When is the image of the functional calculus of a self-adjoint operator also self-adjoint?

Question

Let $A$ be a self-adjoint operator, either bounded or unbounded, and $f$ a Borel function. Using the functional calculus we may define $f(A)$ as a linear operator. Are there any known conditions on $f$ such that $f(A)$ will also be self-adjoint?

Answer 1

The condition is precisely that $f$ is real valued.

If $f$ is real valued then using the Spectral Theorem $$ f(A)^*=\bigg[\int_{\sigma(A)}f(\lambda)\,dE(\lambda)\bigg]^* =\int_{\sigma(A)}\overline{f(\lambda)}\,dE(\lambda)=\int_{\sigma(A)}f(\lambda)\,dE(\lambda)=f(A). $$ Conversely, if $f$ is not real (up to a nullset) on $\sigma(A)$, let $L=f^{-1}(\mathbb C\setminus\mathbb R)$. We have that $E(L)\ne0$. On $L$ necessarily $\overline{f(\lambda)}-f(\lambda)\ne0$. Since $i[\overline{f(\lambda)}-f(\lambda)]\ne0\in\mathbb R$, then at least one of $$ L_+=\{\lambda:\ i[\overline{f(\lambda)}-f(\lambda)]>0\}, \qquad\qquad L_-=\{\lambda:\ i[\overline{f(\lambda)}-f(\lambda)]<0\} $$ is not a nullset. Say it is $L_+$. Then there exists $n$ such that $L_n=\big\{\lambda:\ i[\overline{f(\lambda)}-f(\lambda)]>\frac1n\big\}$ is not a nullset. $$ [f(A)^*-f(A)]\,E(L_n)=\int_{L_n}[\overline{f(\lambda)}-f(\lambda)]\,dE(\lambda) $$ cannot be $0$, because there has to exist $\xi\in E(L_n)H$ such that $E_\xi(L_n)\ne0$ and hence we have $$ i\langle[f(A)^*-f(A)]\xi,\xi\rangle =i\,\int_{L_n}[\overline{f(\lambda)}-f(\lambda)]\,dE_\xi(\lambda) \geq\frac1n\,E_\xi(L_n). $$ Thus $f(A)^*\ne f(A)$ and $f(A)$ is not selfadjoint.