[proofplan] For positive times we use Kirchhoff's formula for the three-dimensional free wave equation, expressing $u(t,x)$ as surface averages of $f$, $\nabla f$, and $g$ over the sphere $\partial B(x,t)$. Since the data are supported in $B(0,R)$, only the spherical patch $\partial B(x,t) \cap B(0,R)$ contributes. For large $t$, this patch has uniformly bounded $\mathcal{H}^2$-measure, so the formula gives a $t^{-1}$ bound. For bounded $t$, continuity and compact support give a uniform bound, and time reversal reduces negative times to the same argument. [/proofplan] [step:Apply Kirchhoff's formula for positive times] Fix $t > 0$ and $x \in \mathbb{R}^3$. By [Kirchhoff's Formula](/theorems/666) for the three-dimensional free wave equation, applied to the classical solution with initial data $u(0,\cdot)=f$ and $\partial_t u(0,\cdot)=g$, we have \begin{align*} u(t,x) = \frac{1}{4\pi t^2}\int_{\partial B(x,t)} f(y)\, d\mathcal{H}^2(y) + \frac{1}{4\pi t}\int_{\partial B(x,t)} \nabla f(y)\cdot \frac{y-x}{t}\, d\mathcal{H}^2(y) + \frac{1}{4\pi t}\int_{\partial B(x,t)} g(y)\, d\mathcal{H}^2(y). \end{align*} Here $\mathcal{H}^2$ denotes two-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on the sphere $\partial B(x,t)$. Since $\operatorname{supp} f \cup \operatorname{supp} g \subset B(0,R)$, all three integrands vanish outside $B(0,R)$. Thus each integral over $\partial B(x,t)$ equals the corresponding integral over \begin{align*} E_{t,x} := \partial B(x,t) \cap B(0,R). \end{align*} Taking absolute values and using $\left|\frac{y-x}{t}\right|=1$ for $y \in \partial B(x,t)$ gives \begin{align*} |u(t,x)| \leq \frac{\|f\|_{L^\infty(\mathbb{R}^3)}}{4\pi t^2}\mathcal{H}^2(E_{t,x}) + \frac{\|\nabla f\|_{L^\infty(\mathbb{R}^3)}}{4\pi t}\mathcal{H}^2(E_{t,x}) + \frac{\|g\|_{L^\infty(\mathbb{R}^3)}}{4\pi t}\mathcal{H}^2(E_{t,x}). \end{align*} [guided] The [representation formula](/theorems/39) is the point where the special role of dimension $3$ enters. For $t > 0$, [Kirchhoff's Formula](/theorems/666) for the [free wave equation](/page/Wave%20Equation) writes the solution with initial data $u(0,\cdot)=f$ and $\partial_t u(0,\cdot)=g$ as a combination of surface integrals over the sphere of radius $t$ centered at $x$: \begin{align*} u(t,x) = \frac{1}{4\pi t^2}\int_{\partial B(x,t)} f(y)\, d\mathcal{H}^2(y) + \frac{1}{4\pi t}\int_{\partial B(x,t)} \nabla f(y)\cdot \frac{y-x}{t}\, d\mathcal{H}^2(y) + \frac{1}{4\pi t}\int_{\partial B(x,t)} g(y)\, d\mathcal{H}^2(y). \end{align*} The hypotheses needed for this formula are satisfied because $f \in C_c^3(\mathbb{R}^3)$ and $g \in C_c^2(\mathbb{R}^3)$, so the corresponding solution is classical and the displayed surface integrals with respect to [Hausdorff measure](/page/Hausdorff%20Measure) are well-defined. These regularity assumptions are used here to justify the classical representation formula; the later estimates use only the displayed sup norms and the support radius. Now define the contributing