If $0\le a\le p$ in a von Neumann algebra then $a \in pMp$

Question

Let $M$ be a von Neumann algebra over a Hilbert space $H$ and $a \in M$ such that $a \ge 0$. Let $p\in M$ be a projection such that $a \le p$. I want to show that $a \in pMp$.
I know that, if $q\le p$ is a projection in $M$, then $q=pqp \in pMp$. But I am unable to do for any general positive element $a \in M$ such that $a \le p$. Please help me to solve this. Thank you for your time.

Answer 1

In general, if $A$ is a C*-algebra and $p$ is a projection then $pAp$ is hereditary.

If $0 \leq a \leq p$, then $0 \leq (1-p)a(1-p) \leq (1-p)p(1-p) = 0$, so $(1-p)a(1-p) = 0$. Hence $\|\sqrt{a}(1-p)\|^2 = \|(1-p)a(1-p)\| = 0$, so $a(1-p) = 0 = (1-p)a$ and $a = pap \in pAp$.