Proof. For the model phase, \[P(y,\xi)=\nabla_y\Phi_{\mathrm{hyp}}(y,\xi)=(\xi_1\xi_2,\xi_1,\xi_2),\] and the second fundamental form \(\text{II}[\omega,\omega]\) is a nonzero multiple of \(\omega_1\omega_2\). Thus it is Carleson–Sjölin but not elliptic. It has the two asymptotic directions parallel to the coordinate axes. The example below stacks wave packets along these two directions.

Let \[J=\left\lfloor \frac{1}{10}\log_2\lambda\right\rfloor .\] Choose a small number \(c>0\), depending only on the amplitude support. For \(1\le j\le J\) set \[R_j= \left\{\xi: 2^j\lambda^{-1/2}\le \xi_1\le (1+c)2^j\lambda^{-1/2}, 0\le \xi_2\le c2^{-j}\lambda^{-1/2} \right\}.\] The rectangles are disjoint, lie in \(\{|\xi|\le c_0\}\) for \(\lambda\) large, and satisfy \(|R_j|\approx\lambda^{-1}\) and \[|\xi_1\xi_2|\lesssim\lambda^{-1},\qquad \xi\in R_j.\] Let \(f_j=|R_j|^{-1/2}{\boldsymbol{1}}_{R_j}\) and \[f_\lambda=J^{-1/2}\sum_{j=1}^J f_j.\] Then \(\|f_\lambda\|_2=1\). For \(0\le s\le1\) and \(|z_1|,|z_2|\le c\lambda^{-1}\), the phase variation of \(\lambda(z\cdot\xi+s\xi_1\xi_2)\) on each \(R_j\) is bounded by a small absolute constant if \(c\) is chosen small. Hence all integrals have the same sign after taking real parts, and \[\label{eq:hyperbolic-low-large} |T_{\lambda,\Phi_{\mathrm{hyp}}} f_\lambda(s,z)| \gtrsim \lambda J^{-1/2}\sum_{j=1}^J |R_j|^{1/2} \gtrsim \lambda^{1/2}\sqrt J .\tag{7}\]

For the high-frequency factor take \[G=\{\eta:0\le\eta_1,\eta_2\le c\mu^{-1/2}\} = \{\eta:0\le\eta_1,\eta_2\le c\lambda^{-1}\}, \qquad g_\mu=|G|^{-1/2}{\boldsymbol{1}}_G .\] Then \(\|g_\mu\|_2=1\) and, on the same box \(0\le s\le1\), \(|z_1|,|z_2|\le c\lambda^{-1}\), the phase variation of \(\mu(z\cdot\eta+s\eta_1\eta_2)\) is also bounded by a small constant. Therefore \[\label{eq:hyperbolic-high-large} |T_{\mu,\Phi_{\mathrm{hyp}}} g_\mu(s,z)| \gtrsim \mu |G|^{1/2} \approx \lambda .\tag{8}\] The box \[Q_\lambda=\{0\le s\le1,\;|z_1|,|z_2|\le c\lambda^{-1}\}\] has measure \(|Q_\lambda|\approx\lambda^{-2}\). Combining 7 and 8 , \[\|T_{\lambda,\Phi_{\mathrm{hyp}}} f_\lambda T_{\mu,\Phi_{\mathrm{hyp}}} g_\mu\|_{L^2} \ge \|T_{\lambda,\Phi_{\mathrm{hyp}}} f_\lambda T_{\mu,\Phi_{\mathrm{hyp}}} g_\mu\|_{L^2(Q_\lambda)} \gtrsim (\lambda^{1/2}\sqrt J)\lambda |Q_\lambda|^{1/2} \gtrsim \lambda^{1/2}\sqrt J.\] Since \(J\approx\log\lambda\), this proves ?? . Replacing the characteristic functions above by nonnegative smooth cutoffs changes only the constants. ◻

Proof. The first two assertions are the standard box packets. For (i), take \(h\) to be a normalized smooth cutoff on a fixed small frequency square. The assumptions make the phase variation bounded on \(Q_{L,R}\), and the integral has size \(\nu\). For (ii), take \(h\) supported in a square of side \(R^{1/2}/\nu\). The normalized integral then has size \[\nu\left(\frac{R}{\nu^2}\right)^{1/2}=R^{1/2},\] and the conditions \(LR\lesssim\nu\) and \(R\lesssim\nu^2\) make the \(s\xi_1\xi_2\) and \(z\cdot\xi\) phase variations bounded.

For (iii), choose dyadic rectangles \[R_m=\Big\{\xi: 2^m(\nu L)^{-1/2}\le \xi_1\le (1+\gamma)2^m(\nu L)^{-1/2},\quad 0\le \xi_2\le \gamma2^{-m}(\nu L)^{-1/2} \Big\},\] with both coordinates \(\lesssim 1\). There are \(J\approx \log(\nu L)\) such rectangles, they are disjoint, and each has area \(\approx(\nu L)^{-1}\). Let \[h=J^{-1/2}\sum_m |R_m|^{-1/2}{\boldsymbol{1}}_{R_m},\] with smooth nonnegative cutoffs in place of characteristic functions if desired. On \(Q_{L,R}\) we have \[\nu |s\xi_1\xi_2|\lesssim 1, \qquad \nu |z\cdot\xi|\lesssim 1,\] for \(\xi\in R_m\). Thus the real parts of the integrals are positive and \[|T_{\nu,\Phi_{\mathrm{hyp}}} h(y)| \gtrsim \nu J^{-1/2}\sum_m |R_m|^{1/2} \approx \Big(\frac{\nu}{L}\Big)^{1/2}J^{1/2}.\] This proves the lemma. ◻

Proof. For (I), use the common box \(Q=Q_{a^{-1},ac}\), so \(|Q|\approx a^{-2}c^{-1}\). For \(j\le k-4\) use Lemma 1(i), giving factors of size \(\lambda_j\) on \(Q\). The \(a\)-factor is also of type (i), hence has size \(a\) on \(Q\). Since \(c\gtrsim b^2/a\), Lemma 1(iii) gives the \(b\)-factor with size \[(ab)^{1/2}\sqrt{\log(b/a)}\] on \(Q\). Finally, Lemma 1(ii) gives the \(c\)- and \(e\)-factors with size \((ac)^{1/2}\) on \(Q\). Multiplying the pointwise lower bounds and then multiplying by \(|Q|^{1/p}\) gives ?? .

For (II), use \(Q=Q_{b^{-1},be}\), so \(|Q|\approx b^{-2}e^{-1}\). The factors \(j\le k-3\) and the \(b\)-factor are of type (i). Since \(e\gtrsim c^2/b\), Lemma 1(iii) gives the \(c\)-factor with size \[(bc)^{1/2}\sqrt{\log(c/b)}\] on \(Q\). Finally, Lemma 1(ii) gives the \(e\)-factor with size \((be)^{1/2}\) on \(Q\). Since \(|Q|^{1/2}\approx b^{-1}e^{-1/2}\), this gives ?? .

For (III), use \(Q=Q_{m^{-1},b^2}\), so \(|Q|\approx m^{-1}b^{-2}\). For \(j\le k-4\) use Lemma 1(i), giving factors of size \(\lambda_j\) on \(Q\). Lemma 1(iii) gives the \(a\)- and \(b\)-factors with sizes \[(am)^{1/2}\sqrt{\log (a/m)}, \qquad (bm)^{1/2}\sqrt{\log (b/m)},\] respectively. Since \(c,e\gtrsim b^2/m\), Lemma 1(ii) gives the \(c\)- and \(e\)-factors with size \(b\) on \(Q\). Since \(|Q|\approx m^{-1}b^{-2}\), this gives ?? . ◻

Proof. Choose \(\chi\in C_0^\infty(W)\) with \(\chi=1\) on \(K_1\). The kernel of \(S_\lambda^*S_\lambda\) is \[K(y,y')=\int_W e^{i\lambda(\psi(x,y')-\psi(x,y))}\chi(x)\overline{c(x,y)}c(x,y')\,dx .\] The trivial estimate gives \(|K(y,y')|\le C\). If \(\lambda|y-y'|>1\), use the vector field \[L=\frac{1}{i\lambda}\frac{\nabla_x(\psi(x,y')-\psi(x,y))}{|\nabla_x(\psi(x,y')-\psi(x,y))|^2}\cdot\nabla_x .\] Then \(L e^{i\lambda(\psi(x,y')-\psi(x,y))}=e^{i\lambda(\psi(x,y')-\psi(x,y))}\). Choose an integration-by-parts order \(N>3\). By ?? –?? and the bounded \(C^\infty\) control of \(c\), each coefficient generated by \((L^*)^N\) is bounded by \(C(\lambda|y-y'|)^{-N}\). Hence \[|K(y,y')|\le C_N(1+\lambda|y-y'|)^{-N}.\] Schur’s test in the three-dimensional \(y\) variable gives \(\|S_\lambda^*S_\lambda\|_{2\to2}\lesssim\lambda^{-3}\), proving the claim. ◻

Proof. It is enough to prove the linearized estimate \[\|B_{\alpha,\beta}F\|_{L^2}\le C\alpha^{1/2}\|F\|_{L^2_{\xi,\eta}},\] where \[B_{\alpha,\beta}F(x)=\alpha\beta\iint e^{i\alpha\varphi(x,\xi)+i\beta\varphi(x,\eta)}b_1(x,\xi)b_2(x,\eta)F(\xi,\eta)\,d\xi d\eta .\] Set \(\sigma=\alpha/\beta\in(0,1]\). At \(\xi_0\), after shrinking the \(x\) patch, the definiteness of the second fundamental form implies that \[M(x,\omega)=\bigl(\partial^2_{\xi\xi}P(x,\xi_0)[\omega,\omega],A_1(x,\xi_0),A_2(x,\xi_0)\bigr),\qquad \omega\in S^1,\] is invertible and there exists \(c_0>0\), independent of \(x\) and \(\omega\), such that \[|M(x,\omega) z| \geq c_0 |z|, \qquad \forall z\in \mathbb{R}^3.\] Choose a small number \(\delta>0\) and concentric balls \[V_0=B(\xi_0, \delta) \Subset V_1=B(\xi_0, 4\delta)\] so that \(A\) and \(\partial^2_{\xi\xi}P\) on \(V_1\) are close to their values at \(\xi_0\).

For \(\xi\ne\eta\), set \[v=\eta+\sigma\xi, \qquad u=\frac{\sigma|\xi-\eta|^2}{2(1+\sigma)}, \qquad \omega=\frac{\xi-\eta}{|\xi-\eta|}\in S^1 .\] The diagonal \(\xi=\eta\) has four-dimensional measure zero and is discarded during the change of variables, and the final identity is recovered by density. The inverse map is \[\rho=\left(\frac{2(1+\sigma)u}{\sigma}\right)^{1/2}, \qquad \xi=\frac{v+\rho\omega}{1+\sigma}, \qquad \eta=\frac{v-\sigma\rho\omega}{1+\sigma},\] and the Jacobian is \[d\xi d\eta=\frac{1}{\sigma(1+\sigma)}\,du\,dv\,d\sigma_{S^1}(\omega).\] For fixed \(\sigma,\omega\), define, for \(u>0\), \[\Psi_{\sigma,\omega}(x,u,v)=\sigma\varphi(x,\xi(u,v))+\varphi(x,\eta(u,v)),\qquad R_{\sigma,\omega}=\nabla_x\Psi_{\sigma,\omega},\] and extend them to \(u=0\) by their one-sided limits. The function \(\Psi_{\sigma,\omega}\) need not be smooth in \(u\) at \(u=0\). This is harmless since Lemma 2 only requires smoothness in the \(x\)-variable, together with the estimates for \(R_{\sigma,\omega}\) as a function of \(y=(u,v)\). These estimates will be verified below.

For each fixed \((u,v)\), the phase and the transformed amplitude are bounded in \(C^\infty\) in \(x\), uniformly in \(\sigma\) and \(\omega\). Let \(K_{\sigma,\omega}\) be the set of \(y=(u,v)\) for which the frequencies \(\xi(y,\omega)\) and \(\eta(y,\omega)\) from the inverse formulas belong to \(V_0\). Extend the transformed amplitude by zero to a fixed bounded set containing all \(K_{\sigma,\omega}\). If \(y=(u,v)\) and \(y'=(u',v')\) lie in \(K_{\sigma,\omega}\), then the segment \(y_t=(1-t)y+ty'\) remains in a region where the frequencies lie in \(V_1\). Indeed, writing \[\rho_t=\left(\frac{2(1+\sigma)u_t}{\sigma}\right)^{1/2},\] one has \[\frac{v_t}{1+\sigma}\in V_0, \qquad \rho_t^2=(1-t)\rho^2+t(\rho')^2\le (2\delta)^2,\] and therefore \[\xi(y_t,\omega)=\frac{v_t+\rho_t\omega}{1+\sigma}, \qquad \eta(y_t,\omega)=\frac{v_t-\sigma\rho_t\omega}{1+\sigma}\] belong to \(V_1\).

For every smooth vector-valued \(G(\xi)\), the following identities follow from the inverse formulas and the chain rule. Since \(\rho\) is independent of \(v\), \[\begin{align} \partial_{v_j}\{\sigma G(\xi)+G(\eta)\} &= \frac{\sigma\partial_{\xi_j}G(\xi)+\partial_{\xi_j}G(\eta)}{1+\sigma}. \label{eq:vder} \end{align}\tag{9}\] For the \(u\)-derivative, using \[\partial_u\rho=\frac{1+\sigma}{\sigma\rho}, \qquad \xi-\eta=\rho\omega,\] we have, for \(u>0\), \[\begin{align} \partial_u\{\sigma G(\xi)+G(\eta)\} &= \frac{\{DG(\xi)-DG(\eta)\}[\omega]}{\rho} \notag\\ &= \int_0^1 \partial^2_{\xi\xi}G(\eta+s\rho\omega)[\omega,\omega]\,ds . \label{eq:uder} \end{align}\tag{10}\] Both derivative formulas have continuous one-sided extensions to \(u=0\). Writing \[\xi_t=\xi(y_t,\omega),\qquad \eta_t=\eta(y_t,\omega),\] we have \[D_{(u,v)}R_{\sigma,\omega}(x,y_t) = \bigl( \partial_u R_{\sigma,\omega}(x,y_t), \partial_{v_1}R_{\sigma,\omega}(x,y_t), \partial_{v_2}R_{\sigma,\omega}(x,y_t) \bigr).\] Applying 9 –10 to \(G=P(x,\cdot)\), we obtain \[\partial_u R_{\sigma,\omega}(x,y_t) = \int_0^1 \partial^2_{\xi\xi}P \bigl(x,(1-r)\eta_t+r\xi_t\bigr)[\omega,\omega]\,dr = \partial^2_{\xi\xi}P(x,\xi_0)[\omega,\omega] + e_0(x,y_t),\] and, for \(j=1,2\), \[\partial_{v_j}R_{\sigma,\omega}(x,y_t) = \frac{\sigma A_j(x,\xi_t)+A_j(x,\eta_t)}{1+\sigma} = A_j(x,\xi_0)+e_j(x,y_t).\] Since \(\xi_t,\eta_t\in V_1\), and since \((1-r)\eta_t+r\xi_t\in V_1\) for \(0\le r\le1\), the choice of \(V_1\) gives \[|e_0(x,y_t)|+|e_1(x,y_t)|+|e_2(x,y_t)|\ll c_0.\] Thus, we get \[D_{(u,v)}R_{\sigma,\omega}(x,y_t) = M(x,\omega)+E(x,y_t),\] where \(E(x,y_t)=\bigl(e_0,e_1,e_2\bigr)\). Hence, by the mean value theorem along the segment \(y_t\), \[R_{\sigma,\omega}(x,y')-R_{\sigma,\omega}(x,y) = \int_0^1D_{(u,v)}R_{\sigma,\omega}(x,y_t)(y'-y)\,dt,\] and therefore \[|R_{\sigma,\omega}(x,y')-R_{\sigma,\omega}(x,y)| \gtrsim c_0|y'-y|,\] which gives the required estimate for Lemma 2.

