This thread is only Q&A!
(See the hint: SE: Q&A)
Given a Hilbert space $\mathcal{H}$.
Consider a spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$
Regard the domain: $$\int|f(\lambda)|^2\mathrm{d}\|E(\lambda)\varphi\|^2<\infty$$
And the calculus: $$\langle f(E)\varphi,\chi\rangle=\int_\mathbb{C} f(\lambda)\mathrm{d}\langle E(\lambda)\varphi,\chi\rangle$$
Then one has: $$f(E)^*=\overline{f}(E)$$
How to prove this?
Regard the sets: $$\Omega_n:=\{|f|\leq n\}\in\mathcal{B}(\mathbb{C})$$
Denote for shorthand: $$1_n:=\chi_{\Omega_n}\quad f_n:=f1_n$$
By dominated convergence: $$\langle f(E)\varphi,\psi\rangle=\lim_n\langle f_n(E)\varphi,\psi\rangle=\lim_n\langle\varphi,\overline{f_n}(E)\psi\rangle=\langle\varphi,\overline{f}(E)\psi\rangle$$
They extend by: $$1_n(E)f(E)^*=1_n(E)^*f(E)^*\subseteq(f(E)1_n(E))^*=f_n(E)^*=\overline{f_n}(E)$$
By monotone convergence: $$\int|\overline{f}|^2\mathrm{d}\nu_\psi=\lim_n\int|\overline{f_n}|^2\mathrm{d}\nu_\psi=\lim_n\|\overline{f_n}(E)\psi\|^2\\ =\|1_n(E)f(E)^*\psi\|^2\leq\|f(E)^*\psi\|^2<\infty$$
Concluding the assertion.