Causal kernels, initial data, and short-time geometry

A wave kernel has two useful descriptions. In frequency space it solves an ordinary differential equation with prescribed initial data. In physical space its singularities sit on a cone. The frequency description fixes constants and initial traces; the cone description explains which parts of a local geometric construction can influence a short time interval. We develop both descriptions before introducing variable coefficients.

The spatial dimension is any integer $n\geq1$. The variable-coefficient construction also permits complex matrix lower-order coefficients, without diagonalizing them. Its principal symbol must be a positive scalar quadratic form times the identity. A general system with several characteristic speeds is a different problem.

1. Conventions and the causal distribution problem

For spatial Fourier transformation use $$\widehat f(\xi)=\int_{\mathbb R^n}e^{-ix\cdot\xi}f(x)\,dx, \qquad f(x)=(2\pi)^{-n}\int e^{ix\cdot\xi}\widehat f(\xi)\,d\xi. \tag{W1}$$ The same convention applies in time. Set $$L=\partial_t^2-\Delta_x,\qquad C_+=\{(t,x):t\geq |x|\},\qquad q(t,x)=t^2-|x|^2. \tag{W2}$$ Distributions supported in $C_+$ will be called causal. Reflection means $\widetilde T(t,x)=T(-t,-x)$; for the radial kernels below it equals $T(-t,x)$.

The Fourier inversion and distribution extension inputs are proved in Sections1–2 of Fourier transforms, finite spectra and convex separation. The elementary integrations, gamma and beta identities, dominated holomorphic derivatives and identity theorem have their complete proof locators CX1–CX11 and CX29–CX40 in Sections16.1–16.10 of Singularities along a submanifold and smooth boundary passage. The finite-pole rectangle case of the residue theorem used here, including its contour orientation and deformation, is proved in Section4.1 from those circle and triangle proofs. The added contour pieces are estimated there after testing, and Section5 supplies the Gaussian spatial regularization. The beta and gamma normalizations used here are displayed in the proof. The distributional inputs are distributional derivatives, multiplication by smooth functions, changes of variables by diffeomorphisms, compact-support distribution estimates, and the wavefront pullback theorem for a submersion with its converse in product coordinates. For wavefront sets we additionally use the exact constant-coefficient microlocal ellipticity implication $$P(D)u\in C^\infty\quad\Longrightarrow\quad WF(u)\subset\{(z,\zeta):p_m(\zeta)=0\}, \tag{W3}$$ where $p_m$ is the principal symbol, and the parameter theorem that absence of covectors normal to the parameter fibers gives smooth distribution-valued dependence. The coordinate, submersion and parameter-smoothness statements are proved in Sections18.3–18.6, WF5–WF17, of Detecting regularity without choosing coordinates. Its Sections18.7–18.8, WF18–WF34, prove the full differential elliptic regularity statement and its independent odd-wave receiving comparison. No propagation theorem or global wave existence theorem is needed.

The jointly smooth exponential chart, radial distance identity and full transformed drift are proved in Sections17.8–17.10, NG0–NG7, of Detecting regularity without choosing coordinates. Section9 below proves their precise wave receiving comparison and the complete ordered matrix transport independently. All estimates for the causal distributions themselves are proved below. The later elliptic and reflected constructions are comparison targets, rather than prerequisites for this proof.

When a time distribution is viewed as a function of a spatial parameter, continuity into distributions of order at most $m\geq0$ means the following local statement: it acts on compactly supported $C^m$ tests, the pairings are continuous in the parameter, and on compact parameter sets they obey a common $C^m$-seminorm bound. This weak finite-order continuity does not assert operator-norm continuity on the dual of $C^m$.

2. A Laplace calculation that constructs the whole family

There is an entire distribution family $\lambda\mapsto R_\lambda$, supported in $C_+$, whose Fourier–Laplace transform is $$\int e^{-st-ix\cdot\xi}R_\lambda(t,x)\,dt\,dx =(s^2+|\xi|^2)^{-\lambda},\qquad s>0. \tag{W4}$$ The integral denotes distributional pairing, equivalently the Fourier transform of the exponentially damped distribution. The power is determined by continuation from positive $s$ and $\xi=0$. The family satisfies $$LR_{\lambda+1}=R_\lambda,\qquad R_0=\delta_{(0,0)},\qquad R_\lambda(at,ax)=a^{2\lambda-n-1}R_\lambda(t,x)\quad(a>0). \tag{W5}$$

Proof. We first justify the gamma normalization for complex parameters. For $\operatorname{Re}z>0$, let $\Gamma(z)=\int_0^\infty e^{-u}u^{z-1}\,du$, using the real logarithm for positive $u$. For an integer $N\geq1$, repeated integration by parts in the beta integral gives $$B(z,N+1)=\int_0^1 t^{z-1}(1-t)^N\,dt =\frac{N!}{z(z+1)\cdots(z+N)},\qquad N^zB(z,N+1)\longrightarrow\Gamma(z). \tag{W4a}$$ For the rational identity, one integration replaces the integral by $N/z$ times the integral with parameters $z+1,N$; the final integral is $1/(z+N)$. No gamma division is used. For the limit put $u=Nt$. On a compact subset of the right half-plane choose $0<\sigma\leq\operatorname{Re}z\leq M$. The resulting integrands are bounded in absolute value by $(u^{\sigma-1}+u^{M-1})e^{-u}$, since $(1-u/N)^N\leq e^{-u}$ for $0<u<N$. Pointwise convergence is uniform for $z$ in that compact set, and the integrable bound makes the integral convergence locally uniform.

