All-order scalar causal extension and finite renormalization

Expand the finite polynomial in the definition of $F$ after substituting $\varphi +{\sum }_j{t}_j{h}_j$ and take the coefficient of $t_1\cdots t_r$. Each resulting term is $\int a(x){\prod }_{j=1}^r{\mathrm{\partial }}^{{\alpha }_j}{h}_j(x)d^{4}x$, with $a\in C_c^{\mathrm{\infty }}(\mathbb{M})$. Write $X=(x_1,\dots ,x_r)$, $i(x)=(x,\dots ,x)$ and $A={\sum }_j\mathrm{\mid }{\alpha }_j\mathrm{\mid }$. Its kernel is $u=(-1{)}^A{\mathrm{\partial }}_{x_1}^{{\alpha }_1}\cdots {\mathrm{\partial }}_{x_r}^{{\alpha }_r}[a(x_r){\prod }_{j<r}\delta (x_j-x_r)]$: pairing it with a product test and using distributional differentiation gives the displayed integral. More generally its value on $v(X)\mathrm{\mid }dX\mathrm{\mid }$ is $\int a(x)({\mathrm{\partial }}_{x_1}^{{\alpha }_1}\cdots {\mathrm{\partial }}_{x_r}^{{\alpha }_r}v)(i(x))d^{4}x$. Thus the complete kernel is a finite sum of finite derivatives of smooth compact diagonal distributions, supported on ${\mathrm{\Delta }}_r$. This includes $r=1$, with the empty delta product equal to $1$, and gives the zero kernel above the finite polynomial degree. Derivatives of the relative deltas are retained; they cannot in general all be transferred to $a$.

Fix an arbitrary smooth ambient chart $\psi :O\to V\subseteq {\mathbb{R}}^{4r}$, a point $q=i(x_0)\in O$, and a nonzero chart covector ${\eta }_0$ with $D(\psi \circ i)(x_0{)}^T{\eta }_0\mathrm{\ne }0$. Put $f=\psi \circ i$ on $i^{-1}(O)$. By continuity, there are a relatively compact neighborhood $B$ of $x_0$, an open cone $\mathcal{C}$ about ${\eta }_0$, and $c>0$ such that $\mathrm{\mid }Df(x{)}^T\eta \mathrm{\mid }\ge c\mathrm{\mid }\eta \mathrm{\mid }$ on $\bar{B}\times \mathcal{C}$. Indeed the left side divided by $\mathrm{\mid }\eta \mathrm{\mid }$ is continuous in $x$ and $\eta \mathrm{/}\mathrm{\mid }\eta \mathrm{\mid }$ and positive at $(x_0,{\eta }_0\mathrm{/}\mathrm{\mid }{\eta }_0\mathrm{\mid })$; restrict to a small closed ball and a small closed spherical cap contained in that positive neighborhood. Choose $\chi \in C_c^{\mathrm{\infty }}(V)$ equal to one near $\psi (q)$, with $i^{-1}({\psi }^{-1}(\mathrm{supp}\chi ))\subset B$. Such a localization is obtained by taking a sufficiently small chart ball inside $\psi (O\cap \{X:x_r\in B\})$. Smooth plateaus in nested balls follow explicitly from $H(t)=0$ for $t\le 0$, $H(t)=e^{-1\mathrm{/}t}$ for $t>0$: for radii $0<b<R$, the ratio $H(R^2-\mathrm{\mid }y-y_0{\mathrm{\mid }}^2)\mathrm{/}(H(R^2-\mathrm{\mid }y-y_0{\mathrm{\mid }}^2)+H(\mathrm{\mid }y-y_0{\mathrm{\mid }}^2-b^2))$ is one on the smaller ball and zero outside the larger, and is smooth because all derivatives of $H$ at zero vanish. No family of charts or localizations is selected. In the test-density convention underlying [F1]–[F2], put $J_{\psi }(X)=\mathrm{\mid }\det D\psi (X)\mathrm{\mid }$. For a summand $u$ from step 1.1 its chart Fourier pairing is $⟨u_{\psi },\chi (y)e^{-2\pi iy\cdot \eta }\mathrm{\mid }dy\mathrm{\mid }⟩=\int a(x){{\mathrm{\partial }}_X^{\alpha }\text{\hspace{0.17em}⁣}[J_{\psi }(X)\chi (\psi (X))e^{-2\pi i\eta \cdot \psi (X)}]\mid }_{X=i(x)}{d}^{4}x\mathrm{.}$ Here ${\mathrm{\partial }}_X^{\alpha }$ denotes the finite ambient derivative from step 1.1. The absolute Jacobian is smooth since the chart determinant never vanishes. Repeated product and chain rules make this a finite sum $\int A_{\beta }(x,\eta )e^{-2\pi i\eta \cdot f(x)}{d}^{4}x$, where each amplitude is a polynomial in $\eta$ of degree at most $m:=\mathrm{\mid }\alpha \mathrm{\mid }$, with smooth coefficients supported in one compact $K\subset B$. The coefficients include every derivative of the Jacobian, localization and chart map. Consequently, for every integer $k\ge 0$, their derivatives satisfy ${\max }_{\mathrm{\mid }\gamma \mathrm{\mid }\le k}{\sup }_x\mathrm{\mid }{\mathrm{\partial }}_x^{\gamma }{A}_{\beta }(x,\eta )\mathrm{\mid }\le C_k(1+\mathrm{\mid }\eta \mathrm{\mid }{)}^m$. This follows by bounding the finitely many coefficient derivatives on $K$; no bound on a varying test function is asserted.

