rho(q)=t$. Suppose otherwise. Let $\epsilon=t-\rho(q)$. Using claim-1, we can find $p\in \text{Proj}(M)$ such that $\rho(p)<t-\rho(q)$. This shows that $\rho(p+q)<t$. However, $p+q\ge q$, contradicting the maximality of $q$.


Now, to answer your question, for any non-zero projection $p$ and $\epsilon\le 1$, we see that $\epsilon\rho(p)\in [0,1]$ and hence, the existence of $e$ is guaranteed by Claim-1 and 2. Clearly $e\le p$. Otherwise, $\rho(e)>\rho(p)\ge \epsilon\rho(p)$ which would lead to a contradiction.