Now using the Jacobian, \(B_{\alpha,\beta}F\) becomes \[B_{\alpha,\beta}F(x)=\frac{\beta^2}{1+\sigma}\int_{S^1}S_{\beta,\sigma,\omega}H_{\sigma,\omega}(x)\,d\sigma_{S^1}(\omega),\] where \[S_{\beta,\sigma,\omega}h(x)=\int e^{i\beta\Psi_{\sigma,\omega}(x,u,v)}c_{\sigma,\omega}(x,u,v)h(u,v)\,du\,dv\] and \(H_{\sigma,\omega}(u,v)=F(\xi(u,v),\eta(u,v))\). Lemma 2 gives \[\|S_{\beta,\sigma,\omega}h\|_2\le C\beta^{-3/2}\|h\|_2 .\] Minkowski’s inequality, Cauchy’s inequality in \(\omega\), and the Jacobian yield \[\begin{align} \|B_{\alpha,\beta}F\|_2 &\le C\frac{\beta^{1/2}}{1+\sigma} \left(\int_{S^1}\int |H_{\sigma,\omega}(u,v)|^2\,du\,dv\,d\sigma_{S^1}(\omega)\right)^{1/2} \\ &= C\frac{\beta^{1/2}}{1+\sigma}\bigl(\sigma(1+\sigma)\bigr)^{1/2}\|F\|_2 = C\alpha^{1/2}\|F\|_2 . \end{align}\] Taking \(F=f\otimes g\) and using density proves the proposition. ◻

Proof. For fixed \(\varepsilon\), interpolate between the Euclidean metric and \(g_\varepsilon\) by \[g_{\varepsilon,u}:=I_3+u h_\varepsilon,\qquad 0\le u\le \varepsilon^2 .\] These metrics are bounded in \(C^\infty\), uniformly in \(\varepsilon\) and \(u\), and are uniformly close to the Euclidean metric on every fixed ball. Hence, on the separated compact set \[|P|,|Q|\le A,\qquad |P-Q|\ge d_0,\] the minimizing geodesic, and therefore its length, depends smoothly on \((P,Q,u)\), with bounds uniform in \(\varepsilon\), since we assume \(B(0,A)\) is strongly geodesically convex.

Because \(g_{\varepsilon,0}=I_3\), \[d_{g_{\varepsilon,0}}(P,Q)=|P-Q|.\] By the fundamental theorem of calculus in \(u\), \[d_{g_\varepsilon}(P,Q)-|P-Q| = \int_0^{\varepsilon^2} \partial_u d_{g_{\varepsilon,u}}(P,Q)\,du = \varepsilon^2E_\varepsilon(P,Q),\] where \[E_\varepsilon(P,Q) := \int_0^1 \partial_u d_{g_{\varepsilon,s\varepsilon^2}}(P,Q)\,ds\] is bounded in \(C^\infty\) on the separated set. Finally, \[d_g(\varepsilon P,\varepsilon Q)=\varepsilon d_{g_\varepsilon}(P,Q),\] which gives \[d_g(\varepsilon P,\varepsilon Q)=\varepsilon|P-Q|+\varepsilon^3E_\varepsilon(P,Q).\] The parametrized version follows from the chain rule. ◻

Proof. The rescaled geodesic equation is \[\ddot Z=\varepsilon^2\mathcal{N}_\varepsilon(Z,\dot{Z}),\] where \(\mathcal{N}_\varepsilon\) is bounded in \(C^\infty\) on bounded sets. Its integral form gives \[Z_\varepsilon(t;P,v)=P+tv+ \varepsilon^2\int_0^t(t-s)\mathcal{N}_\varepsilon (Z_\varepsilon(s;P,v),\dot{Z}_\varepsilon(s;P,v))\,ds .\] Standard ODE estimates on the fixed interval \(|t|\le T\) give \[Z_\varepsilon(t;P,v)=P+tv+\varepsilon^2R_\varepsilon(t;P,v), \qquad \dot{Z}_\varepsilon(t;P,v)=v+\varepsilon^2\dot{R}_\varepsilon(t;P,v),\] with \(R_\varepsilon\) bounded in \(C^\infty\) for \(|P|+|v|\le A\).

For the terminal tangent estimate, consider the time-one endpoint map \[\mathcal{E}_\varepsilon(X,v):=Z_\varepsilon(1;X,v).\] Then Lemma 3 gives \[\mathcal{E}_\varepsilon(X,v)=X+v+O(\varepsilon^2), \qquad D_v\mathcal{E}_\varepsilon=I_3+O(\varepsilon^2).\] By the inverse function theorem, up to reparametrization, the initial velocity of the geodesic from \(X\) to \(Y\) is \[v_\varepsilon(X,Y)=Y-X+O(\varepsilon^2).\] Therefore, we have the terminal velocity \[\dot{Z}_\varepsilon(1;X,v_\varepsilon(X,Y))=Y-X+O(\varepsilon^2).\] Since \(g_\varepsilon=I_3+O(\varepsilon^2)\) and \(|Y-X|\ge d_0\), normalizing this vector with respect to \(g_\varepsilon\) changes the Euclidean normalization only by \(O(\varepsilon^2)\). This proves the stated terminal tangent estimate. ◻

Proof. For \(s=\varepsilon\rho\), put \(P=X\) and \(Q=\rho\Omega(\theta)\). Since \(|X|\le\tau\) and \(\rho\ge1/C_0\), choosing \(\tau\) small gives \[|X-\rho\Omega(\theta)|\ge \frac{1}{2C_0}.\] Thus \[\varepsilon^{-1}\Phi_s(\varepsilon X,\theta) =-|X-\rho\Omega(\theta)|+ \varepsilon^2E_\varepsilon(X,\rho\Omega(\theta)).\] Since \(\nabla_x\Phi_s(\varepsilon X,\theta)\) equals the \(X\)-gradient of the scaled phase, \[\nabla_x\Phi_s(\varepsilon X,\theta) =\frac{\rho\Omega(\theta)-X}{|\rho\Omega(\theta)-X|} +O(\varepsilon^2).\] The first term is \(\Omega(\theta)+O(\tau)\), proving ?? .

Let \[n_{\rho,X}(\theta)=\frac{\rho\Omega(\theta)-X}{|\rho\Omega(\theta)-X|}.\] Let \(\theta=(\theta_1,\theta_2)\) be local coordinates on \(S^2\). Then \[\partial_{\theta_a}n_{\rho,X}(\theta) =\frac{\rho}{|\rho\Omega-X|} (I-n_{\rho,X}\otimes n_{\rho,X})\partial_{\theta_a}\Omega,\quad a=1,2.\] At \(X=0\) this is exactly \(\partial_{\theta_a}\Omega\), since \(\Omega\cdot\Omega=1,\;\Omega\cdot \partial_{\theta_a}\Omega=0\). Hence \[\partial_\theta\nabla_x\Phi_s(\varepsilon X,\theta) =D\Omega(\theta)+O(\tau+\varepsilon^2).\] After shrinking the angular cap and then \(\tau,\varepsilon\), the bi-Lipschitz estimate follows from the corresponding estimate for the chart \(\Omega\). The derivative bound ?? follows from \(\partial_x^\gamma=\varepsilon^{-|\gamma|}\partial_X^\gamma\) and the mean-value theorem in \(\theta\) on the separated set.

For \(s=R\), the surface \(\theta\mapsto\nabla_x\Phi_R(x,\theta)\) is a \(C^2\)-small perturbation of the unit sphere chart \(\theta\mapsto\Omega(\theta)\). The rank condition and definiteness of the second fundamental form therefore persist after the same shrinking. ◻

Proof. Write \(U(\theta,\xi)=\rho_R\Omega(\theta)-\rho\Omega(\xi)\) and \(L=|U|\). For any parameters \(a,b\), \[\partial_{ab}L =\frac{\partial_aU\cdot\partial_bU+U\cdot\partial_{ab}U}{L} -\frac{(U\cdot\partial_aU)(U\cdot\partial_bU)}{L^3}. \label{eq:distance-Hessian-formula}\tag{13}\] At \(\theta=\xi\), \[U=(\rho_R-\rho)\Omega(\xi), \qquad L=\rho_R-\rho.\] Using \[\Omega\cdot\Omega=1,\qquad \Omega\cdot\partial_{\xi_a}\Omega=0, \qquad \Omega\cdot\partial_{\xi_a}\partial_{\xi_b}\Omega=-\partial_{\xi_a}\Omega\cdot\partial_{\xi_b}\Omega,\] formula 13 gives \[\begin{align} \partial_{\theta_a\theta_b}^2G_\rho(\xi,\xi) &=\frac{\rho_R^2\partial_{\xi_a}\Omega\cdot\partial_{\xi_b}\Omega +(\rho_R-\rho)\rho_R\Omega\cdot\partial_{\xi_a}\partial_{\xi_b}\Omega}{\rho_R-\rho} \\ &=\frac{\rho_R\rho}{\rho_R-\rho}\partial_{\xi_a}\Omega\cdot\partial_{\xi_b}\Omega,\\ \partial_{\theta_a\xi_b}^2G_\rho(\xi,\xi) &=\frac{-\rho_R\rho\,\partial_{\xi_a}\Omega\cdot\partial_{\xi_b}\Omega}{\rho_R-\rho} =-\frac{\rho_R\rho}{\rho_R-\rho}\partial_{\xi_a}\Omega\cdot\partial_{\xi_b}\Omega. \end{align}\] This gives the second identity in ?? . Since \(-|0-\rho_R\Omega(\theta)|=-\rho_R\) is constant, the first identity in ?? follows as well.

After shrinking the cap, \(G_S\ge m_SI_2\) on \(V\). Also \[0<\kappa_-\le \frac{\rho_R\rho}{\rho_R-\rho}\le \kappa_+<\infty, \qquad \rho\in[1/C_0,C_0],\] because \(\rho_R-\rho\ge3C_0\). Hence the two matrices in ?? are uniformly positive at \((X,\theta)=(0,\xi)\). The functions are smooth on the shell, and by continuity, the lower bounds persist for \(|X|\le\tau_0\) and \(|\theta-\xi|\le c_1\) after choosing \(V_0\), \(c_1\), and \(\tau_0\) sufficiently small. ◻

Proof. Write \(x=\varepsilon X\) and \(r=\varepsilon\rho\).

(i) and (ii). On the shell, \[\begin{align} \varepsilon^{-1}\Phi_R(\varepsilon X,\theta) &=-|X-\rho_R\Omega(\theta)|+O(\varepsilon^2),\\ \varepsilon^{-1}S_r(\theta,\xi) &=|\rho_R\Omega(\theta)-\rho\Omega(\xi)|+O(\varepsilon^2). \end{align}\] Therefore \[\varepsilon^{-1}\Psi_r(\varepsilon X,\theta,\xi) =F_{\rho,X}(\theta,\xi)+O(\varepsilon^2). \label{eq:Psi-model}\tag{14}\] Differentiating twice and using Lemma 6 gives ?? and ?? for small \(\varepsilon\).

By definition, at \(X=0\), \(\partial_\theta F_{\rho,0}(\xi,\xi)=0\). Hence, for \(|X|\le\tau\), \[|\partial_\theta\Psi_r(\varepsilon X,\xi,\xi)| \le C\varepsilon(\tau+\varepsilon^2).\] Let \(g(\theta)=\partial_\theta\Psi_r(\varepsilon X,\theta,\xi)\). The Hessian lower bound gives the strong monotonicity estimate \[(g(\theta)-g(\eta))\cdot(\theta-\eta) \ge c\varepsilon|\theta-\eta|^2 \label{eq:strong-monotone}\tag{15}\] whenever the segment between \(\theta\) and \(\eta\) lies in \(|\vartheta-\xi|\le c_1\). Choose \(R_0=A_0(\tau+\varepsilon^2)<c_1/2\), with \(A_0\) large. On \(|\theta-\xi|=R_0\), \[\label{eq:outwardpt} g(\theta)\cdot(\theta-\xi) \ge c\varepsilon R_0^2-C\varepsilon(\tau+\varepsilon^2)R_0>0.\tag{16}\]

Let \(B=\overline{B(\xi,R_0)}\), and let \(\pi:\mathbb{R}^2 \to B\) be the closest-point projection onto \(B\). In other words, \(\pi(x)=p\) if and only if \((x-p)\cdot(y-p)\le0\) for all \(y\in B\). Define \[T(\theta)=\pi(\theta-g(\theta)).\] Then \(T\) is continuous and by Brouwer’s fixed point theorem, \(T\) has a fixed point \(\theta_*\in B\). If \(\theta_*\in\partial B\), then \(g(\theta_*)\cdot(\theta_*-\xi)\le0,\) contradicting 16 . Hence \(\theta_*\in B(\xi,R_0)\) and \(g(\theta_*)=0\).

The strict monotonicity 15 gives uniqueness in \(|\theta-\xi|\le c_1\), and ?? follows.

For ?? , integrate the Hessian along the segment from \(\Theta_r(x,\xi)\) to \(\theta\): \[\partial_\theta\Psi_r(x,\theta,\xi) =\left(\int_0^1 \partial_{\theta\theta}^2\Psi_r(x,\Theta_r+t(\theta-\Theta_r),\xi)\,dt \right)(\theta-\Theta_r).\] The lower bound ?? gives the claim.

(iii). For ?? , write \(h=\theta-\theta'\) and \(\theta_t=\theta'+th\). Then \[\nabla_\xi(S_r(\theta,\xi)-S_r(\theta',\xi)) =\left(\int_0^1\partial_{\theta\xi}^2S_r(\theta_t,\xi)^T\,dt\right)h.\] The symmetric part of the averaged matrix is negative definite with size \(\varepsilon\) by ?? , and hence its action on \(h\) has norm at least \(c\varepsilon|h|\), proving ?? .

(iv). Differentiating the critical equation with respect to \(\xi\) gives \[H_r(x,\xi)\partial_\xi\Theta_r(x,\xi)+B_r(x,\xi)=0,\] where \[H_r=\partial_{\theta\theta}^2\Psi_r(x,\Theta_r(x,\xi),\xi), \qquad B_r=\partial_{\theta\xi}^2S_r(\Theta_r(x,\xi),\xi).\] Using ?? , the model identities ?? , and the comparison 14 , \[\begin{align} H_r(\varepsilon X,\xi)&=\varepsilon B_0(\xi,\rho) +O(\varepsilon(\tau+\operatorname{diam}V_0+\varepsilon^2)),\\ B_r(\varepsilon X,\xi)&=-\varepsilon B_0(\xi,\rho) +O(\varepsilon(\tau+\operatorname{diam}V_0+\varepsilon^2)), \end{align}\] where \[B_0(\xi,\rho)=\frac{\rho_R\rho}{\rho_R-\rho}G_S(\xi)\ge c_0I_2.\] Thus \[\partial_\xi\Theta_r=-H_r^{-1}B_r =I_2+O(\tau+\operatorname{diam}V_0+\varepsilon^2),\] which is ?? . After shrinking the parameters, \(\|\partial_\xi\Theta_r-I_2\|\le1/10\), so \(\Theta_r(x,\cdot)\) is injective on the convex set \(V_0\) and is a local diffeomorphism everywhere.

Higher derivatives are obtained by differentiating \(\partial_\theta\Psi_r(x,\Theta_r(x,\xi),\xi)=0\). At each order, the highest new derivative of \(\Theta_r\) appears linearly with coefficient \(H_r\). All other terms involve only derivatives of \(\Psi_r\) and lower-order derivatives of \(\Theta_r\). In scaled variables the derivatives of \(\Psi_r\) are \(O(\varepsilon)\), while \(H_r^{-1}=O(\varepsilon^{-1})\). Induction gives the stated bounded \(C^\infty\) control in the scaled variables. ◻

Proof. Choose once and for all an open set \(W\) satisfying ?? . By Lemma 7, \[\Theta_r(x,\xi)=\xi+O(\tau+\varepsilon^2), \qquad D_\xi\Theta_r(x,\xi)=I_2+O(\tau+\operatorname{diam}V_0+\varepsilon^2), \label{eq:Theta-close-C1}\tag{17}\] uniformly for \(x\in\overline{U_0}\), \(r\in I\), and \(\xi\in V_0\). After shrinking the parameters, the map \(\xi\mapsto\Theta_r(x,\xi)\) is a uniform \(C^1\)-small perturbation of the identity on \(V_0\). Since \(K_2\Subset W\Subset V_0\), the \(C^0\) part of 17 gives \(\Theta_r(x,K_2)\Subset W\).

