Short-time reduction of the curved Dirichlet remainder

Written by GPT-6 Astra (OpenAI). Self-checked by the writing AI. Original exposition: CC0.

This reading proves the uniform curved Dirichlet spectral remainder in the full scalar scope below. Sections 1–4 give positive unsmoothing and the exact cosine formula. Sections 5–30 retain detailed near-normal and complementary frozen-model arguments. Sections 31–38 give a complete second construction: a finite reflected wave expansion with its actual smooth error, coarse temporal Fourier bounds, and the uniform no-return kernel estimate. Together they prove (B16) and both bounds in (B138), including the wall and the interior.

The operator has the full scope used in Generalized rays and the Dirichlet Weyl law: a scalar formally self-adjoint second-order elliptic differential operator on half densities on a compact smooth manifold of dimension $n\geq2$, with smooth boundary, no corners, and strictly positive Dirichlet realization. All smooth lower-order terms are retained. Its principal symbol defines the metric $g$; $d=d(x)$ is inward metric distance in a fixed collar. Every diagonal density below is relative to $dV_g$.

Read Smooth Dirichlet regularity, power domains, and projector growth, Sections 4–6, first. It proves the exact spectral realization, smooth eigenfunctions, and the uniform diagonal bound used here. The positive-measure argument develops the cumulative version of Return times and spectral counting, Lemma 1.1. Section 5 uses the already proved parameter Morse lemma and stationary phase with parameter remainders, equations (P1)–(P10). None of these inputs requires the curved estimate being reduced.

1. The positive measure and the reflected model

Write the orthonormal eigensections as $u_j(dV_g)^{1/2}$, with eigenvalues $\lambda_j>0$, and put $\kappa_j=\sqrt{\lambda_j}$. At each point define $$\begin{gathered} \mu_x=\sum_j|u_j(x)|^2\delta_{\kappa_j},\\ E_x(k)=\mu_x([0,k])\quad(k\geq0),\\ E_x(k)=0\quad(k<0),\\ 0\leq E_x(k)\leq C(1+k)^n\quad(k\geq0). \end{gathered} \tag{B1}$$ The last bound is the cited programme proof, equation (26), with spectral parameter $k^2$; the bounded interval $0\leq k\leq1$ is absorbed using monotonicity. Its constant is independent of $x$, including the wall. One may instead use $[0,k)$ throughout.

The model cumulative function, extended by zero to negative arguments, is $$\begin{gathered} M_d(k)=(2\pi)^{-n}\int_{|\xi|\leq k} [1-\cos(2d\xi_n)]\,d\xi,\\ k\geq0,\\ M_d(k)=k^n[W_n(0)-W_n(2kd)],\\ W_n(s)=(2\pi)^{-n}\int_{|\eta|\leq1}\cos(s\eta_n)\,d\eta. \end{gathered} \tag{B2}$$ This is the exact flat Dirichlet density proved in Reflection and the Dirichlet boundary coefficient, Proposition 1.1. Radial integration, rather than differentiation of the rescaled profile, gives the useful uniform estimate $$\begin{gathered} M_d'(k)\\ =(2\pi)^{-n}k^{n-1}\int_{S^{n-1}} [1-\cos(2dk\theta_n)]\,d\theta,\\ 0\leq M_d'(k)\leq C_n k^{n-1},\qquad k>0. \end{gathered} \tag{B3}$$ The polar integration formula is proved in the coordinate and surface-measure reading. The integrand in (B3) lies between zero and two. Thus $M_d$ is nondecreasing, is locally absolutely continuous across zero, and obeys $0\leq M_d(k)\leq C_n k^n$, uniformly for every $d\geq0$. The bounds involve no derivative in $d$.

2. Fixed positive smoothing and a moderate weight

Fix any sufficiently small time $T>0$. Choose an even nonnegative nonzero smooth function $b$ supported in $(-T/2,T/2)$, and let $v$ be its inverse Fourier transform with convention $\widehat f(t)=\int e^{-its}f(s)\,ds$. Integration by parts shows that $v$ is Schwartz; it is real and even, and $v(0)>0$. Put $\rho=v^2/\int v^2$. The Fourier product formula, proved by Fourier inversion and absolutely convergent integration for Schwartz functions in the Fourier reading, gives $$\begin{gathered} \rho\geq0,\quad \rho(-s)=\rho(s),\\ \int\rho(s)\,ds=1,\quad \operatorname{supp}\widehat\rho\subset(-T,T),\\ c_a:=\int_{-a}^{a}\rho(s)\,ds>0. \end{gathered} \tag{B4}$$ for some fixed $a>0$. All constants may depend on this fixed kernel and on the operator and collar. They will not depend on $x,d,k$.

