Positive energy and vanishing at the far end of space

Decay at infinity can be tested by weights that grow as increasingly high powers of the radius. For a positive-energy equation, the useful estimate must control two scales at once: the power in the weight and the radius at which the solution is being tested. Their competition creates a characteristic frequency shell. An estimate that ignores that shell loses the second derivatives needed to handle rough leading coefficients.

We organize the argument around two complementary estimates. A Fourier calculation, localized on the sphere, recovers derivatives with carefully chosen radius-dependent coefficients. A difference of squared norms supplies the angular and positive-energy terms. Neither estimate closes by itself. Their combination makes the coefficient errors small and gives a weighted uniqueness theorem for complex Lipschitz perturbations of a real elliptic quadratic operator.

1. The exterior equation and its precise assumptions

Let $X\subset\mathbb R^n$, $n\geq1$, be a connected open neighborhood of infinity: $\{|x|>R_*\}\subset X$ for some $R_*$. Let $$p(x,D)=\sum_{j,k=1}^n a_{jk}(x)D_jD_k, \qquad D_j=-i\partial_{x_j}. \tag{E1}$$ The coefficients may be complex. They are locally Lipschitz on $X$, and their symmetric part is elliptic on $X$: $$\sum_{j,k}a_{jk}(x)\xi_j\xi_k\ne0 \quad(x\in X,\ \xi\in\mathbb R^n\setminus0). \tag{E2}$$ Assume that for some $\delta>0$, some $C_a$, and all sufficiently large $|x|$, $$|a_{jk}(x)-\delta_{jk}|\leq C_a|x|^{-1-\delta}, \qquad |\nabla a_{jk}(x)|\leq C_a|x|^{-2-\delta} \quad\text{a.e.} \tag{E3}$$ The two powers in (E3) are distinct. No second derivative of a leading coefficient is assumed. Retain the full matrix, including its skew part. Put $a^s=(a+a^T)/2$ and $a^a=(a-a^T)/2$. The exact comparison is $$\sum_{j,k}a^a_{jk}(x)D_jD_k =\sum_{j<k}\frac{a_{jk}(x)-a_{kj}(x)}2 (D_jD_k-D_kD_j)=0. \tag{E2a}$$ Both derivatives remain to the right of the coefficient. Thus no derivative of a skew coefficient is introduced. Formula (E2a) proves that contribution is zero; it does not erase the original entries from the coefficient data.

Here is the complete formulation for an original real limiting matrix $A_\infty$, which need not be symmetric. Replace the first assumption in (E3) by $$|a_{jk}(x)-(A_\infty)_{jk}|\leq C_a|x|^{-1-\delta}, \qquad |\nabla a_{jk}(x)|\leq C_a|x|^{-2-\delta}. \tag{E3a}$$ Assume its real symmetric part is definite. Choose its actual sign $\sigma\in\{1,-1\}$, and set $$\begin{gathered} A_\infty^s=(A_\infty+A_\infty^T)/2,\qquad A_\infty^a=(A_\infty-A_\infty^T)/2,\\ G=\sigma A_\infty^s>0,\qquad M=G^{1/2},\qquad Y=M^{-1}X,\qquad v(y)=u(My),\\ \widetilde a(y)=\sigma M^{-1}a(My)M^{-T},\quad \widetilde p=\sum_{j,k}\widetilde a_{jk}(y)D_{y_j}D_{y_k},\quad \widetilde\lambda=\sigma\lambda,\\ K_\infty=\sigma M^{-1}A_\infty^aM^{-T},\qquad \widetilde a(\infty)=I+K_\infty,\qquad K_\infty^T=-K_\infty. \end{gathered} \tag{E3b}$$ The original $A_\infty,a,p,u,X,\lambda$ remain fixed. The auxiliary limit is the full $I+K_\infty$, with every skew entry retained. The theorem below holds in this formulation when $\sigma\lambda>0$. For the identity limit, $\sigma=1$, $M=I$, and $K_\infty=0$.

We prove every map needed to apply the exterior argument. The chain rule and its inverse give $$\begin{gathered} D_yv(y)=M^TD_xu(My),\qquad D_xu(My)=M^{-T}D_yv(y),\\ ((\widetilde p-\widetilde\lambda)v)(y) =\sigma((p-\lambda)u)(My),\\ \partial_{y_l}\widetilde a_{jk}(y) =\sigma\sum_{r,s,h}(M^{-1})_{jr}(M^{-1})_{ks}M_{hl} (\partial_{x_h}a_{rs})(My). \end{gathered} \tag{E3c}$$ The derivative identity holds almost everywhere. Constant linear pullback and its inverse map $H^k_{\mathrm{loc}}(X)$ bijectively to $H^k_{\mathrm{loc}}(Y)$, $k=0,1,2$: every weak derivative is the finite chain-rule sum, and changing variables contributes $|\det M|^{-1}$ to its squared norm. Smooth approximation proves these identities at weak regularity. The distributional identity in (E3c) then follows for the weak Lipschitz coefficient product as well; it agrees with the usual expression when the local regularity has reached $H^2$.