We next prove \(W\Subset\Theta_r(x,V_0)\). Fix \(\theta\in W\). Since \(W\Subset V_0\), there is a small ball \(B_\theta\Subset V_0\) centered at \(\theta\), with a radius independent of \(\theta\in W\). On this ball set \[\mathcal{T}(\xi)=\xi+\theta-\Theta_r(x,\xi).\] The \(C^1\)-closeness in 17 makes \(\mathcal{T}\) a contraction, and the \(C^0\)-closeness makes it map \(B_\theta\) into itself. Hence \(\mathcal{T}\) has a unique fixed point, equivalently \(\Theta_r(x,\xi)=\theta\). This proves the right inclusion in ?? . Differentiating \(\Theta_r(x,\Xi_r(x,\theta))=\theta\) gives the derivative bounds for \(\Xi_r\).

It remains to choose \(\chi\). Since \(\Theta_r(x,\xi)=\xi+O(\tau+\varepsilon^2)\) for \(\xi\in K_2\), all critical pairs lie, after shrinking \(\tau\) and \(\varepsilon\), in an arbitrarily small fixed neighborhood of the diagonal over \(K_2\). Choose \(\chi\in C_0^\infty(V_0\times V_0)\) equal to one on such a neighborhood and supported in \(|\theta-\xi|<c_1\). The segment condition follows because the set \(\{|\theta-\xi|<c_1\}\) is convex in the \(\theta\)-variable for fixed \(\xi\). ◻

Proof. Write \[x=\varepsilon X, \qquad r=\varepsilon\rho, \qquad y=Y_r(\xi)=\exp_o(\varepsilon\rho\Omega(\xi)).\] Let \(\gamma_{xy}\) be the minimizing geodesic from \(x\) to \(y\). By Lemma 4, its terminal unit tangent at \(y\), in scaled coordinates, is \[v_y =\frac{\rho\Omega(\xi)-X}{|\rho\Omega(\xi)-X|} +O(\varepsilon^2) =\Omega(\xi)+O(\tau)+O(\varepsilon^2).\] Extend the same \(g_\varepsilon\)-geodesic past \(y\) by solving \[Z(0)=\rho\Omega(\xi), \qquad \dot{Z}(0)=v_y.\] For \(0\le t\le\rho_R-\rho+1\), Lemma 4 gives \[Z(t)=\rho\Omega(\xi)+t\Omega(\xi)+O(\tau)+O(\varepsilon^2), \qquad \dot{Z}(t)=\Omega(\xi)+O(\tau)+O(\varepsilon^2). \label{eq:extended-geodesic}\tag{18}\] Consequently, \[\frac{d}{dt}|Z(t)|=1+O(\tau)+O(\varepsilon^2)\ge\frac{1}{2}\] after shrinking \(\tau\) and \(\varepsilon\). Since \(|Z(0)|=\rho<\rho_R\) while \(|Z(\rho_R-\rho+1)|>\rho_R\), there is a unique transverse hitting time \(t_*\) with \(|Z(t_*)|=\rho_R\). Write \[z=\exp_o(\varepsilon Z(t_*))=Y_R(\theta_*).\] The leading radial expression in 18 implies \[|\theta_*-\xi|\le C(\tau+\varepsilon^2).\] Thus \(\theta_*\) lies in the critical neighborhood where Lemma 7 applies.

The coordinate ball was chosen to be strongly geodesically convex, and \(\varepsilon\) is small so that the unique minimizing geodesic segments connecting \(x\), \(y\), and \(z\) remain in it. See Figure 1. Applying the first variation formula to variations of the endpoint \(z=Y_R(\theta)\) along the outer sphere gives directly \[\partial_\theta\{-d_g(x,Y_R(\theta))+d_g(Y_R(\theta),y)\} \big|_{\theta=\theta_*}=0.\] Since \(y=Y_r(\xi)\), this is exactly \[\partial_\theta\{\Phi_R(x,\theta)+S_r(\theta,\xi)\} \big|_{\theta=\theta_*}=0.\] By uniqueness of the critical point, \[\theta_* = \Theta_r(x,\xi).\] Finally, distance additivity along the minimizing geodesics connecting \(x\), \(y\) and \(z\) gives \[d_g(x,z)=d_g(x,y)+d_g(y,z).\] Therefore \[\begin{align} \Phi_R(x,\theta_*)+S_r(\theta_*,\xi) &=-d_g(x,z)+d_g(z,y)\\ &=-d_g(x,y)=\Phi_r(x,\xi). \end{align}\] Substituting \(\theta_* = \Theta_r(x,\xi)\) proves the identity. ◻

Proof. The kernel of \(U_{\lambda,r}U_{\lambda,r}^*\) is \[K_r(\theta,\theta') =\lambda^2\int_{K_2} e^{i\lambda\{S_r(\theta,\xi)-S_r(\theta',\xi)\}} \chi(\theta,\xi)\overline{\chi(\theta',\xi)}\,d\xi .\] If the integrand is nonzero, the fixed cutoff construction in Section [sec:fixed-margin] implies that the segment \(\theta_t=\theta'+t(\theta-\theta')\) lies in \(|\theta_t-\xi|<c_1\). Hence Lemma 7 gives \[\left\lvert \nabla_\xi\{S_r(\theta,\xi)-S_r(\theta',\xi)\}\right\rvert \ge c\varepsilon|\theta-\theta'| .\] The higher \(\xi\)-derivatives of the scaled phase difference satisfy the corresponding \(O(|\theta-\theta'|)\) bounds on the same support. Integration by parts then gives \[|K_r(\theta,\theta')| \le C_N\lambda^2(1+\lambda\varepsilon|\theta-\theta'|)^{-N}, \qquad N>2.\] Schur’s test on the two-dimensional angular patch yields \[\sup_\theta\int |K_r(\theta,\theta')|\,d\theta' +\sup_{\theta'}\int |K_r(\theta,\theta')|\,d\theta \le C_N\varepsilon^{-2}.\] Thus \(\|U_{\lambda,r}\|_{2\to2}\le C_\varepsilon\). ◻

Proof. The \(L^\infty\) bound follows immediately from Cauchy’s inequality in the compact angular patch. For the \(L^2\) bound, the kernel of \((E^A_{\lambda,R})^*E^A_{\lambda,R}\) is \[K_s(\theta,\eta) =\lambda^2\int_U e^{i\lambda(\Phi_s(x,\eta)-\Phi_s(x,\theta))} \overline{A(x,\theta,\lambda)}A(x,\eta,\lambda)\,dx .\] Lemma 5 gives, on \(K\times V_0\), \[|\nabla_x\Phi_s(x,\eta)-\nabla_x\Phi_s(x,\theta)|\ge c|\eta-\theta|,\] with the corresponding differentiated upper bounds, where the constants may depend on \(\varepsilon\). Repeated integration by parts in \(x\) therefore gives \[|K_s(\theta,\eta)| \le C_{N,\varepsilon}\lambda^2(1+\lambda|\theta-\eta|)^{-N}, \qquad N>2.\] Schur’s test on the two-dimensional angular variables proves ?? . ◻

Proof. Choose a smooth cutoff \(\zeta_r(x,\theta,\xi)\) equal to one near the critical graph and supported where \(\chi=1\). On the support of \((1-\zeta_r)\chi\), 21 gives \[|\partial_\theta\Psi_r(x,\theta,\xi)|\gtrsim\varepsilon.\] For this non-stationary part, repeated integration by parts in \(\theta\) gives, for every \(L\) and fixed \(N\), \[\|I_{\lambda,{\rm ns}}\|_{C^N_{x,\xi}} \le C_{L,N,\varepsilon}\lambda^{-L}.\] Taking \(L\) large allows this term to be absorbed into the remainder.

On the critical part, the variable coefficient version of the stationary phase theorem (see e.g. Sogge [4]) applies uniformly in \(r\). The critical value is 19 and the Hessian is 20 , so the critical contribution is \[e^{i\lambda\Phi_r(x,\xi)} \left[ \lambda^{-1}q_r(x,\xi) B_r(x,\xi,\Theta_r(x,\xi),\lambda) +\lambda^{-2}\mathcal{E}_{\lambda,r}^B(x,\xi) \right],\] with \(\mathcal{E}_{\lambda,r}^B\) bounded in \(C^\infty\). Adding the nonstationary contribution gives ?? . The final variant follows because the proof uses only the listed hypotheses. ◻

Proof. For a trial outer amplitude \(A\), the composition kernel of \(E^A_{\lambda,R}U_{\lambda,r}\) is \[\label{eq:composition-kernel-section4} K^A_{\lambda,r}(x,\xi) =\lambda^2\int e^{i\lambda\Psi_r(x,\theta,\xi)} A(x,\theta,\lambda)\chi(\theta,\xi)\,d\theta .\tag{22}\] Apply Proposition 8, with \(K=\overline{U_0}\), to \(B_r(x,\xi,\theta,\lambda)=A_r(x,\theta,\lambda)\). For \(x\in K_1\) and \(\xi\in K_2\), the cutoff choices and the inverse identity give \[\psi_1(x)=1, \qquad \psi(\Theta_r(x,\xi))=1, \qquad \Xi_r(x,\Theta_r(x,\xi))=\xi.\] Therefore \[\label{eq:leading-match-sogge} q_r(x,\xi)A_r(x,\Theta_r(x,\xi),\lambda)=a_r(x,\xi,\lambda).\tag{23}\] Multiplying the stationary-phase expansion by the outside factor \(\lambda^2\) in 22 , we get, on \(\overline{U_0}\times K_2\), \[K^{A_r}_{\lambda,r}(x,\xi) =\lambda e^{i\lambda\Phi_r(x,\xi)} q_r(x,\xi)A_r(x,\Theta_r(x,\xi),\lambda) +\mathcal{K}_{\lambda,r}(x,\xi),\] where the kernel remainder satisfies \[\|\mathcal{K}_{\lambda,r}\|_{L^\infty(\overline{U_0}\times K_2)}\le C_\varepsilon.\] Together with 23 , this gives the desired original kernel plus the bounded kernel remainder on \(K_1\times K_2\).

Since \(a_r\) is supported in \(K_1\times K_2\), the factor \(a_r(x,\Xi_r(x,\theta),\lambda)\) in \(A_r\) implies the kernel of \(T^a_{\lambda,r}-E^{A_r}_{\lambda,R}U_{\lambda,r}\) is the stationary-phase remainder, hence is uniformly \(O_\varepsilon(1)\) on \(U\times K_2\). The symbol bounds for \(A_r\) follow by differentiating ?? and using the bounds for \(q_r\), \(\Xi_r\), and \(a_r\). ◻

Proof. We only record the changes from the unsigned argument. Write \(x=\varepsilon X\) and \(r=\varepsilon\rho\). The signed Euclidean models are \[F_{\rho,X}^\sigma(\theta,\xi) =-|X-\rho_R\Omega(\theta)| +|\rho_R\Omega(\theta)-\sigma\rho\Omega(\xi)|,\] and \[G_\rho^\sigma(\theta,\xi) =|\rho_R\Omega(\theta)-\sigma\rho\Omega(\xi)|.\] At \(X=0\) and \(\theta=\xi\), the same computation as in Lemma 6, with \(\rho\Omega(\xi)\) replaced by \(\sigma\rho\Omega(\xi)\), gives \[\partial_{\theta\theta}^2F_{\rho,0}^\sigma(\xi,\xi) =\sigma\frac{\rho_R\rho}{\rho_R-\sigma\rho}G_S(\xi), \qquad \partial_{\theta\xi}^2G_\rho^\sigma(\xi,\xi) =-\sigma\frac{\rho_R\rho}{\rho_R-\sigma\rho}G_S(\xi).\] Since \(\rho_R-\sigma\rho\) is uniformly positive for both signs, continuity implies, after shrinking the cap, \(c_1\), and \(\tau\), that \[\sigma\partial_{\theta\theta}^2F_{\rho,X}^\sigma\ge cI_2, \qquad -\sigma\operatorname{Sym}\partial_{\theta\xi}^2G_\rho^\sigma\ge cI_2\] on the region \(|X|\le\tau\), \(|\theta-\xi|\le c_1\).

The scaled distance expansion gives \[\varepsilon^{-1}\Psi_r^\sigma(\varepsilon X,\theta,\xi) =F_{\rho,X}^\sigma(\theta,\xi)+O(\varepsilon^2), \qquad \varepsilon^{-1}S_r^\sigma(\theta,\xi) =G_\rho^\sigma(\theta,\xi)+O(\varepsilon^2).\] Thus \[\sigma\partial_{\theta\theta}^2\Psi_r^\sigma\ge c\varepsilon I_2, \qquad -\sigma\operatorname{Sym}\partial_{\theta\xi}^2S_r^\sigma\ge c\varepsilon I_2.\] Moreover \[\partial_\theta\Psi_r^\sigma(\varepsilon X,\xi,\xi) =O(\varepsilon(\tau+\varepsilon^2)).\] The proof of Lemma 7 now applies verbatim, with \(\partial_{\theta\theta}^2\Psi_r\) replaced by \(\sigma\partial_{\theta\theta}^2\Psi_r^\sigma\) and with \(-\operatorname{Sym}\partial_{\theta\xi}^2S_r\) replaced by \(-\sigma\operatorname{Sym}\partial_{\theta\xi}^2S_r^\sigma\). Indeed, the first of the last two inequalities gives the same strong monotonicity for \(\sigma\partial_\theta\Psi_r^\sigma\), while the second gives the same angular propagation estimate in the \(\xi\) variable. This proves the existence and uniqueness of \(\Theta_r^\sigma\), the bound ?? , and the two nonstationary estimates ?? –?? . The differentiated implicit-equation argument from Lemma 7 also gives \[\partial_\xi\Theta_r^\sigma(\varepsilon X,\xi) =I_2+O(\tau+\operatorname{diam}V_0+\varepsilon^2),\] and the stated \(C^\infty\) boundedness in scaled variables. After shrinking the parameters, \(\Theta_r^\sigma(x,\cdot)\) is therefore a diffeomorphism from \(V_0\) onto its image.

Finally, the common inverse neighborhood and cutoff follow from the same compactness and contraction argument as Lemma 8. For both signs, \[\Theta_r^\sigma(x,\xi)=\xi+O(\tau+\varepsilon^2), \qquad D_\xi\Theta_r^\sigma(x,\xi)=I_2+O(\tau+\operatorname{diam}V_0+\varepsilon^2).\] Since there are only two signs, after shrinking the parameters once more the same open set \(W\) and the same cutoff \(\chi\) work simultaneously for \(\sigma=\pm1\). This gives ?? , the inverse maps \(\Xi_r^\sigma\) on \(\overline{U_0}\times W\), and the common segment condition. ◻

Proof. The case \(\sigma=+1\) is Lemma 9. We prove the case \(\sigma=-1\). Write \[x=\varepsilon X, \qquad r=\varepsilon\rho, \qquad y=Y_r^-(\xi)=\exp_o(-r\Omega(\xi)).\] Let \(\gamma\) be the minimizing geodesic from \(y\) to \(x\). By Lemma 4, in scaled normal coordinates its terminal unit tangent at \(x\), pointing from \(y\) to \(x\), is \[v_x =\frac{X+\rho\Omega(\xi)}{|X+\rho\Omega(\xi)|}+O(\varepsilon^2) =\Omega(\xi)+O(\tau+\varepsilon^2).\]

Continue this geodesic past \(x\), and write the continued scaled curve as \(Z(t)\) with \[Z(0)=X, \qquad \dot{Z}(0)=v_x.\] Again by Lemma 4, for \(0\le t\le\rho_R+2\), \[Z(t)=X+t\Omega(\xi)+O(\tau+\varepsilon^2), \qquad \dot{Z}(t)=\Omega(\xi)+O(\tau+\varepsilon^2).\] Hence \(H(t)=|Z(t)|^2-\rho_R^2\) is strictly increasing on \([\rho_R-1,\rho_R+1]\) and changes sign there. The continued geodesic therefore has a unique transverse first intersection with the positive outer sphere. Write it as \[z=Y_R^+(\theta_*).\] The same expansion gives \[\frac{Z(t_*)}{|Z(t_*)|}=\Omega(\xi)+O(\tau+\varepsilon^2),\] and the bi-Lipschitz property of the angular chart yields \[|\theta_*-\xi|\le C(\tau+\varepsilon^2).\] Thus \(\theta_*\) lies in the signed critical neighborhood.