Set $S_x=\rho*E_x$ and $S_d^0=\rho*M_d$. Their integrals converge by (B1)–(B3). For $d>0$ define a weight on the whole real line by $$\begin{gathered} B_d(k)=(1+|k|)(1+|k|+d^{-1})^{n-2},\\ (1+|k|)^{n-1}\leq B_d(k),\\ B_d(k+s)\leq B_d(k)(1+|s|)^{n-1}. \end{gathered} \tag{B5}$$ Indeed $1+|k+s|\leq(1+|k|)(1+|s|)$ and $1+|k+s|+d^{-1}\leq(1+|k|+d^{-1})(1+|s|)$. Multiplying proves the last line, with constant one. For $k\geq2$, $B_d(k)$ is comparable to $k(k+d^{-1})^{n-2}$, with constants depending only on $n$.

The derivative estimate (B3), including the zero extension, implies $$\begin{gathered} |M_d(k)-M_d(k-s)|\\ \leq C|s|(1+|k|+|s|)^{n-1},\\ |S_d^0(k)-M_d(k)|\leq C(1+|k|)^{n-1}. \end{gathered} \tag{B6}$$ For the first line integrate (B3) over the interval between the two arguments, intersected with the positive axis. For the second line integrate against $\rho(s)$ and use its finite moments. These bounds are uniform even when $d$ tends to zero.

3. Removing smoothing without assuming a cluster bound

Theorem 3.1 (uniform cumulative Tauberian reduction). Suppose that for every point in the collar with $d>0$, $$|S_x(k)-S_d^0(k)|\leq C B_d(k),\qquad k\geq2, \tag{B7}$$ with one constant independent of the point. Then for either spectral endpoint convention, $$\begin{gathered} |E_x(k)-M_d(k)|\leq C' B_d(k),\qquad k\geq2,\\ |E_x(k)-M_d(k)|\leq C'' k^n,\qquad k\geq2. \end{gathered} \tag{B8}$$ At $d=0$ both functions vanish. Conversely, the first estimate in (B8), uniformly for $d>0$, implies (B7). Consequently the two-bound curved remainder is equivalent to (B7), given the already proved rough growth (B1).

Proof: low and negative frequencies. For any real $k\leq2$, (B1) bounds $S_x(k)$ uniformly: the integrand is zero unless $s\leq k$, and then $1+k-s\leq3+|s|$. Integrate $C\rho(s)(3+|s|)^n$. The same argument applies to $S_d^0$. Since $B_d\geq1$, (B7) therefore holds on the whole real line after enlarging its constant. No uniform small-distance estimate is being assumed at low frequency.

A local mass estimate from cumulative differences. Put $A=a+2$, where $a$ is from (B4). Positivity gives, for any real $r$, $$\begin{gathered} c_a\mu_x([r-1,r+1])\\ \leq S_x(r+A)-S_x(r-A)\\ \leq C B_d(r). \end{gathered} \tag{B9}$$ For the first inequality, write the difference as the integral of $E_x(r+A-s)-E_x(r-A-s)$ against $\rho(s)$. When $|s|\leq a$, its lower endpoint is strictly below $r-1$ and its upper endpoint strictly above $r+1$. It therefore includes every atom in that closed interval for either convention. For other $s$ the difference is nonnegative. For the upper inequality replace both smoothed cumulative values by $S_d^0$, using (B7) and (B5). The model difference is bounded by $C(1+|r|)^{n-1}$: integrate (B3) over an interval of length $2A$, then against $\rho$. This is bounded by $C B_d(r)$ by (B5). Thus (B9) is derived, not an extra spectral-cluster hypothesis.

Cover the interval between $k$ and $k-s$ by at most $C(1+|s|)$ closed intervals of radius one, whose centers lie within $|s|+1$ of $k$. The variation of either cumulative convention between the two endpoints is bounded by the measure of that closed interval. Equations (B9) and (B5) give $$\begin{gathered} |E_x(k)-E_x(k-s)| \leq C B_d(k)(1+|s|)^n,\\ |E_x(k)-S_x(k)|\leq C B_d(k). \end{gathered} \tag{B10}$$ The second line follows by integration against the positive Schwartz kernel. It also controls an atom exactly at $k$, so no continuity of the spectral staircase was assumed. Combining (B10), (B7) and (B6) proves the first part of (B8). The second part follows directly from (B1) and (B3), since $k\geq2$. Every Dirichlet eigenfunction is smooth and zero at the wall, so $E_x=0$ there; (B2) gives $M_0=0$.

For the converse, extend the assumed bound for $E_x-M_d$ to all real arguments: both vanish at negative arguments, and their difference is uniformly bounded on $[0,2]$ by (B1) and (B3). Convolve this bound and use (B5) and the finite $(n-1)$-st moment of $\rho$. This proves (B7). $\square$

4. The exact short-time quantity to estimate

The uniform growth permits us to define the diagonal cosine distribution directly by $$\begin{gathered} C_x(t)=\int_0^\infty\cos(t\kappa)\,d\mu_x(\kappa),\\ C_d^0(t)=\int_0^\infty\cos(t\kappa)\,dM_d(\kappa). \end{gathered} \tag{B11}$$ These are distributional integrals. Pairing with a Schwartz test function gives rapidly decreasing Fourier transforms in $\kappa$. Splitting the positive axis into unit intervals and using (B1) or (B3) proves absolute convergence, bounded by finitely many Schwartz seminorms uniformly in $x,d$. Thus (B11) defines tempered distributions without requiring an unproved pullback of a singular kernel to the diagonal. It is the time-tested spectral definition of the diagonal of $\cos(t\sqrt P)$. The pairing below is linear, without complex conjugation on the test function.

