Spectral Measures: Uniqueness

Question

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: http://math.stackexchange.com/questions/1111983/borel-measurable-function-is-pointwise-limit-of-a-sequence-of-continuous-functio — 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-W-star, 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-W-star, Jun 10 '15 at 17:51
Someone wants bounty for free: Post any remark as answer! ;) — C-star-W-star, Jun 16 '15 at 21:11
Hi, just wondering, from which book are you learning stuff about spectral measure? — Cortizol, Jun 16 '15 at 23:20
@Cortizol: I can give you some references anyway; but please open a thread on it. (E.g. "Spectral Measures: References") — C-star-W-star, Jun 16 '15 at 23:23
@Freeze_S I don't need really reference. I was just wondering. For example, Birman and Solomyak have some nice books about it. I know also that in Rudin (Function Analysis) we can find also some stuff, but not too much. — Cortizol, Jun 16 '15 at 23:25
@Cortizol: Yep, but sadly all of them have some deficits... :( — C-star-W-star, Jun 16 '15 at 23:26
@Cortizol: By the way, do you want the bounty? (I'd be happy.) — C-star-W-star, Jun 16 '15 at 23:32
@Cortizol: It would be my first bounty to award. :) (For badges!) — C-star-W-star, Jun 16 '15 at 23:38

C-star-W-star · Accepted Answer · 2015-07-09T00:29:42.363

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.

Spectral Measures: Uniqueness

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