Can the boundedness of $A^2$ imply the boundedness of $A$? 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?
 A: I can give a counterexample where $A:H\to H$ is a linear, bijective, unbounded self map of a Hilbert space with $A^2=\text{id}$.
Something similar can probably be done for many other Banach spaces as well, but I hope this gives a flavor of what might happen.
It might be that no counterexamples exits in somehow reasonable cases, but I don't know what reasonable should mean here.
What I do know is that this construction is not very reasonable, at least from the metric point of view.
Let $H$ be an infinite dimensional separable Hilbert space and $B$ a Hilbert basis (orthonormal, of course).
Write $B$ as a disjoint union of two countable sets: $B=\{a_i;i\in\mathbb N\}\cup\{b_i;i\in\mathbb N\}$.
($0\notin\mathbb N$.)
Now extend $B$ to a Hamel basis $C$ of $H$.
Define $A(a_i)=ib_i$, $A(b_i)=a_i/i$ and $A(x)=x$ for $x\in C\setminus B$ and extend linearly.
It is easy to see that $A:H\to H$ is unbounded and bijective and that $A^2=\text{id}$.
Note: If you restrict $A$ to the span of $B$ (Hamel span, without the closure), you get a more naturally defined operator with dense domain and range.
A: 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).