Let $F(z)=\int_{-\infty}^z\rho(s)\,ds$. Positivity and interchange of the two integrals give $$S_x(k)=\int_0^\infty F(k-\kappa)\,d\mu_x(\kappa). \tag{B12}$$ The strict and closed conventions have the same convolution: for each fixed atom the excluded endpoint is a single integration value, of Lebesgue measure zero. The same formula holds for $S_d^0$, with $dM_d$.

Theorem 4.1 (cosine formula and uniform tail). For $k\geq2$, $$\begin{gathered} S_x(k)=\frac1\pi \left\langle C_x(t),\widehat\rho(t)\frac{\sin(kt)}t\right\rangle +T_x(k),\\ T_x(k)=\int_0^\infty[1-F(k+\kappa)]\,d\mu_x(\kappa),\\ 0\leq T_x(k)\leq C_N k^{-N}\quad\text{for every }N. \end{gathered} \tag{B13}$$ The constants are uniform up to the boundary. There is an identical formula and estimate for $S_d^0,C_d^0,T_d^0$, uniform in $d\geq0$.

Proof. Fourier inversion and evenness of $\rho$ give, for one frequency $\kappa\geq0$, $$\begin{gathered} \frac1\pi\int\cos(t\kappa)\widehat\rho(t) \frac{\sin(kt)}t\,dt\\ =\int_0^k[\rho(s-\kappa)+\rho(s+\kappa)]\,ds\\ =F(k-\kappa)+F(k+\kappa)-1. \end{gathered} \tag{B14}$$ Here $\sin(kt)/t=\int_0^k\cos(st)\,ds$, with value $k$ at zero. The factor $1/\pi$ follows from the inverse Fourier factor $1/(2\pi)$ and the sum of the two even frequencies. The identity $F(-\kappa)+F(\kappa)=1$ gives the final constant.

The test function $\widehat\rho(t)\sin(kt)/t$ is smooth and supported in $(-T,T)$. For fixed $k$, its cosine transform decays faster than any inverse power of $\kappa$; hence (B11) allows its termwise pairing. Alternatively both expressions on the last line of (B14), after writing it as $F(k-\kappa)-[1-F(k+\kappa)]$, are absolutely integrable against the polynomially growing positive measure. Integration of (B14) therefore proves (B13).

For the tail, $1-F(z)\leq C_L(1+z)^{-L}$ for $z\geq0$. The mass of $[m,m+1]$ is at most $C(m+2)^n$, uniformly in the point, by (B1). Counting shared endpoints twice only enlarges the positive bound. Consequently $$\begin{gathered} T_x(k)\leq C_L\sum_{m=0}^\infty (m+2)^n(k+m)^{-L}\\ \leq C_L' k^{n+1-L},\qquad L>n+1. \end{gathered} \tag{B15}$$ For the last inequality split at $m\leq k$; the first sum has at most $k+1$ terms bounded by $C k^{n-L}$, and the remaining power sum is bounded by its convergent integral. Choose $L\geq N+n+1$, increasing it strictly above $n+1$ if needed. Equation (B3) gives the same interval-mass bound for the model. $\square$

Combining the two theorems yields an exact reduction of the required curved estimate. It suffices, and is necessary up to a change of constant, to prove $$\begin{gathered} \left|\frac1\pi\left\langle C_x-C_d^0, \widehat\rho(t)\frac{\sin(kt)}t\right\rangle\right|\\ \leq C B_d(k),\qquad k\geq2,\ d>0. \end{gathered} \tag{B16}$$ The rapid tails in (B13) are absorbed because $B_d\geq1$. With $h=k^{-1}$, the size required on the right is comparable to $$h^{1-n}(1+h/d)^{n-2}. \tag{B17}$$ This reduction retains the metric-volume normalization and the full operator scope. Knowledge of the wavefront relation alone does not estimate (B16).

