TITLE: Can the boundedness of $A^2$ imply the boundedness of $A$?
QUESTION [5 upvotes]: Given an operator $A \in \mathcal L(B)$, $B$ being a Banach space, I came across the following question: assume $\mathrm{dom}(A)=\mathrm{range}(A)$, $\mathrm{dom}(A)$ dense in $B$.
Under which conditions is it possible to obtain the boundedness of $A$ from the boundedness of $A^2$. It is clear that "in most" cases this is not possible but maybe it works for something more than the identity?

REPLY [4 votes]: Your first query has been answered, but not the second.  We make the simple remark that it is the case when $A$ is a self-adjoint (even normal) operator on Hilbert space by the spectral theorem.  Presumably this can be extended to operators on general Banach spaces with good spectral properties (spectral operators, operators of scalar type---see the third volume of Dunford and Schwartz or Dowson's monograph for this).