Given a Hilbert space $\mathcal{H}$.

Consider spectral measures: $$E^{(\prime)}:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$

Denote their operators by: $$M^{(\prime)}:=\int\lambda\mathrm{d}E^{(\prime)}(\lambda)$$ Then one has: $$M=M'\iff E=E'$$

How can I prove this?

  • Use the Stone-Weierstraß Theorem and this: math.stackexchange.com/questions/1111983/… – PhoemueX Jan 20 '15 at 19:45
  • @PhoemueX: You mean building up the whole measurable calculus again? (Note also that the measure may have unbounded support.) – C-Star-Puppy Jan 20 '15 at 19:52
  • Oh, ok... The unbounded support could be a problem for this argument. Apart from that: The existence of the Borel calculus for a spectral measure should be pretty clear, should it not? – PhoemueX Jan 20 '15 at 19:56
  • @PhoemueX: Sure the Borel calculus exists but does equality of the associated operator really imply equality of its spectral measures? – C-Star-Puppy Jun 10 '15 at 17:51
  • Someone wants bounty for free: Post any remark as answer! ;) – C-Star-Puppy Jun 16 '15 at 21:11
up vote 0 down vote accepted

I finally found a proof. :D

Consider the projections: $$N_\Re:=\frac{1}{2}\{N+N^*\}=\frac{1}{2}\{N'+N'^*\}=:N'_\Re$$ $$N_\Im:=\frac{1}{2i}\{N-N^*\}=\frac{1}{2i}\{N'-N'^*\}=:N'_\Im$$

Their resolvents agree: $$R_\alpha(z)=(z-N_\alpha)^{-1}=(z-N'_\alpha)^{-1}=R'_\alpha(z)$$

By Stone's formula: $$E_\alpha(-\infty,\lambda_\alpha]\varphi=\lim_{\delta\to0^+}\lim_{\varepsilon\to0^+}\int_{-\infty}^{\lambda_\alpha+\delta}\Delta R_\alpha(s\pm i\varepsilon)\varphi\mathrm{d}s\\ =\lim_{\delta\to0^+}\lim_{\varepsilon\to0^+}\int_{-\infty}^{\lambda_\alpha+\delta}\Delta R'_\alpha(s\pm i\varepsilon)\varphi\mathrm{d}s=E'_\alpha(-\infty,\lambda_\alpha]\varphi$$

By construction one has: $$E^{(\prime)}_\Re(A)=E^{(\prime)}(A\times\mathbb{R})\quad E^{(\prime)}_\Im(B)=E^{(\prime)}(\mathbb{R}\times B)$$

Therefore one obtains: $$E(-\infty,\lambda]=E\bigg((-\infty,\lambda_\Re]\times\mathbb{R}\bigg)E\bigg(\mathbb{R}\times(-\infty,\lambda_\Im]\bigg)\\ =E'\bigg((-\infty,\lambda_\Re]\times\mathbb{R}\bigg)E'\bigg(\mathbb{R}\times(-\infty,\lambda_\Im]\bigg)=E'(-\infty,\lambda]$$

Concluding the assertion.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.