Let $m_-$ and $m_+$ be the smallest and largest singular values of $M$. Then $0<m_-\leq m_+$ and $m_-|y|\leq|My|\leq m_+|y|$. Assumption (E3a) and the full sums in (E3c) imply, entry by entry, $$\begin{split} |\widetilde a_{jk}(y)-\delta_{jk}-(K_\infty)_{jk}| &\leq C_a m_-^{-1-\delta}|y|^{-1-\delta} \sum_{r,s}|(M^{-1})_{jr}(M^{-1})_{ks}|,\\ |\partial_{y_l}\widetilde a_{jk}(y)| &\leq C_a m_-^{-2-\delta}|y|^{-2-\delta} \sum_{r,s,h}|(M^{-1})_{jr}(M^{-1})_{ks}M_{hl}|. \end{split} \tag{E3d}$$ The image domain contains $\{|y|>R_*/m_-\}$. For the original inequality (E5), its complete pullback is $$|(\widetilde p-\widetilde\lambda)v(y)| \leq C|My|^{-1}\bigl(|v(y)|+|M^{-T}D_yv(y)|\bigr). \tag{E3e}$$ In particular it has the required $|y|^{-1}$ bound with the explicit additional constant $m_-^{-1}\max(1,m_-^{-1})$. The original form of (E3e) is retained when returning the estimate to $x$.

Put $\ell_- =\min(1,m_-)$ and $\ell_+=\max(1,m_+)$. For every real exponent $s$, $$\begin{gathered} \min(\ell_-^s,\ell_+^s)(1+|y|)^s \leq(1+|My|)^s \leq\max(\ell_-^s,\ell_+^s)(1+|y|)^s,\\ \int_Y(1+|My|)^{2s} \bigl(|v|^2+|M^{-T}D_yv|^2\bigr)\,dy =|\det M|^{-1}\int_X(1+|x|)^{2s} \bigl(|u|^2+|D_xu|^2\bigr)\,dx. \end{gathered} \tag{E3f}$$ The bounds on $M^{-T}$, together with (E3f), prove both directions of the all-weight condition (E4), including negative exponents. Ellipticity is also preserved exactly: the auxiliary symbol at a real covector $\xi$ is $\sigma p_{My}(M^{-T}\xi)$, and $M^{-T}\xi\ne0$ for $\xi\ne0$. Equations (E2a) and (E8a) below account for the entire skew operator. Sections 2–12 prove the estimate for the auxiliary expression with these original data present, and (E77)–(E80) return it with its full Jacobian to the original operator. A negative-definite limiting form requires the original $\lambda<0$, since the positive potential is $\widetilde\lambda=\sigma\lambda$. The example in Section 13 shows why that sign cannot be omitted.

Exterior uniqueness theorem. Suppose $\lambda>0$, $u\in H^1_{\mathrm{loc}}(X)$, and $$(1+|x|)^s D^\alpha u\in L^2(X) \quad\text{for every real }s\text{ and }|\alpha|\leq1. \tag{E4}$$ Suppose the distribution $(p-\lambda)u$ has a locally square-integrable representative away from the origin and satisfies $$|(p-\lambda)u|\leq C|x|^{-1}(|u|+|Du|) \quad\text{a.e. on }X\setminus\{0\}. \tag{E5}$$ Then $u=0$ almost everywhere on $X$.

The same conclusion holds if the estimate (E5) is imposed only near infinity and $|pu|\leq C_K(|u|+|Du|)$ is assumed on each compact subset of $X\setminus\{0\}$. The exterior proof uses (E5); propagation through bounded regions uses that local inequality. This more local sufficient formulation does not assert one global constant in bounded regions approaching a finite boundary of $X$. If the displayed equation is imposed on all of $X$, (E5) already supplies the local inequality on every such compact set. A value at the single point zero has no effect on an $H^1$ function as an almost-everywhere equivalence class.

Ellipticity on the whole continuation domain is explicit in (E2). The decay assumptions alone guarantee it sufficiently far out, since $\operatorname{Re}\sum a_{jk}\xi_j\xi_k\geq |\xi|^2/2$ there. They give no ellipticity guarantee in an arbitrary bounded part added to $X$. If only the exterior conditions are known, the conclusion proved below is vanishing on a sufficiently distant exterior region; the extension across a larger connected domain requires (E2) and the local equation there. This makes explicit the elliptic setting of the antecedent theorem.

The following prerequisites specify the current proof boundary.

• Fourier inversion and Plancherel. Sections 1–2 of Fourier transforms, finite spectra and convex separation supply the Fourier convention and Plancherel identity. We prove the parameter inequality used here.
• Weak elliptic regularity. Sections 7–8 of Local inverses and distance-weighted elliptic estimates define weak products with locally Lipschitz coefficients and give $H^1_{\mathrm{loc}},pu\in L^2_{\mathrm{loc}}\Rightarrow H^2_{\mathrm{loc}}$ under second-order ellipticity. Section 6 of Local inverses and distance-weighted elliptic estimates supplies the interior estimate with constants uniform under a common ellipticity bound and coefficient modulus of continuity. Its application on large annuli is proved below.
• Derivative exchange. Section 4 of Curved weights and the directions in which support can end gives the finite integration-by-parts identity which exchanges derivatives across one Lipschitz multiplier. The total order drops by one; individual orders remain at most the larger original order. Its unequal-order refinement is available, but the non-strict version suffices here. The sphere adaptation is detailed in Section 3.
• Polar geometry. Section 3 of Detecting a solution from infinite-order silence at one point establishes the polar vector fields, surface measure, and their adjoints. We state the identities used here and rederive the exterior change of variable.
• Entry tools. Finite-dimensional compactness, real spectral decomposition, positive-definite matrix square roots, smooth coordinate charts, finite smooth partitions on the compact sphere, smooth cutoffs on intervals, the weak chain rule, and density in local integer Sobolev spaces are entry assumptions.
• Support normals. Section 7 of Curved weights and the directions in which support can end gives support-normal exclusion for locally Lipschitz elliptic principal parts at the initial $H^{m-1}_{\mathrm{loc}}$ regularity; Section 8 of Curved weights and the directions in which support can end gives the quadratic root-count argument. We verify its applicability to a complex coefficient field approaching the identity, without assuming that the coefficients become real at a finite point.

