The Product is densely defined?

[smati]

Homework Statement

Hello,
We know that if A and B are two unbounded densely defined operators, it does not mean that AB is also densely defined. But if A is bounded then D (AB) = D (B) ie AB is densely defined.
Is AB densely defined if:
1) B is bounded and A is unbounded densely defined operator.
2) A and B are unbounded densely defined operators such that A or B is invertible of a bounded inverse.
thank you.

The Attempt at a Solution

 

[S.G. Janssens]

But if A is bounded then D (AB) = D (A) ie AB is densely defined.

I think here you mean $D(A B) = D(B)$?

3. The Attempt at a Solution

Yes, $B^{-1}$ has a dense range, which means that $B^{-1}X$ is dense (where $X$ is the underlying Banach space), but $D(A)$ is smaller than $X$ so it is not clear that $B^{-1}D(A)$ is dense as well.