If $p\leq q$ then $pq=p$ where p and q are projections in a Hilbert space

Question

This is from Murphy's C*algebra textbook:

I am trying to understand why 5) implies 2) because the submultiplicative inequality only goes one way. So we have $||p(1-q)||\leq ||p||||(1-q)|| \leq ||q||||1-q||$. But from here , we do not have that $||q||||1-q||\leq ||q(1-q)||$ so how do we get the final inequality?

Answer 1

Let $y= (1-q)(x)$, then we get, using $(5)$

$$ \Vert p(1-q)(x) \Vert = \Vert p (y) \Vert \leq \Vert q(y) \Vert = \Vert q (1-q)(x) \Vert = \Vert (q-q^2)(x) \Vert. $$

Answer 2

If $q(x)=0,$ then $p(x)=0.$ As $q(I-q)=0$ then $p(I-q)=0.$

Answer 3

Yet another way of looking at this: the inequality $\|px\|\leq \|qx\|$ can be written, after squaring and using that $p,q$ are projections, as $$ \langle px,x\rangle\leq\langle qx,x\rangle. $$ As this works of all $x$, this means that $p\leq q$, which is condition $(1)$. To obtain $(2)$, we can multiply on the left and right by $1-q$, to get $$ 0\leq(1-q)p(1-q)\leq (1-q)q(1-q)=0. $$ So $(1-q)p(1-q)=0$. As this can be written as $$ 0=\big[p(1-q)\big]^*\,p(1-q), $$ we get that $p(1-q)=0$.