comparison of projections in $C^*$ algebra

Question

Suppose p is a projection and $vv^*$ is a projection such that $vv^*\leq p\leq Cvv^*$,where $C$ is a constant,how to conclude that $p=vv^*$?

Answer 1

If you have $q\leq p\leq c q$, then $$ 0=(1-q)(cq)(1-q)\geq(1-q)p(1-q)=[p(1-q)]^*p(1-q). $$ So $p(1-q)=0$. Thus $p=pq=qpq\leq q$, and so $p=q$.