Define $H_N=\sum_{k=1}^N k^{-1}$. The numbers $H_N-\log N$ decrease and are bounded below: the decrease follows from $\log(1+1/N)>1/(N+1)$, and comparison with the integral of $1/t$ gives the lower bound. Write their limit as $\gamma$. The exact reciprocal of the scaled beta expression in (W4a) is the entire finite product $$P_N(z)=zN^{-z}\prod_{k=1}^N(1+z/k) =z\exp\bigl(z(H_N-\log N)\bigr) \prod_{k=1}^N(1+z/k)e^{-z/k}. \tag{W4b}$$ Its limit exists locally uniformly on the whole plane. Indeed, on $|z|\leq R$ choose an integer $K>2R$. For $k>K$, use the logarithm defined by its power series at one; then $$\log(1+z/k)-z/k =\sum_{j=2}^{\infty}\frac{(-1)^{j+1}}{j}(z/k)^j, \qquad \bigl|\log(1+z/k)-z/k\bigr|\leq 2R^2/k^2. \tag{W4c}$$ Thus the sum of these tail logarithms converges uniformly on this disk. The derivative of each tail logarithm is $-z/[k(k+z)]$, uniformly $O_R(k^{-2})$; uniform convergence of this derivative series also proves that the tail sum is holomorphic. Exponentiating it, and retaining the first $K$ factors as polynomials times exponentials, proves local uniform convergence to an entire function $$G(z)=z e^{\gamma z}\prod_{k=1}^{\infty}(1+z/k)e^{-z/k}. \tag{W4d}$$ The exponential of the tail sum never vanishes. Consequently the finite initial factors show that the only zeros are $0,-1,-2,\ldots$, each simple: near any of these points precisely one polynomial factor has a simple zero and all the others are nonzero. This argument uses logarithms only for the tail factors close to one, including when the disk contains zeros of the full product.

For $\operatorname{Re}z>0$, the product of $P_N(z)$ and $N^zB(z,N+1)$ is exactly one. Passing to the two locally uniform limits gives $G(z)\Gamma(z)=1$; in particular the gamma integral has no zero there. This identifies $G$ as the entire reciprocal of the meromorphic continuation of gamma, with the constant fixed by the original integral, rather than only up to a nonzero factor. Explicitly $P_N(1)=(N+1)/N$, so $G(1)=1$. Integration by parts in the gamma integral gives $\Gamma(z+1)=z\Gamma(z)$ in the right half-plane. It follows that $G(z)=zG(z+1)$ there, hence everywhere by the identity theorem. Iterating at $z=-m$, for an integer $m\geq0$, gives $$G'(-m)=(-1)^m m!,\qquad \operatorname*{Res}_{z=-m}\Gamma(z)=\frac{(-1)^m}{m!}. \tag{W4e}$$ For this derivative, differentiate $G(z)=z(z+1)\cdots(z+m)G(z+m+1)$ at $-m$ and use $G(1)=1$; its reciprocal has the stated residue. These facts justify both complex gamma normalizations below and the pole cancellation in Section 3.

Initially suppose $\operatorname{Re}\lambda>\max(0,(n-1)/2)$. The locally integrable function $$R_\lambda(t,x)= \frac{2^{1-2\lambda}\pi^{(1-n)/2}} {\Gamma(\lambda)\Gamma(\lambda+(1-n)/2)} \boldsymbol1_{\{t>|x|\}}q(t,x)^{\lambda-(n+1)/2} \tag{W6}$$ has locally integrable cone boundary because the exponent is greater than $-1$. At the vertex, radial integration in $x$ gives a power $t^{2\operatorname{Re}\lambda-1}$, which is integrable by the other inequality. On compact subsets of this half-plane the same estimates hold after inserting any fixed power of $|\log q|$. Thus (W6) is holomorphic as a distribution there.

At $\xi=0$, put $x=ty$. The elementary ball integral, obtained by polar coordinates and $u=|y|^2$, is $$\int_{|y|<1}(1-|y|^2)^{\lambda-(n+1)/2}\,dy =\pi^{n/2}\frac{\Gamma(\lambda+(1-n)/2)}{\Gamma(\lambda+1/2)}. \tag{W7}$$ For $n=1$ the two half-intervals give the same formula. The time integral is $\int_0^\infty e^{-st}t^{2\lambda-1}dt=\Gamma(2\lambda)s^{-2\lambda}$. Their product with the constant in (W6) is $s^{-2\lambda}$, because $$\Gamma(2\lambda)=2^{2\lambda-1}\pi^{-1/2} \Gamma(\lambda)\Gamma(\lambda+1/2). \tag{W8}$$ The full beta and Gaussian calculation CX36–CX38 in the earlier conormal lesson proves (W7) and (W8), retaining their original substitutions, sphere densities and all factors of two and pi. Section2.1 records the exact receiving domains and continuation.

To retain the spatial frequency, first replace $-i\xi$ by a real vector $\eta$ with $|\eta|<s$. The integral with weight $e^{-st+\eta\cdot x}$ is absolutely convergent. Rotate $\eta$ onto the first spatial axis, and in the $(t,x_1)$-plane make a real linear change preserving $t^2-x_1^2$, the future cone, and determinant one. Explicitly its coefficients are $\cosh\theta=s/(s^2-|\eta|^2)^{1/2}$ and $\sinh\theta=|\eta|/(s^2-|\eta|^2)^{1/2}$. It transforms the linear form $st-\eta\cdot x$ into $(s^2-|\eta|^2)^{1/2}t$. The preceding evaluation is therefore $(s^2-\eta\cdot\eta)^{-\lambda}$. Both sides are holomorphic for complex $\eta$ with $|\operatorname{Re}\eta|<s$. Successive one-variable identity theorems extend their equality from a real neighborhood to that tube. Taking $\eta=-i\xi$ proves (W4) in the initial range.

For larger real part, ordinary distributional differentiation, including the boundary where the function and the necessary traces vanish, gives $$Lq^b=2b(2b+n-1)q^{b-1}. \tag{W9}$$ Indeed $\partial_tq=2t$, $\partial_{x_j}q=-2x_j$, $Lq=2(n+1)$, and the difference of the squared gradients is $4q$. The gamma functional equation makes the constants in (W6) satisfy $LR_{\lambda+1}=R_\lambda$. The identity then holds throughout the initial half-plane by holomorphic continuation. For arbitrary $\lambda$, choose an integer $j\geq0$ so that $\lambda+j$ is in that half-plane and define $$R_\lambda=L^jR_{\lambda+j}. \tag{W10}$$ The recursion shows that different choices give the same result. This constructs an entire family, since it constructs it holomorphically on successively larger half-planes. Its support remains in $C_+$. Its members are tempered: (W6) has polynomial growth, and (W10) applies only finitely many derivatives. On the cone, exponential damping makes their transforms meaningful; differentiation proves (W4) for (W10). Fourier injectivity then proves $R_0=\delta$. Homogeneity follows by scaling (W6), then applying (W10); the degree decreases by two at each application of $L$. This also checks the vertex in (W5), not merely points with positive time.

