Is product of two compact operator on a Hilbert space is also compact?

Question

Let me first define the notion of the compact operator
Let $H$ be a Hilbert space over a field $K$ ,An Operator $A$ is said to be compact if for every bounded sequence $x_{n}\in H$ the sequence $A(x_{n})$ contains a convergent subsequence in $H$.

My question is if $A$ and $B$ are compact operators does this implies $AB$ is also a compact Operator? If not can someone please give a counter-example? Or if true please give some hint to prove it.

Answer 1

Hint: Let $(x_n) $ be a bounded sequence.

Let $Tx_{n_k}\subset Tx_n$ convergent subsequence.

Convergent sequence is bounded . Hence $S(T(x_{n_k}) $ has a convergent subsequence. Hence $ST x_n$ has a convergent subsequence.

Is this a hint or a complete solution?

H.W : A compact operator on a Hilbert space can't have bounded inverse. (Hint: Riesz's lemma)