Inequality in von Neumann algebras

Question

U. Krengel. Ergodic Theorems. 1985. Page 278.

Lemma 1.13:

Let $U$ be a von Neumann algebra. If $p\in U$ is a projection and $a,b\in U$ satisfy $0\le a\le b\le 1$, then $||ap||\le \sqrt {||bp||}$.

Proof.

$$||ap||^2 \le ||\sqrt {a}||^2 ||\sqrt {a}p||^2 \le ||pap|| \le ||pbp|| \le ||bp||$$

Everything is clear except the second inequality from left.

Answer 1

The C*-identity says $ \| a^*a\|=\|a\|^2$ so $\|\sqrt{a} p \|^2 = \|p a p\|$. Furthermore $\| \sqrt{a} \| \leq 1$ as $\sqrt{a} \leq 1$.