A question on a lemma in von Neumann algebra. Let $\mathfrak{U}$ be a von Neumann algebra, the lemma says that:
If $p\in \mathfrak{U}$ is a projection and $a,b \in \mathfrak{U}$ s.t $0\leq a \leq b \leq 1$, then: $\| ap \| \leq \| bp\|^{1/2}$
In the proof we have the next two inequalities:
$$\| ap \|^2 \leq \| a^{1/2} \|^2 \| a^{1/2} p \|^2 \leq \| pap \|$$
I don't understand how did they get the second inequality:
$\| a^{1/2} \|^2 \| a^{1/2} p \|^2 \leq \| pap \|$.
Your help is appreciated.

We need to use that $0\leq x\leq 1\Rightarrow ||x||\leq 1$ and that the positive operators are self-adjoint. So, $0\leq a\leq 1\Longrightarrow$ $0\leq a^{1/2}\leq 1\Longrightarrow$ $a^{1/2}=(a^{1/2})^*$ and $||a^{1/2}||\leq 1$. Then the C* identity says $||x||^{2}=||x^{*}x||$, and projections are self-adjoint too. So you have $||a^{1/2}||^{2}||a^{1/2}p||^{2}$ $\leq ||a^{1/2}p||^{2}$ $=||(a^{1/2}p)^{*}a^{1/2}p||$ $=||p^{*}(a^{1/2})^{*}a^{1/2}p||$ $=||pa^{1/2}a^{1/2}p||=||pap||$.
