Suppose $A$ and $B$ are operators on a (separable) Hilbert space $H$ and $A \leq B$. Is it true that if $B$ is compact then $A$ is compact too? If not, could you please show a counterexample?



Does positivity preserve compactness?
What you need to assume is that $|A|\leqslant B$.


Assume first that for all $x$ one has $\Vert A x \Vert \leqslant \Vert B x \Vert$ and $B$ is compact. Then choose any net $P\_i$ of finite rank projections which converge weakly to the identity, one has $\Vert A x - A P\_i x\Vert \leqslant \Vert B x - B P\_i x\Vert$ which converge to $0$ uniformly in $x$ because $B$ is compact. hence $ AP\_i $ converge in norm to $A$ which proves that $A$ is compact.


Now if you have just $|A|\leqslant B$ with $B$ compact then you have $0 \leqslant (x , |A|x) \leqslant (x,Bx)$ which imply that $0 \leqslant \Vert A' x \Vert \leqslant \Vert B' x \Vert$ where $A'$ and $B'$ denote respectively $|A|^{1/2}$ and $B^{1/2}$. $B'$ is also compact, hence by the previous argument $A'$ is compact and hence $|A|$ is compact. Finally, for all $x$, one has $\Vert A x \Vert = \Vert |A|x \Vert$ the previous argument allow to deduce the compactity of $A$ from the compacity of $|A|$.


