If $A$ is self-adjoint operator defined in Hilbert space $\mathscr{H}$ and its resolvent family is $(\mathbb{C}, \mathscr{B}, E)$. If $D\in\mathbb{C}$ is a Borel measuable subset with smooth boundary $\Gamma = \partial D$ and $\Gamma\subset \rho(A)$. How to prove $$E(D)=\frac{1}{2\pi i}\oint_{\Gamma} (zI-A)^{-1}dz.$$

I try to use the $$\frac{1}{2\pi i}\oint_{\Gamma} \frac{1}{\xi-z}d\xi = \chi_{D}(z).$$ Note that $E^{A}(D)=\int_{\sigma(A)}\chi_{D}(z)E(dz)$, we get $$E^{A}(D)=\int_{\sigma(A)}\frac{1}{2\pi i}\Big(\oint_{\Gamma} \frac{1}{\xi-z}d\xi \Big)E^{A}(dz)=\frac{1}{2\pi i}\int_{\sigma(A)}(zI-\xi)^{-1}d\xi.$$

But how to go ahead in the next?

  • $\begingroup$ The question needs editing. It's not clear what the relationship between $D$ and $\Gamma$ is. It looks like the answer below clears this up, but it's really a question for the OP. $\endgroup$ – fourierwho Jun 13 '18 at 21:30
  • $\begingroup$ @fourierwho Sorry, I have edited it. $\endgroup$ – Tinzoe-Yui Jun 14 '18 at 3:12
  • $\begingroup$ @fourierwho smooth boundary $\Gamma = \partial D$ $\endgroup$ – Tinzoe-Yui Jun 14 '18 at 3:14

If $D$ is part of the spectrum, and if $D$ is contained in the interior of a simple closed rectifiable curve $\Gamma$ such that no other part of the spectrum is contained in the interior of or on the curve $\Gamma$, then $$ \frac{1}{2\pi i}\oint_{\Gamma} (\lambda I-A)^{-1}d\lambda \\ = \frac{1}{2\pi i}\oint_{\Gamma} \int_{\sigma(A)} \frac{1}{\lambda-\mu}dE(\mu) d\lambda \\ = \int_{\sigma(A)}\left(\frac{1}{2\pi i}\oint_{\Gamma}\frac{1}{\lambda-\mu}d\lambda\right) dE(\mu) $$ The integral in parentheses is $1$ for $\mu$ in the interior of $\Gamma$, and is $0$ for $\mu$ outside $\Gamma$. And the value of the interior integral does not matter on $\Gamma$ because $E$ is $0$ on a neighborhood of $\Gamma$. So, the above reduces to $$ \int_{\sigma(A)}\chi_{\mbox{Int}(\Gamma)}(\mu)dE(\mu)=\int_{\sigma(A)}\chi_{D}(\mu)dE(\mu) = E(D). $$

  • $\begingroup$ How about the smooth boundary $\Gamma = \partial D$? $\endgroup$ – Tinzoe-Yui Jun 14 '18 at 3:14
  • $\begingroup$ @Tinzoe-Yui : What I wrote applies to your case as well. The only real issue is that $\Gamma$ is in the resolvent set. $\endgroup$ – DisintegratingByParts Jun 15 '18 at 0:32

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