For $\eta \in \mathcal{C}$, $\mathrm{\mid }\eta \mathrm{\mid }\ge 1$, set ${\mathrm{\Phi }}_{\eta }(x)=\eta \cdot f(x)$, $b_{\eta }=\mathrm{\nabla }{\mathrm{\Phi }}_{\eta }\mathrm{/}\mathrm{\mid }\mathrm{\nabla }{\mathrm{\Phi }}_{\eta }{\mathrm{\mid }}^2$, and $L_{\eta }=-(2\pi i{)}^{-1}{b}_{\eta }\cdot {\mathrm{\nabla }}_x$. Then $L_{\eta }{e}^{-2\pi i{\mathrm{\Phi }}_{\eta }}=e^{-2\pi i{\mathrm{\Phi }}_{\eta }}$. Every derivative of $\mathrm{\nabla }{\mathrm{\Phi }}_{\eta }$ is bounded by a constant times $\mathrm{\mid }\eta \mathrm{\mid }$ on $\bar{B}$, while its squared norm is at least $c^2\mathrm{\mid }\eta {\mathrm{\mid }}^2$. Differentiating the quotient therefore gives ${\sup }_{\bar{B}}\mathrm{\mid }{\mathrm{\partial }}_x^{\gamma }{b}_{\eta }\mathrm{\mid }\le C_{\gamma }\mathrm{\mid }\eta {\mathrm{\mid }}^{-1}$: equivalently write $b_{\eta }=\mathrm{\mid }\eta {\mathrm{\mid }}^{-1}{b}_{\eta \mathrm{/}\mathrm{\mid }\eta \mathrm{\mid }}$, and differentiate the smooth rational expression on the compact base ball and closed spherical cap where its denominator is bounded below. The transpose for the bilinear integral is $L_{\eta }^{t}A=(2\pi i{)}^{-1}\mathrm{div}(b_{\eta }A)$. Compact support of each amplitude eliminates boundary terms, so $N$ integrations by parts give $\int A_{\beta }{e}^{-2\pi i{\mathrm{\Phi }}_{\eta }}dx=\int (L_{\eta }^t{)}^N{A}_{\beta }{e}^{-2\pi i{\mathrm{\Phi }}_{\eta }}dx,\mid \int A_{\beta }{e}^{-2\pi i{\mathrm{\Phi }}_{\eta }}dx\mid \le C_N\mathrm{\mid }\eta {\mathrm{\mid }}^{m-N}\mathrm{.}$ For completeness, induction on $N$ expands $(L_{\eta }^t{)}^N{A}_{\beta }$ into finitely many terms, each a derivative of $A_{\beta }$ of order at most $N$ times $N$ factors that are derivatives of components of $b_{\eta }$ of orders at most $N$. The bounds of step 2.1 and the preceding quotient bounds give the estimate, and integration over a fixed rectangle containing $K$ supplies its finite volume factor. Integration by parts itself follows componentwise from the ordinary product rule and the fundamental theorem on that rectangle, with zero boundary products. For $\mathrm{\mid }\eta \mathrm{\mid }\le 1$ the original finite polynomial amplitudes give a uniform integral bound directly. Given any requested decay power $p$, choose a finite integer $N\ge p+\max \mathrm{\mid }\alpha \mathrm{\mid }$ over the finitely many kernel terms; the finite sum of all kernel terms then has $(1+\mathrm{\mid }\eta \mathrm{\mid }{)}^p$ times its localized Fourier pairing bounded throughout $\mathcal{C}$.