Since the whole curve connecting \(y\), \(x\), and \(z\) lies in the fixed strongly convex normal ball, the relevant subsegments are minimizing. The first variation along the positive outer sphere gives \[\partial_\theta\{-d_g(x,Y_R^+(\theta))+d_g(Y_R^+(\theta),y)\} \big|_{\theta=\theta_*}=0,\] or equivalently \[\partial_\theta\{\Phi_R^+(x,\theta)+S_r^-(\theta,\xi)\} \big|_{\theta=\theta_*}=0.\] By the uniqueness part of Lemma 12, \(\theta_* = \Theta_r^-(x,\xi)\).

Finally, distance additivity along the minimizing geodesic segment \(y\to x\to z\) gives \[d_g(z,y)=d_g(z,x)+d_g(x,y).\] Therefore \[\begin{align} \Phi_R^+(x,\Theta_r^-(x,\xi)) +S_r^-(\Theta_r^-(x,\xi),\xi) &=-d_g(x,z)+d_g(z,y)\\ &=d_g(x,y)=\Phi_r^-(x,\xi), \end{align}\] which proves ?? for \(\sigma=-1\). ◻

Proof. The angular estimate is the proof of Lemma 10, with ?? replacing ?? . The same cutoff satisfies the same segment condition for both signs, so the Schur kernel estimate is unchanged.

For the factorization, set \[H_r^\sigma(x,\xi) =\partial_{\theta\theta}^2\Psi_r^\sigma (x,\Theta_r^\sigma(x,\xi),\xi)\] and \[q_r^\sigma(x,\xi) =(2\pi)e^{i\pi\operatorname{sig}(H_r^\sigma(x,\xi))/4} |\det H_r^\sigma(x,\xi)|^{-1/2}.\] By ?? , \(|q_r^\sigma|\approx\varepsilon^{-1}\), and all needed derivatives have bounds depending only on the fixed \(C^\infty\) data and on \(\varepsilon\). Keep the same \(x\)-cutoff \(\psi_1\). Choose \(\psi_{\rm sgn}\in C_0^\infty(W)\) equal to one on a fixed neighborhood of \[\{\Theta_r^\sigma(x,\xi):\sigma=\pm1,\;x\in K_1,\;r\in I,\;\xi\in K_2\}.\] Define \[A_r^\sigma(x,\theta,\lambda) :=\psi_1(x)\psi_{\rm sgn}(\theta) \{q_r^\sigma(x,\Xi_r^\sigma(x,\theta))\}^{-1} a_r(x,\Xi_r^\sigma(x,\theta),\lambda)\] wherever \(\psi_1(x)\psi_{\rm sgn}(\theta)\ne0\), and set it equal to zero elsewhere. This is a smooth order-zero amplitude because \(\operatorname{supp}\psi_1\Subset U_0\), because \(\operatorname{supp}\psi_{\rm sgn}\Subset W\) where \(\Xi_r^\sigma\) is defined, and because \(q_r^\sigma\) is elliptic.

The composition kernel of \(E_{\lambda,R}^{A_r^\sigma}U_{\lambda,r}^\sigma\) is \[\lambda^2\int e^{i\lambda\Psi_r^\sigma(x,\theta,\xi)} A_r^\sigma(x,\theta,\lambda)\chi(\theta,\xi)\,d\theta.\] Apply Proposition 8 in its final general form, with \(\Psi_r,\Theta_r,\Phi_r,H_r\) replaced by \(\Psi_r^\sigma,\Theta_r^\sigma,\Phi_r^\sigma,H_r^\sigma\). The required unique critical point, collar, cutoff, nonstationary estimate, derivative bounds, definite Hessian, and exact critical value are precisely the contents of Lemmas 12 and 13. After multiplying by the outside factor \(\lambda^2\), the composed kernel equals \[\lambda e^{i\lambda\Phi_r^\sigma(x,\xi)} q_r^\sigma(x,\xi)A_r^\sigma(x,\Theta_r^\sigma(x,\xi),\lambda) +\mathcal{K}_{\lambda,r}^\sigma(x,\xi),\] where \[\|\mathcal{K}_{\lambda,r}^\sigma\|_{L^\infty(\overline{U_0}\times K_2)} \le C_\varepsilon.\] For \(x\in K_1\) and \(\xi\in K_2\), \[\psi_1(x)=1, \qquad \psi_{\rm sgn}(\Theta_r^\sigma(x,\xi))=1, \qquad \Xi_r^\sigma(x,\Theta_r^\sigma(x,\xi))=\xi.\] Therefore \[q_r^\sigma(x,\xi)A_r^\sigma(x,\Theta_r^\sigma(x,\xi),\lambda) =a_r(x,\xi,\lambda),\] and the leading stationary-phase term is \[\lambda e^{i\lambda\Phi_r^\sigma(x,\xi)}a_r(x,\xi,\lambda).\] As in Proposition 9, the kernel of \(T_{\lambda,r}^{\sigma,a}-E_{\lambda,R}^{A_r^\sigma}U_{\lambda,r}^\sigma\) is reduced to the stationary-phase remainder, which is uniformly \(O_\varepsilon(1)\) on \(U\times K_2\). This proves ?? – ?? . The stated symbol bounds for \(A_r^\sigma\) follow by differentiating its definition and using the uniform bounds for \(q_r^\sigma\), \((q_r^\sigma)^{-1}\), \(\Xi_r^\sigma\), and \(a_r\). ◻

Proof. Use the signed factorization ?? for both factors. The main-main term has the common positive outer phase \(\Phi_R^+\), so Proposition 7 applies exactly as in Section 5.4. The signed angular bounds ?? control the inputs. All remainders are harmless and only contribute \(O(1)\). This proves Theorem 12. ◻

Proof. Let \(P=\sqrt{-\Delta}\). Choose \(t_0\) with \(\chi_0(t_0)\ne0\), and then choose \(h>0\) so small that \(\chi_0\) is bounded away from zero on \((t_0-2h,t_0+2h)\). Consider a partition of unity of the real line given by \(\rho\in C_0^\infty((-h,h))\) with \[\sum_{m\in h\mathbb{Z}}\rho(t-t_0-m)=1 .\] For \(\widetilde{\chi}\in\mathcal{S}(\mathbb{R})\), set \[b_m(t)= \frac{\rho(t-t_0-m)\widetilde{\chi}(t)}{\chi_0(t-m)}, \qquad m\in h\mathbb{Z} .\] The denominator is harmless on the support of the numerator, and rapid decay of \(\widetilde{\chi}\) gives \[\widetilde{\chi}(t)=\sum_{m\in h\mathbb{Z}}b_m(t)\chi_0(t-m), \qquad \|b_m\|_\infty\le C_N(1+|m|)^{-N}.\]

Expanding the product, fix \(m_1,\ldots,m_k\in h\mathbb{Z}\), and write \[\nu_j=2+|\lambda_j+m_j|,\qquad g_j=b_{m_j}(P-\lambda_j)f_j .\] Up to a rearrangement in the increasing order \(\nu_1\le \dots \le \nu_k\), the assumed estimate for \(\chi_0\) gives \[\left\| \prod_{j=1}^k\chi_0(P-\lambda_j-m_j)g_j \right\|_{L^p} \lesssim \mathcal{E}_0(n,k,p;\nu_1,\ldots,\nu_k) \prod_{j=1}^k\|g_j\|_2 .\] Because \[\frac{2+\lambda_j}{(1+|m_j|)} \leq \nu_j \leq (1+|m_j|) (2+\lambda_j),\] from the explicit formula for \(\mathcal{E}_0\), there is some \(N_0=N_0(n,k,p)\) such that \[\mathcal{E}_0(n,k,p;\nu_1,\ldots,\nu_k) \lesssim \mathcal{E}_0(n,k,p;\lambda_1,\ldots,\lambda_k) \prod_{j=1}^k(1+|m_j|)^{N_0} .\] Therefore, \[\left\| \prod_{j=1}^k\widetilde{\chi}(P-\lambda_j)f_j \right\|_{L^p} \lesssim \mathcal{E}_0(n,k,p;\lambda_1,\ldots,\lambda_k) \prod_{j=1}^k\|f_j\|_2 \left(\sum_{m_1,\ldots,m_k} \prod_{j=1}^k(1+|m_j|)^{N_0-N}\right).\] Choosing \(N>N_0+2\), the last sum is finite. This proves the desired estimate for \(\widetilde{\chi}\). ◻

Proof. Set \[\mathrm{II}_{ij}(r,x,w)=\big\langle \partial_{w_iw_j}^2\nabla_x\varphi_r(x,w),N(x,w)\big\rangle .\] In normal coordinates, as discussed above,Gauss’s lemma gives \(\nabla_x\varphi_r(0,w)=\omega(w), \;\forall r\in[\varepsilon/C,C\varepsilon].\) For a local choice of normal, write \(N(0,w)=\sigma\omega(w),\) where \(\sigma\in\{\pm1\}.\)

Since \(|\omega(w)|^2=1\), differentiating gives \[\mathrm{II}_{ij}(r,0,w) =-\sigma\,\langle\partial_{w_i}\omega,\partial_{w_j}\omega\rangle .\] Thus, for \(\zeta\in\mathbb{R}^{n-1}\), \[-\sigma\sum_{i,j}\mathrm{II}_{ij}(r,0,w)\zeta_i\zeta_j = \sum_{i,j}\langle\partial_{w_i}\omega,\partial_{w_j}\omega\rangle\zeta_i\zeta_j =\left|\sum_i\zeta_i\partial_{w_i}\omega\right|^2 \ge c|\zeta|^2 .\] By smoothness, uniformly for \(r\in[\varepsilon/C,C\varepsilon]\) and \(|x|\le\tau\varepsilon\), \[\|\mathrm{II}(r,x,w)-\mathrm{II}(r,0,w)\|\le c/2.\] Hence \[-\sigma\sum_{i,j}\mathrm{II}_{ij}(r,x,w)\zeta_i\zeta_j \ge \frac{c}{2}|\zeta|^2,\] which is the claimed uniform definiteness. ◻

Proof. It suffices to prove the linearized estimate \[B_{\alpha,\beta}F(x) =\alpha\beta\iint e^{i\alpha\phi_1(x,\xi)+i\beta\phi_2(x,\eta)} b_1(x,\xi,\alpha)b_2(x,\eta,\beta)F(\xi,\eta)\,d\xi d\eta .\] After a finite subdivision of the frequency supports, transversality allows us to write \(\xi=(s,t)\) so that \[\label{eq:transverse-minor} \left|\det\bigl(\partial_sP_1(x,\xi),\partial_{\eta_1}P_2(x,\eta), \partial_{\eta_2}P_2(x,\eta)\bigr)\right|\ge c_1>0\tag{27}\] on the support. Indeed, the two \(\eta\)-columns span the tangent plane of the second covector surface, and ?? says that some tangent direction of the first surface is uniformly transverse to that plane.

Set \(\sigma=\alpha/\beta\) and \(u=\sigma s\). For fixed \(t\), put \(y=(u,\eta)\) and \[\Psi_t(x,u,\eta)=\sigma\phi_1(x,u/\sigma,t)+\phi_2(x,\eta).\] Then \[\nabla_x\Psi_t=\sigma P_1(x,u/\sigma,t)+P_2(x,\eta),\] and differentiation with respect to \((u,\eta_1,\eta_2)\) gives the columns in 27 . Thus, after the subdivision and uniformly in \(\sigma\) and \(t\), \[\begin{align} |\nabla_x\Psi_t(x,y)-\nabla_x\Psi_t(x,y')|&\ge c|y-y'|, \\ |\partial_x^\gamma\{\nabla_x\Psi_t(x,y)-\nabla_x\Psi_t(x,y')\}|&\le C_\gamma |y-y'|, \qquad |\gamma|\le N, \end{align}\] with \(N>3\). The change \(u=\sigma s\) removes the small factor \(\sigma\) from the first column.

For fixed \(t\), let \[S_{\beta,t}h(x)=\int e^{i\beta\Psi_t(x,u,\eta)} c_t(x,u,\eta,\alpha,\beta)h(u,\eta)\,du d\eta,\] where \(c_t=b_1(x,u/\sigma,t,\alpha)b_2(x,\eta,\beta)\). Lemma 2, applied in the three variables \((u,\eta)\), gives \[\|S_{\beta,t}h\|_{L^2_x}\le C\beta^{-3/2}\|h\|_2\] uniformly in \(t\) and \(\sigma\). Since \(ds=\sigma^{-1}du\) and \(\alpha\beta/\sigma=\beta^2\), \[B_{\alpha,\beta}F(x)=\beta^2\int S_{\beta,t}H_t(x)\,dt, \qquad H_t(u,\eta)=F(u/\sigma,t,\eta).\] Minkowski and Cauchy’s inequality in the bounded \(t\)-interval give \[\|B_{\alpha,\beta}F\|_2 \le C\beta^{1/2} \left(\int\|H_t\|_2^2\,dt\right)^{1/2} =C\beta^{1/2}\sigma^{1/2}\|F\|_2 =C\alpha^{1/2}\|F\|_2.\] Taking \(F=f\otimes g\) proves ?? . ◻

Proof. Use geodesic polar coordinates around \(o\) on the \(y\)-annulus. After finite radial and angular partitions of unity, each angular piece has the form \[\label{eq:radial-angular-piece} S_{\lambda,\nu}^{(j)}f_j(x) =\int_I \bigl(T^{(j,\nu)}_{\lambda,r}h^{(j,\nu)}_r\bigr)(x)\,dr,\tag{28}\] where \(I\subset[\varepsilon/C,C\varepsilon]\) is fixed, \(Y_\nu(r,\xi)=\exp_o(r\Omega_\nu(\xi))\), and \[T^{(j,\nu)}_{\lambda,r}h(x) =\lambda\int e^{-i\lambda d_g(x,Y_\nu(r,\xi))} b^{(j,\nu)}_r(x,\xi,\lambda)h(\xi)\,d\xi .\] Splitting the polar Jacobian symmetrically between the amplitude and the input gives \[\label{eq:polar-L2-control} \sum_\nu\int_I\|h^{(j,\nu)}_r\|_{L^2_\xi}^2\,dr \le C_\varepsilon\|f_j\|_{L^2(M)}^2, \qquad f_1=f,\quad f_2=g .\tag{29}\] The frozen amplitudes satisfy the symbol bounds required in the frozen estimates.

We claim that every frozen cap pair satisfies \[\label{eq:frozen-uniform-cap-pair} \|T^{(1,\nu_1)}_{\alpha,r_1}f\, T^{(2,\nu_2)}_{\beta,r_2}g\|_{L^2(U)} \le C\alpha^{1/2}\|f\|_2\|g\|_2,\tag{30}\] for all \(r_1,r_2\in I\). The angular partition is chosen fine enough that each cap pair is either parallel/antipodal or uniformly separated from both of these relations. In the first case there is a common angular chart and there are signs \(\sigma_j\in\{+1,-1\}\) such that, after conjugating the factors with \(\sigma_j=-1\) if necessary, the phases are precisely the signed phases \(\Phi_{r_j}^{\sigma_j}\). Theorem 12 then gives 30 .

In the remaining case, the two direction sets are separated from both \(\omega_1=\omega_2\) and \(\omega_1=-\omega_2\). The scaled distance estimate gives \[\nabla_x[-d_g(x,\exp_o(r\omega))]=\omega+O(\tau+\varepsilon^2)\] with smooth dependence on \(\omega\). Thus, after taking \(\tau\) and \(\varepsilon\) small, the associated covector surfaces have uniformly transverse normal fields. Lemma 17 gives 30 for these cap pairs as well.

