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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: $$\mathcal{N}f(E)=(0)\iff E\{f=0\}=0$$

Especially one has: $$f(E)^{-1}=f^{-1}(E)$$

How to prove this?

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By the calculus: $$f(E)E\{f=0\}=(f1_{\{f=0\}})(E)=0$$

Thus one has: $$\varphi\in\mathcal{R}E\{f=0\}\implies f(E)\varphi=0$$

Conversely regard: $$f^{-1}(\lambda):=\frac{1}{f(\lambda)}\quad(E(\lambda)\neq0)$$

By the calculus: $$f^{-1}(E)f(E)\subseteq(f^{-1}f)(E)=1(E)=1$$

But the domains agree: $$\mathcal{D}f^{-1}(E)f(E)=\mathcal{D}(f^{-1}f)(E)\cap\mathcal{D}f(E)=\mathcal{H}\cap\mathcal{D}f(E)=\mathcal{D}f(E)$$

So for the operator: $$\varphi\in\mathcal{D}f(E):\quad f^{-1}(E)f(E)\varphi=\varphi$$

Hence one obtains: $$f(E)^{-1}\subseteq f^{-1}(E)=(f^{-1}(E)^{-1})^{-1}\subseteq f(E)^{-1}$$

Concluding the assertion.

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