A question about the domain of the product of two unbounded operators As usual, the product $BA$ of two operators is defined on $D(BA)=\{x\in D(A):~Ax\in D(B)\}$.
I have some trouble understanding a notion about that. If $A$ and $B$ are two densely defined operator such that Ran$A \cap D(B)=\{0\}$, then does it follow that $D(BA)=\{0\}$? And why is it in Reed-Simon (Vol. 1) on Page 271 we see that in such case $BA$ does not have a meaning.
Many thanks for your help.
 A: In the standard theory of Banach and Hilbert space it is useful to  compose two linear operator $A:H\to H$ and $B: H\to H$ to get a new linear operator 
$BA: H\to H $. 
There are some problems when you want define the concept of operator ‘composition’ between two operators $B$ and $A$ that have different domains.
If you have two operator 
$A: D(A)\to H$ 
$B: D(B)\to H$ 
in general you can not compose them because if you want for example $BA$ you must have that for each $x \in D(A) $ then $Ax\in D(B)$.
So you want identify the maximal set of $H$ such that the composition $BA$ make sense.  This set is 
$\{x\in D(A) : Ax\in D(B)\}$
that we indicate with $D(BA)$.
So $BA: D(BA)\to H$ is a well defined function, but you want that it is a new operator of your space $H$.
Fortunately you can prove that $D(BA)$ is a subspace of $H$ and that $ BA$ is a linear map, so it make sense define the new operator $BA:D(BA)\to H$. 
There are some particular cases in which 
$D(BA)=\{0\} $, when $Ran(A)\cap D(B)=\{0\}$, so the operator is simply the zero operator.
