Q: Self-adjoint Compact Operators I encountered a problem in functional analysis, as follow.
Let $H$ be a Hilbert space, and $A$ and $B$ be two self-adjoint operators on $H.$ Suppose $B\ge A\ge 0,$ as bilinear forms. If $B$ is compact, then show that $A$ is also compact.
I am not sure how to use the positivity assumption, since this does not imply the relation between $||Bx||$ and $||Ax||.$ I appreciate any idea or comment!

A: Let $x_n$ be a sequence in the unit ball of $H$, then $\sqrt{A}(x_n)$ is a sequence in the ball of length $\|\sqrt A\|$, since $B$ is compact there is a sub-sequence so that $B (\sqrt{A}(x_{n_k}))$ is Cauchy. Now it follows from:
$$\|A(x_{n_k})-A(x_{n_l})\|^2 =\|A(x_{n_k}-x_{n_l})\|^2 =\langle x_{n_k}-x_{n_l}, A^2 (x_{n_k}-x_{n_l})\rangle\\
=\langle \sqrt A(x_{n_k}-x_{n_l}) , A(\sqrt{A}(x_{n_k}-x_{n_l}))\rangle ≤ \langle \sqrt{A}(x_{n_k}-x_{n_l}), B\sqrt{A}(x_{n_k}-x_{n_l})\rangle\\
≤ \|\sqrt{A}(x_{n_k}-x_{n_l})\|\,\|B\sqrt A(x_{n_k}-x_{n_l})\| ≤2\|\sqrt{A}\|\  \|B(\sqrt A(x_{n_k}))-B(\sqrt{A}(x_{n_l}))\|$$
that $A(x_{n_k})$ is also Cauchy. Hence the image of any sequence in the unit ball under $A$ admits a convergent sub-sequence, meaning that the image of the unit ball under $A$ is pre-compact. This is the definition for compactness of an operator, so $A$ is compact.