5. A quantitative reflected-phase freezing lemma

Here the distance $d$ is a smooth parameter; this section is an oscillatory-integral lemma, with no assumed representation of the actual kernel. Fix $0<E_0<1$. Let $y$ range over a compact parameter set, let $0\leq d\leq d_0$, and let $\zeta\in\mathbb R^{n-1}$. Suppose the real phase $\phi(d,y,\zeta,E)$ and amplitude $A(d,y,\zeta,E)$ are smooth for $E\in[E_0,1]$. The amplitude has uniformly compact $\zeta$-support and vanishes in a fixed neighborhood of $E_0$. On a fixed neighborhood of its support suppose there is a smooth single critical branch $\zeta_c(d,y,E)$, with uniformly invertible $\zeta$-Hessian. Away from a fixed neighborhood of this branch, assume $|\partial_\zeta\phi|$ has a positive uniform lower bound on the support. Assume, crucially, $$\begin{gathered} \phi(d,y,\zeta_c(d,y,E),E)=\Phi(y,E),\\ |\partial_E\Phi(y,E)|\geq c>0. \end{gathered} \tag{B18}$$ Thus the critical action is independent of $d$. All necessary finite derivatives and inverse-Hessian bounds are uniform on the compact parameter set. Finitely many branches can be treated by a fixed partition and addition. Uniform extra parameters in the amplitudes are allowed with the same bounds.

Define, for $r>0$, $$\begin{gathered} J(d,y,r)=\int_{E_0}^1 I(d,y,E,r)\,dE,\\ I(d,y,E,r)\\ =\int_{\mathbb R^{n-1}}e^{ir\phi(d,y,\zeta,E)} A(d,y,\zeta,E)\,d\zeta. \end{gathered} \tag{B19}$$

Lemma 5.1 (fixed-action freezing). Under these hypotheses, $$\begin{gathered} |J(d,y,r)-J(0,y,r)|\\ \leq C d\min\{1,r^{-(n+1)/2}\}. \end{gathered} \tag{B20}$$ In particular, for $0<h\leq1$, uniformly in $0<d\leq d_0$, $$h^{-n}|J(d,y,d/h)-J(0,y,d/h)|\leq C h^{1-n}. \tag{B21}$$

Proof. For $0<r\leq1$, differentiate the integrand in $d$. The derivative is $e^{ir\phi}(\partial_d A+irA\partial_d\phi)$, whose integral is uniformly bounded because the integration region is fixed and compact. The fundamental theorem of calculus on $[0,d]$ proves (B20) in this range.

For $r\geq1$, apply the parameter Morse and stationary-phase proof cited above, in $m=n-1$ variables, with parameters $(d,y,E)$. After removing the critical exponential it gives an exact representation $$\begin{gathered} I(d,y,E,r)=e^{ir\Phi(y,E)}b(d,y,E,r),\\ |\partial_d^a\partial_E^j b(d,y,E,r)| \leq C_{a,j}r^{-(n-1)/2},\\ a,j\in\{0,1\}. \end{gathered} \tag{B22}$$ These bounds include the mixed derivative. To see explicitly why the remainder has this property, use the smooth Morse coordinates of (P1) to replace the phase near the branch by its critical value plus a fixed nondegenerate quadratic form. The Jacobian and transformed amplitude have uniform parameter derivatives. Equations (P4)–(P8) give the asserted bounds for this quadratic integral. On the complement, repeated integration by parts using (P10) gives arbitrary decay even after differentiating in $d,E$ and removing $e^{ir\Phi}$; any finitely many extra powers of $r$ are absorbed by additional integrations. A finite parameter cover gives the uniform constants. This also shows that $b$ vanishes near $E_0$. No derivative of a moving critical exponential is hidden in (B22).

Put $D(E)=b(d,y,E,r)-b(0,y,E,r)$. Integration of the $d$-derivative in (B22) gives $|D|+|\partial_E D|\leq C d r^{-(n-1)/2}$. Because the same $\Phi$ occurs at both distances, integration by parts in energy gives $$\begin{gathered} J(d,y,r)-J(0,y,r)\\ =\left[\frac{e^{ir\Phi}D}{ir\partial_E\Phi}\right]_{E_0}^1\\ -\frac1{ir}\int_{E_0}^1 e^{ir\Phi} \partial_E\!\left(\frac D{\partial_E\Phi}\right)dE. \end{gathered} \tag{B23}$$ The lower endpoint is zero; the upper endpoint is retained. The lower bound in (B18) and bounded second energy derivative of $\Phi$ prove the remaining part of (B20).

For (B21), if $d\leq h$, use $h^{-n}d\leq h^{1-n}$. If $d\geq h$, the other part of (B20) gives $$\begin{gathered} h^{-n}d(h/d)^{(n+1)/2}\\ =h^{1-n}(h/d)^{(n-1)/2}\leq h^{1-n}. \end{gathered} \tag{B24}$$ The comparison is zero at $d=0$ by its definition before substitution. This completes the proof. $\square$

The fixed-action assumption is substantial. Even a difference of order $d$ between critical values can produce an additional factor $rd$ when exponentials are subtracted. Smooth coefficients alone therefore do not justify (B20). For the actual boundary kernel one must construct the relevant phase patches, identify the common critical action, verify the parameter bounds, handle all other patches, and control the exact operator remainder in (B16). A formal frozen amplitude is not that construction.

