Dollard integrable remainder

The Hilbert spaces below have fixed countable dense sequences of rational complex simple functions on rational boxes or intervals. Their density follows by simple-function approximation and Lebesgue regularity; CC selects countable approximation sequences for each fixed input. Closed subspaces use their orthogonal projections of these sequences. For a real measurable multiplier $M_a$, truncation to $\mathrm{\mid }a\mathrm{\mid }\le n$ proves dense domain, and testing its adjoint against all vectors supported in $\mathrm{\mid }a\mathrm{\mid }\le n$ identifies the adjoint as $M_a$ on exactly $\{h:ah\in L^2\}$. Thus the explicit Fourier multiplier gives self-adjoint $H_0$ on $H^2$. Hardy's inequality follows by integrating $\mathrm{div}(x\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2)\mathrm{\mid }u{\mathrm{\mid }}^2$ for smooth $u$ away from zero and using Cauchy–Schwarz: $\mathrm{\parallel }u\mathrm{/}r{\mathrm{\parallel }}_2^2\le 2\mathrm{\parallel }u\mathrm{/}r{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$. Smooth cutoffs remove the puncture, and Sobolev density extends $\mathrm{\parallel }u\mathrm{/}r{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$ to $H^1$. Fourier Cauchy–Schwarz gives $\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }u{\mathrm{\parallel }}_2\mathrm{\parallel }{H}_{0}u{\mathrm{\parallel }}_2$. For $z=\pm i\mu$, the explicit $R_0(z)=(H_0-z{)}^{-1}$ has $\mathrm{\parallel }{R}_0\mathrm{\parallel }\le {\mu }^{-1}$ and $\mathrm{\parallel }{H}_0{R}_0\mathrm{\parallel }\le 1$, hence $\mathrm{\parallel }\gamma r^{-1}{R}_0(z)\mathrm{\parallel }\le 2\gamma \mathrm{/}\sqrt{\mu }<1$ when $\mu >4{\gamma }^2$. The factorization $C-z=(I-\gamma r^{-1}{R}_0(z))(H_0-z)$, $C=H_0-\gamma \mathrm{/}r$ on $H^2$, gives onto-ness for both signs by the norm-convergent Neumann inverse. For $y\in D(C^{\ast })$ solve $(C-i\mu )x=(C^{\ast }-i\mu )y$; then $y-x\in \ker (C^{\ast }-i\mu )=\mathrm{ran}(C+i\mu {)}^{\perp }=0$. Thus $C=C^{\ast }$ on exactly $H^2$. Also $⟨Cu,u⟩\ge -{\gamma }^2\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$. This uses the explicit Hardy and Fourier calculation, rather than general Kato–Rellich.

Apply CC spectral calculus on a Hilbert space with a fixed countable dense sequence to the explicit self-adjoint $H_0,C$ from step 1.1, with their rational-box dense sequences. It supplies unique regular PVMs, exact square-integrability domains, bounded Borel calculus and strongly continuous unitary groups with exact generator domains under CC. In particular nonreal inverses are the scalar resolvent multipliers in this representation, and the group derivatives needed below hold on exactly $H^2$. These are the proved local spectral inputs; no stronger hypothesis of the legacy named Coulomb suppliers is borrowed.

For this Coulomb operator, $r^{-1}{R}_0(z)$ is compact under CC. The free resolvent multiplier $m_z(p)=(\mathrm{\mid }p{\mathrm{\mid }}^2-z{)}^{-1}$ belongs to $L^2({\mathbb{R}}^3)$, so Fourier Cauchy–Schwarz gives $\mathrm{\parallel }{R}_0(z)f{\mathrm{\parallel }}_{\mathrm{\infty }}\le (2\pi {)}^{-3\mathrm{/}2}\mathrm{\parallel }{m}_z{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2$. Its convolution kernel is $g_z=(2\pi {)}^{-3\mathrm{/}2}{\mathcal{F}}_p^{\ast }{m}_z\in L^2$. For bounded compactly supported $\chi$, the kernel $\chi (x)g_z(x-y)$ is in product $L^2$. Finite rational rectangular simple kernels approximate it there and give finite-rank operators; Cauchy–Schwarz bounds the operator norm by the kernel norm. Truncate $r^{-1}$ to $\epsilon <r<R$: the near-origin norm error is at most $\sqrt{4\pi \epsilon }(2\pi {)}^{-3\mathrm{/}2}\mathrm{\parallel }{m}_z{\mathrm{\parallel }}_2$, and the tail error at most $R^{-1}\mathrm{\mid }\mathrm{Im}z{\mathrm{\mid }}^{-1}$. Thus $r^{-1}{R}_0$ is a norm limit of compact operators. The resolvent identity makes $(C-z{)}^{-1}-R_0(z)=\gamma (C-z{)}^{-1}{r}^{-1}{R}_0(z)$ compact.