We use the stated predecessor results and prove the exterior estimate below.

For the proof choose once and for all $$0<d<\min(\delta,1),\qquad 0<\eta<d. \tag{E6}$$ The original condition (E3) implies the same bounds with $d$ in place of $\delta$ when $|x|\geq1$. Thus using $d<1$ in an estimate imposes no extra restriction on the theorem. The number $\eta$ controls the curvature of the weight. All constants may depend on these two fixed numbers and on $\lambda,n,C_a$, but never on the subsequently large parameters $T_0$ and $\tau$.

2. A cylindrical operator with a growing potential

Set $x=e^t\omega$, $\omega\in S^{n-1}$, and $Z(t,\omega)=u(e^t\omega)$. Write $\Omega_j$ for the projection of $e_j$ onto the tangent space of the unit sphere. The identities in Section 3 of Detecting a solution from infinite-order silence at one point are $$\begin{gathered} \partial_{x_j}=e^{-t}(\omega_j\partial_t+\Omega_j),\qquad \Omega_j\omega_k=\delta_{jk}-\omega_j\omega_k,\qquad \sum_j\omega_j\Omega_j=0,\\ \Omega_j^*=-\Omega_j+(n-1)\omega_j,\qquad \Delta_S=\sum_j\Omega_j^2, \qquad -\langle\Delta_Sv,v\rangle=\sum_j\|\Omega_jv\|^2. \end{gathered} \tag{E7}$$ The last two formulas use surface measure. Individual $\Omega_j$ are not skew adjoint. Their divergence corrections cancel in the sum of squares because $\sum\omega_j\Omega_j=0$.

Twice applying the first identity, with the derivative of $e^{-t}$ retained, gives $$-e^{2t}pu= \sum_{j,k}a_{jk}(e^t\omega) (\omega_j\partial_t-\omega_j+\Omega_j) (\omega_k\partial_t+\Omega_k)Z. \tag{E8}$$ There is no derivative of $a_{jk}$ in this formula: (E1) has coefficients to the left of the derivatives. When $a_{jk}=\delta_{jk}$, expand (E8) and use (E7) to obtain $$\partial_t^2+(n-2)\partial_t+\Delta_S. \tag{E9}$$ For the full comparison (E3b), apply this calculation to $y=e^t\omega$ and $v(y)$, so the original point is $x=e^tM\omega$. Write $\widetilde a=I+K_\infty+\widetilde r$, retaining each of its three contributions. The exact frozen skew contribution is $$\begin{split} &\sum_{j,k}(K_\infty)_{jk} (\omega_j\partial_t-\omega_j+\Omega_j) (\omega_k\partial_t+\Omega_k)\\ &\qquad=e^{2t}\sum_{j,k}(K_\infty)_{jk} \partial_{y_j}\partial_{y_k}=0. \end{split} \tag{E8a}$$ The equality follows from twice applying the full polar formula (E7). Pairing indices in the last sum proves zero because the second derivatives commute and $K_\infty$ is skew. The surviving expression is (E9) plus every term from $\widetilde r$; (E3d) gives its two required decay bounds. In all subsequent formulas for this case, $U(T,\omega)=u(e^{t(T)}M\omega)$, the coefficient error comes from the complete $\widetilde r$, and the positive potential uses $\widetilde\lambda=\sigma\lambda$. No original matrix entry or sign has been silently replaced.

Every other coefficient in this cylindrical expression, together with its first cylindrical derivatives, is $O(e^{-(1+d)t})$. An angular or $t$ derivative of $a(e^t\omega)$ introduces one factor $e^t$, exactly balancing one power in the derivative bound of (E3). Derivatives of the smooth sphere factors are bounded.

Now introduce the increasing coordinate $$t=T-e^{-\eta T},\qquad h(T)=\frac{dt}{dT}=1+\eta e^{-\eta T}. \tag{E10}$$ For $T>0$, $|T-t|<1$, and $h$ and $h^{-1}$ are uniformly bounded. Put $U(T,\omega)=Z(t(T),\omega)$. Multiply the operator in (E8), with the term $\lambda e^{2t}$ added, by $h^2$. It becomes $$Q=\partial_T^2+c(T)\partial_T+a(T)\Delta_S+B+\Lambda(T), \quad B=\sum_{j+|\alpha|\leq2}b_{\alpha j}(T,\omega) \partial_T^j\Omega^\alpha, \tag{E11}$$ where $$\begin{split} a(T)&=h(T)^2,\\ c(T)&=(n-2)h(T)+\frac{\eta^2e^{-\eta T}}{h(T)},\\ \Lambda(T)&=\lambda e^{2T}e^{-2e^{-\eta T}}h(T)^2,\\ |b_{\alpha j}|+|\partial_Tb_{\alpha j}|+ \sum_l|\Omega_lb_{\alpha j}|&\leq C e^{-(1+d)T}. \end{split} \tag{E12}$$ Here $\Omega^\alpha$ denotes a fixed ordered word; we may sum over all words of the specified length. Indeed $h^2\partial_t^2=\partial_T^2-(h'/h)\partial_T$, with $h'=-\eta^2e^{-\eta T}$, which proves the formula for $c$. Smooth factors involving $h,h^{-1},h'$ preserve the last bound of (E12), including its weak first derivatives.