6. Constructing the two phases from the metric

Write a boundary normal chart as $(a,z)$, where $a\geq0$ is distance and $z\in\mathbb R^{n-1}$. Reserve $(d,y)$ for the source point. On a compact chart inside a larger one the actual principal symbol is $$\begin{gathered} p(a,z,s,\zeta)=s^2+r(a,z,\zeta),\\ r(a,z,\zeta)=\zeta^{\mathsf T}G(a,z)\zeta,\\ \lambda_\sigma(a,z,\zeta,E)\\ =\sigma\sqrt{E-r(a,z,\zeta)},\qquad\sigma=\pm1. \end{gathered} \tag{B25}$$ Here $G$ is real symmetric and uniformly positive. Fix $0<E_0<1$. For $E\in[E_0,1]$, choose a small tangential frequency ball and a short fixed collar so that $E-r\geq E_0/2$ on a slightly larger region. Every derivative of both roots is bounded there, uniformly in all parameters. Smoothness up to the wall permits an auxiliary smooth extension for local inverse arguments at zero; none of the asserted equations uses a physical point outside the collar.

We construct an incident phase $S^i_\sigma$ from the source normal slice, and a reflected phase $S^r_\sigma$ from the wall: $$\begin{gathered} \partial_a S^i_\sigma =\lambda_\sigma(a,z,\partial_z S^i_\sigma,E),\\ S^i_\sigma(d,z;d,y,\eta,E)=(z-y)\cdot\eta,\\ \partial_a S^r_\sigma =\lambda_\sigma(a,z,\partial_z S^r_\sigma,E),\\ S^r_\sigma(0,z;d,y,\eta,E) =S^i_{-\sigma}(0,z;d,y,\eta,E). \end{gathered} \tag{B26}$$ Both phases are real and smooth jointly in all variables for $0\leq a,d\leq d_0$, $y,z$ in retained chart neighborhoods, small $\eta$, and $E\in[E_0,1]$. The same neighborhood works at $d=0$. The matching phases at the wall are equal; Dirichlet cancellation will put a minus sign in the amplitudes.

Here is the construction and its uniformity. For any smooth initial function $f(w)$ on the slice $a=a_0$, solve $$\begin{gathered} Z'=-\partial_\zeta\lambda_\sigma(a,Z,\Xi,E),\\ \Xi'=\partial_z\lambda_\sigma(a,Z,\Xi,E),\\ Z(a_0)=w,\qquad \Xi(a_0)=df(w),\\ s(a)=f(w)+\int_{a_0}^{a}\Xi(v)\cdot Z'(v)\,dv\\ +\int_{a_0}^{a}\lambda_\sigma(v,Z(v),\Xi(v),E)\,dv. \end{gathered} \tag{B27}$$ The contraction and differentiated integral-equation proof in Existence and compactness of generalized reflected curves, equation (G13), supplies this smooth flow with smooth dependence on the initial slice and every displayed parameter. Its constants are uniform on the larger compact region. At $a=a_0$, the derivative $Z_w$ is the identity. Bounded differentiated flow equations therefore give $Z_w=I+O(|a-a_0|)$ for the bounded initial families used here. Choose the collar once so that this matrix stays invertible. The parameter inverse proof then solves $Z(a,w)=z$ smoothly for $w$.

Define $S(a,z)=s(a,w(a,z))$. Varying $w$ under the action integral shows $d_ws(a)=\Xi(a)d_wZ(a)$: the variation of its integrand is the derivative of $\Xi\cdot\delta Z$, because $Z'=-\lambda_\zeta$ and $\Xi'=\lambda_z$; its initial endpoint cancels $df=\Xi(a_0)dw$. It follows that $$\begin{gathered} \partial_zS=\Xi(a,w(a,z)),\\ \partial_aS=\lambda_\sigma(a,z,\partial_zS,E). \end{gathered} \tag{B28}$$ The second identity follows by differentiating $Z(a,w(a,z))=z$ at fixed $z$, so the action's $\Xi\cdot Z'$ term cancels. Conversely, the gradient of any solution of this initial problem follows (B27), by differentiating its equation. Flow uniqueness and the initial value give uniqueness of $S$.

First apply this construction to $f(w)=(w-y)\cdot\eta$ and $a_0=d$. Its parameter derivatives are bounded. Then take the resulting smooth boundary value $f(w)=S^i_{-\sigma}(0,w;d,y,\eta,E)$ and $a_0=0$. Its first two spatial derivatives are uniformly bounded and its gradient stays in the larger root region after reducing the collar. The construction applies again. Finite covers of a compact boundary chart give one common collar and frequency radius. This proves (B26), including all parameter derivatives and the normal-root gap; no formal eikonal solution is being assumed.