For every integer $\nu\geq0$, define $$E_\nu=\nu!R_{\nu+1}. \tag{W11}$$ If $\sigma>0$, (W4), with the time frequency on the line $\tau\in\mathbb R-i\sigma$, gives the contour formula $$E_\nu(t,x)=\frac{\nu!}{(2\pi)^{n+1}} \int_{\operatorname{Im}\tau=-\sigma}\int_{\mathbb R^n} \frac{e^{i(x\cdot\xi+t\tau)}}{(|\xi|^2-\tau^2)^{\nu+1}} \,d\xi\,d\tau. \tag{W12}$$ The right side is a distributional inverse Fourier–Laplace transform. It is not an absolutely convergent integral at arbitrary dimension. For fixed $\sigma$, the denominator is nowhere zero on the real integration variables and its inverse and derivatives have polynomial growth, so the inverse transform exists. Formula (W4) identifies it with (W11) for every $\sigma>0$, proving independence of the contour height. In particular the support is the full causal cone condition $t\geq|x|$, including its vertex, rather than merely $t\geq0$.

2.1. The full damping, branch and continuation maps

The integration and complex analysis used above have complete earlier proofs: Singularities along a submanifold and smooth boundary passage, CX1–CX11 proves the circle formula, identity theorem and every dominated complex derivative; CX29–CX40 proves the gamma product, beta substitution, duplication, sphere density and Laplace branch. In particular CX36 retains the substitution $x=ut,\ y=u(1-t)$ and its Jacobian $u$. CX37 evaluates $B(z,z)$ by $t=(1+v)/2$ and then $v^2$, giving exactly (W8); CX38 gives exactly (W7), including dimension one. These are proof inputs with specified receiving formulas, rather than citations offered in place of those calculations.

Here are the additional analytic details needed for the distribution construction. On a compact set of complex parameters choose one integer $j$ so that every $\lambda+j$ is in the original convergence region. Write $b=\lambda+j-(n+1)/2$. On its compact parameter set $\operatorname{Re}b>-1$ and $\operatorname{Re}(\lambda+j)>0$, with positive margins. In the integral (W6) use $x=tu$. Its absolute spatial integral, with a bounded test inserted, is bounded by a constant times $t^{2\operatorname{Re}(\lambda+j)-1}$. The constants are uniform on that compact parameter set: the positive power margin bounds the integral at the sphere boundary, and the reciprocal gamma factors are bounded there. For $0<t\leq1$ use the smallest exponent, and for $t\geq1$ the largest. Choosing a Schwartz decay exponent larger than the latter exponent plus one proves a fixed Schwartz seminorm estimate. For (W10), apply that estimate to the full derivative $L^j\phi$, retaining its finite expansion and all coefficients. This proves temperedness and locally uniform finite-order bounds for the entire family. Logarithmic parameter derivatives have the same estimates after reducing those positive margins; for every fixed integer $m$, $u^\epsilon|\log u|^m$ is bounded on $0<u\leq1$. The dominated holomorphic integral proof just cited therefore applies to every compact parameter set and every Schwartz test.

In (W9) one may initially take $\operatorname{Re}b>2$. The cone expression, extended by zero, is then $C^2$ away from the vertex, with its first two boundary jets zero. At the vertex its derivatives of order at most two are bounded by constants times $|(t,x)|^{2\operatorname{Re}b-2}$, times bounded annular derivatives, and the corresponding lower-order bounds prove their actual derivatives there by difference quotients. Thus it is $C^2$ there as well. Ordinary differentiation consequently gives the distributional recursion with no boundary or vertex term in this region. Holomorphic continuation of each tested identity supplies the recursion in the original half-plane and makes the definitions (W10) agree on every overlap. Derivatives preserve support; applying this to every test outside $C_+$ proves the complete causal support claim for every parameter.

Exponential damping is an actual Schwartz multiplier after a harmless extension off the support. Choose a smooth scalar cutoff $\theta(t)$ equal to one for $t\geq-1$ and zero for $t\leq-2$. For $s>0$, $e^{-st}\theta(t)$ and all its derivatives are bounded, and multiplication preserves every Schwartz seminorm by the full product rule. Its value and all derivatives coincide with $e^{-st}$ on a neighborhood of $C_+$, so the damped distribution is independent of this cutoff. This defines the damping of every derivative in (W10), without treating an exponentially growing function on negative time as a Schwartz multiplier.

The boost calculation also fixes the complex branch. Write $\eta=A+iB$, where $A,B$ are real and $|A|<s$. Then

$$\operatorname{Re}(s^2-\eta\cdot\eta) =s^2-|A|^2+|B|^2>0. \tag{WA1}$$

Thus the power in that tube is the holomorphic right-half-plane branch, agreeing with the positive real value at $\eta=0$. On a compact sub-tube the integrand is bounded by its original cone power times $e^{-(s-|A|)t}$; all fixed $\eta$-derivatives add only powers of $x$, bounded by powers of $t$ on the cone. This is an integrable common envelope at infinity and at zero. Successive one-variable identity theorems on a small product of disks extend the equality from a real ball to a complex neighborhood. Equality then extends over the connected tube: along a polygonal path inside it, use a finite chain of overlapping product disks and their convergent local power series. This proves the stated equality on the whole tube and the substitution $\eta=-i\xi$.

Now retain a complex damping parameter $p$ with $\operatorname{Re}p>0$. For real $\xi$, the number $p^2+|\xi|^2$ never lies on the nonpositive real axis: its imaginary part is $2\operatorname{Re}p\,\operatorname{Im}p$, and if this vanishes then $p$ is positive real. The integral is holomorphic in $p$, with the same exponential envelopes on compact subsets. Continuation from real $p=s>0$ gives

$$\mathcal F_{t,x}(e^{-st}R_\lambda)(\sigma,\xi) =\big((s+i\sigma)^2+|\xi|^2\big)^{-\lambda}. \tag{WA2}$$

For initial parameters this follows by absolutely integrable testing. Multiplication by the polynomial $(s+i\sigma)^2+|\xi|^2$ is precisely the Fourier transform of the damped operator $L$, since $$e^{-st}LT=\big((\partial_t+s)^2-\Delta_x\big)(e^{-st}T) \tag{WA3}$$ on distributions supported in $C_+$; the cutoff derivatives vanish near their support. This proves (WA2) for the continuation (W10). At $\lambda=0$ its full spacetime Fourier transform is one. The Fourier inverse proved in the prerequisite lesson gives $e^{-st}R_0=\delta_{(0,0)}$; multiplication on compact tests by $e^{st}$ gives $R_0=\delta_{(0,0)}$. This uses the full time frequency as well as the spatial frequency, and fixes the point source exactly.

