Positive scaling bounds for products transverse at the scaling center

Fix one factor $u$ and a positive exponent $s>\max (0,{\mathrm{sd}}_0(u))$. Multiply $u$ by a compact cutoff equal to one near zero and denote the result by $v$. Multiplication by this cutoff preserves the scaling upper bound: the tests acquire the factor $\psi (\rho x)$, a bounded smooth family on every fixed compact, and the uniform estimate established next applies to it. For any fixed compact test support $K$, the continuous linear functionals $L_{\rho }={\rho }^s{v}_{\rho }$, $0<\rho \le T<\mathrm{\infty }$, are pointwise bounded: convergence at zero follows from [F1], and away from zero the dilation tests vary continuously on compact scale intervals. To justify the uniform estimate without assuming it, apply [F2] first to ${\rho }^s{u}_{\rho }$ on a compact containing all small-scale supports. The closed sets $E_N=\{\phi :{\sup }_{\rho }\mathrm{\mid }{L}_{\rho }(\phi )\mathrm{\mid }\le N\}$ cover the fixed-support space. One has interior, so subtracting two elements of that interior and rescaling gives $\mathrm{\mid }{L}_{\rho }(\phi )\mathrm{\mid }\le C{p}_m(\phi )$ uniformly in $\rho$, with $p_m={\max }_{\mathrm{\mid }\alpha \mathrm{\mid }\le m}\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}$. The same argument then applies to $v$ and to every fixed compact scale interval.

Choose slightly enlarged closed frequency cones $K_j$ around the $C_j$ such that the joint no-cancellation condition still holds for the $K_j$. This is possible by compactness of tuples with ${\sum }_j\mathrm{\mid }{\xi }_j\mathrm{\mid }=1$: the set of tuples in ${\prod }_j(C_j\cup \{0\})$ is compact there and its sum stays away from zero. If $C_j$ is empty use the empty cone. Closedness of the wavefront set permits localization to a sufficiently small common neighborhood so that all singular directions of each localized $v_j$ lie strictly inside $K_j$. Choose a smooth Fourier cutoff $q_j$ supported in $K_j$ outside a bounded ball and equal to one around those singular directions at large frequencies. Then $(1-q_j){\widehat{v}}_j$ is Schwartz. In fact [F3] gives rapid decay on its unbounded support, while each frequency derivative is a localized Fourier transform with a coordinate monomial inserted, so has the same regular-cone bounds by [F3]–[F4]. Bounded frequencies cause no problem. Thus $v_j=b_j+c_j$, with $b_j$ Schwartz and the Fourier transform of $c_j$ supported in $K_j$ together with a bounded ball.

The family $L_{\rho }={\rho }^s{v}_{\rho }$ is uniformly bounded on Schwartz space by one Schwartz seminorm. Indeed, decompose a Schwartz test into a fixed ball and smooth dyadic annuli of radii $R=2^k\ge 1$. On an annulus write ${\phi }_R(x)=\theta (x\mathrm{/}R)\phi (x)$ with one fixed annular cutoff. Dilation gives $⟨{\rho }^s{v}_{\rho },{\phi }_R⟩=R^{D-s}⟨(\rho R{)}^s{v}_{\rho R},\theta (y)\phi (Ry)⟩\mathrm{.}$ If $\mathrm{supp}v\subset B(0,M)$, annuli with $\rho R>2M$ give zero, after increasing $M$ to accommodate the fixed annular radii. For $0<\rho R\le 2M$, step 1.1 on the one fixed annulus bounds this by $C{R}^{D-s+m-L}{P}_{m,L}(\phi )$, where $P_{m,L}(\phi )={\max }_{\mathrm{\mid }\alpha \mathrm{\mid }\le m}{\sup }_x(1+\mathrm{\mid }x\mathrm{\mid }{)}^L\mathrm{\mid }{\mathrm{\partial }}^{\alpha }\phi (x)\mathrm{\mid }$. Choose $L>D-s+m+1$; the geometric sum over $R=2^k$ is finite uniformly in $\rho$. The fixed ball is controlled by step 1.1. This proves $\mathrm{\mid }⟨L_{\rho },\phi ⟩\mathrm{\mid }\le C^{\mathrm{\prime }}{P}_{m^{\mathrm{\prime }},L^{\mathrm{\prime }}}(\phi )$ for fixed $m^{\mathrm{\prime }},L^{\mathrm{\prime }},C^{\mathrm{\prime }}$ independent of $\rho$. By transposition [F5] the Fourier transforms are likewise uniformly tempered.

