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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?

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

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