Let $A$ be a closed (densely defined) operator on a Hilbert space $H$.


We define for a natural number $k$, the operator $A^k$ with its natural domain.


Is $A^k$ closed?



If $A$ is a closed operator, is $A^k$ closed?
Here's a counterexample (subject perhaps to what you consider "natural").
Take a separable Hilbert space with orthonormal basis $\{u\_n : n = 1, 2, \ldots\}$ and the operator $A$ defined by
$$ A u\_n = \cases{ n u\_{n+1} & if $n$ is odd\cr 0 & if $n$ is even}$$
with domain $D = \{x = \sum\_n c\_n x\_n : \sum\_n |c\_n|^2 < \infty, \sum\_{n \text{ odd}} n^2 |c\_n|^2 < \infty\}$. Then $A$ is closed and densely defined. For integers $k > 1$ I would expect the "natural" domain of $A^k$ to be contained in $D$. However, $A^k x = 0$ where it is defined, so that $A^k$ is not closed.