The potential is retained as a main term. Direct differentiation gives $$\frac{\Lambda'}{\Lambda} =2+2\eta e^{-\eta T} -\frac{2\eta^2e^{-\eta T}}{h(T)}\longrightarrow2, \quad \Lambda\asymp\lambda e^{2T},\quad \Lambda'\geq\lambda e^{2T} \tag{E13}$$ after increasing a fixed lower threshold for $T$. The sign $\lambda>0$ is used here.

In these coordinates the differential inequality becomes $$|QU|\leq C\left(e^T|U|+|\partial_TU|+ \sum_j|\Omega_jU|\right). \tag{E14}$$ To check the powers, before changing $t$, multiplication of (E5) by $e^{2t}$ gives a zeroth-order term $e^t|Z|$ and first-order terms $|(\omega_j\partial_t+\Omega_j)Z|$. The bounded factors in (E10) then give (E14). The norm convention on the cylinder is $dT\,d\omega$; it is not Euclidean measure. The latter is $$dx=e^{nt(T)}h(T)\,dT\,d\omega. \tag{E15}$$

For $\tau\geq1$, write $$V=e^{\tau T}U,\qquad Q_\tau=e^{\tau T}Qe^{-\tau T}. \tag{E16}$$ Thus every $\partial_T$ in (E11) is replaced by $\partial_T-\tau$, whereas angular derivatives are unchanged. Our target is $$\tau^2\|V\|^2+\|\partial_TV\|^2+ \sum_j\|\Omega_jV\|^2+\|e^TV\|^2 \leq C\left(e^{-T_0/2}+\tau^{-1}\right)\|Q_\tau V\|^2 \tag{E17}$$ for compactly supported smooth $V$ in $(T_0,\infty)\times S^{n-1}$. Since $(\tau+e^T)^2$ is comparable to $\tau^2+e^{2T}$, this is also the estimate with weight $(\tau+e^T)^{1-k}$ on each derivative of order $k\leq1$.

3. The error budget and one-derivative algebra

Let $\mathcal D^k V$ denote the finite family of all words of length $k$ in $\partial_T,\Omega_1,\ldots,\Omega_n$. A squared norm of this family means the sum of the squared norms of its members; $\mathcal D^0V=V$. On a fixed sphere chart these norms through order two are equivalent to coordinate Sobolev norms, with lower orders included. Define $$\begin{split} W(V)&=\tau^{-1}\|e^{-T}\mathcal D^2V\|^2 +\|\mathcal D^1V\|^2+\tau^2\|V\|^2,\\ F(V)&=\tau^3\|e^{-\eta T/2}V\|^2,\\ R(V)&=W(V)+\tau^3\|e^{-dT}V\|^2 +\tau\|e^{-dT}\mathcal D^1V\|^2. \end{split} \tag{E18}$$ The highest-order term uses exactly order two; it carries both $e^{-T}$ and $\tau^{-1}$. We will not replace it by an unweighted lower-order error.

Two elementary product estimates explain this choice. Consider a term $$I=\int e^{-(1+d)T}\tau^j |\mathcal D^rV|\,|\mathcal D^sV|\,dT\,d\omega, \quad j+r+s\leq3,\quad r,s\leq2,\quad j\geq0. \tag{E19}$$ All the indices here are nonnegative integers. If $r=2$, then $j+s\leq1$, and the integrand is the product $$\left(\tau^{-1/2}e^{-T}|\mathcal D^2V|\right) \left(\tau^{j+1/2}e^{-dT}|\mathcal D^sV|\right). \tag{E20}$$ The square of the second factor is bounded by the order-$s$ term in the decaying part of $R$, because $2j+1\leq3-2s$. Cauchy–Schwarz and $2ab\leq a^2+b^2$ bound (E19) by $CR$. The case $s=2$ is identical with the two factors reversed. If $r,s\leq1$, use $d<1$, hence $e^{-(1+d)T}\leq e^{-2dT}$ for $T>0$, and $j\leq3-r-s$. The product of $\tau^{3/2-r}e^{-dT}|\mathcal D^rV|$ and $\tau^{3/2-s}e^{-dT}|\mathcal D^sV|$ dominates the integrand. Their squares again occur in $R$. This proves a uniform bound for every monomial in (E19), not just a representative highest-order term.