Finally combine 28 , 30 , Minkowski’s inequality, Cauchy’s inequality in the two radial variables, and 29 . For each cap pair, \[\begin{align} \|S_{\alpha,\nu_1}^{(1)}f\,S_{\beta,\nu_2}^{(2)}g\|_{L^2(U)} &\le \int_I\int_I \|T^{(1,\nu_1)}_{\alpha,r_1}h^{(1,\nu_1)}_{r_1}\, T^{(2,\nu_2)}_{\beta,r_2}h^{(2,\nu_2)}_{r_2}\|_2\,dr_1dr_2 \\ &\le C\alpha^{1/2} \left(\int_I\|h^{(1,\nu_1)}_{r_1}\|_2^2\,dr_1\right)^{1/2} \left(\int_I\|h^{(2,\nu_2)}_{r_2}\|_2^2\,dr_2\right)^{1/2}. \end{align}\] Summing over the finitely many cap pairs proves ?? . ◻

Proof. The assertion is local in the product support. Fix \[y_0\in \mathbb{R}^{d+1},\qquad \eta_j^0\in \mathbb{R}^d,\quad j=1,\ldots,k,\] with \((y_0,\eta_j^0)\in\operatorname{supp}a_j\). By the rank condition, the image \[\operatorname{Image}\partial_\xi\nabla_y\phi_j(y_0,\eta_j^0) \subset \mathbb{R}^{d+1}\] is a \(d\)-dimensional subspace for each \(j\). Choose a unit vector \(\ell\) outside the finite union of these subspaces. Let \(E:\mathbb{R}^d\to \ell^\perp\subset\mathbb{R}^{d+1}\) be an isometry, and write \[y=y_0+s\ell+Ez,\qquad s\in\mathbb{R},\quad z\in\mathbb{R}^d .\] Then \[\partial_{z\xi}^2\phi_j = E^T\partial_\xi\nabla_y\phi_j .\] Since \(\ker E^T=\mathbb{R}\ell\) and \(\ell\notin \operatorname{Image}\partial_\xi\nabla_y\phi_j(y_0,\eta_j^0)\), the projection \(E^T\) is injective on this image. Hence, using the rank condition again, \[\det \partial_{z\xi}^2\phi_j(0,0,\eta_j^0)\ne0 .\] After shrinking the supports, this remains true throughout the localized piece. Compactness gives the required finite decomposition.

On each such piece, the map \(\xi\mapsto \zeta=\partial_z\phi_j(s,z,\xi)\) is a local diffeomorphism. Using \(\zeta\) as the new frequency variable, we may write \[\partial_s\phi_j(s,z,\xi)=H^{(j)}_{s,z}(\zeta).\] Therefore \[S_{\phi_j}(s,z) = \{(H^{(j)}_{s,z}(\zeta),\zeta)\}.\] So the curvature condition is equivalent to \[\det\partial_{\zeta\zeta}^2H^{(j)}_{s,z}\ne0.\] ◻

Proof. We prove the estimate on one normal-form patch. The finite sum of patches only changes the constant. Write \[U_\lambda(s)u(z):=T_{\lambda,\Phi}u(s,z).\] The nondegeneracy \(\det\partial_{z\xi}^2\Phi\ne 0\) gives the fixed-time \(L^2\) bound uniform in \(s\) and \(\lambda\): \[\|U_\lambda(s)u\|_{L_z^2}\le C\|u\|_{L_\xi^2}.\]

We also need the usual dispersive estimate. The kernel of \(U_\lambda(s)U_\lambda(s')^*\) is \[K_\lambda(s,z;s',z')= \lambda^d\int e^{i\lambda(\Phi(s,z,\xi)-\Phi(s',z',\xi))} b(s,z,\xi)\overline{b(s',z',\xi)}\,d\xi .\] If \(|s-s'|\lesssim\lambda^{-1}\), the trivial bound gives \(|K_\lambda|\lesssim\lambda^d\). If \(|s-s'|\gtrsim\lambda^{-1}\), stationary phase in \(\xi\), using the graph curvature condition, gives \[|K_\lambda(s,z;s',z')|\le C\lambda^d(\lambda |s-s'|)^{-d/2}.\] Combining the two regions, \[\|U_\lambda(s)U_\lambda(s')^*F\|_{L_z^\infty} \le C\lambda^d(1+\lambda |s-s'|)^{-d/2}\|F\|_{L_z^1}.\]

Now rescale both time and space to put this in the normalized Keel–Tao form. Set \[\widetilde{U}_\lambda(t)u(x) :=\lambda^{-d/2}U_\lambda(t/\lambda)u(x/\lambda).\] Then \[\|\widetilde{U}_\lambda(t)u\|_{L_x^2}\le C\|u\|_{L^2}\] and the previous kernel bound becomes \[\|\widetilde{U}_\lambda(t)\widetilde{U}_\lambda(t')^*F\|_{L_x^\infty} \le C(1+|t-t'|)^{-d/2}\|F\|_{L_x^1}.\] Thus the Keel–Tao theorem applies with dispersive exponent \(\sigma=d/2\) in the truncated-decay case. Since Keel–Tao admissibility is \[\frac{1}{a}+\frac{d/2}{b}\le \frac{d/2}{2},\] it is exactly the condition \[\frac{2}{a}+\frac{d}{b}\le \frac{d}{2}.\] The only excluded Keel–Tao endpoint \((a,b,\sigma)=(2,\infty,1)\) is precisely \((a,b,d)=(2,\infty,2)\). Therefore, for every other admissible pair, \[\|\widetilde{U}_\lambda(t)u\|_{L_t^aL_x^b}\le C\|u\|_{L^2}.\] Undoing the rescaling gives \[\|\widetilde{U}_\lambda u\|_{L_t^aL_x^b} =\lambda^{-d/2+d/b+1/a}\|T_{\lambda,\Phi} u\|_{L_s^aL_z^b},\] with the usual interpretation when \(a=\infty\) or \(b=\infty\). Hence \[\|T_{\lambda,\Phi} u\|_{L_s^aL_z^b} \le C\lambda^{d/2-1/a-d/b}\|u\|_{L^2},\] which proves ?? .

When \((a,b,d)=(2,\infty,2)\), the same dyadic \(TT^*\) argument is borderline: the time kernel has size \((1+\lambda |s-s'|)^{-1}\), whose dyadic summation over \(\lambda^{-1}\lesssim |s-s'|\lesssim1\) contributes \(\log\lambda\) at the \(TT^*\) level. Taking the square root gives \[\|T_{\lambda,\Phi} u\|_{L_s^2L_z^\infty}\le C\lambda^{1/2}\sqrt{\log\lambda}\|u\|_{L^2}.\] This is the asserted endpoint estimate. ◻

Proof. We first check that the beams are spherical harmonics. Since \[a_\eta\cdot a_\eta =1+(i\sqrt{1-|\eta|^2})^2+\sum_{j=1}^d(i\eta_j)^2=0,\] one has \[\Delta_{\mathbb{R}^{n+1}}(a_\eta\cdot x)^\nu =\nu(\nu-1)(a_\eta\cdot a_\eta)(a_\eta\cdot x)^{\nu-2}=0.\] Thus \((a_\eta\cdot x)^\nu\) is a homogeneous harmonic polynomial of degree \(\nu\). Its restriction to \(S^n\) is therefore an eigenfunction of \(-\Delta_{S^n}\) with eigenvalue \(\nu(\nu+n-1)\), hence belongs to \(\mathcal{H}_\nu(S^n)\).

It remains to compute the normalization and the overlaps. For \(c\in\mathbb{C}^{n+1}\) we use the standard entire expansion \[\int_{S^n}e^{c\cdot x}\,dS(x) =|S^n|\Gamma\!\left(\frac{n+1}{2}\right) \sum_{m\ge0}\frac{(c\cdot c/4)^m}{m!\Gamma(m+(n+1)/2)} .\] Indeed, this is the usual Bessel-function formula for real \(c\), continued analytically to complex \(c\). Apply it with \(c=s a_\eta+t\overline{a_\zeta}\). Since \(a_\eta\cdot a_\eta=\overline{a_\zeta}\cdot\overline{a_\zeta}=0\), \[c\cdot c=2st\,a_\eta\cdot\overline{a_\zeta} =2st\bigl(1+\omega(\eta)\cdot\omega(\zeta)\bigr).\] On the other hand, \[\int_{S^n}e^{s a_\eta\cdot x+t\overline{a_\zeta}\cdot x}\,dS(x) =\sum_{j,k\ge0}\frac{s^j t^k}{j!k!} \int_{S^n}(a_\eta\cdot x)^j (\overline{a_\zeta}\cdot x)^k\,dS(x).\] Comparing the coefficient of \(s^\nu t^\nu\) gives \[\int_{S^n}(a_\eta\cdot x)^\nu(\overline{a_\zeta}\cdot x)^\nu\,dS(x) =|S^n|\Gamma\!\left(\frac{n+1}{2}\right) \frac{\nu!}{\Gamma(\nu+(n+1)/2)} \left(\frac{1+\omega(\eta)\cdot\omega(\zeta)}{2}\right)^\nu.\] Taking \(\zeta=\eta\) shows that the constant \(\kappa_{\nu,n}\) makes \(\|q_{\nu,\eta}\|_2=1\), and the same identity gives the stated overlap formula. Stirling’s formula yields \[\frac{\Gamma(\nu+(n+1)/2)}{\nu!}\approx \nu^{(n-1)/2}=\nu^{d/2},\] so \(\kappa_{\nu,n}\approx \nu^{d/4}\).

Finally, \[\frac{1+\omega(\eta)\cdot\omega(\zeta)}{2} =1-\frac{|\omega(\eta)-\omega(\zeta)|^2}{4}.\] After restricting to a sufficiently small ball \(|\eta|,|\zeta|\le\delta_0\), the map \(\eta\mapsto\omega(\eta)\) is bi-Lipschitz and \(|\omega(\eta)-\omega(\zeta)|^2/4\le1/2\). Therefore \[1-C|\eta-\zeta|^2 \le \frac{1+\omega(\eta)\cdot\omega(\zeta)}{2} \le 1-c|\eta-\zeta|^2 .\] Raising to the \(\nu\)-th power and using \((1-u)^\nu\le e^{-\nu u}\) and \((1-u)^\nu\ge e^{-2\nu u}\) for \(0\le u\le1/2\) gives the two Gaussian overlap bounds. ◻

Proof. Let \(\delta=(\nu r)^{-1/2}\), and choose a fixed nonnegative \(\psi\in C_0^\infty(\{|\theta|\le 2\varepsilon\})\), equal to one for \(|\theta|\le\varepsilon\). Define \[U_{\nu,r}(x)=\int \psi(\eta/\delta)q_{\nu,\eta}(x)\,d\eta, \qquad u_{\nu,r}=U_{\nu,r}/\|U_{\nu,r}\|_2 .\] By the Gaussian overlap estimate in Lemma 19, \[\|U_{\nu,r}\|_2^2\approx \delta^d\nu^{-d/2}, \qquad \|U_{\nu,r}\|_2^{-1}\approx \nu^{d/4}\delta^{-d/2}.\] On the box \(|t|\le cr\), \(|z|\le c(r/\nu)^{1/2}\), the factors \((a_\eta\cdot x(t,z))^\nu\), \(|\eta|\le2\varepsilon\delta\), have modulus \(\gtrsim1\) and phases differing by at most a sufficiently small absolute constant, provided \(c\) and \(\varepsilon\) are small. Hence \[|U_{\nu,r}(x(t,z))|\gtrsim \nu^{d/4}\delta^d,\] and normalization gives ?? .

For the upper bound, put \(\eta=\delta\theta\). Then \[u_{\nu,r}(x(t,z)) =B_{\nu,r}\int \psi(\theta)\exp\{\nu\log w_{\delta\theta}(t,z)\}\,d\theta, \qquad B_{\nu,r}\approx \nu^{d/4}r^{-d/4},\] where \[w_\eta(t,z)=a_\eta\cdot x(t,z) =\cos\rho\cos t +i\bigl(\cos\rho\sin t\,(1-|\eta|^2)^{1/2}+\beta_\rho z\cdot\eta\bigr), \quad \beta_\rho=\frac{\sin\rho}{\rho},\quad \rho=|z|.\] On the region \(|z|\le K(r/\nu)^{1/2}\), the logarithm is taken on a fixed branch and \(|w_{\delta\theta}|\approx1\). Write \[\Phi(\theta)=\operatorname{Im}(\nu\log w_{\delta\theta}(t,z)), \qquad A(\theta)=\psi(\theta)\exp\{\operatorname{Re}(\nu\log w_{\delta\theta}(t,z))\}, \qquad \Lambda=1+\frac{|t|}{r}.\] A Taylor expansion of the explicit formula for \(w_\eta\), using \(\nu\delta^2=1/r\), gives uniformly on the support of \(\psi\) \[\partial_{\theta_i}\partial_{\theta_j}\Phi(\theta) =-\frac{\sin t\cos t}{r}\delta_{ij} +O_K\!\left(1+\varepsilon\frac{|t|}{r}\right),\] and the scaled symbol bounds \[|\partial_\theta^\alpha \Phi(\theta)|\le C_{\alpha,K}\Lambda\quad(|\alpha|\ge2), \qquad |\partial_\theta^\alpha A(\theta)|\le C_{\alpha,K}\Lambda^{|\alpha|/2}.\] If \(|t|\lesssim r\), the integral is \(O(1)\), which is the desired bound. If \(|t|\ge C_0r\), then \(|\sin t\cos t|\approx |t|\) on the fixed small coordinate interval. Taking \(C_0\) large and \(\varepsilon\) small, the phase has Hessian of size \(\Lambda\). The stationary phase estimate with the above scaled symbol bounds gives \[\left|\int A(\theta)e^{i\Phi(\theta)}\,d\theta\right| \le C_K\Lambda^{-d/2}.\] Multiplying by \(B_{\nu,r}\) proves ?? .

Finally, when \(r=\nu^{-1}\), the lower-bound box contains a ball of radius \(c\nu^{-1}\) after reducing \(c\). Rotating the construction gives the stated zonal function. ◻

Proof. Use the chart \[X(s,y)=(\cos s,\sin s\,\omega(y)), \qquad s\in[1/4,1/2],\quad |y|\ll1,\] where \(\omega(y)=(\sqrt{1-|y|^2},y)\in S^d\). In this chart the beams from Lemma 19 satisfy, after shrinking the chart, \[\label{eq:beam-belt-decay} |q_{\nu,\eta}(X(s,y))| \le C\nu^{d/4}e^{-c\nu|y-\eta|^2}, \qquad |q_{\nu,\eta}(X(s,y))| \ge c\nu^{d/4}\tag{35}\] whenever \(|y-\eta|\le c\nu^{-1/2}\). This follows directly from \[q_{\nu,\eta}(X(s,y)) =\kappa_{\nu,n}\bigl(\cos s+i\sin s\,\omega(\eta)\cdot\omega(y)\bigr)^\nu\] and the bi-Lipschitz property of \(\omega\) near the origin.

Choose a subinterval \(I_L\subset[1/4,1/2]\) with \(|I_L|\approx L\). Let \(D\gg1\) be fixed, and choose a maximal \(D\nu^{-1/2}\)-separated set \[\mathcal{A}\subset\{\eta: |\eta|\le c_0R^{-1/2}\}.\] Then \[N:=|\mathcal{A}|\approx (\nu/R)^{d/2}.\] For \(\eta\in\mathcal{A}\), set \[T_\eta=\{X(s,y):s\in I_L,\;|y-\eta|\le c\nu^{-1/2}\}, \qquad E=\bigcup_{\eta\in\mathcal{A}}T_\eta .\] The tubes \(T_\eta\) are disjoint if \(c\ll D\), and therefore \[|E|\approx N L\nu^{-d/2}\approx LR^{-d/2}.\]

Set \[Q=\sum_{\eta\in\mathcal{A}}q_{\nu,\eta}.\] By the Gaussian overlap bound in Lemma 19, \[\|Q\|_2^2 \le \sum_{\eta,\zeta\in\mathcal{A}}e^{-c\nu|\eta-\zeta|^2} \le C_D N.\] If \(x\in T_\eta\), then the \(\eta\)-beam is \(\gtrsim\nu^{d/4}\) by 35 , while the remaining beams contribute at most \[C\nu^{d/4}\sum_{\zeta\in\mathcal{A}\setminus\{\eta\}} e^{-c\nu|\eta-\zeta|^2}.\] Taking \(D\) sufficiently large makes this tail smaller than half of the main term. Hence \(|Q|\gtrsim\nu^{d/4}\) on \(E\). With \(u=Q/\|Q\|_2\), \[|u(x)|\gtrsim \nu^{d/4}N^{-1/2}\approx R^{d/4}, \qquad x\in E.\] ◻

Proof. Let \(R_a\) be the rotation along the reference geodesic, so that \(R_a x(t,z)=x(t+a,z)\), and write \(P_a=R_a u_{\nu,r}\). The lower bound on a single core and the tail estimate away from that core are exactly ?? and ?? , after replacing \(t\) by \(t-a\). It remains to prove the almost-orthogonality estimate.

