Resolvent-tail short-range completeness

For $z=\pm i\mu$, $\mu >0$, write $m_z(p)=(\mathrm{\mid }p{\mathrm{\mid }}^2-z{)}^{-1}$. Its $L^2$ norm is $C{\mu }^{-1\mathrm{/}4}$ by $p=\sqrt{\mu }q$ in $\int (\mathrm{\mid }p{\mathrm{\mid }}^4+{\mu }^2{)}^{-1}dp$, and its supremum is ${\mu }^{-1}$. Thus $\mathrm{\parallel }{V}_2{R}_0(z)\mathrm{\parallel }\le C^{\mathrm{\prime }}\mathrm{\parallel }{V}_2{\mathrm{\parallel }}_2{\mu }^{-1\mathrm{/}4}$ and $\mathrm{\parallel }{V}_{\mathrm{b}}{R}_0(z)\mathrm{\parallel }\le \mathrm{\parallel }{V}_{\mathrm{b}}{\mathrm{\parallel }}_{\mathrm{\infty }}{\mu }^{-1}$, by [F1]. Choose a large integer $\mu$ so their sum is less than one. Multiplication by $V$ maps $H^2$ to $L^2$: Fourier Cauchy–Schwarz with weight $(1+\mathrm{\mid }p{\mathrm{\mid }}^2{)}^{-1}$ gives $H^2\subset L^{\mathrm{\infty }}$, and the two parts of $V$ have the stated bounds. $H$ on $H^2$ is symmetric because $V$ is real. The factorization $H-z=(I+V{R}_0(z))(H_0-z)$ is onto at both signs by the norm-convergent Neumann series. If $y\in D(H^{\ast })$, solve $(H-i\mu )x=(H^{\ast }-i\mu )y$; then $y-x\in \ker (H^{\ast }-i\mu )=\mathrm{ran}(H+i\mu {)}^{\perp }=0$. Hence the exact operator $H$ is self-adjoint on $H^2$, and [F3] constructs its dynamics without an imported AC perturbation theorem.

Put $K=V{R}_0(i)$, which is bounded by step 1.1. The function $m_i$ belongs to $L^2$, so $g_i=(2\pi {)}^{-3\mathrm{/}2}{\mathcal{F}}^{-1}{m}_i\in L^2$ is the free convolution kernel. Its kernel realization follows first on smooth compact test functions from Fourier convolution, then by density. For finite $R$, the $V_2$ part of $K{\chi }_R$ has kernel-square integral at most $\mathrm{\parallel }{V}_2{\mathrm{\parallel }}_2^2\mathrm{\parallel }{g}_i{\mathrm{\parallel }}_2^2$; the $V_{\mathrm{b}}$ part has integral at most $\mathrm{\parallel }{V}_{\mathrm{b}}{\mathrm{\parallel }}_{\mathrm{\infty }}^2\mathrm{\mid }{B}_R\mathrm{\mid }\mathrm{\parallel }{g}_i{\mathrm{\parallel }}_2^2$, integrating $x$ first and then $y\in B_R$. Thus $K{\chi }_R$ is compact by [F4]; boundedness and density identify this kernel operator with the original composition. Since $\mathrm{\parallel }(K-K{\chi }_R)\mathrm{\parallel }=h(R)$, $K$ is compact: $h$ is nonincreasing, because a farther-tail projection is a subprojection of an earlier one, and nonnegative monotone integrability implies $h(R)\to 0$. This proof cuts off the INPUT variable $y$ of the resolvent, and handles globally noncompact $V_2$ and $V_{\mathrm{b}}$ without falsely calling the full potential compact.

Let $f\in C_c^{\mathrm{\infty }}((0,\mathrm{\infty }))$ have support in $(a^2,b^2)$ and set $g(E)=(E-i)f(E)$. For $t\ge 1$ define $B(t)=e^{-it{H}_0}g(H_0)P_{+}$. The exact input split of $K$ gives $\mathrm{\parallel }V{e}^{-it{H}_0}f(H_0)P_{+}\mathrm{\parallel }\le \mathrm{\parallel }K\mathrm{\parallel }\mathrm{\parallel }{\chi }_{2at}B(t)\mathrm{\parallel }+h(2at)\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }K\mathrm{\parallel }{C}_{N}t(1+t{)}^{-N}+h(2at)\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ Take $N=3$. The first term is integrable, and the second has integral at most $(2a{)}^{-1}\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}h(r)dr$. The finite interval $[0,1]$ has bound $\mathrm{\parallel }K\mathrm{\parallel }\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$. The incoming estimate is identical with $\mathrm{\mid }t\mathrm{\mid }$ and $P_{-}$. This proves operator-norm integrability, not merely vectorwise Cook integrability, for the FULL resolvent-tail class. In particular it transfers the all-$N$ growing-ball bound to the required norm integral with no stationary input.

