Let $A,B$ be two bounded operators on a Hilbert space $\mathcal{H}$ and suppose there is a densely defined injective operator $(\Lambda,\mathcal{D}(\Lambda))$ such that $$A\Lambda x=\Lambda Bx$$ for all $x\in\mathcal{G}$ where $\mathcal{G}\subsetneq\mathcal{D}(\Lambda)$ is a dense linear subspace of $\mathcal{H}$. Further assume that the range of $\Lambda$ is also dense in $\mathcal {H}$. In this case, $(\Lambda,\mathcal{D}(\Lambda))$ admits an inverse $(L,\mathcal{D}(L))$ with $\mathcal{D}(L)=\operatorname{Range}(\Lambda)$ and $\operatorname{Range}(L)=\mathcal{D}(\Lambda)$. So, we can write $$LA\Lambda x=Bx$$ for all $x\in\mathcal{G}$. $B$ is bounded, hence closable and therefore $(L A\Lambda,\mathcal{G})$ is closable. Can we say $$L A\Lambda x=Bx$$ for all $x\in\mathcal{D}(\Lambda)$?
My belief is NO because I can't see why $A\Lambda x\in\mathcal{D}(L)$ for all $x\in\mathcal{D}(\Lambda)$. I don't have a counterexample though.