7. The exact critical action and Hessian on the diagonal

Put $\Theta_\sigma(d,y,\eta,E)=S^r_\sigma(d,y;d,y,\eta,E)$. Since $\Theta_\sigma(0,y,\eta,E)=0$, its normalized phase extends smoothly through zero: $$\begin{gathered} \psi_\sigma(d,y,\eta,E) =\int_0^1\partial_d\Theta_\sigma(vd,y,\eta,E)\,dv,\\ \Theta_\sigma=d\psi_\sigma,\\ \psi_\sigma(0,y,\eta,E) =2\sigma\sqrt{E-\eta^{\mathsf T}G(0,y)\eta}. \end{gathered} \tag{B29}$$ The last equality follows by differentiating (B26) when both normal slices are zero. The target-slice derivative contributes $\lambda_\sigma$. The derivative of the incident initial slice contributes $-\lambda_{-\sigma}$, because differentiation of $S^i_{-\sigma}(d,z;d,y,\eta,E)=(z-y)\cdot\eta$ gives $\partial_d S^i_{-\sigma}=-\partial_a S^i_{-\sigma}$ there. Their sum is $2\lambda_\sigma$. The first formula in (B29), which is Taylor's integral identity in the single distance variable, also proves every uniform derivative bound after division by $d$.

At $\eta=0$, the incident and reflected solutions and their first frequency derivatives are particularly simple: $$\begin{gathered} S^i_\sigma(a,z;d,y,0,E)=\sigma(a-d)\sqrt E,\\ S^r_\sigma(a,z;d,y,0,E)=\sigma(a+d)\sqrt E,\\ \left.\partial_\eta S^i_\sigma\right|_{\eta=0}=z-y, \qquad \left.\partial_\eta S^r_\sigma\right|_{\eta=0}=z-y. \end{gathered} \tag{B30}$$ Indeed the first two expressions solve (B26), since the root at zero tangential momentum is $\sigma\sqrt E$, independent of $a,z$. Uniqueness proves them. Differentiating the eikonal equation once in $\eta$ gives $\partial_a S_\eta=\lambda_\zeta S_{z\eta}$. At $\eta=0$ its right side vanishes, so its prescribed initial value $z-y$ remains unchanged on each leg. Thus $$\begin{gathered} \psi_\sigma(d,y,0,E)=2\sigma\sqrt E,\\ \partial_\eta\psi_\sigma(d,y,0,E)=0,\\ \partial_E\psi_\sigma(d,y,0,E)=\sigma/\sqrt E. \end{gathered} \tag{B31}$$ These identities include $d=0$ by (B29). In particular the critical action is exactly independent of distance, rather than just constant to first order.

The Hessian is also explicit. Define the positive matrix $\overline G(d,y)=\int_0^1G(vd,y)\,dv$. Differentiating the eikonal equation twice and using $S_{z\eta}=I$ from (B30) gives $\partial_a S_{\eta\eta}=\lambda_{\zeta\zeta}(a,z,0,E)=-\sigma G(a,z)/\sqrt E$ on the $\sigma$ leg. The incident $-\sigma$ leg starts with zero Hessian on $a=d$; its backward integral from $d$ to zero is $-\sigma E^{-1/2}\int_0^dG(v,z)\,dv$. The reflected $\sigma$ leg adds the same integral on its forward journey to $a=d$. Consequently $$\begin{gathered} \partial_\eta^2\psi_\sigma(d,y,0,E) =-\frac{2\sigma}{\sqrt E}\,\overline G(d,y),\\ \overline G(0,y)=G(0,y),\qquad \overline G(d,y)\geq c_G I. \end{gathered} \tag{B32}$$ This calculation differentiates at fixed target coordinate $z$ before setting $z=y$; it does not identify a moving coordinate with that fixed derivative.

Smoothness of $\psi_\sigma$, compactness of the parameter range and (B32) give, on one sufficiently small frequency ball, $$\begin{gathered} -\sigma\partial_\eta^2\psi_\sigma\geq cI,\\ |\partial_\eta\psi_\sigma(d,y,\eta,E)|\geq c|\eta|. \end{gathered} \tag{B33}$$ For the second inequality, integrate the Hessian on the segment from zero to $\eta$, pair with $-\sigma\eta$, and use (B31) and Cauchy–Schwarz. Hence zero is the unique critical point in that ball, the Hessian has signature $-\sigma(n-1)$, and the phase is uniformly nonstationary outside any smaller ball. Since $|\sigma/\sqrt E|\geq1$ on the chosen energy interval, all hypotheses (B18) of the fixed-action freezing lemma are now verified for these actual metric phases. This assertion concerns these near-normal phases, not a representation of every part of the spectral kernel.

8. Transport with the actual lower-order terms

Represent a half density by its scalar coefficient relative to $|da\,dz|^{1/2}$. In this representation the actual differential expression has the form $$\begin{gathered} P=\sum_{j,k=1}^n g^{jk}(a,z)D_jD_k\\ +\sum_{j=1}^n b_j(a,z)D_j+c(a,z),\\ D_j=-i\partial_j,\qquad (g^{jk})=\begin{pmatrix}1&0\\0&G\end{pmatrix}. \end{gathered} \tag{B34}$$ The smooth coefficients $b_j,c$ include all ordering and lower-order terms; they need not be real in left quantization. Nothing in the following construction discards them or assumes a Laplace operator.

