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 $T$.

  • $\begingroup$ But $A^{-1} A$ is bounded even if $A^{-1}$ isn't no? $\endgroup$ – SecretlyAnEconomist Jan 28 '18 at 20:59
  • $\begingroup$ Sorry I meant $AA^{-1}$ is bounded? $\endgroup$ – SecretlyAnEconomist Jan 28 '18 at 21:55
  • $\begingroup$ Thanks for the edit! Sorry to keep pestering you but I'm still confused. Doesn't the Hahn-Banach theorem require the densely defined linear operator be bounded? And the densely defined inverse is not necessarily bounded no? $\endgroup$ – SecretlyAnEconomist Jan 30 '18 at 19:31
  • $\begingroup$ The answer to the 1st question is no. Densely defined means unbounded, in general. $\endgroup$ – rytis_j Jan 30 '18 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.