The essential-spectrum inference also has a specified countable construction. A real $\lambda$ is essential exactly when every spectral neighborhood has infinite rank. If it does, project the fixed dense sequence onto $E_A((\lambda -1\mathrm{/}n,\lambda +1\mathrm{/}n))H$ intersected with the orthogonal complement of previously constructed vectors, take the least nonzero projection and normalize. This yields an orthonormal sequence $u_n$ with $\mathrm{\parallel }(A-\lambda )u_n\mathrm{\parallel }\le 1\mathrm{/}n$, weakly null by the elementary finite Bessel inequality. Conversely, for a unit weakly null sequence with $(A-\lambda )u_n\to 0$, a finite-rank spectral neighborhood would give $E_{\epsilon }{u}_n\to 0$ and $\mathrm{\parallel }(I-E_{\epsilon })u_n\mathrm{\parallel }\le \mathrm{\parallel }(A-\lambda )u_n\mathrm{\parallel }\mathrm{/}\epsilon \to 0$, a contradiction. If $K=(C-z{)}^{-1}-(A-z{)}^{-1}$ is compact, $v_n=(C-z{)}^{-1}(A-z)u_n=u_n+K(A-z)u_n$ satisfies $v_n-u_n\to 0$ and $(C-\lambda )v_n=(A-\lambda )u_n+(z-\lambda )(v_n-u_n)\to 0$. Compact operators send bounded weakly null sequences to norm-null sequences: otherwise a subsequence with image norm bounded below has a norm-convergent image subsequence by compactness and CC, but its weak limit is zero, a contradiction. Normalize $v_n$ and reverse roles to get essential-spectrum equality. Explicit momentum shells give infinite-rank neighborhoods at every $\lambda \ge 0$ for $H_0$ and zero projections below zero, so ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(C)=[0,\mathrm{\infty })$. Below zero finite-rank neighborhoods have finite-dimensional self-adjoint restrictions and finite sets of eigenvalues; every negative spectral point is isolated of finite multiplicity. Least rational isolating intervals make these points countable. The other negative points have zero-projection neighborhoods and a countable rational subcover, so the negative PVM is exhausted by the countable eigenprojections.

If $a_0=b_0$, the smooth profile is zero and all conclusions are immediate; otherwise $a_0<b_0$. Under the explicit Fourier multiplier, $U_D(t)$ for $\tau \ge 1$ is the multiplier $e^{-it\mathrm{\mid }p{\mathrm{\mid }}^2+i\sigma a(p)L}$ and is unitary. It agrees at $t=\pm 1$ with the free branch, so strong continuity follows from dominated convergence on the Fourier classes. For $\mathrm{\mid }t\mathrm{\mid }>1$, $d(\sigma \log \mathrm{\mid }t\mathrm{\mid })\mathrm{/}dt=1\mathrm{/}\mathrm{\mid }t\mathrm{\mid }$, so on the annular class differentiation gives $i{U}_D^{\mathrm{\prime }}(t)=[H_0-a(P)\mathrm{/}\mathrm{\mid }t\mathrm{\mid }]U_D(t)$; the exterior derivatives at the join have the same value. The annular Fourier class stays Schwartz, with the same momentum support, for each fixed time.

For $\epsilon >0$, the Fourier multiplier $e^{-(\epsilon +it)\mathrm{\mid }p{\mathrm{\mid }}^2}$ has kernel $(4\pi (\epsilon +it){)}^{-3\mathrm{/}2}{e}^{-\mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}[4(\epsilon +it)]}$. To obtain it, continue the real Gaussian Fourier formula into the right half-plane: the integral is holomorphic by a Gaussian majorant on compact parameter sets, and the identity theorem gives the same principal-power formula there (Complex Gaussian continuation, Euclidean Gaussian transform with the 2π normalization). For Schwartz $g$, dominated convergence against $\mathrm{\mid }g(y)\mathrm{\mid }$ gives the pointwise limit $(4\pi it{)}^{-3\mathrm{/}2}\int e^{i\mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}(4t)}g(y)dy$, since the kernels have modulus bounded by $(4\pi \mathrm{\mid }t\mathrm{\mid }{)}^{-3\mathrm{/}2}$. Fourier dominated convergence also gives $L^2$ convergence to the explicit $e^{-it{H}_0}g$; an a.e.-convergent subsequence of this $L^2$ limit identifies the pointwise kernel limit. The damped multipliers are contractions, not isometries; the limit is unitary because its momentum multiplier has modulus one. All principal powers use $\exp (-\frac{3}{2}\mathrm{Log}(4\pi it))$. Define $m_t(x)=e^{i\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(4t)}$, $M_{t}g=m_{t}g$, and $(D_{t}h)(x)=(2it{)}^{-3\mathrm{/}2}{e}^{i\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(4t)}h(x\mathrm{/}(2t))$, using the branch of the free Gaussian formula. $D_t$ is unitary because its modulus is $(2\tau {)}^{-3\mathrm{/}2}$ and its dilation has absolute Jacobian $(2\tau {)}^3$. Expanding $\mathrm{\mid }x-y{\mathrm{\mid }}^2$ in the free Gaussian formula gives the exact identity $e^{-it{H}_0}=D_t{\mathcal{F}}_p{M}_t$ on Schwartz functions and then on $L^2$. For the bounded multiplier $F(P)={\mathcal{F}}_p^{\ast }{M}_F{\mathcal{F}}_p$, it follows that $[F(x\mathrm{/}(2t))-F(P)]e^{-it{H}_0}g=D_t{\mathcal{F}}_p[F(P),M_t]g$.