We also need products containing the potential. If $k$ and its first derivatives are $O(e^{-(1+d)T})$, then $$|\Lambda k|+|\partial_T(\Lambda k)|+ \sum_l|\Omega_l(\Lambda k)| \leq C e^{(1-d)T}. \tag{E21}$$ Consequently, for every fixed $\theta>0$, $$\begin{split} C\int e^{(1-d)T} \left(\tau|V|^2+|\mathcal D^1V|\,|V|\right) &\leq \theta\|e^TV\|^2 +C_\theta\left(\tau^2\|V\|^2+\|\mathcal D^1V\|^2\right). \end{split} \tag{E22}$$ For the first integrand apply Young’s inequality to $e^T|V|$ and $\tau e^{-dT}|V|$; for the second use $e^T|V|$ and $e^{-dT}|\mathcal D^1V|$. The factors $e^{-dT}\leq1$ give the displayed right side. Constants in (E21) may contain the fixed energy $\lambda$.

Here is the exact algebra allowing only first coefficient derivatives. In a Euclidean chart, for $D=-i\partial$, the difference $$\int k\left(D^\alpha v\,\overline{D^\beta v} -D^\beta v\,\overline{D^\alpha v}\right) \tag{E23}$$ is a finite sum with multiplier $D_lk$, total derivative order $|\alpha|+|\beta|-1$, and each individual order no greater than $\max(|\alpha|,|\beta|)$. This is the derivative-exchange result in Section 4 of Curved weights and the directions in which support can end. It follows by transferring an unmatched derivative once; when the two factors have equal orders, interchange one derivative from each factor before continuing. Each transfer differentiates $k$ once in its own integral. One never differentiates that error again.

For a sphere chart with smooth density $\rho\,dy$, replace $k$ by $\rho k$. Its derivative is $\rho D_lk+kD_l\rho$, so the estimate contains both $|dk|$ and $|k|$, with bounded chart constants. Expanding a word of sphere fields into coordinate derivatives produces its leading coordinate monomials and lower-order terms with bounded smooth coefficients. Reordering two sphere fields costs their commutator, a smooth first-order field. These costs also lower the total derivative count by one. The finite atlas has uniform bounds on these smooth factors.

For later use, attach degree one both to $\tau$ and to each derivative. A differential polynomial of degree at most two and its formal complex conjugate have equal squared principal symbols. Expanding their squared norms, pairing terms with indices $(\alpha,\beta)$ and $(\beta,\alpha)$, and using (E23) therefore removes every term of total degree four without differentiating a coefficient twice. The remaining terms have the form (E19). If one factor was multiplication by $\Lambda$, the degree-two exchange leaves at most one derivative on $V$, or one power of $\tau$ and no derivative, and (E21) gives exactly (E22). Terms of degree below two already satisfy the same counts. This proves the assertion for every pair of monomials; linearity and the finite number of monomials prove it for their sums.

In this argument “formal complex conjugate” is an algebraic operation on the displayed polynomial, not the assertion that the resulting operator is the Hilbert adjoint. We never expand the actual adjoint of a rough second-order monomial: doing so could introduce distributional second derivatives of its coefficient. Instead each paired integral exchange differentiates its combined multiplier once. Reversing the order of smooth sphere fields and using their divergence corrections costs only the lower-degree geometric terms described above. This distinction is essential for Lipschitz coefficients.

4. A Fourier inequality across a characteristic shell

Let $y=(y',y_n)\in\mathbb R^n$, $A,\tau\geq1$, $K>0$, $M\geq0$, and $$L_\tau=\Delta_{y'}+(\partial_{y_n}-\tau)^2.$$ Assume the four conditions $$\begin{gathered} A^{-\eta}\tau^2\leq\tau^3/K+A^2, \qquad \tau^3/K\leq A^{-\eta}\tau^2+A^2,\\ M^2K\leq\tau A^2, \qquad M\leq A^{1-\eta/2}. \end{gathered} \tag{E24}$$ Then, for $v\in C_c^\infty(\mathbb R^n)$, $$\begin{split} &K^{-1}\sum_{|\alpha|\leq2}\tau^{4-2|\alpha|}\|D^\alpha v\|^2 +A^{-\eta}\tau^3\|v\|^2 +M\sum_{|\alpha|\leq1}\tau^{2-2|\alpha|}\|D^\alpha v\|^2\\ &\quad\leq C\left(K^{-1}\|L_\tau v\|^2 +\tau A^{-\eta}\sum_{j<n}\|D_jv\|^2 +\tau A^2\|v\|^2\right). \end{split} \tag{E25}$$ The constant depends only on dimension. The same proof works in dimension one, with the angular sum empty.

Proof. Put $S=|\xi|^2+\tau^2$ and $$q(\xi)=|\xi|^2-\tau^2+2i\tau\xi_n.$$ The Fourier symbol of $L_\tau$ is $-q$, so its modulus is $|q|$. The finite monomial sums on the left of (E25) are comparable, with dimensional constants, to $$K^{-1}S^2+A^{-\eta}\tau^3+MS. \tag{E26}$$ It suffices to bound this by $C(K^{-1}|q|^2+\tau A^{-\eta}|\xi'|^2+\tau A^2)$.