In (W12) put $\tau=\sigma-i s$, retaining the original time variable in its exponential. For $\omega=|\xi|$, $$|\omega^2-(\sigma-is)^2|^2 =(\omega^2-\sigma^2+s^2)^2+4s^2\sigma^2\geq s^4. \tag{WA4}$$ If $|\sigma|<s$, the first square is at least $(s^2-\sigma^2)^2$, whose sum with the second term is $(s^2+\sigma^2)^2$; otherwise the second term alone is at least $4s^4$. Thus the denominator has a uniform positive bound. Every derivative of its reciprocal is its original polynomial numerator divided by a positive power of that denominator, so it has polynomial growth. The inverse Fourier transform is therefore a tempered distribution before undoing damping, and (WA2) identifies the resulting distribution on compact tests with (W11). This proves the contour formula and its height independence without an untested integral rearrangement.

The Taylor subtraction used for (W13) can likewise be written with all endpoint terms. For an integer $M\geq1$, put $p_{M-1}(s)=\sum_{j=0}^{M-1}\phi^{(j)}(0)s^j/j!$. The exact pairing is $$\langle\chi_+^a,\phi\rangle=\frac1{\Gamma(a+1)}\left[ \int_0^1s^a(\phi(s)-p_{M-1}(s))\,ds +\int_1^\infty s^a\phi(s)\,ds +\sum_{j=0}^{M-1}\frac{\phi^{(j)}(0)}{j!(a+j+1)}\right]. \tag{WA5}$$ The first integral converges for $\operatorname{Re}a>-M-1$, the second is entire, and the displayed simple poles are exactly canceled by the proved simple zeros of $1/\Gamma(a+1)$. On overlapping half-planes the expression agrees with the original integral, hence agrees everywhere by its tested identity theorem. Taylor’s integral remainder bounds it by a finite test seminorm on each compact parameter set. This proves the entire distribution family, its support and its finite-order estimates. Integration by parts on the initial half-plane gives the derivative identity; analytic continuation and differentiation of the Heaviside distribution give every delta derivative in (W13). No divergent endpoint integral is assigned a value.

3. Cone formulas and exact recursion

Let $\chi_+^a(s)=s_+^a/\Gamma(a+1)$ for $\operatorname{Re}a>-1$. Its entire continuation is determined by $$\partial_s\chi_+^{a+1}=\chi_+^a,\qquad \chi_+^0=\boldsymbol1_{\{s>0\}},\qquad \chi_+^{-1}=\delta_0. \tag{W13}$$ The entire reciprocal gamma function and its simple zeros were proved in Section 2. For completeness, integrate the initial definition against a compactly supported smooth test after subtracting its Taylor polynomial at zero. The remainder is $O(s^M)$, so its integral continues to $\operatorname{Re}a>-M-1$; the removed monomials give simple fractions in $a$, whose poles are canceled by $1/\Gamma(a+1)$. Increasing $M$ gives the entire family. Integration by parts proves the first identity initially, hence everywhere. This construction also gives $\chi_+^{-j-1}=\delta^{(j)}$ for $j\geq0$.

Write $$A_\nu=2^{-2\nu-1}\pi^{(1-n)/2},\qquad a_\nu=\nu+(1-n)/2. \tag{W14}$$ On the open set $t>0$, $$E_\nu(t,x)=A_\nu\chi_+^{a_\nu}(q(t,x)). \tag{W15}$$ At nonzero points of $q=0$, the map $q$ is a submersion, so this is a legitimate distributional pullback. At the vertex (W15) is interpreted by the construction of Section 2, not by substituting into a singular one-dimensional distribution at a critical point. To prove (W15), use (W6) in its initial range, restrict to $t>0$, and apply the compatible differential continuations (W10) and (W13). The chain-rule calculation in (W9), now in normalized form, reads $$L\chi_+^a(q)=(4a+2n-2)\chi_+^{a-1}(q) \quad\text{away from the vertex}. \tag{W16}$$ There is no claim that (W16) discards the vertex term in the fundamental solution. That term was fixed separately by (W5).

The global recursions, including every vertex contribution, are $$LE_0=\delta_{(0,0)},\qquad LE_\nu=\nu E_{\nu-1}\quad(\nu\geq1), \qquad -2\partial_{x_j}E_\nu=x_jE_{\nu-1}\quad(\nu\geq1). \tag{W17}$$ The first two follow from (W5). For the last, apply spatial Fourier–Laplace transformation. Multiplication by $x_j$ becomes $i\partial_{\xi_j}$, and $$i\partial_{\xi_j}\bigl((\nu-1)!(|\xi|^2-\tau^2)^{-\nu}\bigr) =-2i\xi_j\nu!(|\xi|^2-\tau^2)^{-\nu-1}. \tag{W18}$$ The identity follows by injectivity; it therefore has no possible omitted distribution supported at the origin. The degree of homogeneity of $E_\nu$ is $2\nu+1-n$.

4. Initial traces from a Volterra problem

Each $E_\nu$, restricted to nonnegative time, is a one-sided $C^\infty$ function of $t$ with values in $\mathcal D'(\mathbb R^n)$. Its initial traces are $$\partial_t^jE_\nu(0+,\cdot)=0\quad(0\leq j\leq2\nu),\qquad \partial_t^{2\nu+1}E_\nu(0+,\cdot)=\nu!\delta_0. \tag{W19}$$ These are traces of the one-sided function. Repeatedly differentiating its extension by zero to negative time can also produce distributions at $t=0$.

