# Showing that $0\leq A\leq B$ and $B \in \mathcal{L}_c(H)$ implies that $A \in \mathcal{L}_c(H)$.

### Solution:

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.