Prove that the domain of unbounded self-adjoint operator is strictly larger than that of its square

176 Views Asked by At

This is the problem from the textbook Unbounded Self-adjoint Operators on Hilbert Space by Konrad Schmudgen

Let $A$ be a self-adjoint operator which is not bounded. Show that there exists a vector $x \in \mathcal{D}(A)$ such that $x \notin \mathcal{D}\left(A^2\right)$.

If $A$ is bounded, it seems like $x \in \mathcal{D}(A)$ implies $x \notin \mathcal{D}\left(A^2\right)$. However, I don't know how to solve this problem. Should I find a counterexample or show this using the standard inner product technique?