Proof. Put $\omega=|\xi|$, and define $$S_\omega(t)=\begin{cases}\sin(t\omega)/\omega,&\omega\ne0,\\t,&\omega=0.\end{cases} \tag{W20}$$ For $t\geq0$, its Volterra convolution with a function $f$ is $(S_\omega*f)(t)=\int_0^tS_\omega(t-s)f(s)\,ds$. Differentiating twice proves $(\partial_t^2+\omega^2)(S_\omega*f)=f$, with zero initial value and derivative. Its Laplace transform is $(s^2+\omega^2)^{-1}$. Hence (W4) implies $$\widehat E_\nu(t,\xi)=h_\nu(t,\omega),\qquad h_\nu=\nu!\underbrace{S_\omega*\cdots*S_\omega}_{\nu+1\text{ factors}} \quad(t\geq0). \tag{W21}$$ All interchanges here can first be made after multiplying the spatial transform by a Schwartz cutoff; the bounds below then permit its removal.

There is a useful version on a fixed simplex. For $\nu\geq1$, integrate over $u_i\geq0$, $\sum_{i=1}^{\nu}u_i\leq1$, and put $u_{\nu+1}=1-\sum_{i=1}^{\nu}u_i$. With the evident single factor interpretation for $\nu=0$, $$h_\nu(t,\omega)=\nu!t^{2\nu+1} \int_{\Sigma_\nu}\prod_{i=1}^{\nu+1} u_i\,\frac{\sin(t\omega u_i)}{t\omega u_i}\,du_1\cdots du_\nu. \tag{W22}$$ This follows by changing the $\nu$ time variables in the convolution to $tu_i$. The removable quotients have value one at zero. Their derivatives in real $t$ obey polynomial bounds in $\omega$ on bounded time intervals; for example $\sin z/z=\int_0^1\cos(sz)ds$. Thus each $\partial_t^jh_\nu$ has a bound $C_{j,T}(1+\omega)^j$ times a fixed polynomial in $T$. Every compactly supported smooth spatial test has a rapidly decreasing Fourier transform, so inversion and dominated differentiation yield all one-sided distributional derivatives. Uniformity over bounded sets of tests follows from their common Fourier decay seminorms, giving the strong distribution topology as well.

At zero, the simplex integral is $$\int_{\Sigma_\nu}\prod_i u_i\,du =\frac1{(2\nu+1)!}. \tag{W23}$$ One proves this by successively applying $\int_0^1v^{p-1}(1-v)^{q-1}dv=\Gamma(p)\Gamma(q)/\Gamma(p+q)$, beginning with parameters all equal to two. Therefore the first nonzero Taylor term in (W22) is $\nu!t^{2\nu+1}/(2\nu+1)!$. Its inverse spatial Fourier transform is exactly (W19). The same expression extends to an odd smooth distribution-valued function on all real $t$, namely $$W_\nu=E_\nu-\widetilde E_\nu. \tag{W24}$$ The distinction between this odd extension and the causal extension will be essential when time distributions are restricted at a spatial point.

4.1. The simplex factor and the orientation of the tested contour

For completeness the simplex constant in (W23) follows by a finite induction with every factor present. Let $I_\nu$ denote its integral and $I_0=1$. Fix $u_1=a$; in the remaining simplex write $u_i=(1-a)v_i$, $2\leq i\leq\nu+1$. The $\nu$ remaining factors contribute $(1-a)^\nu$, their $\nu-1$ independent integration variables contribute $(1-a)^{\nu-1}$, and the first factor contributes $a$. Hence $$I_\nu=I_{\nu-1}\int_0^1a(1-a)^{2\nu-1}\,da =I_{\nu-1}\frac{(2\nu-1)!}{(2\nu+1)!} =\frac1{(2\nu+1)!}. \tag{WA6}$$ The beta identity CX36 applies with its positive real parameters $2,2\nu$. This gives all initial jets in (W19) and the odd extension (W24), including $\nu=0$, whose simplex is a point of mass one.

Here is the complete contour estimate used in Section 5. If $\phi$ is supported in $[a,b]$ with $0<a<b$, put $\Phi(z)=\int e^{itz}\phi(t)\,dt$. Integration by parts $M$ times, with zero endpoint jets, gives $$|\Phi(u+iv)|\leq C_M e^{-av}(1+|u+iv|)^{-M}\quad(v\geq0); \tag{WA7}$$ on the fixed lower strip $-s\leq v\leq0$, the same bound holds with $e^{sb}$ instead of $e^{-av}$. Take a rectangle with bottom on $\operatorname{Im}z=-s$, vertical sides at $\pm R$, top at $\operatorname{Im}z=R$, and $R>2(\omega+s)+1$. On its added sides the functions $1/(\omega^2-z^2)$ and $iz/(\omega^2-z^2)$ are bounded by constants times $R^{-2}$ and $R^{-1}$, respectively. Their side lengths are at most $2(R+s)$; (WA7) makes all added integrals tend to zero. The bottom runs from left to right, so this closure is positively oriented. To make the finite-pole step explicit, expand the entire tested exponential at each of the two poles, subtract its complete principal parts, and use CX7 to see that the remainder extends holomorphically there. Its rectangle integral is zero by a two-triangle subdivision and CX3. A principal part of order greater than one has its actual single-valued power primitive and contributes zero. For the simple part inside the rectangle, remove a small polygonal disk around the pole and triangulate the region between the two polygons; all interior edges cancel. Let the inner polygons tend to the positively oriented circle: continuity of the integrand on that compact annulus and bounded polygon lengths give convergence of their parametrized integrals. CX6 gives exactly $2\pi i$ for the inner circle integral of $1/(z-z_0)$. A pole outside the rectangle instead leaves a holomorphic integrand inside, so its integral is zero. Thus the circle/triangle proof CX1–CX9 gives the contour integral as $2\pi i$ times the sum of residues. The first rational function has residues $-e^{it\omega}/(2\omega)$ and $e^{-it\omega}/(2\omega)$. Its inverse-transform coefficient produces $S_\omega(t)$. The second has the two residues displayed in Section 5 and produces $\cos(t\omega)$.