First suppose $$\left||\xi'/\tau|^2-1\right|\geq\tfrac12 \quad\text{or}\quad |\xi_n/\tau|\geq\tfrac12. \tag{E27}$$ On this set $S\leq C|q|$. For completeness, divide by $\tau^2$ and put $z=\xi/\tau$. For $|z|\geq2$, the real part $|z|^2-1$ bounds a fixed multiple of $1+|z|^2$. On $|z|\leq2$, the closed subset (E27) is compact and contains no zero of $|z|^2-1+2iz_n$: such a zero has $z_n=0$ and $|z'|=1$, excluded by (E27). Its modulus therefore has a positive minimum. These two regions prove the claimed uniform bound. The first condition in (E24), multiplied by $\tau$, controls $A^{-\eta}\tau^3$ by $\tau^4/K+\tau A^2$. Since $S\geq\tau^2$, its first term is bounded by $CS^2/K$, and hence by $C|q|^2/K$. Finally $$MS\leq\tfrac12K^{-1}S^2+\tfrac12M^2K \leq\tfrac12K^{-1}S^2+\tfrac12\tau A^2. \tag{E28}$$ This proves the bound away from the shell.

On the complementary set, $|\xi'|^2\geq\tau^2/2$, $|\xi'|^2\leq3\tau^2/2$, and $|\xi_n|\leq\tau/2$. Thus $S\asymp\tau^2$. The second condition in (E24), multiplied by $\tau$, controls $\tau^4/K$ by $\tau^3A^{-\eta}+\tau A^2$. The angular term already controls $A^{-\eta}\tau^3$. The last condition gives $$M\tau^2\leq A^{1-\eta/2}\tau^2 =\sqrt{(\tau^3A^{-\eta})(\tau A^2)} \leq\tfrac12\tau^3A^{-\eta}+\tfrac12\tau A^2. \tag{E29}$$ Every term in (E26) is now bounded. The two sets may share their boundaries; the constants are uniform there. Multiply the pointwise inequality by $|\widehat v|^2$, integrate, and apply Plancherel. Its fixed normalization factor cancels from both sides. This proves (E25). $\square$

5. Choosing the radius-dependent coefficient

The potential in (E11) is of size $A^2$ on a slab where $e^T\asymp A$. To place its square in the $\tau A^2\|V\|^2$ term of (E25), we also require $$A^2/K\leq\tau. \tag{E30}$$ An admissible choice, of the smallest order allowed by the lower bounds, is $$K= \begin{cases} A^\eta\tau,& A^{2+\eta}\leq\tau^2,\\ \tau^3/A^2,& A^2\leq\tau^2\leq A^{2+\eta},\\ A^2/\tau,& \tau^2\leq A^2. \end{cases} \tag{E31}$$ Both formulas agree at each transition. This is an optimization up to absolute constants, not a claim of an exact unique minimizer.

Here are all the checks. In the first region $\tau^3/K=A^{-\eta}\tau^2$, so the first two conditions in (E24) hold, and (E30) follows from $A^{2-\eta}\leq\tau^2$. In the middle region $\tau^3/K=A^2$, while $A^{-\eta}\tau^2\leq A^2$; the first two conditions again hold. Moreover $A^2/K=A^4/\tau^3\leq\tau$ since $A^2\leq\tau^2$. In the last region $A^2/K=\tau$ and $\tau^3/K=\tau^4/A^2\leq A^2$; also $A^{-\eta}\tau^2\leq A^2$, proving both conditions.

In these three regions, respectively, $$\frac{\tau A^2}{K} =A^{2-\eta},\quad A^4/\tau^2\geq A^{2-\eta}, \quad\tau^2. \tag{E32}$$ Therefore the remaining two conditions in (E24) hold whenever $$M\leq\min(A^{1-\eta/2},\tau). \tag{E33}$$ The inequality $K\geq A$ also holds in every region. In the first, $\tau\geq A^{1+\eta/2}$ and hence $K\geq A^{1+3\eta/2}$. In the middle, $\tau\geq A$, giving $\tau^3/A^2\geq A$. In the last, $\tau\leq A$, giving $A^2/\tau\geq A$.

For the eventual estimate set $$M=\left(e^{-T_0/2}+\tau^{-1}\right)^{-1}. \tag{E34}$$ This has $M\leq\tau$ and $M\leq e^{T_0/2}$. On a slab $|T-\log A|<1$ meeting $T>T_0$, one has $A>e^{T_0-1}$. Since $\eta<1$, increasing the fixed threshold for $T_0$ ensures $$e^{T_0/2}\leq e^{(1-\eta/2)(T_0-1)}\leq A^{1-\eta/2}.$$ Thus (E33) holds uniformly for every slab. Also $M\geq1$ when both $T_0$ and $\tau$ are sufficiently large. There is no factor $\eta$ in the exponent $-T_0/2$ in (E34).

To verify the asserted scale optimality, the second condition in (E24) and (E30) imply $$K\geq\max\left(\frac{A^2}{\tau}, \frac{\tau^3}{A^{-\eta}\tau^2+A^2}\right). \tag{E35}$$ When $A^{-\eta}\tau^2\geq A^2$, the second term is between $A^\eta\tau/2$ and $A^\eta\tau$, and dominates the first up to constants. When $A^{-\eta}\tau^2\leq A^2$, it is between $\tau^3/(2A^2)$ and $\tau^3/A^2$; comparison with $A^2/\tau$ divides this range at $\tau^2=A^2$. This proves that (E31) is within a factor two of the necessary lower bound.

