**Proof plan.** We prove the contrapositive: if the initial data vanish on $B(x_0, t_0)$, then $u(x_0, t_0) = 0$. The key is to compute the energy not over all of $\mathbb{R}^n$ but over the spatial cross-section $B(x_0, t_0 - t)$ of the backward cone at time $t$. This **local energy** is non-increasing in $t$ (proved by the same [integration](/page/Integral)-by-parts argument, now with a [boundary](/page/Boundary) flux term that has the correct sign). Since it starts at zero, it remains zero. **Step 1: Define the local energy on the cone.** The backward cone with vertex $(x_0, t_0)$ and unit speed is \begin{align*} K := \{(x, t) : 0 \leq t \leq t_0, \; |x - x_0| \leq t_0 - t\}. \end{align*} For $0 \leq t \leq t_0$, the time-$t$ cross-section is $B(x_0, t_0 - t)$. Define \begin{align*} e(t) := \frac{1}{2}\int_{B(x_0, t_0 - t)} \bigl((\partial_t u)^2 + |\nabla u|^2\bigr) \, d\mathcal{L}^n. \end{align*} **Step 2: Differentiate the local energy.** The domain of integration depends on $t$, so the Leibniz rule for moving domains gives \begin{align*} e'(t) = \int_{B(x_0, t_0 - t)} \bigl(\partial_t u \cdot \partial_t^2 u + \nabla u \cdot \nabla \partial_t u\bigr) \, d\mathcal{L}^n - \frac{1}{2}\int_{\partial B(x_0, t_0 - t)} \bigl((\partial_t u)^2 + |\nabla u|^2\bigr) \, d\mathcal{H}^{n-1}, \end{align*} where the second term arises because the radius $t_0 - t$ shrinks at rate $-1$. Integrating by parts in the first integral: \begin{align*} \int_{B} \nabla u \cdot \nabla \partial_t u \, d\mathcal{L}^n = -\int_{B} \Delta u \cdot \partial_t u \, d\mathcal{L}^n + \int_{\partial B} \partial_t u \, \frac{\partial u}{\partial \nu} \, d\mathcal{H}^{n-1}, \end{align*} where $\nu$ is the outward unit normal to $\partial B(x_0, t_0 - t)$. Using $\partial_t^2 u = \Delta u$: \begin{align*} e'(t) = \int_{\partial B(x_0, t_0 - t)} \partial_t u \, \frac{\partial u}{\partial \nu} \, d\mathcal{H}^{n-1} - \frac{1}{2}\int_{\partial B(x_0, t_0 - t)} \bigl((\partial_t u)^2 + |\nabla u|^2\bigr) \, d\mathcal{H}^{n-1}. \end{align*} **Step 3: Estimate the boundary flux.** By the Cauchy–Schwarz inequality applied pointwise: \begin{align*} \partial_t u \, \frac{\partial u}{\partial \nu} \leq |\partial_t u| \, |\nabla u| \leq \frac{1}{2}\bigl((\partial_t u)^2 + |\nabla u|^2\bigr), \end{align*} since $|\partial u/\partial \nu| \leq |\nabla u|$ and $ab \leq (a^2 + b^2)/2$. Therefore: \begin{align*} e'(t) \leq \frac{1}{2}\int_{\partial B} \bigl((\partial_t u)^2 + |\nabla u|^2\bigr) \, d\mathcal{H}^{n-1} - \frac{1}{2}\int_{\partial B} \bigl((\partial_t u)^2 + |\nabla u|^2\bigr) \, d\mathcal{H}^{n-1} = 0. \end{align*} **Step 4: Conclude.** We have shown $e'(t) \leq 0$, so $e$ is non-increasing. The hypothesis states that $g = 0$ and $h = 0$ on $B(x_0, t_0)$. At $t = 0$, the cross-section is $B(x_0, t_0)$, and \begin{align*} e(0) = \frac{1}{2}\int_{B(x_0, t_0)} \bigl(h^2 + |\nabla g|^2\bigr) \, d\mathcal{L}^n = 0. \end{align*} Therefore $e(t) = 0$ for all $t \in [0, t_0]$, which forces $\partial_t u = 0$ and $\nabla u = 0$ inside the cone $K$. Combined with $u(\cdot, 0) = g = 0$ on $B(x_0, t_0)$, we conclude $u \equiv 0$ on $K$. In particular, $u(x_0, t_0) = 0$.