Exercise :
Let $H$ be a Hilbert space and $A,B \in \mathcal{L}(H)$ be self-adjoint operators with $0 \leq A \leq B$ and $B \in \mathcal{L}_c(H)$. Show that $A \in \mathcal{L}_c(H)$.
Thoughts :
Relying only on the definition of a compact operator, we essentialy need to conclude that $A$ transfers bounded sets to relatively compact sets (compact closure).
Now, since $B$ is compact and self adjoint, I know that also $B^*B$ is compact. This may be of use since the property of $A$ and $B$ being self adjoint is noted in the exercise.
I think that $A \leq B \implies \|A\| \leq \|B\|$ since they are both bounded and we could take $\mathbf{1} \in H$ which yields that
$$\|A(\mathbf{1})\| \leq \|A\|\|1\| \equiv \|A\| \quad \text{and} \quad \|B(\mathbf{1})\| \leq \|B\|\|1\| \equiv \|B\|$$
and since $0 \leq A \leq B$ implies that their values follow the inequality for any $x \in H$ thus the implied result.
Request : Beyond these points, I sadly do not have an intuition for a head-start, so I would really appreciate any hints or elaborations.
Best Answer
Because $B-A \ge 0$, then $[x,y]=\langle (B-A)x,y\rangle$ is a pseudo inner product, lacking only positive definiteness. As such, the Cauchy-Schwarz inequality holds $$ |[x,y]|^2 \le [x,x][y,y] \\ |\langle (B-A)x,y\rangle|^2 \le \langle (B-A)x,x\rangle\langle (B-A)y,y\rangle. $$ Now set $y=(B-A)x$ in the above to obtain $$ \|(B-A)x\|^4 \le \langle (B-A)x,x\rangle\langle(B-A)(B-A)x,(B-A)x\rangle \\ \|(B-A)x\|^4 \le \langle (B-A)x,x\rangle\|B-A\|\|(B-A)x\|^2 \\ \|(B-A)x\|^2 \le \|B-A\|\langle(B-A)x,x\rangle \\ \|(B-A)x\|^2 \le \|B-A\|\langle Bx,x\rangle. $$ Suppose $\{ x_n \}$ is a bounded sequence. Because $B$ is compact, there is a subsequence $\{ x_{n_k} \}$ such that $\{ Bx_{n_k} \}$ converges. By the above, $\{ (B-A)x_{n_k}\}$ is a Cauchy sequence and, hence, converges to some $y$. But $\{ Bx_{n_k} \}$ also converges. Therefore $\{ Ax_{n_k} \}$ converges, leading to the conclusion that $A$ is compact.