For any one of the phases in (B26), direct differentiation of the degree-two differential expression gives the exact identity $$\begin{gathered} e^{-iS/h}(h^2P-E)(e^{iS/h}A)\\ =h\mathcal T_S A+h^2PA. \end{gathered} \tag{B35}$$ where $$\begin{aligned} \mathcal T_S A={}&\frac2i\sum_{j,k}g^{jk}S_j\partial_kA\\ &+\frac1i\sum_{j,k}g^{jk}S_{jk}A\\ &+\left(\sum_j b_jS_j\right)A. \end{aligned} \tag{B36}$$ The eikonal equation cancels $(p(dS)-E)A$. The coefficient of $\partial_a A$ in $\mathcal T_S$ is $2S_a/i$, uniformly separated from zero. After division by that coefficient, the transport equation is an ordinary scalar equation along the base characteristics with velocity $G S_z/S_a=-\partial_\zeta\lambda_\sigma(a,z,S_z,E)$. These are precisely the projected characteristics used above.

For every integer $L\geq0$, choose smooth bounded initial coefficient functions $g_{\sigma,j}(z;d,y,\eta,E)$, $0\leq j\leq L$, supported in the retained frequency and energy region. Prescribe incident coefficients on $a=d$, and reflected coefficients on $a=0$, by $$\begin{gathered} \mathcal T_S a_0=0,\\ \mathcal T_S a_j=-P a_{j-1}\quad(1\leq j\leq L),\\ a^i_{\sigma,j}(d,z;d,y,\eta,E)\\ =g_{\sigma,j}(z;d,y,\eta,E),\\ a^r_{\sigma,j}(0,z;d,y,\eta,E)\\ =-a^i_{-\sigma,j}(0,z;d,y,\eta,E). \end{gathered} \tag{B37}$$ Solve the incident recursion first and then the reflected one. To justify every step, write its divided equation along a characteristic as $u'+q u=f$. Multiplication by $e^{\int q}$ gives $u(a)=e^{-\int_{a_0}^a q}[u(a_0)+\int_{a_0}^a e^{\int_{a_0}^v q}f(v)\,dv]$. Differentiation verifies the equation and initial value, with oriented integrals for a backward incident leg. The intervals are uniformly bounded, their coefficients and all parameter derivatives are bounded, and the flow inverse is uniformly smooth. Differentiating this formula proves all parameter bounds. At each recursion step $P a_{j-1}$ uses only the already obtained smooth derivatives. This proves existence, uniqueness and uniform bounds for all finite coefficient families, including mixed derivatives in the source distance $d$ through zero. No expansion or division in that distance is used in the transport equations.

Set $A^i_{\sigma,L}=\sum_{j=0}^Lh^j a^i_{\sigma,j}$, and define $A^r_{\sigma,L}$ in the same way. Equations (B35)–(B37) telescope to $$\begin{gathered} (h^2P-E)(e^{iS/h}A_L) =h^{L+2}e^{iS/h}P a_L,\\ \left.e^{iS^i_{-\sigma}/h}A^i_{-\sigma,L} +e^{iS^r_\sigma/h}A^r_{\sigma,L}\right|_{a=0}=0. \end{gathered} \tag{B38}$$ The boundary cancellation is exact for every $h>0$. In the frozen flat case the phases are $(z-y)\eta-\sigma(a-d)\sqrt{E-r_0(\eta)}$ and $(z-y)\eta+\sigma(a+d)\sqrt{E-r_0(\eta)}$; they agree at $a=0$. This also checks the reflection sign directly.

Frequency and energy supports of the coefficients are retained by the recursion: its differentiations are in $(a,z)$, and the characteristic parameter $(\eta,E)$ is fixed. A fixed energy cutoff vanishing near $E_0$ can therefore be built into the initial coefficients. Initial base supports can be chosen strictly inside the chart so that their two transported supports stay inside the larger chart. Alternatively every estimate can be restricted to a compact subchart where an auxiliary base cutoff is one. Derivatives of a subsequently imposed base cutoff are separate commutator errors on its transition region; (B38) does not call those errors small.

