Coulomb relative compactness

[The resolvent as a Fourier multiplier] Fix $z\mathrm{\notin }\mathbb{R}$ and define $m_z(\xi ):=(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2-z{)}^{-1}$. Since $\mathrm{\mid }4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2-z\mathrm{\mid }\ge \mathrm{\mid }\mathrm{Im}z\mathrm{\mid }$ one has $\mathrm{\mid }{m}_z(\xi )\mathrm{\mid }\le \mathrm{\mid }\mathrm{Im}z{\mathrm{\mid }}^{-1}$ everywhere; and whenever $4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2\ge 2\mathrm{\mid }\mathrm{Re}z\mathrm{\mid }$ one has $\mathrm{\mid }4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2-z\mathrm{\mid }\ge 4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2-\mathrm{\mid }\mathrm{Re}z\mathrm{\mid }\ge \frac{1}{2}4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2$, so that $\mathrm{\mid }{m}_z(\xi )\mathrm{\mid }\le \frac{1}{2{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2}$. With $\rho :=\max \{1,\sqrt{2\mathrm{\mid }\mathrm{Re}z\mathrm{\mid }}\mathrm{/}(2\pi )\}>0$, so that $4{\pi }^2{\rho }^2\ge 2\mathrm{\mid }\mathrm{Re}z\mathrm{\mid }$ this yields ${\int }_{{\mathbb{R}}^3}\mathrm{\mid }{m}_z{\mathrm{\mid }}^2\le \mathrm{\mid }\mathrm{Im}z{\mathrm{\mid }}^{-2}\mathrm{\mid }B(0,\rho )\mathrm{\mid }+\frac{1}{4{\pi }^4}{\int }_{\mathrm{\mid }\xi \mathrm{\mid }>\rho }\mathrm{\mid }\xi {\mathrm{\mid }}^{-4}d\xi <\mathrm{\infty }$, because ${\int }_{\mathrm{\mid }\xi \mathrm{\mid }>\rho }\mathrm{\mid }\xi {\mathrm{\mid }}^{-4}d\xi =4\pi {\int }_{\rho }^{\mathrm{\infty }}{r}^{-2}dr$; so $m_z\in L^2({\mathbb{R}}^3)$; moreover $m_z(\xi )(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2-z)=1$ for a.e. $\xi$. Define $R_0(z):={\mathcal{F}}_2^{\ast }{M}_{m_z}{\mathcal{F}}_2$; then for $f\in L^2$ the vector $R_0(z)f$ lies in $D(H_0)$, because $\mathrm{\mid }\xi {\mathrm{\mid }}^2{m}_z(\xi ){\mathcal{F}}_{2}f(\xi )=({\mathcal{F}}_{2}f(\xi )+z{m}_z(\xi ){\mathcal{F}}_{2}f(\xi ))\mathrm{/}(4{\pi }^2)$ is a sum of two $L^2$ functions, and $(H_0-z)R_0(z)f=f$ since ${\mathcal{F}}_2((H_0-z)R_0(z)f)=(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2-z)m_z(\xi ){\mathcal{F}}_{2}f={\mathcal{F}}_{2}f$; conversely for $g\in D(H_0)$ the function $(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2-z){\mathcal{F}}_{2}g$ is in $L^2$, so $R_0(z)(H_0-z)g$ is defined and its transform is $m_z(\xi )(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2-z){\mathcal{F}}_{2}g={\mathcal{F}}_{2}g$. Hence $R_0(z)=(H_0-z{)}^{-1}$ and $\mathrm{\parallel }{R}_0(z)\mathrm{\parallel }\le \mathrm{\mid }\mathrm{Im}z{\mathrm{\mid }}^{-1}$.

[$L^{\mathrm{\infty }}$ bound of the free resolvent] For Schwartz $f$ the function $m_z{\mathcal{F}}_{2}f$ lies in $L^1\cap L^2$, so $R_0(z)f$ has the everywhere-defined representative $x\mapsto \int m_z(\xi ){\mathcal{F}}_{2}f(\xi )e^{2\pi ix\cdot \xi }d\xi$, whose modulus is at most $\mathrm{\parallel }{m}_z{\mathrm{\parallel }}_2\mathrm{\parallel }{\mathcal{F}}_{2}f{\mathrm{\parallel }}_2=\mathrm{\parallel }{m}_z{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2$ by [F7]. For general $f\in L^2$ choose Schwartz $f_n\to f$ in $L^2$; then $R_0(z)f_n\to R_0(z)f$ in $L^2$ and $\mathrm{\parallel }{R}_0(z)f_n-R_0(z)f_m{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{m}_z{\mathrm{\parallel }}_2\mathrm{\parallel }{f}_n-f_m{\mathrm{\parallel }}_2$, so the representatives converge uniformly to a bounded function $h$ with $\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{m}_z{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2$; uniform convergence gives $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^2$ convergence and hence $h=R_0(z)f$ a.e., so $\mathrm{\parallel }{R}_0(z)f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{m}_z{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2$.