spherical patch by \begin{align*} E_{t,x} := \partial B(x,t) \cap B(0,R). \end{align*} The support assumption says that $f(y)=0$, $\nabla f(y)=0$, and $g(y)=0$ whenever $y \notin B(0,R)$. Therefore the parts of the sphere outside $E_{t,x}$ contribute nothing to the three integrals. For $y \in \partial B(x,t)$, the vector $(y-x)/t$ has Euclidean norm $1$. Hence \begin{align*} \left|\nabla f(y)\cdot \frac{y-x}{t}\right| \leq |\nabla f(y)| \leq \|\nabla f\|_{L^\infty(\mathbb{R}^3)}. \end{align*} Taking absolute values in Kirchhoff's formula and estimating each integrand by its supremum norm gives \begin{align*} |u(t,x)| \leq \frac{\|f\|_{L^\infty(\mathbb{R}^3)}}{4\pi t^2}\mathcal{H}^2(E_{t,x}) + \frac{\|\nabla f\|_{L^\infty(\mathbb{R}^3)}}{4\pi t}\mathcal{H}^2(E_{t,x}) + \frac{\|g\|_{L^\infty(\mathbb{R}^3)}}{4\pi t}\mathcal{H}^2(E_{t,x}). \end{align*} This reduces the analytic decay estimate to a geometric estimate on the area of the patch $E_{t,x}$. [/guided] [/step] [step:Bound the contributing spherical patch for large times] Assume $t \geq 4R$. If $E_{t,x}=\varnothing$, then $\mathcal{H}^2(E_{t,x})=0$. Otherwise choose $y_0 \in E_{t,x}$. Since $|y_0|<R$ and $|y_0-x|=t$, the triangle inequality gives \begin{align*} |x| \geq t-|y_0| > t-R \geq \frac{3t}{4}. \end{align*} In particular $x \neq 0$. Define the unit vector $e \in \mathbb{R}^3$ by \begin{align*} e := \frac{x}{|x|}. \end{align*} Let $P := e^\perp$ be the plane through the origin perpendicular to $e$, and let $\pi: \mathbb{R}^3 \to P$ be the [orthogonal projection](/theorems/437). For $y \in E_{t,x}$, the outward unit normal to $\partial B(x,t)$ at $y$ is \begin{align*} \nu(y) := \frac{y-x}{t}. \end{align*} We first justify that the projection has one sheet on the contributing patch. If $y,y' \in E_{t,x}$ and $\pi(y)=\pi(y')$, then there is a scalar $a \in \mathbb{R}$ such that $y=z+ae$ for the common projected point $z:=\pi(y)=\pi(y') \in P$. Since $x=|x|e$ and $z\cdot e=0$, the sphere equation along this line is \begin{align*} |z+ae-x|^2 = |z|^2+(a-|x|)^2=t^2. \end{align*} Thus the possible values of $a$ are the roots of $(a-|x|)^2=t^2-|z|^2$, so there are at most two roots and, when two occur, their arithmetic midpoint is $|x|$. Both points lie in $B(0,R)$, so any admissible root must satisfy $|a|=|e\cdot y|<R$. Since $|x|>3t/4\geq 3R$, two numbers both lying in $(-R,R)$ cannot have midpoint $|x|$. Hence at most one root can lie in $B(0,R)$, and $\pi|_{E_{t,x}}$ is injective onto its image. The [area formula](/theorems/3075) for orthogonal projection from the smooth surface $\partial B(x,t)$ to $P$ gives \begin{align*} \mathcal{H}^2(E_{t,x}) = \int_{\pi(E_{t,x})} \frac{1}{|\nu(y_z)\cdot e|}\, d\mathcal{H}^2_P(z), \end{align*} where $d\mathcal{H}^2_P$ denotes two-dimensional Hausdorff measure on $P$ and $y_z$ is the unique point of $E_{t,x}$ with $\pi(y_z)=z$ on the visible graph over $P$. For every $y \in E_{t,x}$, \begin{align*} |\nu(y)\cdot e| = \frac{|x|-e\cdot