For each sign the linear span of all free translates $e^{-is{H}_0}f(H_0)P_{\pm }\xi$, where $s\in \mathbb{R}$, $\xi \in L^2$ and smooth positive cutoffs $f$ vary, is dense. Indeed, if $u$ is orthogonal to the outgoing span, then $P_{+}\bar{f}(H_0)e^{is{H}_0}u=0$ for all $s$. Set $s=-t$ and use $P_{-}\bar{f}(H_0)e^{-it{H}_0}u\to 0$ as $t\to +\mathrm{\infty }$ from [F2]. Since $P_{+}+P_{-}=I$, this forces $\mathrm{\parallel }\bar{f}(H_0)u\mathrm{\parallel }=0$. Positive smooth cutoffs exhaust the free spectrum, and the free projection at zero is zero because $\{p=0\}$ is null, so $u=0$. The incoming case uses $t\to -\mathrm{\infty }$. Each generating vector is in $H^2$. The Duhamel derivative is $i{e}^{itH}V{e}^{-it{H}_0}\eta$; domain continuity follows from free graph continuity, boundedness of $V:H^2\to L^2$, and the exact generators of [F3]. Step 3.1 gives its integrability at the relevant infinity, also after any fixed time translation, while boundedness on finite intervals follows from the graph estimate. Its vector integral is Cauchy, so both wave limits exist on their respective dense spans. Uniform unitarity extends them to all $L^2$, giving isometries. Changing the time variable proves group intertwining, and [F3] gives PVM intertwining and reducing closed ranges. Their scalar spectral measures are transported free measures, which are absolutely continuous by polar momentum coordinates (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma). Thus ${\mathcal{R}}_{\pm }=\mathrm{ran}{\mathrm{\Omega }}_{\pm }\subseteq {\mathcal{H}}_{\mathrm{a}\mathrm{c}}(H)$.

The integrand $F_{+}(t)=e^{itH}V{e}^{-it{H}_0}f(H_0)P_{+}=e^{itH}KB(t)$ is compact and norm-continuous. $B(t)$ is norm-continuous since its compact-energy scalar multiplier has Lipschitz time differences; $t\mapsto e^{itH}K$ is norm-continuous by finite-rank approximation of $K$ and strong continuity on each finite-dimensional image. Step 3.1 makes $F_{+}$ norm-integrable, so domain Duhamel and step 4.1 give $C_{+}=({\mathrm{\Omega }}_{+}-I)f(H_0)P_{+}=i{\int }_0^{\mathrm{\infty }}{F}_{+}(t)dt$ in operator norm. Compact Riemann sums and the norm-small tail show $C_{+}$ compact. The reversed integral proves the same for $C_{-}$. ONLY smooth strictly positive cutoffs are used here; no claim for arbitrary cutoffs nonzero at zero is made.

The identity $R_H(z)-R_0(z)=-R_H(z)V{R}_0(z)$ makes the resolvent difference compact for nonreal $z$: $V{R}_0(z)=K(H_0-i)R_0(z)$, whose last factor is bounded by its explicit scalar multiplier. For the Cayley unitaries $Z_A=I-2i{R}_A(-i)$, both $Z_H-Z_0$ and their adjoint difference are compact. Telescoping makes differences of their Laurent polynomials compact. For $f\in C_0(\mathbb{R})$, the circle function $F(z)=f(i(1+z)\mathrm{/}(1-z))$ off $z=1$, with $F(1)=0$, is continuous. Fejer polynomials approximate it uniformly by [F3]; the multiplier norm estimate implies $f(H)-f(H_0)$ is the norm limit of the compact polynomial differences, hence compact.