For $s_j>a_j\ge 0$, step 2.1 makes ${\rho }^{s_j}(v_j{)}_{\rho }$ uniformly tempered. The Schwartz part ${\rho }^{s_j}(b_j{)}_{\rho }$ is also uniformly tempered, since $b_j(\rho x)$ and each of its derivatives are bounded uniformly for $0<\rho \le 1$, and its local smooth seminorms tend to zero. Consequently the Fourier family of ${\rho }^{s_j}(c_j{)}_{\rho }$ is uniformly tempered and supported in $K_j$ together with a fixed bounded ball: dilation shrinks that ball in frequency. Let $V$ be a closed cone disjoint from a slightly larger cone around $K_j$, and let $\chi$ be any fixed compact smooth test cutoff. Evaluate the localized Fourier transform by pairing this Fourier family with $\widehat{\chi }(\xi -\eta )$ times a smooth cutoff equal to one on its support. On the unbounded conic support, separation gives $\mathrm{\mid }\xi -\eta \mathrm{\mid }\ge c(\mathrm{\mid }\xi \mathrm{\mid }+\mathrm{\mid }\eta \mathrm{\mid })$ for $\xi \in V$; on the bounded part use $1+\mathrm{\mid }\xi -\eta \mathrm{\mid }\ge c^{\mathrm{\prime }}(1+\mathrm{\mid }\xi \mathrm{\mid })$ for large $\xi$. Every Schwartz seminorm of this test kernel is therefore $O((1+\mathrm{\mid }\xi \mathrm{\mid }{)}^{-N})$ for arbitrary $N$. The uniform tempered bound gives ${\sup }_{0<\rho \le 1}{\sup }_{\xi \in V}(1+\mathrm{\mid }\xi \mathrm{\mid }{)}^N\mid \widehat{\chi {\rho }^{s_j}(v_j{)}_{\rho }}(\xi )\mid <\mathrm{\infty }\mathrm{.}$ The Schwartz part has the same estimate by integration by parts on the fixed support of $\chi$. An intermediate exponent between $a_j$ and $s_j$ even makes each displayed seminorm tend to zero. Ordinary weak convergence follows from [F1]. Since $v_j=u_j$ near zero, every fixed output localization uses these same dilates for all sufficiently small $\rho$. This proves the required directional control rather than inferring it from pointwise wavefront inclusion of the rescaled family.

Shrink the cones and localization neighborhoods as in step 1.2, retaining a little angular margin. The product exists there by [F3], iterated over the finitely many factors. For a fixed compactly supported test $\phi$, its pairing with the product of the localized scaled factors is the absolutely convergent Fourier integral of $\widehat{\phi }(-{\sum }_j{\eta }_j)$ times the factor transforms. Here is a uniform integrable bound. Compactness of normalized frequency tuples and the joint no-cancellation condition imply that either $\mathrm{\mid }{\sum }_j{\eta }_j\mathrm{\mid }\ge c{\sum }_j\mathrm{\mid }{\eta }_j\mathrm{\mid }$, or some ${\eta }_j$ lies in a regular cone and $\mathrm{\mid }{\eta }_j\mathrm{\mid }\ge c{\sum }_k\mathrm{\mid }{\eta }_k\mathrm{\mid }$, for a fixed $c>0$, apart from bounded frequencies. In the first region the Schwartz decay of $\widehat{\phi }$ controls the total frequency size; in the second region step 3.1 supplies arbitrarily high decay from that factor. All other localized factor transforms have fixed uniform polynomial bounds by step 1.1 and [F4]. Choosing the decay power larger than the total polynomial degree plus $Dr+1$ gives an integrable majorant independent of $\rho$. Each localized factor transform tends pointwise to zero by weak convergence. Dominated convergence therefore proves ${\rho }^{{\sum }_j{s}_j}{\prod }_j(u_j{)}_{\rho }\to 0$. Dilation commutes with the locally defined product by the pullback formula [F3], so this is precisely the scaling estimate for the product. Let every $s_j$ decrease to $a_j$ to obtain the claimed bound.