y}{t} \geq \frac{|x|-|y|}{t} \geq \frac{3t/4-R}{t} \geq \frac{1}{2}. \end{align*} Also $\pi(E_{t,x}) \subseteq \pi(B(0,R)) \subseteq P \cap B(0,R)$, so \begin{align*} \mathcal{H}^2_P(\pi(E_{t,x})) \leq \pi R^2. \end{align*} Therefore \begin{align*} \mathcal{H}^2(E_{t,x}) \leq 2\pi R^2. \end{align*} [guided] The geometric problem is to bound the area of the part of the sphere that lies inside the fixed ball $B(0,R)$, uniformly in $x$, once $t$ is large. Assume $t \geq 4R$. If $E_{t,x}=\varnothing$, then $\mathcal{H}^2(E_{t,x})=0$, so suppose $E_{t,x}$ is nonempty and choose $y_0 \in E_{t,x}$. Since $|y_0|<R$ and $|y_0-x|=t$, the triangle inequality gives \begin{align*} |x| \geq t-|y_0| > t-R \geq \frac{3t}{4}. \end{align*} Thus $x\neq 0$, and we define $e:=x/|x|$, the plane $P:=e^\perp$, and the orthogonal projection $\pi:\mathbb{R}^3\to P$. Why project onto $P$? The sphere is nearly transverse to the direction $e$ on the small patch inside $B(0,R)$, so the patch is a single graph over $P$ with controlled slope. To prove the single-sheet property, take $y,y'\in E_{t,x}$ with $\pi(y)=\pi(y')$. Writing the common projected point as $z\in P$, both points have the form $z+ae$ for some scalar $a\in\mathbb{R}$. Since $x=|x|e$ and $z\cdot e=0$, the sphere equation becomes \begin{align*} |z+ae-x|^2=|z|^2+(a-|x|)^2=t^2. \end{align*} The possible values of $a$ are therefore roots of $(a-|x|)^2=t^2-|z|^2$. If two roots occur, their arithmetic midpoint is $|x|$. But points in $B(0,R)$ satisfy $|a|=|e\cdot y|<R$, while $|x|>3t/4\geq 3R$. Two numbers in $(-R,R)$ cannot have midpoint $|x|$. Hence at most one point of $E_{t,x}$ lies over each $z\in P$, so $\pi|_{E_{t,x}}$ is injective. Let $\nu:E_{t,x}\to\mathbb{R}^3$ be the outward unit normal field defined by $\nu(y)=(y-x)/t$. The area formula for orthogonal projection from the smooth surface $\partial B(x,t)$ to $P$ gives \begin{align*} \mathcal{H}^2(E_{t,x})=\int_{\pi(E_{t,x})}\frac{1}{|\nu(y_z)\cdot e|}\,d\mathcal{H}^2_P(z), \end{align*} where $\mathcal{H}^2_P$ is two-dimensional Hausdorff measure on $P$ and $y_z$ is the unique point of $E_{t,x}$ with $\pi(y_z)=z$. For every $y\in E_{t,x}$, \begin{align*} |\nu(y)\cdot e|=\frac{|x|-e\cdot y}{t}\geq \frac{|x|-|y|}{t}\geq \frac{3t/4-R}{t}\geq \frac{1}{2}. \end{align*} Also $\pi(E_{t,x})\subseteq \pi(B(0,R))\subseteq P\cap B(0,R)$, so $\mathcal{H}^2_P(\pi(E_{t,x}))\leq \pi R^2$. Substituting these two estimates into the area formula gives \begin{align*} \mathcal{H}^2(E_{t,x})\leq 2\pi R^2. \end{align*} [/guided] [/step] [step:Derive the large-time decay estimate] For $t \geq 4R$, the previous two steps give \begin{align*} |u(t,x)| \leq \frac{R^2}{2t^2}\|f\|_{L^\infty(\mathbb{R}^3)} + \frac{R^2}{2t}\|\nabla f\|_{L^\infty(\mathbb{R}^3)} + \frac{R^2}{2t}\|g\|_{L^\infty(\mathbb{R}^3)}. \end{align*} If $R=0$, then $B(0,R)=\varnothing$, the support condition forces $f=0$ and $g=0$, and the estimate is immediate with $C_1=0$. Assume henceforth in