By rotation invariance it suffices to estimate \(\langle P_0,P_\tau\rangle\), where \(\tau=b-a\). Put \(\delta=(\nu r)^{-1/2}\) and \(\alpha_\eta=(1-|\eta|^2)^{1/2}\). Lemma 19, applied after rotating one beam by \(\tau\), gives \[\langle q_{\nu,\eta},R_\tau q_{\nu,\zeta}\rangle =\Gamma_\tau(\eta,\zeta)^\nu,\] where \[\Gamma_\tau(\eta,\zeta) =\frac{1}{2}\Bigl( \cos\tau\,(1+\alpha_\eta\alpha_\zeta)+\eta\cdot\zeta +i\sin\tau\,(\alpha_\eta+\alpha_\zeta) \Bigr).\] Consequently, \[\langle P_0,P_\tau\rangle =c_{\nu,r}\iint \psi(\theta)\overline{\psi(\vartheta)} \Gamma_\tau(\delta\theta,\delta\vartheta)^\nu \,d\theta\,d\vartheta, \qquad c_{\nu,r}\approx r^{-d/2}.\] For \(|\tau|\lesssim r\), Cauchy’s inequality gives \(|\langle P_0,P_\tau\rangle|\le1\). Assume \(|\tau|\ge C_0r\). Taylor expansion of the explicit kernel gives \[\operatorname{Re}\bigl(\nu\log\Gamma_\tau(\delta\theta,\delta\vartheta)\bigr) \le -c\frac{|\theta-\vartheta|^2}{r}+C.\] With \[u=\frac{\theta+\vartheta}{2}, \qquad v=\frac{\theta-\vartheta}{\sqrt r},\] the Jacobian is \(d\theta\,d\vartheta=r^{d/2}\,du\,dv\), which cancels \(c_{\nu,r}\). The same expansion gives, in the \(u\)-variables, \[\partial_{u_i}\partial_{u_j} \operatorname{Im}\bigl(\nu\log\Gamma_\tau(\delta\theta,\delta\vartheta)\bigr) =-\frac{\sin\tau}{r}\delta_{ij} +O\!\left(1+\varepsilon\frac{|\tau|}{r}\right).\] On the fixed coordinate interval, \(|\sin\tau|\approx |\tau|\). Therefore the inner \(u\)-integral has a nondegenerate Hessian of size \(|\tau|/r\), while the \(v\)-dependence is Gaussian. The same scaled stationary-phase estimate used in Lemma 20 yields \[\label{eq:short-packet-overlap} |\langle P_0,P_\tau\rangle| \le C\left(1+\frac{|\tau|}{r}\right)^{-d/2}.\tag{36}\] Rotating gives the same bound for \(P_a\) and \(P_b\).

Now \[\Big\|\sum_j P_{a_j}\Big\|_2^2 \le \sum_{j,\ell}|\langle P_{a_j},P_{a_\ell}\rangle| \le C\sum_j\sum_{m\in\mathbb{Z}}(1+D|m|)^{-d/2} \lesssim N,\] because the points \(a_j\) are \(Dr\)-separated on a line and \(d/2>1\). This proves ?? .

Finally, fix \(x\in F_j\). The main packet satisfies \(|P_{a_j}(x)|\gtrsim\nu^{d/4}r^{-d/4}\) by ?? . For \(\ell\ne j\), the upper bound ?? gives \[|P_{a_\ell}(x)| \le C\nu^{d/4}r^{-d/4} \left(1+\frac{|a_\ell-a_j|}{r}\right)^{-d/2},\] after decreasing the core constant \(c\). Hence \[\sum_{\ell\ne j}|P_{a_\ell}(x)| \le C\nu^{d/4}r^{-d/4} \sum_{m\ge1}(1+Dm)^{-d/2}.\] Choosing the fixed separation constant \(D\) large makes this tail smaller than the main term. This proves ?? . ◻

Proof. Since \(\nu\le R\le\nu^2\), the number \(r=\nu/R\) satisfies \(\nu^{-1}\le r\le1\), and the transverse radius in Lemma 20 is \[(r/\nu)^{1/2}=R^{-1/2}.\] If \(L\le r\), take one envelope \(u_{\nu,r}\) from Lemma 20 and restrict its core to \(|t|\lesssim L\). On this set \[|u_{\nu,r}| \gtrsim \nu^{d/4}r^{-d/4}=R^{d/4},\] and the measure is \(\approx LR^{-d/2}\). This is the first alternative in ?? .

Assume now that \(L>r\) and \(d\ge3\). Choose times \(a_j\) in an interval of length \(\approx L\), separated by \(Dr\), with \[N\approx L/r=LR/\nu.\]

Let \(P_{a_j}\) be the corresponding rotated envelopes, set \[Q=\sum_{j=1}^N P_{a_j}, \qquad u=Q/\|Q\|_2,\] and let \[E=\bigcup_{j=1}^N \{x(t,z): |t-a_j|\le cr, \;|z|\le cR^{-1/2}\}.\] The pieces are disjoint, so \[|E|\approx N r R^{-d/2}\approx LR^{-d/2}.\] By Lemma 22, \(\|Q\|_2\lesssim N^{1/2}\) and \(|Q|\gtrsim R^{d/4}\) on \(E\). Therefore \[|u(x)|\gtrsim R^{d/4}N^{-1/2} \approx R^{d/4}(L/r)^{-1/2}, \qquad x\in E.\] ◻

Proof. This is the endpoint case \(r=\nu^{-1}\) of the envelope train. Since \((r/\nu)^{1/2}=\nu^{-1}\) and \(R^{-1/2}\le\nu^{-1}\), a single zonal function from Lemma 20 gives the result when \(L\nu\lesssim1\), after restricting its core to \[E=\{x(t,z): |t|\le cL,\;|z|\le cR^{-1/2}\}.\] Indeed \(|u|\gtrsim\nu^{d/2}\) there and \(|E|\approx LR^{-d/2}\).

Assume now that \(L\nu\gg1\) and \(d\ge3\). Choose times \(a_j\) in an interval of length \(\approx L\), separated by \(D\nu^{-1}\), with \[N\approx L\nu.\]

Let \(P_{a_j}\) be the corresponding rotations of \(u_{\nu,\nu^{-1}}\), and set \[Q=\sum_{j=1}^N P_{a_j}, \qquad u=Q/\|Q\|_2.\] Let \[E=\bigcup_{j=1}^N \{x(t,z): |t-a_j|\le c\nu^{-1},\;|z|\le cR^{-1/2}\}.\] The pieces are disjoint and \[|E|\approx N\nu^{-1}R^{-d/2}\approx LR^{-d/2}.\] Lemma 22, with \(r=\nu^{-1}\), gives \(\|Q\|_2\lesssim N^{1/2}\) and \(|Q|\gtrsim\nu^{d/2}\) on \(E\). Therefore \[|u(x)|\gtrsim \nu^{d/2}N^{-1/2} \approx \nu^{d/2}(L\nu)^{-1/2}, \qquad x\in E.\] ◻

Proof. Work in one box \(B_{L,R}=\{x(t,z):t\in I,\;z\in Q_R\}\). A shift means the origin of a packet grid. For a time scale \(\ell\le L\) and shift \(\tau\in[0,D\ell)\), set \[S_{\ell,\tau}=I\cap\bigcup_{j\in\mathbb{Z}} \{t:|t-(\tau+jD\ell)|\le c\ell\}.\] For a transverse scale \(\rho\le R^{-1/2}\) and shift \(\sigma\in[0,D\rho)^d\), set \[G_{\rho,\sigma}=Q_R\cap\bigcup_{k\in\mathbb{Z}^d} \bigl(\sigma+D\rho k+[-c\rho,c\rho]^d\bigr).\] Thus \(\tau\) translates a time train and \(\sigma\) translates a transverse lattice. The required shifts are chosen by averaging. For each fixed \(t\in I\) and \(z\in Q_R\), \[\mathbb{E}_\tau {\boldsymbol{1}}_{S_{\ell,\tau}}(t):={1\over D\ell}\int_0^{D\ell}{\boldsymbol{1}}_{S_{\ell,\tau}}(t)\,d\tau={2c\over D},\] \[\mathbb{E}_\sigma {\boldsymbol{1}}_{G_{\rho,\sigma}}(z):= {1\over (D\rho)^d}\int_{[0,D\rho)^d}{\boldsymbol{1}}_{G_{\rho,\sigma}}(z)\,d\sigma =\left({2c\over D}\right)^d .\] Then for any fixed \(A,B\le m\) \[\mathbb{E}_{\tau_1,...,\tau_A}\left|I\cap\bigcap_{a=1}^A S_{\ell_a,\tau_a}\right|=\int_{I}\prod_{a=1}^A\mathbb{E}_{\tau_a}{\boldsymbol{1}}_{S_{\ell_a,\tau_a}}(t)dt=\Big(\frac{2c}{D}\Big)^A|I|\approx L,\] \[\mathbb{E}_{\sigma_1,...,\sigma_B}\left|Q_R\cap\bigcap_{b=1}^B G_{\rho_b,\sigma_b}\right|=\int_{Q_R}\prod_{b=1}^B\mathbb{E}_{\sigma_b}{\boldsymbol{1}}_{G_{\rho_b,\sigma_b}}(z)dz=\Big(\frac{2c}{D}\Big)^{dB}|Q_R|\approx R^{-d/2},\] Thus there are shifts \(\tau_1,...,\tau_A\) and \(\sigma_1,...,\sigma_B\) so that \[\label{eq:grid-shift-choice} \left|I\cap\bigcap_{a=1}^A S_{\ell_a,\tau_a}\right|\gtrsim L, \qquad \left|Q_R\cap\bigcap_{b=1}^B G_{\rho_b,\sigma_b}\right|\gtrsim R^{-d/2}.\tag{38}\]

For each frequency choose a good product set \(T_i\times Z_i\subset I\times Q_R\). Full factors \(I\) and \(Q_R\) impose no restriction. If \(1\le R\le\nu_i\), take \(\rho_i=\nu_i^{-1/2}\) and, by Lemma 21, \[T_i=I, \qquad Z_i=G_{\rho_i,\sigma_i}, \qquad |u_i|\gtrsim R^{d/4}\gtrsim A_{\nu_i}(L,R) \quad onT_i\times Z_i .\] If \(\nu_i\le R\le\nu_i^2\), set \(r_i=\nu_i/R\). For \(L\le r_i\), Lemma 23 gives the bound on all of \(I\times Q_R\). For \(L>r_i\), admissibility gives \(d\ge3\), and the train with centers \(\tau_i+jDr_i\) gives \[T_i=S_{r_i,\tau_i}, \qquad Z_i=Q_R, \qquad |u_i|\gtrsim R^{d/4}(L/r_i)^{-1/2}=A_{\nu_i}(L,R) \quad onT_i\times Z_i .\] If \(R\ge\nu_i^2\), Lemma 24 gives the bound on all of \(I\times Q_R\) when \(L\nu_i\lesssim1\). For \(L\nu_i\gg1\), use the endpoint train with centers \(\tau_i+jD\nu_i^{-1}\), obtaining \[T_i=S_{\nu_i^{-1},\tau_i}, \qquad Z_i=Q_R, \qquad |u_i|\gtrsim \nu_i^{d/2}(L\nu_i)^{-1/2}=A_{\nu_i}(L,R) \quad onT_i\times Z_i .\]

Choose all nontrivial shifts by 38 and set \[T=I\cap\bigcap_i T_i, \qquad Z=Q_R\cap\bigcap_i Z_i,\] omitting full factors. Then \(|T|\gtrsim L\) and \(|Z|\gtrsim R^{-d/2}\); after discarding subsets we may take comparable upper bounds as well. With \(E=x(T,Z)\), the coordinate density gives \(|E|\approx LR^{-d/2}\). Since \(T\times Z\subset T_i\times Z_i\) for every \(i\), the pointwise bounds above imply ?? . ◻

Proof. The one-function statement is Lemma 21, Lemma 23, or Lemma 24, according as \((\mathrm A_1)\), \((\mathrm A_2)\), or \((\mathrm A_3)\) holds. The common-set statement is Lemma 25. ◻

Proof. If \(M=\prod_iM_i\), then \(\prod_i|u_i|\ge M\) on \(E\), and hence \[\left\|\prod_i u_i\right\|_{L^{p}} \ge \|M\|_{L^p(E)}=|E|^{\frac{1}{p}}M.\] ◻

Proof. Let \(1\ll\lambda\le \mu\) and \(d=n-1\). We first use the box \(L\approx1\), \(R=\lambda\). The \(\lambda\)-factor is a beam that contributes \(A_\lambda(1,\lambda)= \lambda^{d/4}\), while the \(\mu\)-factor is a beam block that contributes \(A_\mu(1,\lambda)= \lambda^{d/4}.\) Since \(|E_{L,R}|\approx \lambda^{-d/2}\), Lemma 26 gives \[\|uv\|_p\gtrsim \lambda^{d/2}\lambda^{-d/(2p)} =\lambda^{\frac{d}{2}(1-1/p)}.\] This agrees with the powers in Theorem 1 for \(d=1,2\), and the first range for \(d\ge3\).

Assume \(d\ge3\). If \(\mu\le\lambda^2\), use first \(L\approx1\), \(R=\mu\). The \(\lambda\)-factor is an envelope train that contributes \[A_\lambda(1,\mu)=\mu^{d/4}(\mu/\lambda)^{-1/2} =\lambda^{1/2}\mu^{d/4-1/2},\] while the \(\mu\)-factor is a beam that contributes \(A_\mu(1,\mu)=\mu^{d/4}\). Since \(|E_{L,R}|\approx\mu^{-d/2}\), \[\|uv\|_p\gtrsim \lambda^{1/2}\mu^{\frac{d-1}{2}-\frac{d}{2p}}.\] In the same frequency regime, use the box \(L=\mu/\lambda^2\), \(R=\lambda^2\). The \(\lambda\)-factor is a zonal train that contributes \[A_\lambda(\mu/\lambda^2,\lambda^2)=\lambda^{d/2}(\mu/\lambda)^{-1/2}=\lambda^{(d+1)/2}\mu^{-1/2},\] while the \(\mu\) factor is an envelope train that contributes \[A_\mu(\mu/\lambda^2,\lambda^2)=\lambda^{d/2}.\] Since \(|E_{L,R}|\approx \mu\lambda^{-(d+2)}\), \[\|uv\|_p\gtrsim \lambda^{\frac{2d+1}{2}-\frac{d+2}{p}}\mu^{\frac{1}{p}-\frac{1}{2}}.\] If \(\mu\ge\lambda^2\), use \(L\approx1\), \(R=\lambda^2\). Then the \(\lambda\)-factor is a zonal train that contributes \(A_\lambda(1,\lambda^2)=\lambda^{(d-1)/2},\) while the \(\mu\)-factor is a beam block that contributes \(A_\mu(1,\lambda^2)=\lambda^{d/2}\). Since \(|E_{L,R}|\approx\lambda^{-d}\), \[\|uv\|_p\gtrsim \lambda^{d-\frac{1}{2}-\frac{d}{p}}.\] These are the two minimum branches in the remaining ranges. ◻

Proof. For \(k\ge3\), use Propositions 17, 18, 19, and 20 below, according as \(n\ge5\), \(n=4\), \(n=3\), and \(n=2\). ◻

Proof. Let \(d=n-1.\) Then \(d\ge 4\). The lower-frequency factors that are not named explicitly below are chosen to be zonal functions from Lemma 20, with scale \(r=\lambda_j^{-1}\), all centered at the same point as the packet box. The boxes used below are contained in the region where these zonal functions have size \(\gtrsim \lambda_j^{d/2}\). Hence such filler factors contribute the product of the corresponding powers \(\lambda_j^{d/2}\).