For a test supported in $[-b,-a]$, close below with bottom at $-R$. Integration by parts now gives $e^{-a|v|}$ in the lower half-plane. The rectangle is clockwise and contains no poles, so both tested inverses vanish. At $\omega=0$ the first function is $-z^{-2}$, whose residue after multiplication by $e^{itz}$ is $-it$; the derivative integrand is $-i/z$, with residue $-i$. Their inverses are respectively $t$ and one on positive time. Finally the directly convergent Laplace integral of $\boldsymbol1_{\{t>0\}}S_\omega(t)$ is $(p^2+\omega^2)^{-1}$ for $\operatorname{Re}p>0$: two integrations of its differential equation retain its zero initial value and first derivative one. Formula (WA2) and Fourier injectivity therefore identify the inverse also at time zero. There is no additional point-supported term. Its distributional first derivative has no delta because its initial value is zero. The Gaussian spatial factor in Section 5 supplies absolute spatial bounds, and its removal is the original Schwartz multiplier convergence. This closes all contour, zero-frequency and initial-trace cases with their stated orientations.

4.2. Every one-sided flat initial jet

The simplex proof supplies all higher jets as well, with the full original coefficients. For nonnegative integers $m_1,\ldots,m_{\nu+1}$, put $m=\sum_i m_i$. Successive beta substitutions on that same simplex give $$\int_{\Sigma_\nu}\prod_{i=1}^{\nu+1}u_i^{2m_i+1}\,du =\frac{\prod_{i=1}^{\nu+1}(2m_i+1)!} {(2m+2\nu+1)!}. \tag{WJ1}$$ Indeed fix the first coordinate, rescale the remaining ones by its complement as in (WA6), and retain their full sum of exponents and the $\nu-1$ Jacobian power. The resulting beta integral has parameters $2m_1+2$ and $\sum_{i=2}^{\nu+1}(2m_i+2)$. CX36 gives their gamma quotient; it cancels precisely the denominator in the lower-dimensional induction, leaving the displayed product. The case $\nu=0$ is the point simplex and its equal numerator and denominator. This proves the formula for every dimension and every listed exponent.

Expand each original sinc by its convergent series at $t=0$. For fixed $\omega$, those series and all fixed derivatives converge uniformly on a bounded time interval and the compact simplex. Their absolute product is bounded by the product of finitely many convergent exponential series, so integration and coefficient extraction are justified. The term indexed by $(m_i)$ has sign $(-1)^m$, frequency factor $\omega^{2m}$, and denominator $\prod_i(2m_i+1)!$. Formula (WJ1) retains and then evaluates exactly those factors. The number of nonnegative tuples with sum $m$ is $\binom{m+\nu}{\nu}$: arrange $m$ marked slots and $\nu$ dividers in a row of length $m+\nu$; the divider positions uniquely determine the lengths of all $\nu+1$ successive groups, including zero groups. Thus $$h_\nu(t,\omega)=\nu!\sum_{m=0}^\infty (-1)^m\binom{m+\nu}{\nu}\, \omega^{2m}\frac{t^{2m+2\nu+1}}{(2m+2\nu+1)!}. \tag{WJ2}$$ This equality is a fixed-frequency Taylor calculation of the original Volterra expression, not an assumption of an untested termwise spatial Fourier integral. For each fixed derivative at zero, (W22)’s polynomial bounds already justify its original Fourier inversion on spatial Schwartz tests. Consequently $$\begin{split} \partial_t^{2\ell}E_\nu(0+,\cdot)&=0\quad(\ell\geq0),\\ \partial_t^{2\ell+1}E_\nu(0+,\cdot)&=0\quad(0\leq\ell<\nu),\\ \partial_t^{2\ell+1}E_\nu(0+,\cdot) &=\nu!(-1)^{\ell-\nu}\binom{\ell}{\nu} (-\Delta_x)^{\ell-\nu}\delta_0 \quad(\ell\geq\nu). \end{split} \tag{WJ3}$$ The highest binomial top index is the original $\ell$, the coefficient includes $\nu!$, and the original Fourier multiplier of $-\Delta_x$ is $|\xi|^2$. Formula (W19) is its first nonzero instance. The smooth odd family has the same jets from either side, whereas the causal extension retains its separate spacetime point-source terms.

5. The cosine kernel and its spectral measure

The derivative of the odd fundamental kernel satisfies $$\mathcal F_x(\partial_tW_0)(t,\xi)=\cos(t|\xi|). \tag{W25}$$ Define the flat spectral cutoff kernel for $\lambda\in\mathbb R$ by $$e(x,\lambda^2)=(2\pi)^{-n}\int_{|\xi|<|\lambda|}e^{ix\cdot\xi}\,d\xi. \tag{W26}$$ Let $d_\lambda e(x,\lambda^2)$ mean its distributional differential in $\lambda$, not differentiation with respect to $\lambda^2$. The measure-valued tempered distribution $$d\mu_x(\lambda)=\tfrac12 \operatorname{sgn}\lambda\,d_\lambda e(x,\lambda^2) \tag{W27}$$ satisfies $$\partial_tW_0(t,x)=\int_{\mathbb R}e^{it\lambda}\,d\mu_x(\lambda). \tag{W28}$$ Here the transform of the measure has no additional $(2\pi)^{-1}$ factor. In the normalized inverse-transform convention, its time Fourier transform is $2\pi\mu_x$.

Proof. Differentiating (W20) proves (W25). For a fixed $\omega>0$, $$d_\lambda\boldsymbol1_{\{|\lambda|>\omega\}} =\delta_\omega-\delta_{-\omega}. \tag{W29}$$ Multiplication by $\tfrac12\operatorname{sgn}\lambda$ makes this $\tfrac12(\delta_\omega+\delta_{-\omega})$. Its transform in (W28) is $\cos(t\omega)$. The set $\xi=0$ has Lebesgue measure zero since $n\geq1$, and alternatively the two-point measures converge to $\delta_0$ as $\omega\downarrow0$. Pairing with Schwartz tests justifies integration in $\xi$ and proves (W28) with the stated constants.