this step that $R>0$. For $t \geq 4R$, direct differentiation shows that the maps $t \mapsto (1+t)/t$ and $t \mapsto (1+t)/t^2$ are decreasing on $(0,\infty)$. Hence \begin{align*} \frac{1}{t} \leq \left(1+\frac{1}{4R}\right)(1+t)^{-1} \end{align*} and \begin{align*} \frac{1}{t^2} \leq \frac{1+4R}{16R^2}(1+t)^{-1}. \end{align*} Substituting these two bounds into the previous estimate gives \begin{align*} |u(t,x)| \leq C_1(1+t)^{-1}, \end{align*} where one may take \begin{align*} C_1 := \frac{1+4R}{32}\|f\|_{L^\infty(\mathbb{R}^3)} + \frac{R^2}{2}\left(1+\frac{1}{4R}\right)\left(\|\nabla f\|_{L^\infty(\mathbb{R}^3)} + \|g\|_{L^\infty(\mathbb{R}^3)}\right). \end{align*} [guided] For $t\geq 4R$, the first step reduces $|u(t,x)|$ to the patch area times the three coefficients in Kirchhoff's formula, and the previous step gives $\mathcal{H}^2(E_{t,x})\leq 2\pi R^2$. Substitution gives \begin{align*} |u(t,x)|\leq \frac{R^2}{2t^2}\|f\|_{L^\infty(\mathbb{R}^3)}+\frac{R^2}{2t}\|\nabla f\|_{L^\infty(\mathbb{R}^3)}+\frac{R^2}{2t}\|g\|_{L^\infty(\mathbb{R}^3)}. \end{align*} If $R=0$, then $B(0,R)=\varnothing$, so the support condition forces $f=0$ and $g=0$, and the estimate holds with $C_1=0$. Now assume $R>0$. Since $t\mapsto (1+t)/t$ and $t\mapsto (1+t)/t^2$ are decreasing on $(0,\infty)$, evaluating them at $t=4R$ gives \begin{align*} \frac{1}{t}\leq \left(1+\frac{1}{4R}\right)(1+t)^{-1} \end{align*} and \begin{align*} \frac{1}{t^2}\leq \frac{1+4R}{16R^2}(1+t)^{-1}. \end{align*} Applying these two one-variable estimates to the preceding bound yields \begin{align*} |u(t,x)|\leq C_1(1+t)^{-1}, \end{align*} where \begin{align*} C_1:=\frac{1+4R}{32}\|f\|_{L^\infty(\mathbb{R}^3)}+\frac{R^2}{2}\left(1+\frac{1}{4R}\right)\left(\|\nabla f\|_{L^\infty(\mathbb{R}^3)}+\|g\|_{L^\infty(\mathbb{R}^3)}\right). \end{align*} [/guided] [/step] [step:Absorb the bounded positive-time regime] First consider $t=0$. Since $u(0,x)=f(x)$ for every $x \in \mathbb{R}^3$, \begin{align*} |u(0,x)| \leq \|f\|_{L^\infty(\mathbb{R}^3)}. \end{align*} Now let $0<t\leq 4R$. In the Kirchhoff estimate from the first step, the contributing set $E_{t,x}$ is contained in the whole sphere $\partial B(x,t)$, whose two-dimensional Hausdorff measure is $4\pi t^2$. Therefore \begin{align*} \mathcal{H}^2(E_{t,x}) \leq 4\pi t^2. \end{align*} Substituting this bound into the first-step estimate gives \begin{align*} |u(t,x)| \leq \|f\|_{L^\infty(\mathbb{R}^3)} + t\|\nabla f\|_{L^\infty(\mathbb{R}^3)} + t\|g\|_{L^\infty(\mathbb{R}^3)}. \end{align*} Since $0<t\leq 4R$, we have \begin{align*} |u(t,x)| \leq \|f\|_{L^\infty(\mathbb{R}^3)} + 4R\|\nabla f\|_{L^\infty(\mathbb{R}^3)} + 4R\|g\|_{L^\infty(\mathbb{R}^3)}. \end{align*} Define \begin{align*} C_2 := (1+4R)\left(\|f\|_{L^\infty(\mathbb{R}^3)} + 4R\|\nabla f\|_{L^\infty(\mathbb{R}^3)} + 4R\|g\|_{L^\infty(\mathbb{R}^3)}\right). \end{align*} Because $(1+t)^{-1}\geq (1+4R)^{-1}$ for $0\leq t\leq 4R$, the preceding estimates imply \begin{align*} |u(t,x)| \leq C_2(1+t)^{-1} \end{align*} for all $0\leq t\leq 4R$ and all $x \in \mathbb{R}^3$. [guided] The large-time argument used transversality of the sphere with respect to a fixed projection direction. For bounded times, no geometry is needed: we can bound the whole sphere. At $t=0$, the initial condition gives $u(0,x)=f(x)$, so \begin{align*} |u(0,x)|\leq \|f\|_{L^\infty(\mathbb{R}^3)}. \end{align*} For $0<t\leq 4R$, the set $E_{t,x}$ is contained in $\partial B(x,t)$, and the surface measure of this sphere is $4\pi t^2$. Therefore \begin{align*} \mathcal{H}^2(E_{t,x})\leq 4\pi t^2. \end{align*} Substituting this into the Kirchhoff estimate from the first step gives \begin{align*} |u(t,x)|\leq \|f\|_{L^\infty(\mathbb{R}^3)}+t\|\nabla f\|_{L^\infty(\mathbb{R}^3)}+t\|g\|_{L^\infty(\mathbb{R}^3)}. \end{align*} Since $0<t\leq 4R$, this implies \begin{align*} |u(t,x)|\leq \|f\|_{L^\infty(\mathbb{R}^3)}+4R\|\nabla f\|_{L^\infty(\mathbb{R}^3)}+4R\|g\|_{L^\infty(\mathbb{R}^3)}. \end{align*} Define \begin{align*} C_2:=(1+4R)\left(\|f\|_{L^\infty(\mathbb{R}^3)}+4R\|\nabla f\|_{L^\infty(\mathbb{R}^3)}+4R\|g\|_{L^\infty(\mathbb{R}^3)}\right). \end{align*} Because $(1+t)^{-1}\geq (1+4R)^{-1}$ for $0\leq t\leq 4R$, this uniform bound becomes \begin{align*} |u(t,x)|\leq C_2(1+t)^{-1}. \end{align*} [/guided] [/step] [step:Use time reversal to handle negative times] Let $v: [0,\infty)\times \mathbb{R}^3 \to \mathbb{R}$ be the map defined by \begin{align*} v(s,x) := u(-s,x). \end{align*} Then $v$ solves the same free wave equation for $s\geq 0$ with initial data \begin{align*} v(0,x)=f(x), \qquad \partial_s v(0,x)=-g(x). \end{align*} The support assumptions are unchanged, and \begin{align*} \|-g\|_{L^\infty(\mathbb{R}^3)}=\|g\|_{L^\infty(\mathbb{R}^3)}. \end{align*} Applying the positive-time estimate to $v$ gives \begin{align*} |u(t,x)| = |v(-t,x)| \leq C(1+|t|)^{-1} \end{align*} for every $t<0$ and every $x \in \mathbb{R}^3$. Combining this with the estimates for $t\geq 0$ proves the theorem. [guided] It remains to transfer the positive-time estimate to negative time. Define the map $v:[0,\infty)\times\mathbb{R}^3\to\mathbb{R}$ by \begin{align*} v(s,x):=u(-s,x). \end{align*} Differentiating in $s$ gives $\partial_s v(s,x)=-\partial_t u(-s,x)$ and $\partial_s^2 v(s,x)=\partial_t^2 u(-s,x)$. Since $u$ solves the free wave equation, $v$ solves the same free wave equation for $s\geq 0$. Its initial data are \begin{align*} v(0,x)=f(x), \qquad \partial_s v(0,x)=-g(x). \end{align*} The support condition is unchanged, and the relevant velocity sup norm is unchanged because \begin{align*} \|-g\|_{L^\infty(\mathbb{R}^3)}=\|g\|_{L^\infty(\mathbb{R}^3)}. \end{align*} Applying the positive-time estimate already proved to $v$ gives, for $t<0$ with $s=-t>0$, \begin{align*} |u(t,x)|=|v(-t,x)|\leq C(1+|t|)^{-1}. \end{align*} Together with the $t\geq 0$ estimate, this proves the bound for all $t\in\mathbb{R}$ and $x\in\mathbb{R}^3$. [/guided] [/step]