We first consider \[1\le p\le p_2=\frac{d+2}{d+1}.\] Set \[a=\lambda_{k-3},\qquad b=\lambda_{k-2},\qquad c=\lambda_{k-1},\qquad e=\lambda_k .\] The filler factors contribute \(\prod_{j=1}^{k-4}\lambda_j^{d/2}\).

Assume first that \(ac\le b^2.\) Choose \(L=c/b^2,\; R=b^2.\) Then \(L\lesssim 1\), and \[|E_{L,R}|\approx \frac{c}{b^2}(b^2)^{-d/2} = c b^{-(d+2)}.\] We now list the sizes of the four active factors on \(E_{L,R}\).

The \(a\)-factor is zonal and \(A_a(L,R)=a^{d/2}.\)

The \(b\)-factor is a zonal train and \(A_b(L,R)=b^{d/2}\left(\frac{c}{b}\right)^{-1/2} =b^{(d+1)/2}c^{-1/2}.\)

The \(c\)-factor is either a beam block or an envelope, and in both cases \(A_c(L,R)=R^{d/4}=b^{d/2}.\)

The \(e\)-factor is again a beam block or an envelope, and \(A_e(L,R)=R^{d/4}=b^{d/2}.\)

Therefore, Lemma 26 gives \[\begin{align} L^{(k)}_{p,n}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-4}\lambda_j^{d/2}\right) a^{d/2} b^{(3d+1)/2} c^{-1/2} \left(c b^{-(d+2)}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-4}\lambda_j^{d/2}\right) a^{d/2} b^{\frac{3d+1}{2}-\frac{d+2}{p}} c^{\frac{1}{p}-\frac{1}{2}}. \end{align}\] This is the first branch in the first range of Theorem 2.

Assume next that \(ac\ge b^2.\) Choose \(L=a^{-1},\;R=ac .\) Then \[|E_{L,R}|\approx a^{-1}(ac)^{-d/2} =a^{-1-d/2}c^{-d/2}.\]

The \(a\)-factor is zonal and \(A_a(L,R)=a^{d/2}.\)

The \(b\)-factor is a zonal train and \(A_b(L,R)=b^{d/2}\left(\frac{b}{a}\right)^{-1/2} =a^{1/2}b^{(d-1)/2}.\)

The \(c\)-factor is an envelope and \(A_c(L,R)=R^{d/4}=(ac)^{d/4}.\)

The \(e\)-factor is a beam block or an envelope, and \(A_e(L,R)=R^{d/4}=(ac)^{d/4}.\)

Therefore, Lemma 26 gives \[\begin{align} L^{(k)}_{p,n}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-4}\lambda_j^{d/2}\right) a^{(2d+1)/2}b^{(d-1)/2}c^{d/2} \left(a^{-1-d/2}c^{-d/2}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-4}\lambda_j^{d/2}\right) a^{\frac{2d+1}{2}-\frac{d+2}{2p}} b^{\frac{d-1}{2}} c^{\frac{d}{2}-\frac{d}{2p}}. \end{align}\] This is the second branch in the first range of Theorem 2. Moreover, the ratio of the first branch to the second branch is \[\left(\frac{b^2}{ac}\right)^{{(d+1)}/{2}-{(d+2)}/{(2p)}}.\] Since \(1\le p\le (d+2)/(d+1)\), the exponent is nonpositive, so the two subcases above select the smaller of the two branches.

We now turn to the remaining ranges. Put \[a=\lambda_{k-2},\qquad b=\lambda_{k-1},\qquad c=\lambda_k .\] The filler factors contribute \(\prod_{j=1}^{k-3}\lambda_j^{d/2}.\)

First suppose \[p_2\le p\le p_1=\frac{d}{d-1}.\] Choose \(L=a^{-1},\; R=ab .\) Then \[|E_{L,R}|\approx a^{-1}(ab)^{-d/2} =a^{-1-d/2}b^{-d/2}.\]

The \(a\)-factor is zonal and \(A_a(L,R)=a^{d/2}.\)

The \(b\)-factor is an envelope and \(A_b(L,R)=R^{d/4}=(ab)^{d/4}.\)

The \(c\)-factor is a beam block or an envelope, and in either case \(A_c(L,R)=R^{d/4}=(ab)^{d/4}.\)

Hence, Lemma 26 yields \[\begin{align} L^{(k)}_{p,n}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-3}\lambda_j^{d/2}\right) a^d b^{d/2} \left(a^{-1-d/2}b^{-d/2}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-3}\lambda_j^{d/2}\right) a^{d-\frac{d+2}{2p}} b^{\frac{d}{2}-\frac{d}{2p}}. \end{align}\] This is precisely the branch in Theorem 2 for \(p_2\le p\le p_1\).

Next assume \[p_1\le p\le p_0=\frac{d+2}{d}.\] There are two frequency regimes.

If \(ac\le b^2,\) choose \(L=a^{-1},\; R=ac .\) Then \[|E_{L,R}|\approx a^{-1}(ac)^{-d/2} =a^{-1-d/2}c^{-d/2}.\] The \(a\)-factor is zonal and \(A_a(L,R)=a^{d/2}.\) The \(b\)-factor is an envelope train and contributes \[A_b(L,R)=R^{d/4}\left(\frac{LR}{b}\right)^{-1/2} =(ac)^{d/4}\left(\frac{c}{b}\right)^{-1/2} =a^{d/4}b^{1/2}c^{d/4-1/2}.\] The \(c\)-factor is an envelope and contributes \(A_c(L,R)=R^{d/4}=(ac)^{d/4}.\)

Consequently, Lemma 26 gives \[\begin{align} L^{(k)}_{p,n}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-3}\lambda_j^{d/2}\right) a^d b^{1/2}c^{(d-1)/2} \left(a^{-1-d/2}c^{-d/2}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-3}\lambda_j^{d/2}\right) a^{d-\frac{d+2}{2p}} b^{1/2} c^{\frac{d-1}{2}-\frac{d}{2p}}. \end{align}\] This is the first branch in the range \(p_1\le p\le p_0\).

If instead \(ac\ge b^2,\) choose \(L=a^{-1},\; R=b^2.\) Then \[|E_{L,R}|\approx a^{-1}b^{-d}.\] The \(a\)-factor is zonal and \(A_a(L,R)=a^{d/2}.\) The \(b\)-factor is a zonal train and contributes \[A_b(L,R)=b^{d/2}\left(\frac{b}{a}\right)^{-1/2} =a^{1/2}b^{(d-1)/2}.\] The \(c\)-factor is a beam block or an envelope, and \(A_c(L,R)=R^{d/4}=b^{d/2}.\) Thus Lemma 26 yields \[\begin{align} L^{(k)}_{p,n}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-3}\lambda_j^{d/2}\right) a^{(d+1)/2}b^{d-1/2} \left(a^{-1}b^{-d}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-3}\lambda_j^{d/2}\right) a^{\frac{d+1}{2}-\frac{1}{p}} b^{d-\frac{1}{2}-\frac{d}{p}}. \end{align}\] This is the second branch in the range \(p_1\le p\le p_0\). The ratio of the first branch to the second branch is \[\left(\frac{ac}{b^2}\right)^{{(d-1)}/{2}-{d}/{(2p)}}.\] Since \(p\ge p_1=d/(d-1)\), the exponent is nonnegative, and the two subcases again select the smaller branch.

It remains to prove the lower bounds for \[p_0\le p\le 2.\] We keep the notation \(a=\lambda_{k-2}\), \(b=\lambda_{k-1}\), and \(c=\lambda_k\), and the same filler contribution \(\prod_{j=1}^{k-3}\lambda_j^{d/2}\).

If \(ac\le b^2,\) choose \(L=c/b^2,\;R=b^2 .\) Then \[|E_{L,R}|\approx c b^{-(d+2)}.\] The \(a\)-factor is zonal and \(A_a(L,R)=a^{d/2}.\) The \(b\)-factor is a zonal train, and \[A_b(L,R)=b^{d/2}\left(\frac{c}{b}\right)^{-1/2} =b^{(d+1)/2}c^{-1/2}.\] The \(c\)-factor is a beam block or an envelope, and contributes \(A_c(L,R)=R^{d/4}=b^{d/2}.\)

Hence, Lemma 26 gives \[\begin{align} L^{(k)}_{p,n}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-3}\lambda_j^{d/2}\right) a^{d/2} b^{(2d+1)/2} c^{-1/2} \left(c b^{-(d+2)}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-3}\lambda_j^{d/2}\right) a^{d/2} b^{\frac{2d+1}{2}-\frac{d+2}{p}} c^{\frac{1}{p}-\frac{1}{2}}. \end{align}\] This is the first branch in the range \(p_0\le p\le 2\).

If \(ac\ge b^2,\) we use the same box as in the second subcase of the preceding range, namely \(L=a^{-1},\; R=b^2 .\) Then \[|E_{L,R}|\approx a^{-1}b^{-d}.\] The \(a\)-factor is zonal and \(A_a(L,R)=a^{d/2}\). The \(b\)-factor is a zonal train and \[A_b(L,R)=b^{d/2}(b/a)^{-1/2}=a^{1/2}b^{(d-1)/2}.\] The \(c\)-factor is a beam block or an envelope, and \(A_c(L,R)=b^{d/2}\). Therefore Lemma 26 gives \[\begin{align} L^{(k)}_{p,n}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-3}\lambda_j^{d/2}\right) a^{(d+1)/2}b^{d-1/2} \left(a^{-1}b^{-d}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-3}\lambda_j^{d/2}\right) a^{\frac{d+1}{2}-\frac{1}{p}} b^{d-\frac{1}{2}-\frac{d}{p}}. \end{align}\] This is the second branch in the range \(p_0\le p\le 2\). In this range, the ratio of the first branch to the second branch is \[\left(\frac{ac}{b^2}\right)^{{1}/{p}-{1}/{2}},\] and the exponent is nonnegative because \(p\le 2\). Hence the two subcases select the smaller branch.

Combining the four ranges, and recalling that \(d=n-1\), we obtain exactly the powers in Theorem 2 for \(n\ge 5\). This proves the proposition. ◻

Proof. Here \(d=3\). As in the preceding proof, unnamed lower-frequency factors are chosen to be zonal functions from Lemma 20, centered at the same packet box.

First assume \[1\le p\le \frac{10}{9}.\] Set \[a=\lambda_{k-3},\qquad b=\lambda_{k-2},\qquad c=\lambda_{k-1},\qquad e=\lambda_k .\] The filler factors contribute \(\prod_{j=1}^{k-4}\lambda_j^{3/2}\).

Suppose first that \(ac\le b^2\). Choose \(L=a^{-1},\;R=ac .\) Then \[|E_{L,R}|\approx a^{-1}(ac)^{-3/2}=a^{-5/2}c^{-3/2}.\] The \(a\)-factor is zonal and contributes \(A_a(L,R)=a^{3/2}\). The \(b\)-factor is an envelope train and contributes \[A_b(L,R)=(ac)^{3/4}\left(\frac{c}{b}\right)^{-1/2} =a^{3/4}b^{1/2}c^{1/4}.\] The \(c\)- and \(e\)-factors are beam blocks or envelopes, and each contributes \[A_c(L,R)=A_e(L,R)=(ac)^{3/4}\] Hence \[\begin{align} \mathcal{L}_{p,4}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-4}\lambda_j^{3/2}\right) a^{3/2}\big(a^{3/4}b^{1/2}c^{1/4}\big)(ac)^{3/4}(ac)^{3/4} \left(a^{-5/2}c^{-3/2}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-4}\lambda_j^{3/2}\right) a^{\frac{15}{4}-\frac{5}{2p}} b^{1/2} c^{\frac{7}{4}-\frac{3}{2p}} . \end{align}\] This is the first branch in the first range of Theorem 2.

Suppose next that \(ac\ge b^2\). Use the same box \(L=a^{-1}\), \(R=ac\). The \(a\)-, \(c\)-, and \(e\)-factors are as above. Now the \(b\)-factor is a zonal train and contributes \[A_b(L,R)=b^{3/2}\left(\frac{b}{a}\right)^{-1/2}=a^{1/2}b .\] Thus \[\begin{align} \mathcal{L}_{p,4}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-4}\lambda_j^{3/2}\right) a^{3/2}(a^{1/2}b)(ac)^{3/4}(ac)^{3/4} \left(a^{-5/2}c^{-3/2}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-4}\lambda_j^{3/2}\right) a^{\frac{7}{2}-\frac{5}{2p}} b c^{\frac{3}{2}-\frac{3}{2p}} . \end{align}\] This is the second branch in the first range. The ratio of the first branch to the second branch is \[\left(\frac{ac}{b^2}\right)^{1/4},\] so the two subcases select the smaller branch.

We next assume \[\frac{10}{9}\le p\le \frac{5}{4},\] and keep the same notation \(a,b,c,e\). If \(ac\le b^2\), choose \(L=c/b^2,\;R=b^2 .\) Then \(L\lesssim1\) and \[|E_{L,R}|\approx \frac{c}{b^2}(b^2)^{-3/2}=cb^{-5}.\] The \(a\)-factor is zonal and contributes \(A_a(L,R)=a^{3/2}\). The \(b\)-factor is a zonal train and contributes \[A_b(L,R)=b^{3/2}\left(\frac{c}{b}\right)^{-1/2}=b^2c^{-1/2}.\] The \(c\)- and \(e\)-factors are beam blocks or envelopes, and each contributes \[A_c(L,R)=A_e(L,R)=b^{3/2}.\] Therefore \[\begin{align} \mathcal{L}_{p,4}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-4}\lambda_j^{3/2}\right) a^{3/2}(b^2c^{-1/2})b^{3/2}b^{3/2} \left(cb^{-5}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-4}\lambda_j^{3/2}\right) a^{3/2} b^{5-\frac{5}{p}} c^{\frac{1}{p}-\frac{1}{2}} . \end{align}\] This is the first branch in the second range.

If \(ac\ge b^2\), use again \(L=a^{-1}\), \(R=ac\). Then \[|E_{L,R}|\approx a^{-5/2}c^{-3/2}.\] The \(a\)-factor is zonal and \(A_a(L,R)=a^{3/2}\). The \(b\)-factor is a zonal train and \(A_b(L,R)=a^{1/2}b\). The \(c\)-factor is an envelope and \(A_c(L,R)=(ac)^{3/4}\). The \(e\)-factor is a beam block or an envelope, and \(A_e(L,R)=(ac)^{3/4}\).

Hence \[\begin{align} \mathcal{L}_{p,4}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-4}\lambda_j^{3/2}\right) a^{\frac{7}{2}}bc^{3/2} \left(a^{-5/2}c^{-3/2}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-4}\lambda_j^{3/2}\right) a^{\frac{7}{2}-\frac{5}{2p}} b c^{\frac{3}{2}-\frac{3}{2p}} . \end{align}\] This is the second branch in the second range. The ratio of the first branch to the second branch is \[\left(\frac{ac}{b^2}\right)^{\frac{5}{2p}-2}.\] Since \(10/9\le p\le5/4\), the exponent \(5/(2p)-2\) is nonnegative. Thus the two subcases again select the smaller branch.

For the remaining three ranges, set \[a=\lambda_{k-2},\qquad b=\lambda_{k-1},\qquad c=\lambda_k,\] and let the lower-frequency factors contribute \(\prod_{j=1}^{k-3}\lambda_j^{3/2}\).

