Unbounded self-adjoint "nth-root"operators

Question

Suppose we have a self-adjoint operator on a separable Hilbert space $H$ such that there exist $n>1$ that $T^n=1$ on $\mathrm{Dom}(T^n)$. Does it mean that $T$ is bounded ? If not, what are the minimal assumptions that implies this ?

Answer 1

The operator $T$ would be closed and densely-defined to qualify as selfadjoint. If $T^n=I$ on $\mathcal{D}(T^n)$, then the following holds on $\mathcal{D}(T^n)$, which is a dense subspace for any selfadjoint $T$: $$ (1-\lambda^n)I=T^n - \lambda^n I = (T-\lambda I)(T^{n-1}+\lambda T^{n-2}+\cdots + \lambda^{n-1}I). $$ Because $T$ is selfadjont, then $T-\lambda I$ is invertible for non-real $\lambda$, and the resolvent $R(\lambda)=(T-\lambda I)^{-1}$ must satisfy $$ (1-\lambda^n)R(\lambda)=T^{n-1}+\lambda T^{n-2}+\cdots +\lambda^{n-1}I, \;\; \mbox{on}\; \mathcal{D}(T^n). $$ Differentiating $n-2$ times gives $$ \frac{d^{n-2}}{d\lambda^{n-2}}((1-\lambda^n)R(\lambda)x)=(n-2)!Tx+\lambda(n-1)!x,\;\;\; x\in\mathcal{D}(T^{n}). $$ The left side is a bounded operator for any non-real $\lambda$ because the resolvent and all of its derivatives are bounded operators. So $T$ is bounded on $\mathcal{D}(T^n)$. Because $T$ is closed and densely-defined, then $T$ is bounded with domain $H$.