Suppose $\mathscr{H}_{1}, \mathscr{H}_{2}$ are both Hilbert spaces, $T: \mathscr{H}_{1} \to \mathscr{H}_{1}$ and $S: \mathscr{H}_{2} \to \mathscr{H}_{2}$ are both linear self-adjoint (possibly unbounded). Suppose there is a unitary map $U: \mathscr{H}_{1} \to \mathscr{H}_{2}$ such that $T = U^{*}SU$, that is, $T$ is unitarily equivalent to $S$. My question is: given a continuous function $f: \mathbb{R} \to \mathbb{C}$, does it follow that $f(T) = U^{*}f(S)U$? I know this result holds for bounded Borel measurable functions, but I don't know about continuous functions.
This seems just an easy application of the functional calculus, but I am not sure. For instance, if $f$ is continuous then it is measurable and one can approximate it by simple functions, which are bounded and measurable. Let $\{f_{n}\}_{n\in \mathbb{N}}$ be such simple functions, such that $f_{n} \to f$, with $f$ continuous. Then, by the functional calculus $f_{n}(T) \to f(T)$, $f_{n}(S) \to f(S)$ and, because $f_{n}(T) = U^{*}f_{n}(S)U$, we get $f(T) = f(S)$. Is my argument correct?
Let $E$ be the spectral measure of $T$ and $F$ the spectral measure of $S$. Then we have $E(A) = U^* F(A) U$ for any measurable $A\subset \mathbb{R}$.
This is because for any $x$ in the domain of $T$ and $y \in \mathcal{H}_1$: $$ \int \lambda d \langle U^* F(\lambda) U x ,y \rangle = \int \lambda d \langle F(\lambda) Ux, Uy \rangle = \langle SU x , Uy \rangle = \langle U^* SU x , y \rangle = \langle Tx, y\rangle = \int \lambda d \langle E(\lambda) x ,y \rangle . $$ The assertion follows since the above uniquely specifies the spectral measure.