For $\tau$ large, the inverse Fourier representative is $(2\pi {)}^{-3\mathrm{/}2}\int e^{i{\theta }_{t,x}(p)}\varphi (p)dp$, with ${\theta }_{t,x}(p)=p\cdot x-t\mathrm{\mid }p{\mathrm{\mid }}^2+\sigma a(p)L$. If $\mathrm{\mid }x\mathrm{\mid }<a_0\tau$ or $\mathrm{\mid }x\mathrm{\mid }>4b_0\tau$, then on a fixed neighborhood of the compact support of $\varphi$, $\mathrm{\mid }{\mathrm{\nabla }}_p\theta \mathrm{\mid }=\mathrm{\mid }x-2tp+\sigma L\mathrm{\nabla }a(p)\mathrm{\mid }\ge c(\tau +\mathrm{\mid }x\mathrm{\mid })$ for sufficiently large $\tau$, since $L\mathrm{/}\tau \to 0$. Every derivative of order at least two of $\theta$ is bounded by $C_{\beta }\tau$ there. The quotient rule consequently gives $\mathrm{\mid }{\mathrm{\partial }}_p^{\beta }(\mathrm{\nabla }\theta \mathrm{/}\mathrm{\mid }\mathrm{\nabla }\theta {\mathrm{\mid }}^2)\mathrm{\mid }\le C_{\beta }\mathrm{/}(\tau +\mathrm{\mid }x\mathrm{\mid })$ for each fixed multi-index: each differentiated numerator or denominator contributes at most a factor $C\tau$, and $\tau \mathrm{/}(\tau +\mathrm{\mid }x\mathrm{\mid })\le 1$. The integration-by-parts operator $\mathcal{L}=(\mathrm{\nabla }\theta \mathrm{/}(i\mathrm{\mid }\mathrm{\nabla }\theta {\mathrm{\mid }}^2))\cdot {\mathrm{\nabla }}_p$ satisfies $\mathcal{L}{e}^{i\theta }=e^{i\theta }$. Repeating its formal adjoint $N$ times on the fixed smooth compact amplitude gives an $L^1$ norm at most $C_{N,\psi }(\tau +\mathrm{\mid }x\mathrm{\mid }{)}^{-N}$, by the derivative bounds and induction on the product rule. No boundary term occurs by the cutoff construction. Thus $\mathrm{\mid }{U}_D(t)\psi (x)\mathrm{\mid }\le C_{N,\psi }(\tau +\mathrm{\mid }x\mathrm{\mid }{)}^{-N}$ on this complement for both signs, with no extra logarithmic factor.

The Schwartz convolution kernel of $F(P)$ is $k_F(y)=(2\pi {)}^{-3}\int e^{ip\cdot y}F(p)dp$, by the Fourier convolution identity. Thus $[F(P),M_t]g(x)=\int k_F(y)[m_t(x-y)-m_t(x)]g(x-y)dy$. The phase difference satisfies $\mathrm{\mid }{m}_t(x-y)-m_t(x)\mathrm{\mid }\le [2\mathrm{\mid }x-y\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }+\mathrm{\mid }y{\mathrm{\mid }}^2]\mathrm{/}(4\tau )$. If $w\in L^1$ is nonnegative and $h\in L^2$, Cauchy–Schwarz gives $\mathrm{\mid }\int w(y)h(x-y)dy{\mathrm{\mid }}^2\le \mathrm{\parallel }w{\mathrm{\parallel }}_1\int w(y)\mathrm{\mid }h(x-y){\mathrm{\mid }}^{2}dy$, and Tonelli then gives the convolution norm bound $\mathrm{\parallel }w\ast h{\mathrm{\parallel }}_2\le \mathrm{\parallel }w{\mathrm{\parallel }}_1\mathrm{\parallel }h{\mathrm{\parallel }}_2$. Apply this with $w=\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }{k}_F(y)\mathrm{\mid }$ and $w=\mathrm{\mid }y{\mathrm{\mid }}^2\mathrm{\mid }{k}_F(y)\mathrm{\mid }$, both integrable, and use the $L^2$ triangle inequality to obtain $\mathrm{\parallel }[F(P),M_t]g{\mathrm{\parallel }}_2\le C_F(\mathrm{\parallel }\mathrm{\mid }x\mathrm{\mid }g{\mathrm{\parallel }}_2+\mathrm{\parallel }g{\mathrm{\parallel }}_2)\mathrm{/}\tau$. Every exchange is absolutely integrable by these same bounds.