If $q\mathrm{\notin }{\mathrm{\Delta }}_r$, a localization disjoint from the closed diagonal makes the kernel zero by step 1.1. At a diagonal point, steps 2.1–3.1 exclude every chart direction for which $Df(x_0{)}^T\eta \mathrm{\ne }0$. The intrinsic covector represented by $\eta$ is $\xi =D{\psi }_q^T\eta$; the ordinary chain rule gives $Df(x_0{)}^T\eta =D{i}_{x_0}^T\xi ={\sum }_j{\xi }_j$. Thus, in every arbitrary ambient chart, the only possible singular directions annihilate the tangent to the diagonal. Taking the all-chart union in [F4]'s wavefront convention therefore proves precisely $\mathrm{WF}(F^{(r)}(\varphi ))\subseteq N^{\ast }{\mathrm{\Delta }}_r\setminus 0$. No chart-invariance theorem is invoked. For $r=1$, $i$ is the identity and $D\psi$ is invertible, so every nonzero direction has been excluded and the wavefront set is empty.

At a diagonal point, if ${\xi }_1,\dots ,{\xi }_r$ are all future causal, their metric-dual time components have the same weak sign, and each nonzero causal covector has a strictly nonzero time component. Hence ${\sum }_j{\xi }_j=0$ forces every ${\xi }_j=0$; the same holds for the past cone. Step 4.1 therefore excludes every nonzero tuple in the forbidden cones of [F3]. The finite Taylor expansion of the original polynomial at $\varphi =0$ gives its symmetric compact coefficient kernels $f_m=F^{(m)}(0)\mathrm{/}m!$ (and its constant term), so it is also a polynomial functional in the kernel sense of [F3]. Its derivative kernels have the conormal bound just proved for every smooth $\varphi$ and every $r\ge 1$. This proves the asserted membership in ${\mathcal{F}}_{\mu c}^{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{y}}(\mathbb{M})$. All localizations, finite expansions and integrations above were constructed for a fixed chart and direction; this proof adds no choice principle.

Polarize each homogeneous field-degree-$r$ jet polynomial into $r$ field slots. Commutativity of the scalar fields identifies it with a polynomial in the formal derivative variables $p_a$, invariant under slot permutations. The invertible linear change $p_a=q_a+P\mathrm{/}r$ identifies the full polynomial algebra with $\mathbb{C}[P]\otimes \mathbb{C}[q_1,\dots ,q_r]\mathrm{/}({\sum }_a{q}_a)$. Slot permutations fix $P$ and act only on the relative factor. Expanding in the independent monomials of $P$, invariance holds if and only if every relative coefficient is invariant. Thus the invariant space is exactly $\mathbb{C}[P]\otimes {\mathcal{B}}_r$. In each finite degree choose a basis by ordinary finite Gaussian elimination on the real coefficient polynomials; the monomial ordering gives compatible choices across degrees. The Leibniz rule identifies multiplication by $P_{\mu }$ with the total derivative ${\mathrm{\partial }}_{\mu }$ of the local field polynomial. This is a direct tensor decomposition, so has no hidden relations between total and balanced derivatives.