The contour prescription can also be checked without an interchange of unregularized oscillatory integrals. Insert $e^{-\varepsilon|\xi|^2}$, $\varepsilon>0$, in (W12) for $\nu=0$. For fixed $\xi$, inverse transformation of $(\omega^2-\tau^2)^{-1}$ on a line below the poles gives $\boldsymbol1_{t>0}S_\omega(t)$. For $t>0$, closing upward, the residues of $i\tau e^{it\tau}/(\omega^2-\tau^2)$ at $\tau=\omega,-\omega$ are $-ie^{it\omega}/2$ and $-ie^{-it\omega}/2$. Multiplication by $2\pi i$, then by $(2\pi)^{-1}$, gives $\cos(t\omega)$. For $t<0$, closing downward contains no poles and gives zero. At $\omega=0$, the limiting residue gives the same value. One may justify the closures first with a compact time test supported away from zero; its transform decays on the closing contour after integration by parts. The initial value $S_\omega(0)=0$ then shows that differentiating its causal extension adds no Dirac term.

Subtract the reflected causal kernel. Its derivative is now the cosine for either sign of time. The Gaussian makes the remaining spatial integral absolutely convergent, and it also gives the version of (W27) with $e^{-\varepsilon\lambda^2}$. Finally $e^{-\varepsilon|\xi|^2}\to1$ on Schwartz tests with all required dominated seminorm bounds. Thus the regularized distributions converge, proving the contour and spectral computations agree. This argument records both the time-contour signs and the role of the Gaussian; neither is hidden in a formal residue integral.

6. Testing at the spatial center with finite regularity

Put $a=a_\nu=\nu-(n-1)/2$ and $c_{\nu,n}=A_\nu$, as in (W14). For nonzero $x$, the odd kernel is $c_{\nu,n}\operatorname{sgn}(t)\chi_+^a(t^2-|x|^2)$. This expression uses a submersion near its nonzero cone points; its value at $x=0$ requires the construction that follows. The finite-order topology is the weak topology on compactly supported $C^m$ tests specified in Section 1.

For an even function $\phi\in C_c^{j+2}(\mathbb R)$, define $$(T\phi)(t)=\frac{\phi'(t)}{2t}\quad(t\ne0),\qquad (T\phi)(0)=\frac{\phi''(0)}2. \tag{OE3}$$ This defines an even $C^j$ function with support contained in the support of $\phi$. To see both regularity and the quantitative bound, use $\phi'(0)=0$ and write $$(T\phi)^{(j)}(t) =\frac12\int_0^1 s^j\phi^{(j+2)}(st)\,ds, \qquad \|(T\phi)^{(j)}\|_\infty \leq\frac{\|\phi^{(j+2)}\|_\infty}{2(j+1)}. \tag{OE4}$$ Differentiation under this integral is allowed for the displayed regularity, and it also proves the formula at zero. Away from the support, the original quotient vanishes. Iteration therefore gives a continuous operation from even $C_c^{2k}$ functions to $C_c^0$, and $$\|T^k\phi\|_\infty \leq\frac{k!}{(2k)!}\|\phi^{(2k)}\|_\infty, \qquad (T^k\phi)(0)=\frac{k!}{(2k)!}\phi^{(2k)}(0). \tag{OE5}$$ The norm bound follows by applying (OE4) with derivative indices $0,2,\ldots,2k-2$. The value at zero follows from the Taylor polynomial of the even function through degree $2k$, because $T(t^{2j})=j t^{2j-2}$. For $k=0$, both formulas mean the identity operation.

Positive measures and finite-order distributions.

Fix $\delta\geq0$. For $b>0$, define the even positive measure $e_{\delta,b}$ by $$\langle e_{\delta,b},\phi\rangle =\frac1{\Gamma(b)}\int_0^\infty s^{b-1}\bigl[\phi(\sqrt{s+\delta})+ \phi(-\sqrt{s+\delta})\bigr] \,ds. \tag{OE6}$$ At $b=0$, use $$e_{\delta,0}=\delta_{\sqrt\delta}+\delta_{-\sqrt\delta}; \quad e_{0,0}=2\delta_0. \tag{OE7}$$ Thus (OE6) agrees, when it is an ordinary function, with $2|t|\chi_+^{b-1}(t^2-\delta)$. For tests supported in $[-R,R]$, the mass bounds are $$|\langle e_{\delta,b},\phi\rangle| \leq M_{b,R}\|\phi\|_\infty, \quad M_{b,R}= \begin{cases} 2,&b=0,\\ 2R^{2b}/\Gamma(b+1),&b>0. \end{cases} \tag{OE8}$$ They hold for every $\delta\geq0$; if $\delta>R^2$, the pairing vanishes. For a fixed continuous test, (OE6) varies continuously with $\delta$: its integrand converges pointwise, and a constant times $s^{b-1}$ on $[0,R^2]$ dominates it. Formula (OE7) proves the same assertion for $b=0$. In particular these are weakly continuous measures even at $\delta=0$.

For even smooth tests, integration by parts in $s$ gives $$\langle e_{\delta,b},\phi\rangle =-\langle e_{\delta,b+1},T\phi\rangle, \qquad b\geq0. \tag{OE9}$$ For $b>0$, differentiate $s^b/\Gamma(b+1)$ and use $d\phi(\sqrt{s+\delta})/ds=(T\phi)(\sqrt{s+\delta})$. The boundary at zero vanishes. For $b=0$, the boundary term is precisely the two evaluations in (OE7). This argument includes $\delta=0$, because (OE3) supplies the continuous derivative of the even function of $\sqrt s$.

Now let $a$ be any real number, choose $k\in\mathbb N_0$ with $a+k\geq0$, and set $\phi_{\mathrm e}(t)=(\phi(t)+\phi(-t))/2$. Define $$\langle e_{\delta,a},\phi\rangle =(-1)^k\langle e_{\delta,a+k},T^k\phi_{\mathrm e}\rangle. \tag{OE10}$$ Applying (OE9) once proves that increasing $k$ does not change this definition. The definition kills odd tests. It agrees for $\delta>0$ with the distribution $2|t|\chi_+^{a-1}(t^2-\delta)$, by the same integration by parts on each side of the two nonzero roots, or equivalently by the differential recurrence for $\chi_+^a$. At $\delta=0$, (OE10), rather than an undefined pullback, specifies its extension.