The position difference $\gamma \mathrm{/}\mathrm{\mid }x\mathrm{\mid }-\gamma F(x\mathrm{/}(2t))\mathrm{/}(2\tau )$ vanishes on $a_0\tau \le \mathrm{\mid }x\mathrm{\mid }\le 4b_0\tau$, by the cutoff construction. On the complement, the nonstationary-phase calculation bounds the squared norm of its $\mathrm{\mid }x{\mathrm{\mid }}^{-1}$ part by $C_N{\int }_0^{\mathrm{\infty }}(\tau +r{)}^{-2N}dr\le C_N{\tau }^{1-2N}$, since the three-dimensional volume factor $r^2$ cancels $r^{-2}$. The bounded $F$ part has support $\mathrm{\mid }x\mathrm{\mid }\le 8b_0\tau$, and its squared norm is at most $C_N{\tau }^{-2}{\int }_0^{8b_0\tau }(\tau +r{)}^{-2N}{r}^{2}dr\le C_N{\tau }^{1-2N}$. Taking square roots and choosing $N=3$ gives a norm at most $C_{\psi }{\tau }^{-5\mathrm{/}2}$.

Put $g_t={\mathcal{F}}_p^{\ast }(e^{i\sigma aL}\varphi )$. Integration by parts in the compact momentum support gives ${\mathcal{F}}_p(x_j{g}_t)=i{\mathrm{\partial }}_{p_j}(e^{i\sigma aL}\varphi )$; hence $\mathrm{\parallel }\mathrm{\mid }x\mathrm{\mid }{g}_t{\mathrm{\parallel }}_2\le \mathrm{\parallel }\mathrm{\nabla }\varphi {\mathrm{\parallel }}_2+L\mathrm{\parallel }\varphi \mathrm{\nabla }a{\mathrm{\parallel }}_2\le C_{\psi }(1+L)$. Also $\mathrm{\parallel }{g}_t{\mathrm{\parallel }}_2=\mathrm{\parallel }\psi {\mathrm{\parallel }}_2$. On its support $F(P)g_t=\mathrm{\mid }P{\mathrm{\mid }}^{-1}{g}_t$, and this equality is preserved by free evolution. The factorization and commutator calculations therefore give $\mathrm{\parallel }[F(x\mathrm{/}(2t))-\mathrm{\mid }P{\mathrm{\mid }}^{-1}]U_D(t)\psi {\mathrm{\parallel }}_2\le C_{\psi }(1+L)\mathrm{/}\tau$. This uses the convolution identity, not a functional-calculus assertion about noncommuting position and momentum.

Split $[\gamma \mathrm{/}\mathrm{\mid }x\mathrm{\mid }-\gamma \mathrm{/}(2\tau \mathrm{\mid }P\mathrm{\mid })]U_D(t)\psi$ into the position complement of the singular-complement calculation and $\gamma [F(x\mathrm{/}(2t))-\mathrm{\mid }P{\mathrm{\mid }}^{-1}]U_D(t)\psi \mathrm{/}(2\tau )$. The moment and singular-complement calculations give norm at most $C_{\psi }(1+\log \tau )\mathrm{/}{\tau }^2+C_{\psi }{\tau }^{-5\mathrm{/}2}\le C_{\psi }^{\mathrm{\prime }}(1+\log \tau )\mathrm{/}{\tau }^2$ for both signs and large $\tau$. On each compact exterior time interval $1\le \tau \le T$, the Fourier profile is Schwartz with bounded Schwartz seminorms, so its spatial representative is bounded near zero and rapidly decreasing at infinity, uniformly in that interval; $\mathrm{\mid }x{\mathrm{\mid }}^{-1}$ is locally square integrable in three dimensions, and $\mathrm{\mid }P{\mathrm{\mid }}^{-1}$ is bounded by $a_0^{-1}$ on the momentum support. Thus the remainder norm is bounded there. Integration over both half-lines is finite because ${\int }_T^{\mathrm{\infty }}(1+\log \tau ){\tau }^{-2}d\tau <\mathrm{\infty }$. This proves (a) and (b), including the original full annular class and the required single logarithm.