Compactness of the product of a compact operator and an unbounded one

Question

I remember the following result that many years ago I was able to show that the following claim is valid:

Let $X$ be some Hilbert space with norm denoted by $\left\|\;\!\cdot\;\!\right\|_{X}$. Let the operators $A:{D}\left(A\right)\rightarrow X$ and $B:X\rightarrow X$, where $B$ is compact, possibly of finite rank, and $A$ is boundedly invertible, but, by definition, not defined on the whole of the underlying space. Then $BA$ is compact if and only it is bounded in the sense that, for all $x\in{D}\left(A\right)$, \begin{equation} \left\|BAx\right\|_{X}\leq c\left\|x\right\|_{X} \end{equation} for some real number $c$.

I cannot reproduce the proof, so I am not even sure anymore if the claim is entirely true. The result played a role in analysing the spectral properties of ordinary differential operators.

Answer 1

False as stated. Let $X$ be $\ell^2$, and let $B$ be the multiplication by $1/n$; that is $Bx = (x_1/1, x_2/2, x_3/3,\dots)$. This is an injective compact operator. Let $A=B^{-1}$, defined on $\operatorname{ran}B$ (which is a dense subspace of $\ell^2$). All of the assumptions are satisfied, and in particular $BAx=x$ for all $x\in D(A)$. But $BA$ is not compact, since it's just the identity restricted to $D(A)$.