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Assumptions: Suppose $T$ is a self-adjoint operator (possibly unbounded) from the (dense) domain $D(T)$ on a Hilbert space $H$, hence $T:D(T)\rightarrow H$. Assume that $f$ is a continuous function on the spectrum of $T$, i.e. $f\in C(\sigma(T))$.

Question: We can define $f(T)$ (via spectral theorems), but what is the domain of $D(f(T))$, what is a good definition here. In particular, if $f$ is real valued, then $f(T)$ is self-adjoint (also by spectral theorems) but on which domain?

Thanks a lot.

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  • $\begingroup$ $D(T^2)$ will be a common core for $f(T)$ and $T$. The general description is given in the answer below by Yurii. $\endgroup$ – voldemort Feb 26 '15 at 0:27
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$D(f(T))$ consists of those vectors $\varphi\in H$ such that $\int|f|^2d\mu_\varphi<\infty,$ where $\mu_\varphi(\cdot)=\langle E_T(\cdot)\varphi,\varphi\rangle$ is the spectral measure of associated with $\varphi$ and $E_T$ is the projection-valued measure of $T.$

Note, that $\varphi\in D(f(T))$ iff $$||f(T)\varphi||^2=\langle f(T)\varphi,f(T)\varphi\rangle=\langle f^*f(T)\varphi,\varphi\rangle=\int|f(\lambda)|^2d\langle E_T(\lambda)\varphi,\varphi\rangle<\infty$$

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  • $\begingroup$ Can one avoid the spectral measures? $\endgroup$ – Arnetta Barberio Feb 25 '15 at 10:12
  • $\begingroup$ How do you define $f(T)$ then? $\endgroup$ – Yurii Savchuk Feb 25 '15 at 10:16

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