(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?

• What exactly is the purpose of this? It is not a frequently asked question, is it? – wythagoras Jun 21 '15 at 17:56
• @wythagoras: Yeaah but I need these lemmata for other real problems. – C-Star-W-Star Jun 21 '15 at 17:59
• How does that answer my question? Math Stack Exchange is not a storage place for your results. – wythagoras Jun 21 '15 at 18:06
• Mhhh yaa I know :/ But see the hint: SE: Q&A – C-Star-W-Star Jun 21 '15 at 18:11
• Oh, in that case, nevermind my comments. – wythagoras Jun 21 '15 at 18:21

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: $$\int|\overline{f}|^2\mathrm{d}\nu_\psi=\lim_n\int|\overline{f_n}|^2\mathrm{d}\nu_\psi=\lim_n\|\overline{f_n}(E)\psi\|^2\\ =\|1_n(E)f(E)^*\psi\|^2\leq\|f(E)^*\psi\|^2<\infty$$