9. Quantitative residuals and the frozen diagonal family

For a fixed choice $\varepsilon=\pm1$, form the finite wave-phase integral $$\begin{gathered} \mathcal V_{\sigma,L}=e^{iS^i_{-\sigma}/h}A^i_{-\sigma,L} +e^{iS^r_\sigma/h}A^r_{\sigma,L},\\ K_{h,L}(t,a,z;d,y)\\ =(2\pi h)^{-n}\sum_{\sigma=\pm1}\int_{E_0}^1\int e^{it\varepsilon\sqrt E/h}\mathcal V_{\sigma,L}\,d\eta\,dE. \end{gathered} \tag{B39}$$ The amplitudes have uniformly compact $\eta$-support. They may be nonzero at the upper energy endpoint. This causes no problem when applying a differential operator or taking a fixed ordinary parameter derivative under the finite integral. With $Q_h=(hD_t)^2-h^2P$, (B38) gives, on the retained compact charts and any fixed bounded time interval, $$\begin{gathered} K_{h,L}|_{a=0}=0,\\ |\partial^\alpha Q_h K_{h,L}| \leq C_{L,\alpha}h^{L+2-n-|\alpha|}. \end{gathered} \tag{B40}$$ Here $\partial^\alpha$ may differentiate time, target coordinates or source coordinates. Each derivative of an exponential costs at most one power of $h^{-1}$; all phase derivatives and amplitude derivatives are uniformly bounded. The un-differentiated residual is exactly the integral of $-h^{L+2}e^{i(t\varepsilon\sqrt E+S)/h}Pa_L$. Its absolute integral, and the same calculation after differentiating, prove (B40). Thus any prescribed finite number of derivatives and any prescribed residual power can be attained by increasing $L$. The construction has not yet imposed the physical Cauchy data that identify this family with the actual cosine kernel.

For density normalization write $$\begin{gathered} dV_g=\gamma(a,z)\,da\,dz,\\ \gamma(a,z)=(\det G(a,z))^{-1/2}. \end{gathered} \tag{B41}$$ A coordinate half-density kernel has scalar diagonal relative to $dV_g$ equal to its coordinate diagonal divided by $\gamma$. Indeed $v|da\,dz|^{1/2}=u(dV_g)^{1/2}$ means $u=\gamma^{-1/2}v$; apply this to both kernel factors. The smooth positive factors $\gamma^{\pm1}$ and all their derivatives are bounded on the retained collar.

Consider the reflected diagonal family $$\begin{gathered} a^{\rm diag}_{\sigma,L}(d,y,\eta,E)\\ =\gamma(d,y)^{-1}A^r_{\sigma,L}(d,y;d,y,\eta,E),\\ F_{h,L}(d,y)\\ =(2\pi h)^{-n}\sum_\sigma\int_{E_0}^1\int e^{id\psi_\sigma/h}a^{\rm diag}_{\sigma,L}\,d\eta\,dE. \end{gathered} \tag{B42}$$ Let $F_h^0(d,y)$ be the same expression keeping only $j=0$, and replacing the smooth distance parameters in $\psi,\gamma,a^r_0$ by zero, while retaining $d/h$ in the exponential. Then $$\begin{gathered} |F_{h,L}(d,y)-F_h^0(d,y)|\\ \leq C_Lh^{1-n},\qquad 0\leq d\leq d_0. \end{gathered} \tag{B43}$$ To prove this, apply Lemma 5.1 to each actual metric phase $\psi_\sigma$. Its hypotheses were verified in (B29)–(B33), and (B37) supplies all required distance and energy derivatives of the amplitude, including the factor $\gamma^{-1}$. The lower energy cutoff makes that amplitude vanish near $E_0$. This gives the claimed bound for the leading term. Every higher term is bounded directly by $C_jh^{j-n}$, using its compact integration region, so their finite sum is $O(h^{1-n})$. At $d=0$ the leading difference vanishes and the same estimate for the higher terms applies. This proves a uniform coefficient-freezing bound for the constructed family, including arbitrary smooth lower-order coefficients.

Its frozen normalization can be fixed explicitly. Suppose the leading incident seeds on the zero source slice satisfy, at $z=y$, $$g_{\sigma,0}(y;0,y,\eta,E) =\frac{\chi(E)\beta(y,\eta,E)} {2\sqrt{E-r(0,y,\eta)}}. \tag{B44}$$ Here $\chi$ vanishes near $E_0$, and $\beta$ is supported in the small near-normal frequency ball. Smooth such seed families exist because the denominator is separated from zero; multiply by an initial base cutoff equal to one near $z=y$ and extend in $d$ using the same positive root. By (B37), the reflected leading coefficient at zero is the negative of (B44). Thus $$\begin{gathered} \lambda_0(y,\eta,E)=\sqrt{E-r(0,y,\eta)},\\ F_h^0(d,y)=-\frac{(2\pi h)^{-n}}{\gamma(0,y)} \sum_{\sigma=\pm1}\int_{E_0}^1\int\\ \quad e^{2i\sigma d\lambda_0/h} \frac{\chi(E)\beta(y,\eta,E)}{2\lambda_0}\,d\eta\,dE. \end{gathered} \tag{B45}$$ This is precisely the corresponding part of the frozen reflected phase-volume integral. For each fixed $\eta$, use the two branches $s=\sigma\sqrt{E-r(0,y,\eta)}$; the positive Jacobian is $|ds/dE|=1/(2\sqrt{E-r})$. This proves (B45)'s normalization and its negative Dirichlet sign without assigning the spectral interpretation to an arbitrary initial seed.