Dominated by compact operator implies compact?

Question

Given a Hilbert space $H$, a projection $P\in B(H)$ and a compact operator $T\in B(H)$ how does one conclude that $P\leq T$ implies $P$ compact?

I seem to recall that one can apply the fact that ideals in $B(H)$ are hereditary, but I have forgotten the steps in this reasoning.

Answer 1

Notice that if $P \le T$ then $$ \|Pu\|^2 = (Pu,Pu)= (Pu,u) \le (Tu,u) \le \|Tu\| \|u\|, \tag 1 $$ for all $u \in H$. Let $(u_n)$ be a bounded sequence in $H$. Since $T$ is compact, there is an increasing sequence of indices $(k_n)$ such that $(Tu_{k_n})$ converges. It follows from (1) that $(Pu_{k_n})$ is Cauchy in $H$ and thus converges.

Answer 2

More general statement is true. If $0\le B\le A$ and $A$ is compact, then $B$ is compact. Indeed, the assumption is equivalent to $$\|B^{1/2}x\|^2\le \langle Ax,x\rangle $$ This implies compactness of $B^{1/2}$. Hence $B$ is compact as well.