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

Consider a Hamiltonian: $$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$

And its spectral measure: $$E:\mathcal{B}(\mathbb{R})\to\mathcal{B}(\mathcal{H}):\quad H=\int\lambda\mathrm{d}E(\lambda)$$

Denote its resolvent: $$R(z):=(z-H)^{-1}\in\mathcal{B}(\mathcal{H})$$

And singular spectrum: $$\sigma_\perp(H):=\sigma(H_\perp):=\sigma(J_\perp^*HJ_\perp)$$

Denote for shorthand: $$\|\chi\|:=\|\chi_\lambda\|_{\lambda\in\Lambda}:=\sup_{\lambda\in\Lambda}\|\chi_\lambda\|$$

Regard an open subset: $$\Delta\in\mathcal{B}(\mathbb{R}):\quad\Delta=\Delta^\circ$$

That bounds resolvent: $$\|\langle R(s+i\varepsilon)\varphi,\varphi\rangle\|_{s\in\Delta}^{|\varepsilon|<1}<\infty\quad(\varphi\in\mathcal{H})$$

Then it is regular: $$\sigma_\perp(H)\cap\Delta=\varnothing$$

How can I prove this?

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  • $\begingroup$ Please avoid too many edits unless absolutely necessary. $\endgroup$ – Pedro Tamaroff Jul 16 '15 at 0:19
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Meanwhile I got it...

Preparation

By reducibility one has: $$\sigma_\perp(H)\cap\Delta=\varnothing\iff E_\perp(\Delta)=0\iff P_\perp E(\Delta)=0$$

For the Borel measures: $$\nu_{E(\Delta)\varphi}(A)=\langle E(A\cap\Delta)\varphi,\varphi\rangle=\nu_\varphi(A\cap\Delta)$$

So one must check that: $$\lambda(N)=0\implies\nu_\varphi(N\cap\Delta)=0$$

Construction

Denote for readability: $$R_\Delta^\varphi:=\|\langle R(s+i\varepsilon)\varphi,\varphi\rangle\|_{s\in\Delta}^{|\varepsilon|<1}<\infty$$

By regularity one finds: $$N\cap\Delta\subseteq U_\varepsilon\subseteq\Delta:\quad\lambda(U_\varepsilon)<\varepsilon$$

By second countability:* $$U_\varepsilon=\bigcup_{k=1}^\infty\overline{B_k}:\quad B_k\cap B_{k'}\stackrel{k\neq k'}{=}\varnothing$$

By Stone's formula one has: $$\langle E(\overline{B_k})\varphi,\varphi\rangle\leq\frac{1}{2\pi}R_\Delta^\varphi\lambda(\overline{B_k})=\frac{1}{2\pi}R_\Delta^\varphi\lambda(B_k)$$

Summing all these up gives: $$\nu_\varphi(N\cap\Delta)\leq\langle E(U_\varepsilon)\varphi,\varphi\rangle\leq\sum_{k=1}^\infty\langle E(\overline{B_k})\varphi,\varphi\rangle\\ \leq\frac{1}{2\pi}R_\Delta^\varphi\sum_{k=1}^\infty\lambda(B_k)=\frac{1}{2\pi}R_\Delta^\varphi\lambda(U_\varepsilon)<\frac{1}{2\pi}R_\Delta^\varphi\varepsilon$$

Concluding the assertion.

*Real line: Disjunction gives intervals!

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