# Spectral Measures: Uniqueness

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-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

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.