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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: $$E:\mathcal{B}(\mathbb{R})\to\mathcal{B}(\mathcal{H}):\quad H=\int\lambda\operatorname{dE(\lambda)}$$

Denote its resolvent by: $$R(z):=(H-z)^{-1}:=\int\frac{1}{z-\lambda}\operatorname{dE(\lambda)}$$

Regard the strong integral: $$\frac{1}{2\pi i}\int_{-\infty}^\infty\{R(s+i\varepsilon)-R(s-i\varepsilon)\}\varphi\operatorname{ds}$$ How can I check integrability?

The weak version is definitely integrable: $$\int_{-\infty}^\infty|\langle\{R(s+i\varepsilon)-R(s-i\varepsilon)\}\varphi,\chi\rangle|\operatorname{ds}=\int_{-\infty}^\infty\left|\frac{2\varepsilon}{(s-\lambda)^2+\varepsilon^2}\right|\operatorname{ds}\operatorname{d|\mu_{\varphi\chi}|(\lambda)}\leq2\pi\|\varphi\|\cdot\|\chi\|$$

The strong version is possibly integrable: $$\int_{-\infty}^\infty\|\{R(s+i\varepsilon)-R(s-i\varepsilon)\}\varphi\|\operatorname{ds}=\int_{-\infty}^\infty\sqrt{\int_{\sigma E}\left|\frac{2\varepsilon}{(s-\lambda)^2+\varepsilon^2}\right|^2\operatorname{d|\mu_{\varphi\chi}|(\lambda)}}\operatorname{ds}<\infty?$$

The uniform version is generally not integrable: $$\int_{-\infty}^\infty\|\{R(s+i\varepsilon)-R(s-i\varepsilon)\}\|\operatorname{ds}=\int_{-\infty}^\infty\left\|\frac{2\varepsilon}{(s-\lambda)^2+\varepsilon^2}\right\|_{\lambda\in\sigma E=\mathbb{R}}\operatorname{ds}=\int_{-\infty}^\infty\frac{1}{\varepsilon}\operatorname{ds}=\infty$$

Still missing the strong version?

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Let $(Mf)(x)=xf(x)$ be the multiplication operator on its natural domain in $L^2(\mathbb{R})$. Then $$ \|(\lambda I-M)^{-1}f\|^2 = \int_{\mathbb{R}}\frac{1}{|\lambda -s|^2}|f(s)|^2ds. $$ As an example, take $f(s)=1/\sqrt{1+s^2}$. For this $f$ and for $\Im\lambda > 0$, \begin{align} \|(\lambda I-M)^{-1}f\|^2 & =\frac{1}{2i\Im\lambda}\int_{-\infty}^{\infty} \frac{2i\Im\lambda}{|s-\lambda|^2}\frac{1}{1+s^2}ds \\ & = \frac{1}{(2i)^2\Im\lambda}\int_{-\infty}^{\infty}\left[\frac{1}{s-\lambda}-\frac{1}{s-\overline{\lambda}}\right]\left[\frac{1}{s-i}-\frac{1}{s+i}\right]ds \\ & = \frac{1}{4\Im\lambda}\int_{-\infty}^{\infty}\frac{1}{(s-\lambda)(s+i)}+\frac{1}{(s-\overline{\lambda})(s-i)}ds \\ & = \frac{2\pi i}{4\Im\lambda}\left[\frac{1}{\lambda+i}+\frac{1}{i-\overline{\lambda}}\right] \\ & = -\frac{2\pi i}{4\Im\lambda}\frac{2i(1+\Im\lambda)}{|\lambda+i|^2} \\ & = \frac{\pi}{\Im\lambda}\frac{1+\Im\lambda}{\Re\lambda^2+(1+\Im\lambda)^2} \end{align} For this example, and for this $f$, $$ \int_{-\infty}^{\infty}\|(\lambda I-M)^{-1}f\|d\Re\lambda = \infty. $$

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  • $\begingroup$ I needed some time to think about the proof - I'm sorry for the late reply. :/ $\endgroup$ – C-Star-W-Star Apr 15 '16 at 10:59
  • $\begingroup$ Now, it proves that $R(s\pm i\varepsilon)\varphi$ fail to be integrable. But it doesn't disprove (yet) the integrability of $\{R(s+i\varepsilon)-R(s-i\varepsilon)\}\varphi$. As a (little stupid) example take $f(z):=\Re\{z\}$. Then $f(s\pm i\varepsilon)$ both fail to be integrable but $f(s+i\varepsilon)-f(s-i\varepsilon)\equiv0$ is trivially integrable. $\endgroup$ – C-Star-W-Star Apr 15 '16 at 11:01
  • $\begingroup$ Conversely, I used your example operator together with the integrable vector. Mathematica reveals $\int_{-\infty}^\infty \left|\frac{1}{\lambda-(s+i1)}-\frac{1}{\lambda-(s-i1)}\right|^2 \frac{1}{\lambda^2+1}\operatorname{d\lambda}=\frac{\pi}{2} \frac{s^2+12}{(s^2+4)^2}$. Thus $\frac{\sqrt{s^2+12}}{(s^2+4)}\geq\frac{|s|}{s^2+4}\approx\frac{1}{|s|}$ is not integrable. $\endgroup$ – C-Star-W-Star Apr 15 '16 at 11:20

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