Let $A$ be a bounded and boundedly invertible operator and $B$ an unbounded operator such that $BAB^{-1}$ is bounded and boundedly invertible. Can I conclude that $BA^{-1}B^{-1}$ is bounded?
Basically, the question seems to boil down to whether $(BAB^{-1})^{-1} = BA^{-1}B^{-1}$ still holds under the given assumptions.
$$(BA^{-1}B^{-1}) \cdot (BAB^{-1}) = BA^{-1}AB^{-1} = B B^{-1} = 1 $$ $$(BAB^{-1}) \cdot (BA^{-1}B^{-1}) = BAA^{-1}B^{-1} = B B^{-1} = 1 $$ Since it satisfies the definition, $BA^{-1} B^{-1}$ is the inverse of $BAB^{-1}$.