[proofplan] We use the precise Kirchhoff-Hadamard formula in odd spatial dimension, written in terms of spherical means and the radial operator $r^{-1}\partial_r$. This formula expresses $u(t,x)$ for positive time using only finitely many radial derivatives of spherical means over $\partial B(x,t)$. Expanding those radial derivatives shows that only finitely many spatial derivatives of $f$ and $g$ on that sphere occur. The support conclusion follows because every derivative of a smooth function vanishes off its closed support, and negative times are reduced to positive times by time reversal. [/proofplan] [step:State the odd-dimensional Kirchhoff-Hadamard formula with all notation fixed] Let $S^{n-1}:=\partial B(0,1)$ denote the unit sphere in $\mathbb{R}^n$, and let $\mathcal{H}^{n-1}$ denote $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on $S^{n-1}$. Define the surface measure constant \begin{align*} \sigma_{n-1}:=\mathcal{H}^{n-1}(S^{n-1}) \end{align*} and, for every $h\in C_c^\infty(\mathbb{R}^n)$, define the spherical mean map $M h: (0,\infty)\times\mathbb{R}^n\to\mathbb{R}$ by \begin{align*} (M h)(r,x):=\frac{1}{\sigma_{n-1}}\int_{S^{n-1}} h(x+r\omega)\,d\mathcal{H}^{n-1}(\omega). \end{align*} Let \begin{align*} m:=\frac{n-3}{2} \end{align*} which is a non-negative integer because $n\geq 3$ is odd. Let $L$ denote the radial differential operator acting on smooth functions of $r\in(0,\infty)$ by \begin{align*} L\Phi(r):=\frac{1}{r}\frac{\partial \Phi}{\partial r}(r). \end{align*} Finally define \begin{align*} c_n:=\frac{1}{1\cdot 3\cdot 5\cdots (n-2)}. \end{align*} The classical Kirchhoff-Hadamard formula for odd spatial dimension states that, for every $t>0$ and $x\in\mathbb{R}^n$, \begin{align*} u(t,x)=c_n\frac{\partial}{\partial t}\left[L^m\left(r\mapsto r^{n-2}(M f)(r,x)\right)(t)\right]+c_n L^m\left(r\mapsto r^{n-2}(M g)(r,x)\right)(t). \end{align*} This is the standard odd-dimensional spherical mean representation for the smooth Cauchy problem for the free [wave equation](/page/Wave%20Equation). Its hypotheses are satisfied here: the spatial dimension is odd and at least $3$, the initial data $f$ and $g$ are in $C_c^\infty(\mathbb{R}^n)$, and the solution $u$ is the smooth solution with those Cauchy data. The compact support and smoothness of $f$ and $g$ ensure that every spherical mean above is smooth in $(r,x)$ for $r>0$, so the displayed differential expression is well-defined. In dimension $n=3$ this gives \begin{align*} u(t,x)=\frac{\partial}{\partial t}\left[t(M f)(t,x)\right]+t(M g)(t,x), \end{align*} which is exactly the usual three-dimensional Kirchhoff formula. [/step] [step:Expand radial derivatives into finitely many spatial derivatives on the sphere] Fix $h\in C_c^\infty(\mathbb{R}^n)$ and $x\in\mathbb{R}^n$. For each integer $q\geq 0$, repeated differentiation under the surface integral gives functions $P_{q,\alpha}:S^{n-1}\to\mathbb{R}$, polynomial in $\omega$, such that \begin{align*} \frac{\partial^q}{\partial r^q}(M h)(r,x)=\frac{1}{\sigma_{n-1}}\sum_{|\alpha|=q}\int_{S^{n-1}}P_{q,\alpha}(\omega)D^\alpha h(x+r\omega)\,d\mathcal{H}^{n-1}(\omega). \end{align*} For $q=0$ this is the definition of $M h$ with $P_{0,0}=1$. For $q\geq 1$, the identity follows by applying the chain rule to the smooth map $r\mapsto h(x+r\omega)$ and using compact support to justify differentiation under the finite measure integral over $S^{n-1}$. The expression $L^m(r^{n-2}(M h)(r,x))$ is obtained by composing $m$ copies of the operator $r^{-1}\partial_r$ with multiplication by $r^{n-2}$. Therefore it is a finite linear combination, with coefficients depending only on $n$ and on powers of $r$, of derivatives $\partial_r^q(M h)(r,x)$ for $0\leq q\leq m$. After applying the additional outer derivative $\partial_t$ in the $f$ term, the highest derivative order of $M f$ is $m+1=(n-1)/2$. In the $g$ term, the highest derivative order of $M g$ is $m=(n-3)/2$. Hence the Kirchhoff-Hadamard formula expresses $u(t,x)$ for $t>0$ using only values of $D^\alpha f$ with $|\alpha|\leq (n-1)/2$ and $D^\beta g$ with $|\beta|\leq (n-3)/2$ at points of the form $x+t\omega$, where $\omega\in S^{n-1}$. Since \begin{align*} \{x+t\omega:\omega\in S^{n-1}\}=\partial B(x,t), \end{align*} the asserted finite-derivative dependence on $\partial B(x,t)$ follows for $t>0$. [guided] Fix $h\in C_c^\infty(\mathbb{R}^n)$ and $x\in\mathbb{R}^n$. The first point is that differentiating a spherical mean in the radius never moves the