Equations (OE5), (OE8) and (OE10) imply $$|\langle e_{\delta,a},\phi\rangle| \leq M_{a+k,R}\frac{k!}{(2k)!} \|\phi^{(2k)}\|_\infty. \tag{OE11}$$ For $k=0$, the last norm is $\|\phi\|_\infty$. The estimate is valid on every fixed compact test support and is uniform in $\delta\geq0$. Since $T^k\phi_{\mathrm e}$ is continuous for $\phi\in C_c^{2k}$, weak measure continuity of the base family proves continuity of the pairing in (OE10) for every $C_c^{2k}$ test. It also proves joint continuity when $\delta_j\to\delta$ and $\phi_j\to\phi$ in $C^{2k}$ on one compact support: use (OE11) for the changing test and pointwise continuity for the fixed test. This establishes the finite-order topology directly, including equality in $a+k\geq0$.

No norm continuity is hidden here. For $\delta>0$, the measure $\delta_{\sqrt\delta}+\delta_{-\sqrt\delta}-2\delta_0$ has total variation $4$ on an interval containing all three points. Nevertheless its pairing with every continuous test tends to zero as $\delta\downarrow0$.

The same formulas yield smooth dependence on $x$ with values in ordinary distributions, without asserting a fixed finite order for all parameter derivatives. For smooth even tests, differentiate (OE6) in $\delta$; the derivative of the test is $T\phi$, and all derivatives are dominated on a fixed bounded $s$ interval. Equations (OE9)–(OE10) then give, including one-sided derivatives at zero, $$\partial_\delta^j\langle e_{\delta,a},\phi\rangle =(-1)^j\langle e_{\delta,a-j},\phi\rangle. \tag{OE12}$$ The same conclusion for the base $b=0$ follows from the smoothness of $\phi(\sqrt\delta)$ for even smooth $\phi$, whose successive derivatives are $(T^j\phi)(\sqrt\delta)$. Composing with $\delta=|x|^2$ gives smooth functions of $x$: differentiation produces polynomials in $x$ multiplying the continuous one-sided derivatives in (OE12), and induction extends every such derivative through $x=0$. Each parameter derivative has an estimate of the form (OE11), at an order that may increase with the derivative. This is the requisite distribution-valued smoothness.

The odd primitive retains the order.

For $x\ne0$, differentiation of the cone expression (W15) and its odd reflection gives $$\partial_tW_\nu(t,x)=c_{\nu,n}e_{|x|^2,a}(t). \tag{OE13}$$ To recover an odd primitive while keeping its finite order, let $\phi_{\mathrm o}(t)=(\phi(t)-\phi(-t))/2$, and set $$(J\phi)(t)=\int_{-\infty}^t\phi_{\mathrm o}(s)\,ds. \tag{OE14}$$ The total integral of an odd compactly supported function is zero, so $J\phi$ is compactly supported. It is even, and its support lies in $[-R,R]$ if that of $\phi$ does. Moreover, $$\|J\phi\|_\infty\leq2R\|\phi\|_\infty, \qquad \|(J\phi)^{(j)}\|_\infty\leq \|\phi^{(j-1)}\|_\infty\quad(j\geq1). \tag{OE15}$$ For an even distribution $h$, define its odd primitive by $$\langle Ah,\phi\rangle=-\langle h,J\phi\rangle. \tag{OE16}$$ It is odd because $J$ kills even tests. Further, $J(\phi')=\phi_{\mathrm e}$, which gives $\langle\partial_tAh,\phi\rangle=\langle h,\phi\rangle$. An odd primitive is unique: a distribution with derivative zero is constant, and a constant distribution is even. For completeness, the first fact follows by writing any integral-zero test as the derivative of its compactly supported primitive; a distribution killing all such tests depends only on their integral.

Define $W_\nu(\cdot,x)=A(c_{\nu,n}e_{|x|^2,a})$ also at $x=0$. For $x\ne0$, uniqueness identifies it with the odd reflection of (W15). Estimate (OE15) proves that $A$ preserves every nonnegative finite order; it is not necessary to differentiate the test one extra time. It also preserves the weak continuity just proved. Consequently, $$x\longmapsto W_\nu(\cdot,x),\quad x\longmapsto\partial_tW_\nu(\cdot,x) \quad\hbox{are continuous in order at most }2k \tag{OE17}$$ for every $$k\in\mathbb N_0,\qquad k\geq\frac{n-1}{2}-\nu. \tag{OE18}$$ This includes the threshold whenever the right side is a nonnegative integer. Equations (OE12) and (OE16) also give their smooth dependence on $x$ with values in ordinary distributions. The restriction at $x=0$ is obtained by this continuity. It agrees with the restriction of the full spacetime odd kernel; one may either use its already established distribution-valued smoothness or the wavefront criterion proved below.

7. The exact endpoint and the meaning of the integer parameter

Suppose $a=-k$ with $k\in\mathbb N_0$. Equations (OE7), (OE10) and (OE5) give, for every smooth test, $$\langle e_{0,-k},\phi\rangle =2(-1)^k(T^k\phi_{\mathrm e})(0) =\frac{2(-1)^kk!}{(2k)!}\phi^{(2k)}(0).$$ The derivative order is even, so the sign in the action of $\delta^{(2k)}$ is positive. Hence $$e_{0,-k}=\frac{2(-1)^kk!}{(2k)!}\delta^{(2k)}, \qquad \partial_tW_\nu(t,0) =2^{2k-n+1}\pi^{(1-n)/2} \frac{(-1)^kk!}{(2k)!}\delta^{(2k)}(t). \tag{OE19}$$ In the last equality we used $2\nu=n-1-2k$. For $k=0$, this states that the derivative is $2c_{\nu,n}\delta_0$, with odd primitive $c_{\nu,n}\operatorname{sgn}(t)$. For $k\geq1$, replace $\delta^{(2k)}$ by $\delta^{(2k-1)}$ in the second formula to obtain $W_\nu(t,0)$.

The endpoint order $2k$ for the derivative is exact when $k>0$. Indeed, a nonzero multiple of $\delta^{(2k)}$ cannot have order at most $2k-1$: choose a compactly supported smooth $\psi$ with $\psi^{(2k)}(0)\ne0$, and test against $\epsilon^{2k-1}\psi(t/\epsilon)$. Derivatives through order $2k-1$ stay bounded, while the pairing grows like $\epsilon^{-1}$. The case $k=0$ is a nonzero measure.