Assume first that \[\frac{5}{4}\le p\le \frac{3}{2}.\] Choose \(L=a^{-1},\; R=ab .\) Then \[|E_{L,R}|\approx a^{-1}(ab)^{-3/2}=a^{-5/2}b^{-3/2}.\] The \(a\)-factor is zonal and \(A_a(L,R)=a^{3/2}\). The \(b\)- and \(c\)-factors are beam blocks or envelopes, and each contributes \[A_b(L,R)=A_c(L,R)=(ab)^{3/4}.\] Therefore \[\begin{align} \mathcal{L}_{p,4}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-3}\lambda_j^{3/2}\right) a^{3/2}(ab)^{3/4}(ab)^{3/4} \left(a^{-5/2}b^{-3/2}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-3}\lambda_j^{3/2}\right) a^{3-\frac{5}{2p}} b^{\frac{3}{2}-\frac{3}{2p}} . \end{align}\] This is the branch in the range \(5/4\le p\le3/2\).

Next suppose that \[\frac{3}{2}\le p\le \frac{5}{3}.\] If \(ac\le b^2\), choose \(L=a^{-1}\), \(R=ac\). Then \[|E_{L,R}|\approx a^{-5/2}c^{-3/2}.\] The \(a\)-factor is zonal and contributes \(A_a(L,R)=a^{3/2}\). The \(b\)-factor is an envelope train and contributes \[A_b(L,R)=(ac)^{3/4}\left(\frac{c}{b}\right)^{-1/2} =a^{3/4}b^{1/2}c^{1/4}.\] The \(c\)-factor is an envelope and contributes \(A_c(L,R)=(ac)^{3/4}\). Hence \[\begin{align} \mathcal{L}_{p,4}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-3}\lambda_j^{3/2}\right) a^{3/2}\big(a^{3/4}b^{1/2}c^{1/4}\big)(ac)^{3/4} \left(a^{-5/2}c^{-3/2}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-3}\lambda_j^{3/2}\right) a^{3-\frac{5}{2p}} b^{1/2} c^{1-\frac{3}{2p}} . \end{align}\] This is the first branch in the range \(3/2\le p\le5/3\).

If \(ac\ge b^2\), choose \(L=a^{-1}\), \(R=b^2\). Then \[|E_{L,R}|\approx a^{-1}b^{-3}.\] The \(a\)-factor is zonal and contributes \(A_a(L,R)=a^{3/2}\). The \(b\)-factor is a zonal train and contributes \[A_b(L,R)=b^{3/2}\left(\frac{b}{a}\right)^{-1/2}=a^{1/2}b.\] The \(c\)-factor is a beam block or an envelope, and contributes \(A_c(L,R)=b^{3/2}\). Therefore \[\begin{align} \mathcal{L}_{p,4}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-3}\lambda_j^{3/2}\right) a^{3/2}(a^{1/2}b)b^{3/2} \left(a^{-1}b^{-3}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-3}\lambda_j^{3/2}\right) a^{2-\frac{1}{p}} b^{\frac{5}{2}-\frac{3}{p}} . \end{align}\] This is the second branch in the range \(3/2\le p\le5/3\). The ratio of the first branch to the second branch is \[\left(\frac{ac}{b^2}\right)^{1-\frac{3}{2p}}.\] Since \(p\ge3/2\), the exponent is nonnegative, and the two subcases select the smaller branch.

It remains to consider \[\frac{5}{3}\le p\le 2.\] If \(ac\le b^2\), choose \(L=c/b^2,\; R=b^2 .\) Then \(|E_{L,R}|\approx cb^{-5}\). The \(a\)-factor is zonal and contributes \(A_a(L,R)=a^{3/2}\). The \(b\)-factor is a zonal train and contributes \[A_b(L,R)=b^{3/2}\left(\frac{c}{b}\right)^{-1/2}=b^2c^{-1/2}.\] The \(c\)-factor is a beam block or an envelope, and contributes \(A_c(L,R)=b^{3/2}\). Thus \[\begin{align} \mathcal{L}_{p,4}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-3}\lambda_j^{3/2}\right) a^{3/2}(b^2c^{-1/2})b^{3/2} \left(cb^{-5}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-3}\lambda_j^{3/2}\right) a^{3/2} b^{\frac{7}{2}-\frac{5}{p}} c^{\frac{1}{p}-\frac{1}{2}} . \end{align}\] This is the first branch in the range \(5/3\le p\le2\).

If \(ac\ge b^2\), use the same box as in the second subcase of the preceding range, namely \(L=a^{-1}\), \(R=b^2\). The \(a\)-factor is zonal and contributes \(A_a(L,R)=a^{3/2}\). The \(b\)-factor is a zonal train and contributes \[A_b(L,R)=b^{3/2}(b/a)^{-1/2}=a^{1/2}b.\] The \(c\)-factor is a beam block or an envelope, and contributes \(A_c(L,R)=b^{3/2}\).

Thus, \[\begin{align} \mathcal{L}_{p,4}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-3}\lambda_j^{3/2}\right) a^{2}b^{5/2} \left(a^{-1}b^{-3}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-3}\lambda_j^{3/2}\right) a^{2-\frac{1}{p}} b^{\frac{5}{2}-\frac{3}{p}} . \end{align}\] This is the second branch in the range \(5/3\le p\le2\). The ratio of the first branch to the second branch is \[\left(\frac{ac}{b^2}\right)^{\frac{1}{p}-\frac{1}{2}}.\] Since \(p\le2\), the exponent is nonnegative, so the two subcases select the smaller branch.

Combining the five ranges gives exactly the \(n=4\) powers in Theorem 2. This proves the proposition. ◻

Proof. Here \(d=2\). We use only the admissible non-train subcases from the packet profile: in the envelope range \(\nu\le R\le \nu^2\) we always have \(L\le \nu/R\), and in the zonal range \(R\ge \nu^2\) we always have \(L\nu\lesssim1\). Thus the geometric objects which occur below are zonals, beam blocks, and envelopes, but not trains. The unnamed lower-frequency factors are chosen to be zonal functions from Lemma 20, centered at the same point as the packet boxes.

First suppose that \[1\le p\le \frac{4}{3} .\] Set \[a=\lambda_{k-3},\qquad b=\lambda_{k-2},\qquad c=\lambda_{k-1},\qquad e=\lambda_k .\] We use \(L=a^{-1},\;R=ac\). Then \[|E_{L,R}|\approx a^{-2}c^{-1}.\] The filler factors contribute \(\prod_{j=1}^{k-4}\lambda_j\). The \(a\)-factor is zonal and \(A_a(L,R)=a\). The \(b\)-factor is an envelope at the coarser transverse scale \(R'=ab\le R\). Indeed \(b\le ab\le b^2\) and \[L=a^{-1}=\frac{b}{ab}=b/R'.\] Thus this is the endpoint envelope case, and \[A_b(L,R')=(ab)^{1/2}.\] Since \((ac)^{-1/2}\le (ab)^{-1/2}\), the same lower bound holds after restricting to the smaller set \(E_{L,R}\). The \(c\)-factor is an envelope and \(A_c(L,R)=(ac)^{1/2}.\) Finally, the \(e\)-factor is a beam block or an envelope, and \(A_e(L,R)=(ac)^{1/2}\). Hence, Lemma 26 gives \[\begin{align} \mathcal{L}_{p,3}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-4}\lambda_j\right) a^{5/2}b^{1/2}c\, \left(a^{-2}c^{-1}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-4}\lambda_j\right) a^{\frac{5}{2}-\frac{2}{p}} b^{1/2} c^{1-\frac{1}{p}}. \end{align}\] Substituting back \(a=\lambda_{k-3}\), \(b=\lambda_{k-2}\), and \(c=\lambda_{k-1}\), this is exactly the power \(\mathcal{E}_0\) in Theorem 2 for \(1\le p\le4/3\).

We now suppose that \[\frac{4}{3}\le p\le 2 .\] Set \[a=\lambda_{k-2},\qquad b=\lambda_{k-1},\qquad c=\lambda_k .\] We use \(L=a^{-1},\;R=ab.\) Thus \[|E_{L,R}|\approx a^{-1}(ab)^{-1}=a^{-2}b^{-1}.\] The filler factors contribute \(\prod_{j=1}^{k-3}\lambda_j\). The \(a\)-factor is zonal and \(A_a(L,R)=a.\) The \(b\)-factor is an envelope and \(A_b(L,R)=(ab)^{1/2}.\) The \(c\)-factor is a beam block or an envelope, and \(A_c(L,R)=(ab)^{1/2}.\) Consequently, Lemma 26 gives \[\begin{align} \mathcal{L}_{p,3}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{k-3}\lambda_j\right) a^2b\, \left(a^{-2}b^{-1}\right)^{1/p} \\ &= \left(\prod_{j=1}^{k-3}\lambda_j\right) a^{2-\frac{2}{p}} b^{1-\frac{1}{p}}. \end{align}\] Since \(a=\lambda_{k-2}\) and \(b=\lambda_{k-1}\), this is exactly the power \(\mathcal{E}_0\) in Theorem 2 for \(4/3\le p\le2\).

Combining the two ranges proves the sharp lower bounds for the powers in the case \(n=3\). This proves the proposition. ◻

Proof. Here \(d=1\). We use only the admissible non-train cases from the packet profile. Thus the geometric objects which occur below are zonals, envelopes, and beam blocks; no envelope trains or zonal trains are used. All packets are centered on the same geodesic, and the unnamed lower-frequency factors are chosen to be zonal functions from Lemma 20.

Choose \(m\in\{4,5,6\}\) by \[m=6\quad \left(1\le p\le \frac{6}{5}\right), \qquad m=5\quad \left(\frac{6}{5}\le p\le \frac{3}{2}\right), \qquad m=4\quad \left(\frac{3}{2}\le p\le2\right).\] Set \[a=k-m+1, \qquad \alpha=\lambda_a, \qquad \omega=\lambda_{k-1}.\] We use \(L=\alpha^{-1}, R=\alpha\omega.\) Thus, \[|E_{L,R}|\approx \alpha^{-1}(\alpha\omega)^{-1/2} =\alpha^{-3/2}\omega^{-1/2}.\] The filler factors are zonal and contribute \(\prod_{j=1}^{a-1}\lambda_j\). The \(\alpha\)-factor is also zonal and \(A_\alpha(L,R)=\alpha^{1/2}.\)

For the intermediate factors \(a+1\le j\le k-2\), we use the larger tube with the same longitudinal length and transverse parameter \[R_j=\alpha\lambda_j.\] Since \(\alpha\le\lambda_j\), we have \[\lambda_j\le R_j=\alpha\lambda_j\le\lambda_j^2, \qquad L=\alpha^{-1}=\frac{\lambda_j}{\alpha\lambda_j}=\frac{\lambda_j}{R_j}.\] Hence each such factor is an endpoint envelope, and \[A_{\lambda_j}(L,R_j) =(\alpha\lambda_j)^{1/4}, \qquad a+1\le j\le k-2.\] Because \(\lambda_j\le\omega\), the tube defining \(E_{L,R}\) is contained in this larger envelope tube. Therefore the same lower bound holds on \(E_{L,R}\).

The \(\omega\)-factor is an endpoint envelope, and \(A_\omega(L,R)=(\alpha\omega)^{1/4}.\) The \(\lambda_k\)-factor is a beam block or an envelope, and \(A_{\lambda_k}(\alpha^{-1},\alpha\omega)=(\alpha\omega)^{1/4}.\)

Thus, Lemma 26 gives \[\begin{align} \mathcal{L}_{p,2}^{(k)}(\lambda_1,\ldots,\lambda_k) &\gtrsim \left(\prod_{j=1}^{a-1}\lambda_j^{1/2}\right) \alpha^{1/2} \left(\prod_{j=a+1}^{k-2}(\alpha\lambda_j)^{1/4}\right) (\alpha\omega)^{1/4}(\alpha\omega)^{1/4} \left(\alpha^{-3/2}\omega^{-1/2}\right)^{1/p} \\ &= \left(\prod_{j=1}^{a-1}\lambda_j^{1/2}\right) \alpha^{\frac{m+1}{4}-\frac{3}{2p}} \left(\prod_{j=a+1}^{k-2}\lambda_j^{1/4}\right) \omega^{\frac{1}{2}-\frac{1}{2p}}. \end{align}\]

Taking \(m=6\), so that \(a=k-5\) and \(\alpha=\lambda_{k-5}\), gives for \(1\le p\le6/5\) \[\mathcal{L}_{p,2}^{(k)} \gtrsim \left(\prod_{j=1}^{k-6}\lambda_j^{1/2}\right) \lambda_{k-5}^{\frac{7}{4}-\frac{3}{2p}} \lambda_{k-4}^{1/4} \lambda_{k-3}^{1/4} \lambda_{k-2}^{1/4} \lambda_{k-1}^{\frac{1}{2}-\frac{1}{2p}}.\] Taking \(m=5\), so that \(a=k-4\) and \(\alpha=\lambda_{k-4}\), gives for \(6/5\le p\le3/2\) \[\mathcal{L}_{p,2}^{(k)} \gtrsim \left(\prod_{j=1}^{k-5}\lambda_j^{1/2}\right) \lambda_{k-4}^{\frac{3}{2}-\frac{3}{2p}} \lambda_{k-3}^{1/4} \lambda_{k-2}^{1/4} \lambda_{k-1}^{\frac{1}{2}-\frac{1}{2p}}.\] Taking \(m=4\), so that \(a=k-3\) and \(\alpha=\lambda_{k-3}\), gives for \(3/2\le p\le2\) \[\mathcal{L}_{p,2}^{(k)} \gtrsim \left(\prod_{j=1}^{k-4}\lambda_j^{1/2}\right) \lambda_{k-3}^{\frac{5}{4}-\frac{3}{2p}} \lambda_{k-2}^{1/4} \lambda_{k-1}^{\frac{1}{2}-\frac{1}{2p}}.\] These are exactly the three powers in Theorem 2 for \(n=2\). This proves the proposition. ◻

Proof. The proof of sharpness for \(p>2\) only needs the envelopes in Lemma 20. For \(\nu^{-1}\le r\ll1\), there is an \(L^2\)-normalized spherical harmonic satisfying \[|u_{\nu,r}|\gtrsim (\nu/r)^{d/4}\] on a tube of length \(r\), transverse radius \((r/\nu)^{1/2}\), and measure \(\approx r(r/\nu)^{d/2}\). Put all envelopes on the same geodesic. If \(r_1\ge\cdots\ge r_k\) and \(r_j\ge\lambda_j^{-1}\), then the smallest tube gives \[\label{eq:large-p-master-lower} \mathcal{L}_{p,n}^{(k)}(\lambda_1,\ldots,\lambda_k) \gtrsim \prod_{j=1}^k\left(\frac{\lambda_j}{r_j}\right)^{d/4} \left[r_k\left(\frac{r_k}{\lambda_k}\right)^{d/2}\right]^{1/p}.\tag{40}\]

For \(n=2\) and \(2\le p\le3\), take \[r_j=\lambda_j^{-1}\;(j\le k-2), \qquad r_{k-1}=r_k=\lambda_{k-2}^{-1}.\] Then 40 gives \[\mathcal{L}_{p,2}^{(k)} \gtrsim \Big(\prod_{j=1}^{k-3}\lambda_j^{1/2}\Big) \lambda_{k-2}^{1-\frac{3}{2p}} \lambda_{k-1}^{1/4} \lambda_k^{1/4-\frac{1}{2p}}.\]

Next take \[r_j=\lambda_j^{-1}\;(j\le k-1), \qquad r_k=\lambda_{k-1}^{-1}.\] For every \(d\ge1\), this yields \[\mathcal{L}_{p,n}^{(k)} \gtrsim \Big(\prod_{j=1}^{k-2}\lambda_j^{d/2}\Big) \lambda_{k-1}^{\frac{3d}{4}-\frac{d+2}{2p}} \lambda_k^{\frac{d}{4}-\frac{d}{2p}}.\] Finally, taking \(r_j=\lambda_j^{-1}\) for all \(j\) gives \[\mathcal{L}_{p,n}^{(k)} \gtrsim \Big(\prod_{j=1}^{k-1}\lambda_j^{d/2}\Big) \lambda_k^{d/2-\frac{d+1}{p}}.\] ◻