About what happens to eigenspace under functional calculus for Unbounded Operator

Question

Let $T$ be an unbounded self adjoint positive operator on a Hilbert Space $\mathcal{H}$. Let $x \in \mathcal{H}$ be a vector such that $Tx = x$. Is it true that $T^{\frac{1}{2}} x = x$. For what $f$ do we have that $f(T)x = x $ in general. Also, what about the same question when $T$ is just self-adjoint or normal?

Thanks a lot in advance.

Answer 1

If you know the spectral theorem $T = \int_{\sigma} \lambda dE(\lambda)$, then you should not have any problem showing that $\mathcal{N}(T-\lambda I)=E(\{\lambda\})X$. That is $T$ has an eigenvalue $\lambda$ iff $\lambda$ is an atom of the spectral measure. Then $f(T)E(\{\lambda\})=\int_{\sigma}f(t)dE(t)E(\{\lambda\})=f(\lambda)E(\{\lambda\})$ because $E(U)E(V)=E(U\cap V)$. So $Tx=\lambda x$ implies $f(T)x=f(\lambda)x$.