[Convolution kernel] Put $g_z:={\mathcal{F}}_2^{\ast }{m}_z\in L^2$ by [F3]. We first prove $R_0(z)f=g_z\ast f$ for Schwartz $f$. Choose Schwartz functions ${\phi }_n$ with ${\phi }_n\to g_z$ in $L^2$ ([F10]). For every $n$ both ${\phi }_n$ and $f$ are Schwartz, so [F4] gives ${\mathcal{F}}_2({\phi }_n\ast f)=\widehat{{\phi }_n}\widehat{f}$; here $\widehat{{\phi }_n}\to {\mathcal{F}}_2{g}_z=m_z$ in $L^2$ by the isometry [F3], while $\mathrm{\mid }\widehat{f}(\xi )\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_1<\mathrm{\infty }$ for every $\xi$ by the defining integral of the $L^1$ transform [F3] and the integrability of the Schwartz function $f$; hence ${\mathcal{F}}_2({\phi }_n\ast f)=\widehat{{\phi }_n}\widehat{f}\to m_z\widehat{f}={\mathcal{F}}_2(R_0(z)f)$ in $L^2$. Since ${\mathcal{F}}_2$ is an isometry, ${\phi }_n\ast f\to R_0(z)f$ in $L^2$. On the other hand ${\phi }_n\ast f\to g_z\ast f$ in $L^2$: for $h:={\phi }_n-g_z$ and a.e. $x$, $\int \mathrm{\mid }h(x-y)\mathrm{\mid }\mathrm{\mid }f(y)\mathrm{\mid }dy=\int \mathrm{\mid }h(x-y)\mathrm{\mid }\sqrt{\mathrm{\mid }f(y)\mathrm{\mid }}\sqrt{\mathrm{\mid }f(y)\mathrm{\mid }}dy\le (\int \mathrm{\mid }h(x-y){\mathrm{\mid }}^2\mathrm{\mid }f(y)\mathrm{\mid }dy{)}^{1\mathrm{/}2}\mathrm{\parallel }f{\mathrm{\parallel }}_1^{1\mathrm{/}2}$ by [F7], so Tonelli [F6] gives $\mathrm{\parallel }h\ast f{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }f{\mathrm{\parallel }}_1\iint \mathrm{\mid }h(x-y){\mathrm{\mid }}^2\mathrm{\mid }f(y)\mathrm{\mid }dydx=\mathrm{\parallel }f{\mathrm{\parallel }}_1^2\mathrm{\parallel }h{\mathrm{\parallel }}_2^2\to 0$. Therefore $R_0(z)f=g_z\ast f$ for Schwartz $f$. This identity also gives $\mathrm{\parallel }{g}_z\ast f{\mathrm{\parallel }}_2=\mathrm{\parallel }{R}_0(z)f{\mathrm{\parallel }}_2=\mathrm{\parallel }{m}_z{\mathcal{F}}_{2}f{\mathrm{\parallel }}_2\le \mathrm{\parallel }{m}_z{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }f{\mathrm{\parallel }}_2$ for Schwartz $f$, so $f\mapsto g_z\ast f$ is bounded on the dense subspace of Schwartz classes and extends uniquely to a bounded operator on $L^2$ by [F10], and this extension agrees with the bounded operator $R_0(z)$ on the dense subspace; hence the extension equals $R_0(z)$ on every $f\in L^2$. It also equals the ordinary convolution representative: for fixed $x$, Cauchy–Schwarz gives $\int \mathrm{\mid }{g}_z(x-y)f(y)\mathrm{\mid }dy\le \mathrm{\parallel }{g}_z{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2$, and for Schwartz $f_n\to f$ it gives ${\sup }_x\mathrm{\mid }{g}_z\ast (f_n-f)(x)\mathrm{\mid }\le \mathrm{\parallel }{g}_z{\mathrm{\parallel }}_2\mathrm{\parallel }{f}_n-f{\mathrm{\parallel }}_2\to 0$. This is uniform convergence, while $R_0(z)f_n\to R_0(z)f$ in $L^2$. On every bounded set both limits agree by $L^2$ convergence and uniform convergence, proving $R_0(z)f=g_z\ast f$ almost everywhere for the ordinary convolution as well.