For completeness we give the CC negative-spectrum argument explicitly. A real $\lambda$ has infinite-rank spectral neighborhoods for a self-adjoint $A$ precisely when there is a weakly-null unit sequence $u_n\in D(A)$ with $(A-\lambda )u_n\to 0$. Forward: recursively take a nonzero projection of the least eligible member of the fixed dense sequence into $E_A((\lambda -1\mathrm{/}n,\lambda +1\mathrm{/}n))L^2$ orthogonal to the previous finite family, and normalize; the resulting orthonormal sequence is weakly null. Reverse: a finite-rank $E_A((\lambda -\epsilon ,\lambda +\epsilon ))$ sends $u_n$ to zero, while the complementary norm is bounded by $\mathrm{\parallel }(A-\lambda )u_n\mathrm{\parallel }\mathrm{/}\epsilon$, a contradiction. A compact resolvent difference $L=R_B(i)-R_A(i)$ transfers such a sequence by $v_n=R_B(i)(A-i)u_n=u_n+L(A-i)u_n$: the compact term tends to zero and $(B-\lambda )v_n=(A-\lambda )u_n+(i-\lambda )(v_n-u_n)\to 0$. Reverse the roles for the converse. Therefore step 5.2 transfers the free set of infinite-rank-neighborhood points, exactly $[0,\mathrm{\infty })$, to $H$. Every negative spectral point of $H$ has a finite-rank bounded neighborhood; its restriction is a finite self-adjoint matrix, hence negative points are isolated eigenvalues of finite multiplicity. Rational isolating intervals enumerate them countably; rational zero-projection neighborhoods cover the remaining negative points. Countable PVM additivity proves $E_H((-\mathrm{\infty },0))L^2\subseteq {\mathcal{H}}_{\mathrm{p}\mathrm{p}}$. Also $E_H(\{0\})L^2=\ker H$ is pure point. Consequently ${\mathcal{H}}_{\mathrm{c}}\subseteq E_H((0,\mathrm{\infty }))L^2$, exhausted by $J_n=[1\mathrm{/}n,n]$.

Set $\mathcal{R}=\overline{{\mathcal{R}}_{+}+{\mathcal{R}}_{-}}$ and let $Q$ project onto ${\mathcal{R}}^{\perp }$. All these spaces reduce the PVM by step 4.1; $\mathcal{R}$ lies in the closed absolutely continuous subspace and $Q$ preserves ${\mathcal{H}}_{\mathrm{c}}$. For a compact $J\subset (0,\mathrm{\infty })$ choose real smooth $f=1$ on $J$ and put $f_0=f(H_0)$. The exact identity $C_{+}+C_{-}={\mathrm{\Omega }}_{+}{f}_0{P}_{+}+{\mathrm{\Omega }}_{-}{f}_0{P}_{-}-f_0$ gives $Q{f}_0=-Q(C_{+}+C_{-})$ compact. Step 5.2 implies $Qf(H)$ compact, hence $Q{E}_H(J)$ is a compact orthogonal projection. Its finite-dimensional reducing range consists of eigenvectors, so $Q{E}_H(J)u=0$ for $u\in {\mathcal{H}}_{\mathrm{c}}$. Exhausting with $J_n$ gives ${\mathcal{H}}_{\mathrm{c}}\subseteq \mathcal{R}\subseteq {\mathcal{H}}_{\mathrm{a}\mathrm{c}}$. The absolutely continuous subspace is closed (its vectors are annihilated by every null-set projection) and orthogonal to each eigenspace, so ${\mathcal{H}}_{\mathrm{c}}={\mathcal{H}}_{\mathrm{a}\mathrm{c}}$ and ${\mathcal{H}}_{\mathrm{s}\mathrm{c}}=0$: an atomless singular vector lies in ${\mathcal{H}}_{\mathrm{c}}$ and is orthogonal to ${\mathcal{H}}_{\mathrm{a}\mathrm{c}}$. No abstract AC spectral-type-decomposition theorem is needed for this direct null-set argument.