6. Derivative recovery in one chart and one slab

Rotate the sphere to center a chart at its north pole and use $y=(\omega_1,\ldots,\omega_{n-1})$ with $|y|<r<1$. The last coordinate is $\sqrt{1-|y|^2}$. The induced metric and its inverse give $$\begin{split} d\omega&=(1-|y|^2)^{-1/2}\,dy,\\ \Delta_S&=\sum_{i,j<n}(\delta_{ij}-y_iy_j)\partial_i\partial_j -(n-1)\sum_{i<n}y_i\partial_i,\\ \sum_j|\Omega_jv|^2 &=\sum_{i,j<n}(\delta_{ij}-y_iy_j) \partial_iv\,\overline{\partial_jv}. \end{split} \tag{E36}$$ For example, the density is the square root of the determinant of $I+yy^t/(1-|y|^2)$, and the inverse matrix is $I-yy^t$. Substituting these into the divergence formula for the Laplace–Beltrami operator gives the middle identity. Thus the coordinate gradient norm and the angular gradient norm are comparable with fixed constants on a fixed small chart. The same is true through order two after adding lower orders, by expanding each field and expressing each coordinate derivative in the spanning fields.

Suppose $V$ is supported in this chart and in $|T-\log A|<1$, $T>T_0$. In these coordinates write $$Q_\tau=L_\tau+E_2+E_1+\Lambda(T), \quad L_\tau=\Delta_y+(\partial_T-\tau)^2. \tag{E37}$$ The polynomial $E_2$ has total degree at most two in derivatives and $\tau$, with coefficients bounded by $C(r+e^{-\eta T_0})$. It includes the variable principal coefficients and all their conjugated $\tau$-terms. The polynomial $E_1$ has total degree at most one, with uniformly bounded coefficients; it includes $c(\partial_T-\tau)$ and the bounded first-order sphere terms. To see these bounds directly, subtract the leading constant matrix in (E36), use $a-1=O(e^{-\eta T})$, and use (E12). No derivative of a rough coefficient is used in this step.

Put $$H_K(V)=K^{-1}\sum_{|\alpha|\leq2} \tau^{4-2|\alpha|}\|D^\alpha V\|^2$$ in the coordinate chart. Expanding the finite polynomials in (E37) gives $$K^{-1}\|E_2V\|^2\leq C(r+e^{-\eta T_0})^2H_K(V),\qquad K^{-1}\|E_1V\|^2\leq C\tau^{-2}H_K(V). \tag{E38}$$ Use (E25) and $L_\tau V=Q_\tau V-E_2V-E_1V-\Lambda V$. Choose the fixed chart radius small first, then choose fixed sufficiently large lower thresholds for $T_0$ and $\tau$, so the contributions (E38) consume at most half of $H_K$. The equivalence of the smooth density norms changes only fixed constants. The potential contributes $$K^{-1}\|\Lambda V\|^2\leq C A^4K^{-1}\|V\|^2 \leq C\tau A^2\|V\|^2, \tag{E39}$$ where (E30) is used. This is a bound by the designated potential term on the right, with no smallness assertion about $\Lambda$.

On the slab, powers of $A$ and $e^T$ are comparable by constants determined by its fixed width. After (E38) has been absorbed, one may drop the factors $\tau^{4-2|\alpha|}\geq1$ in the retained first sum. We obtain $$\begin{split} &K^{-1}\sum_{k=0}^2\|\mathcal D^kV\|^2+F(V) +M\bigl(\tau^2\|V\|^2+\|\mathcal D^1V\|^2\bigr)\\ &\quad\leq C\left(K^{-1}\|Q_\tau V\|^2 +\tau\sum_j\|e^{-\eta T/2}\Omega_jV\|^2 +\tau\|e^TV\|^2\right). \end{split} \tag{E40}$$ The first sum here deliberately has no remaining powers of $\tau$. Its stronger predecessor was needed for the coefficient absorption; (E40) alone would not justify that absorption.

Now use (E31) and (E34). The inequality $K\geq A$ gives $K^{-1}\leq A^{-1}\asymp e^{-T}$. The third condition in (E24) gives $$K^{-1}\geq\frac{M^2}{\tau A^2} \geq\frac{M}{\tau A^2}, \tag{E41}$$ where $M\geq1$ is used in the second step. Retain this lower bound only for order two in (E40), and keep the displayed $M$-weighted lower orders. It follows that $$\begin{split} M W(V)+F(V)\leq C\bigg(&\|e^{-T/2}Q_\tau V\|^2 +\tau\sum_j\|e^{-\eta T/2}\Omega_jV\|^2 +\tau\|e^TV\|^2\bigg). \end{split} \tag{E42}$$ This proves the required derivative estimate on each chart and slab, with constants independent of their location.

7. Localization with constants independent of support

We now prove (E42) for every $V\in C_c^\infty((T_0,\infty)\times S^{n-1})$. Choose finitely many real smooth functions $\phi_j$, supported in charts of the fixed size used above, with $\sum_j\phi_j^2=1$. Such a square partition is obtained from any nonnegative subordinate partition by dividing its members by the square root of the sum of their squares. The denominator is smooth and strictly positive on the compact sphere.