[Localized resolvent is Hilbert–Schmidt] Let $\chi$ be bounded with compact support. By step 2.2, $M_{\chi }{R}_0(z)f=\chi \cdot (g_z\ast f)$, so the operator $M_{\chi }{R}_0(z)$ has the kernel $k(x,y):=\chi (x)g_z(x-y)$. The function $k$ is product-measurable and $\int \int \mathrm{\mid }k(x,y){\mathrm{\mid }}^{2}dxdy=\int \mathrm{\mid }\chi (x){\mathrm{\mid }}^2(\int \mathrm{\mid }{g}_z(x-y){\mathrm{\mid }}^{2}dy)dx=\mathrm{\parallel }\chi {\mathrm{\parallel }}_2^2\mathrm{\parallel }{g}_z{\mathrm{\parallel }}_2^2<\mathrm{\infty }$ by [F6]; hence $M_{\chi }{R}_0(z)=T_k$ is Hilbert–Schmidt in the sense of L^2 kernel operators and Hilbert-Schmidt operators on R^3 and compact by [F5].

[Short-range and tail estimates] For $\epsilon >0$ let $B_{\epsilon }:=\{x:\mathrm{\mid }x\mathrm{\mid }<\epsilon \}$. By step 2.1, $\mathrm{\parallel }{M}_{\gamma \mathrm{\mid }\cdot {\mathrm{\mid }}^{-1}{\mathbf{1}}_{B_{\epsilon }}}{R}_0(z)f{\mathrm{\parallel }}_2\le \gamma ({\int }_{B_{\epsilon }}\mathrm{\mid }x{\mathrm{\mid }}^{-2}dx{)}^{1\mathrm{/}2}\mathrm{\parallel }{R}_0(z)f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \gamma \sqrt{4\pi \epsilon }\mathrm{\parallel }{m}_z{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2$, so the operator norm is at most $\gamma \sqrt{4\pi \epsilon }\mathrm{\parallel }{m}_z{\mathrm{\parallel }}_2$. For $R>0$, $\mathrm{\parallel }{M}_{\gamma \mathrm{\mid }\cdot {\mathrm{\mid }}^{-1}{\mathbf{1}}_{\{\mathrm{\mid }x\mathrm{\mid }>R\}}}{R}_0(z)\mathrm{\parallel }\le \gamma R^{-1}\mathrm{\parallel }{R}_0(z)\mathrm{\parallel }\le \gamma R^{-1}\mathrm{\mid }\mathrm{Im}z{\mathrm{\mid }}^{-1}$.

[Compactness of $M{R}_0(z)$] Fix $R>\epsilon >0$ and choose a bounded measurable $\chi$ with $0\le \chi \le 1$, $\chi =1$ on $\{\epsilon \le \mathrm{\mid }x\mathrm{\mid }\le R\}$ and $\chi =0$ off $\{\epsilon \mathrm{/}2\le \mathrm{\mid }x\mathrm{\mid }\le 2R\}$; this is possible with compact support. Since $\chi =1$ on the closed annulus $\{\epsilon \le \mathrm{\mid }x\mathrm{\mid }\le R\}$, the function $1-\chi$ vanishes there and satisfies the pointwise domination $0\le \gamma \frac{1-\chi (x)}{\mathrm{\mid }x\mathrm{\mid }}\le \gamma \frac{{\mathbf{1}}_{B_{\epsilon }}(x)}{\mathrm{\mid }x\mathrm{\mid }}+\gamma \frac{{\mathbf{1}}_{\{\mathrm{\mid }x\mathrm{\mid }>R\}}(x)}{\mathrm{\mid }x\mathrm{\mid }}$. Hence $\gamma M{R}_0(z)=M_{\gamma \mathrm{\mid }\cdot {\mathrm{\mid }}^{-1}\chi }{R}_0(z)+M_{\gamma \mathrm{\mid }\cdot {\mathrm{\mid }}^{-1}(1-\chi )}{R}_0(z)$, where the first summand is compact by step 3.1 and the second has operator norm at most $\gamma \sqrt{4\pi \epsilon }\mathrm{\parallel }{m}_z{\mathrm{\parallel }}_2+\gamma R^{-1}\mathrm{\mid }\mathrm{Im}z{\mathrm{\mid }}^{-1}$ by step 3.2, because its multiplier is pointwise dominated by the sum of the two multipliers of step 3.2. Given $\delta >0$ choose $\epsilon$ and $R$ making this bound smaller than $\delta$. Thus $\gamma M{R}_0(z)$ lies within $\delta$ of a compact operator for every $\delta >0$, hence is compact by [F9].

[Conclusion] By step 4.1 and [F8], $V{R}_0(z)=-\gamma M{R}_0(z)$ is compact; the graph-norm boundedness of $V$ with respect to $H_0$ is [F2]. Hence $V$ is $H_0$-compact in the sense of Relative compactness with respect to an operator, and the equivalent formulations in (a)–(c) are steps 1.1, 2.1, 2.2 and 3.1.