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This thread is only 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 equivalence holds: $$\|f(E)\|<\infty\iff\|f\|_\infty<\infty$$

Especially one has: $$\mathcal{D}f(E)=\mathcal{H}:\quad\|f(E)\|=\|f\|_\infty$$

How to prove this?

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Suppose one has: $$\vartheta:=\|f(E)\|<\infty$$

Regard the sets: $$\Omega_n:=\{\vartheta+\frac{1}{n}\leq|f|\leq\vartheta+n\}\in\mathcal{B}(\mathbb{C})$$

And their union: $$\Omega:=\bigcup_{n=1}^\infty\Omega_n=\{|f|>\vartheta\}\in\mathcal{B}(\mathbb{C})$$

Regard the functions: $$1_n:=\chi_{\Omega_n}:\quad 1_n(E)=E(\Omega_n)$$

Denote for readability: $$f_n:=f1_n:\quad\|f_n\|_\infty<\infty$$

For them one has: $$\mathcal{D}f_n(E)=\mathcal{H}:\quad f_n(E)=f(E)1_n(E)$$

So one obtains inequality: $$\left(\vartheta+\tfrac{1}{n}\right)^2\|E(\Omega_n)\varphi\|^2=\int\left(\vartheta+\tfrac{1}{n}\right)^21_n^2\mathrm{d}\nu_\varphi\\ \leq\int|f|^21_n^2\mathrm{d}\nu_\varphi=\|f_n(E)\varphi\|^2=\|f(E)1_n(E)\varphi\|^2\leq\vartheta^2\|E(\Omega_n)\varphi\|^2$$

That only holds for: $$E(\Omega_n)=0\implies E(\Omega)=0$$

So it was bounded: $$E\{|f|>\vartheta\}=E(\Omega)=0$$

Conversely one has: $$\|f(E)\varphi\|^2=\int|f|^2\mathrm{d}\nu_\varphi\leq\|f\|_\infty\|\varphi\|<\infty$$

Therefore it holds: $$\mathcal{D}f(E)=\mathcal{H}:\quad\|f(E)\|\leq\|f\|_\infty\leq\|f(E)\|$$

Concluding the assertion.

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