Expand a density using step 1.1 as a finite sum $p(\mathrm{\partial })B_b$. In its integral against $f$, move $p(\mathrm{\partial })$ onto the compact coefficient by integration by parts, obtaining $p(-\mathrm{\partial })f$ multiplying $B_b$. This gives existence of the displayed representation and preserves compact support. Degree zero gives the scalar $c=F(0)$ and is separated from the positive field degrees. To prove uniqueness, suppose a balanced expression represents zero. Separate its field degrees by replacing $\varphi$ by $t\varphi$, polarize, and evaluate its $r$ slots on the smooth exponential fields $e^{2\pi i{k}_a\cdot x}$; complex exponentials are permitted by complex multilinear extension from real configurations. The resulting identity is a finite sum of ${\widehat{f}}_{r,b}(-{\sum }_a{k}_a)$ times the balanced polynomial $b(2\pi i{k}_1,\dots ,2\pi i{k}_r)$. For fixed total momentum, vary freely the relative momenta. Independence of the basis polynomials on ${\sum }_a{q}_a=0$ makes every coefficient Fourier transform zero at that total momentum. Since that momentum is arbitrary, [F2] gives every $f_{r,b}=0$; also $c=0$.

The same uniqueness is local. In center and relative coordinates for the polarized variational kernel on the thin diagonal, balanced differential operators differentiate only the relative delta variables; their coefficients are the $f_{r,b}$ in the center variable, up to fixed nonzero coordinate constants. Test with a center test times a relative monomial times a cutoff equal to one near the relative origin. Starting at the highest relative order and descending isolates every independent relative delta-jet coefficient. Thus vanishing of the kernel over an open center set forces each $f_{r,b}$ to vanish there. If $F$ is unchanged by all variations supported in an open set $U$, its polarized kernels vanish where any one field slot lies in $U$; near their thin diagonal this is precisely the preceding situation. Conversely, if every coefficient vanishes on $U$, no variation supported in $U$ changes $F$. This proves equality with the functional support. Under the action defined in the Statement, a Lorentz transformation substitutes $p_a\mapsto {\mathrm{\Lambda }}^{\mathsf{T}}{p}_a$ and transforms each coefficient by $f(x)\mapsto f({\mathrm{\Lambda }}^{-1}(x-a))$; $\det \mathrm{\Lambda }=1$ preserves $d^{4}x$. Thus $P$ and all $q_a$ transform by the same linear map and the tensor decomposition is preserved. Translations act on coefficients by the same formula with $\mathrm{\Lambda }=I$, and, because the chosen basis is real, complex conjugation sends each coefficient to its complex conjugate. Hence the decomposition is equivariant as stated.

For a general polynomial label write its unique tensor decomposition as ${\sum }_{\beta ,b}{c}_{\beta ,b}{\mathrm{\partial }}^{\beta }{B}_b$, and define its coefficient by ${\sum }_{\beta ,b}{c}_{\beta ,b}{\mathrm{\partial }}_{x_i}^{\beta }K(\dots ,B_b,\dots )$, independently in every entry. Commutativity of coordinate derivatives and the direct decomposition in step 1.1 make this consistent, linear, and the unique extension satisfying the displayed derivative identity. Smearing and integrating by parts gives exactly the coefficient normal form of step 2.1, so equivalent functional presentations yield the same map. The construction respects the equivariance and support statements of step 3.1. Finally, dilation of a distributional derivative gives $({\mathrm{\partial }}^{\beta }T{)}_{\rho }={\rho }^{-\mathrm{\mid }\beta \mathrm{\mid }}{\mathrm{\partial }}^{\beta }{T}_{\rho }$, so the added derivative increases a scaling upper bound by at most $\mathrm{\mid }\beta \mathrm{\mid }$. All sums and derivative orders here are finite.

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.