# Composition of bounded linear operator and an inverse of a linear operator bounded?

I've tried and failed to find an answer to this on google and elsewhere on this site.

Suppose the inverse of the linear operator $A$, $A^{-1}$ is unbounded (in the norm associated with the relevant Hilbert space e.g. L2). $B$ is a bounded linear operator. Are there simple conditions under which the composition $BA^{-1}$ is bounded? In particular I'm interested in the case where $A$ and $B$ are both integral operators with the latter having a continuous kernel.

Since $A^\prime:=A^{-1}:D(A^\prime)\to H$ (from the domain $D(A^\prime)\subseteq H$ to the Hilbert space $H$) and since $B\in \mathbb{B}(H)$ is bounded, the product $BA^\prime$ is defined on $D(BA^\prime)=D(A^\prime)$; hence it is bounded iff so is $A^\prime$.
Now, if you put $B=A\in\mathbb{B}(H)$, then, since according to your hypothesis the inverse $A^\prime$ exists, $A^\prime$ necessarily extends to a (unique) bounded operator, and the identity operator $AA^\prime=I=A^\prime A$ is everywhere defined in $H$. Indeed, by hypothesis (that the null space $N(A)=\{0\}$), the adjoint $A^*$ is also invertible, ie the null space $N(A^*)=\{0\}$ is trivial. Then it follows from $H=N(A^*)\oplus \overline{R(A)}$ that the closure of the range $\overline{R(A)}=\overline{D(A^\prime)}=H$, ie a densely defined $A^\prime$ extends uniquely to a bounded operator by Hahn-Banach thm. As an example take $A=(T-z)^{-1}$ as the resolvent, for $z$ in the resolvent set of a closed operator $T$.
• But $A^{-1} A$ is bounded even if $A^{-1}$ isn't no? – SecretlyAnEconomist Jan 28 '18 at 20:59
• Sorry I meant $AA^{-1}$ is bounded? – SecretlyAnEconomist Jan 28 '18 at 21:55