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Assume $u$ is a solution of the wave equation in $\mathbb R^n \times [0,\infty)$ with $x \mapsto u(x,0)$ smooth with compact support. Why then must the support of $u(\cdot,t)$ be compact for all $t \geq 0$ ? Does this follow from the following theorem?

Domain of dependence: Assume suppose $u \in \mathcal C^2(\overline \Omega \times[0,T])$ solves the wave equation in $\Omega \times (0,T)$ and $u=g$ on $\Omega \times \{0\} \cup \partial \Omega \times[0,T)$ and $u_t(\cdot,0)=h$ on $\Omega$. Then if for some point $x_0 \in \Omega$ and $t_0 >0$ and $B_{t_0}(x_0) \subset \Omega$ we have

$u = u_t = 0$ on $B_{t_0}(x_0)\times\{0\}$, then we already have

$u=0$ on the cone $K = \{(x,t):0 \leq t \leq t_0, |x-x_0|\leq t_0-t\}$

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    I think it follows from the fact that the wave equation has a finite velocity of propagation. So I would say you just need D'Alembert's solution formula. The theorem you propose seems to refer to bounded domains rather than to the problem on the whole space. – 2017-02-24
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    But I want to consider also the case of more than one space dimension. – 2017-02-24
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    Sorry you are definitely right. In many dimensions what you wrote is exactly the statement for the finite speed of propagation. So yes: having compact support at any time follows directly from this theorem. In fact if $u(0)$ is supported in $[-R, R]$ we get $u(t)$ supported in $[-R -t, R+t]$ – 2017-02-24

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