integration away from the same sphere; it only differentiates the integrand along the radial direction. The map $r\mapsto h(x+r\omega)$ is smooth for every $\omega\in S^{n-1}$, and $S^{n-1}$ has finite $\mathcal{H}^{n-1}$ measure. Since $h$ is smooth with compact support, all derivatives appearing below are continuous and bounded on the relevant compact sets, so differentiation under the $\mathcal{H}^{n-1}$ integral is justified. For one derivative, the chain rule gives \begin{align*} \frac{\partial}{\partial r}h(x+r\omega)=\sum_{i=1}^n \omega_i\partial_{x_i}h(x+r\omega). \end{align*} Therefore \begin{align*} \frac{\partial}{\partial r}(M h)(r,x)=\frac{1}{\sigma_{n-1}}\sum_{i=1}^n\int_{S^{n-1}}\omega_i\partial_{x_i}h(x+r\omega)\,d\mathcal{H}^{n-1}(\omega). \end{align*} Repeating the same chain-rule computation gives, for each integer $q\geq 0$, polynomial coefficient functions $P_{q,\alpha}:S^{n-1}\to\mathbb{R}$ such that \begin{align*} \frac{\partial^q}{\partial r^q}(M h)(r,x)=\frac{1}{\sigma_{n-1}}\sum_{|\alpha|=q}\int_{S^{n-1}}P_{q,\alpha}(\omega)D^\alpha h(x+r\omega)\,d\mathcal{H}^{n-1}(\omega). \end{align*} Thus every radial derivative of the spherical mean is still an integral over $S^{n-1}$ of spatial derivatives of $h$ evaluated only at the points $x+r\omega$. Now apply this observation to the [representation formula](/theorems/39). The operator $L$ was defined by $L\Phi(r)=r^{-1}\partial_r\Phi(r)$. Applying $L$ repeatedly to $r^{n-2}(M h)(r,x)$ differentiates only in $r$ and multiplies by powers of $r$. It cannot introduce values of $h$ away from the points $x+r\omega$. After $m$ applications of $L$, only radial derivatives of $M h$ up to order $m$ occur. The $g$ contribution in the Kirchhoff-Hadamard formula has precisely this form, so it uses only derivatives $D^\beta g$ with $|\beta|\leq m=(n-3)/2$. The $f$ contribution has one additional outside derivative $\partial_t$. This raises the maximum radial derivative order by at most one. Hence the $f$ contribution uses only derivatives $D^\alpha f$ with $|\alpha|\leq m+1=(n-1)/2$. Finally, the set of all integration points is \begin{align*} \{x+t\omega:\omega\in S^{n-1}\}=\partial B(x,t). \end{align*} Therefore, for $t>0$, the value $u(t,x)$ is determined entirely by the stated finite jets of $f$ and $g$ restricted to the sphere $\partial B(x,t)$. [/guided] [/step] [step:Use vanishing of all derivatives off the support] Assume that $\operatorname{supp}(f)\cup\operatorname{supp}(g)\subset K$ for a compact set $K\subset\mathbb{R}^n$, and let $t>0$ satisfy \begin{align*} \partial B(x,t)\cap K=\varnothing. \end{align*} For every $y\in\partial B(x,t)$, the point $y$ lies outside $K$, hence outside $\operatorname{supp}(f)$ and outside $\operatorname{supp}(g)$. Since supports are closed supports, there is an open neighbourhood of $y$ on which $f$ is identically zero and an open neighbourhood of $y$ on which $g$ is identically zero. Therefore every spatial derivative of $f$ and every spatial derivative of $g$ vanishes at $y$. All derivatives of $f$ and $g$ appearing in the spherical representation from the previous step therefore vanish at every point $x+t\omega$ with $\omega\in S^{n-1}$. Every surface integral in the Kirchhoff-Hadamard formula is zero, and consequently \begin{align*} u(t,x)=0. \end{align*} [/step] [step:Reduce negative times to the positive-time result] Let $t<0$ and define the time-reversed function $v:\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}$ by \begin{align*} v(s,x):=u(-s,x). \end{align*} Because $u\in C^\infty(\mathbb{R}\times\mathbb{R}^n)$ and solves $\partial_t^2u-\Delta u=0$, the function $v$ belongs to $C^\infty(\mathbb{R}\times\mathbb{R}^n)$ and satisfies $\partial_s^2v-\Delta v=0$. Its Cauchy data are \begin{align*} v(0,x)=f(x) \end{align*} and \begin{align*} \partial_s v(0,x)=-g(x). \end{align*} The functions $f$ and $-g$ are smooth and compactly supported, and their supports are contained in the same compact set $K$ whenever the support hypothesis holds for $f$ and $g$. Applying the positive-time conclusions to $v$ at time $|t|>0$ shows that $v(|t|,x)$ is determined by the corresponding finite derivatives of $f$ and $-g$ on $\partial B(x,|t|)$ and that $v(|t|,x)=0$ if this sphere is disjoint from $K$. Since $v(|t|,x)=u(t,x)$, the dependence statement and the support conclusion hold for negative $t$ as well. The theorem only concerns $t\neq 0$, so the positive-time and negative-time cases complete the proof. [/step]