Each wave range is individually complete, as follows. Suppose $u\in {\mathcal{H}}_{\mathrm{a}\mathrm{c}}\cap {\mathcal{R}}_{+}^{\perp }$ and put $\psi =E_H(J)u$ on a compact positive $J$, with real smooth $f=1$ on $J$. Then ${\psi }_t=e^{-itH}\psi$ has fixed norm, $f(H){\psi }_t={\psi }_t$ and ${\psi }_t\perp {\mathcal{R}}_{+}$. It is weakly null as $t\to +\mathrm{\infty }$ by [F5]: cyclic scalar PVM representation for an absolutely continuous vector expresses its pairing with a fixed test as the Fourier transform of an $L^1$ density, by Cauchy–Schwarz, so Riemann–Lebesgue lemma applies. The compact operators in steps 5.1–5.2 send ${\psi }_t$ to zero in norm. The identity in step 7.1 therefore gives $\mathrm{\parallel }\psi {\mathrm{\parallel }}^2={\lim }_{t\to +\mathrm{\infty }}⟨{\psi }_t,{\mathrm{\Omega }}_{+}{f}_0{P}_{+}{\psi }_t+{\mathrm{\Omega }}_{-}{f}_0{P}_{-}{\psi }_t⟩$. The outgoing summand is zero. The remaining summand has magnitude at most $\mathrm{\parallel }\psi \mathrm{\parallel }\mathrm{\parallel }{P}_{-}\bar{f}(H_0)e^{-it{H}_0}{\mathrm{\Omega }}_{-}^{\ast }\psi \mathrm{\parallel }$, by adjoint intertwining, and tends to zero by [F2] for the FIXED vector ${\mathrm{\Omega }}_{-}^{\ast }\psi$. Thus $\psi =0$; exhaustion proves $u=0$ and ${\mathcal{R}}_{+}={\mathcal{H}}_{\mathrm{a}\mathrm{c}}$. For $u\in {\mathcal{H}}_{\mathrm{a}\mathrm{c}}\cap {\mathcal{R}}_{-}^{\perp }$ run the same proof at $t\to -\mathrm{\infty }$, eliminate the incoming summand, and use $P_{+}\bar{f}(H_0)e^{-it{H}_0}{\mathrm{\Omega }}_{+}^{\ast }\psi \to 0$. This proves ${\mathcal{R}}_{-}={\mathcal{H}}_{\mathrm{a}\mathrm{c}}$. Finally ${\mathrm{\Omega }}_{\pm }^{\ast }{\mathrm{\Omega }}_{\pm }=I$ and ${\mathrm{\Omega }}_{\pm }{\mathrm{\Omega }}_{\pm }^{\ast }=P_{\mathrm{a}\mathrm{c}}$, so $S^{\ast }S=S{S}^{\ast }=I$.

We verify the concrete decay example, rather than postulating its resolvent estimate. The free kernel at $i$ is $g_i(x)=e^{-\kappa \mathrm{\mid }x\mathrm{\mid }}\mathrm{/}(4\pi \mathrm{\mid }x\mathrm{\mid })$ with $\kappa =\sqrt{-i}$ chosen with positive real part. One direct verification is the radial Fourier integral: $(2\pi {)}^{3\mathrm{/}2}\mathcal{F}{g}_i(p)=(\mathrm{\mid }p{\mathrm{\mid }}^2+{\kappa }^2{)}^{-1}$, since angular integration reduces it to $\mathrm{\mid }p{\mathrm{\mid }}^{-1}{\int }_0^{\mathrm{\infty }}{e}^{-\kappa s}\sin (\mathrm{\mid }p\mathrm{\mid }s)ds=(\mathrm{\mid }p{\mathrm{\mid }}^2+{\kappa }^2{)}^{-1}$; $g_i\in L^1\cap L^2$ legitimizes the calculation. Thus ${\int }_{\mathrm{\mid }z\mathrm{\mid }\ge d}\mathrm{\mid }{g}_i(z){\mathrm{\mid }}^{2}dz=C{e}^{-2cd}$, $c=\mathrm{Re}\kappa >0$. If $\mathrm{supp}{V}_{\mathrm{l}\mathrm{o}\mathrm{c}}\subset B_R$ and $r\ge 2R$, its resolvent-input tail norm is at most its kernel norm, hence at most $C\mathrm{\parallel }{V}_{\mathrm{l}\mathrm{o}\mathrm{c}}{\mathrm{\parallel }}_2{e}^{-c(r-R)}$. For $V_{\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{l}}$ split the OUTPUT variable into $\mathrm{\mid }x\mathrm{\mid }\ge r\mathrm{/}2$ and $\mathrm{\mid }x\mathrm{\mid }<r\mathrm{/}2$. The outer multiplier has norm at most $C(1+r\mathrm{/}2{)}^{-1-\delta }$, and $\mathrm{\parallel }{R}_0(i)\mathrm{\parallel }\le 1$. On the inner output region and input $\mathrm{\mid }y\mathrm{\mid }\ge r$, $\mathrm{\mid }x-y\mathrm{\mid }\ge r\mathrm{/}2$, so the kernel-square integral is at most $C^2\mathrm{\mid }{B}_{r\mathrm{/}2}\mathrm{\mid }{\int }_{\mathrm{\mid }z\mathrm{\mid }\ge r\mathrm{/}2}\mathrm{\mid }{g}_i(z){\mathrm{\mid }}^{2}dz\le C^{\mathrm{\prime }}{r}^3{e}^{-cr}$. Taking square roots and adding gives the stated bound with $c^{\mathrm{\prime }}=c\mathrm{/}2$ after adjusting constants. For bounded $r$, the graph bound of step 1.1 makes $h(r)\le \mathrm{\parallel }V{R}_0(i)\mathrm{\parallel }<\mathrm{\infty }$. Therefore $h$ is integrable for every $\delta >0$, proving the full unbounded-local-plus-decaying-tail example together with every asserted range and spectral conclusion.