Choose a real nonnegative smooth function supported in $(-1,1)$, positive on $[-3/4,3/4]$. Normalize it by the square root of the sum of the squares of its integer translates. The result is a smooth $\psi$, with $\operatorname{supp}\psi\subset(-1,1)$, such that $$\sum_{k\in\mathbb Z}\psi(T-k)^2=1.$$ The denominator is smooth, periodic, and bounded away from zero. All derivatives of $\psi(T-k)$ of any fixed order are bounded uniformly in $k$. Put $$\chi_{jk}(T,\omega)=\phi_j(\omega)\psi(T-k),\qquad \sum_{j,k}\chi_{jk}^2=1. \tag{E43}$$ Only a bounded number of these functions meet a point. A slab meeting $\operatorname{supp}V$ has $k>T_0-1$, so (E34) is admissible with $A=e^k$, including slabs whose lower endpoint is below $T_0$. The function $\chi_{jk}V$ itself still has compact support above $T_0$.

Apply (E42) to these functions and sum. Zeroth-order terms sum exactly. For each real field $Y$ in $\mathcal D$, expansion and $\sum\chi_{jk}Y\chi_{jk}=0$ give $$\sum_{j,k}|Y(\chi_{jk}V)|^2 =|YV|^2+\sum_{j,k}|Y\chi_{jk}|^2|V|^2. \tag{E44}$$ For order two, the product rule shows that $\chi_{jk}Y_1Y_2V$ equals $Y_1Y_2(\chi_{jk}V)$ minus terms bounded by $C(|\mathcal D^1V|+|V|)$ on the cutoff support. Squaring, summing, and using bounded overlap yields $$W(V)\leq C\sum_{j,k}W(\chi_{jk}V) +C\tau^{-1}e^{-2T_0} (\|\mathcal D^1V\|^2+\|V\|^2).$$ The last term is absorbed into $W(V)$ after increasing the parameter thresholds. Thus the summed left sides control $MW(V)+F(V)$, up to fixed constants.

The growing potential commutes with all cutoffs. In the other terms of $Q_\tau$, two derivatives may hit $\chi$, or one may hit $\chi$ and one may remain on $V$. Conjugation supplies at most one $\tau$ in a nonzero commutator, since a $\tau^2$ term is multiplication. Explicitly, $$[\partial_T^2,\chi]V=2\chi_TV_T+\chi_{TT}V,\qquad [-2\tau\partial_T,\chi]V=-2\tau\chi_TV,$$ and $[Y_1Y_2,\chi]V=(Y_1Y_2\chi)V+(Y_2\chi)Y_1V+(Y_1\chi)Y_2V$. Coefficients stay on the left and are uniformly bounded; none is differentiated. It follows that $$\sum_{j,k}\|e^{-T/2}[Q_\tau,\chi_{jk}]V\|^2 \leq C e^{-T_0} (\|\mathcal D^1V\|^2+\tau^2\|V\|^2) \leq C e^{-T_0}W(V). \tag{E45}$$ The angular right-side terms in (E42) contribute, besides a fixed multiple of the original angular norm, at most $C\tau\|e^{-\eta T/2}V\|^2=C\tau^{-2}F(V)$. The potential norm sums exactly. Therefore the only additional right-side errors are $Ce^{-T_0}W+C\tau^{-2}F$. They are absorbed by $MW+F$ for large thresholds, since $M\geq1$. This proves the global estimate (E42). The number of slabs occupied by $V$ never enters its constant.

8. A norm identity revealing the two positive terms

For the moment set $B=0$ in (E11). Define $$\begin{split} H&=\partial_T^2+a(T)\Delta_S+\tau^2-c(T)\tau+\Lambda(T),\\ J&=-\beta(T)\partial_T,\qquad \beta(T)=2\tau-c(T). \end{split} \tag{E46}$$ Then $Q_{0,\tau}=H+J$, and its algebraically reversed companion is $Q_{0,\tau}^-=H-J$. With respect to $dT\,d\omega$, $H^*=H$ and $J^*=-J-c'$. Hence $$\|Q_{0,\tau}V\|^2-\|Q_{0,\tau}^-V\|^2 =4\operatorname{Re}\langle HV,JV\rangle =2\operatorname{Re}\langle[H,J]V,V\rangle -2\operatorname{Re}\langle c'HV,V\rangle. \tag{E47}$$ The real part makes this identity independent of which slot of the complex inner product is chosen linear.

We compute it completely. Since $\beta'=-c'$, $$\begin{split} [\partial_T^2,-\beta\partial_T]&=2c'\partial_T^2+c''\partial_T,\\ [a\Delta_S,-\beta\partial_T]&=\beta a'\Delta_S,\\ [\tau^2-c\tau+\Lambda,-\beta\partial_T] &=\beta(-\tau c'+\Lambda'). \end{split} \tag{E48}$$ These derivatives concern the smooth scalar functions $a,c,\Lambda$, not a rough principal coefficient. Insert (E48) in (E47). Integration by parts in $T$ combines $2c'\partial_T^2+2c''\partial_T$ into $-2c'|V_T|^2$; the angular integration uses (E7). Thus the exact identity is $$\begin{split} &\|Q_{0,\tau}V\|^2-\|Q_{0,\tau}^-V\|^2\\ &=\int\bigg[ -2c'|V_T|^2 -2(\beta a'-c'a)\sum_j|\Omega_jV|^2\\ &\hspace{25mm} +\Big((-6\tau^2+4c\tau)c' +2\beta\Lambda'-2c'\Lambda\Big)|V|^2 \bigg]\,dT\,d\omega. \end{split} \tag{E49}$$ This identity is also a useful sign check on the direction of the curved radial adjustment.