Since I haven't been able to track down Selberg's lecture notes since he moved to Bergen, and since the proof of the result I mentioned in this comment is super-short anyway, let me just include a proof here.

**Theorem** Let $\psi\in \mathscr{D}'(\mathbb{R}^{n+1})$ be a distributional solution of $\Box \psi = 0$. Suppose further that $\mathrm{supp}(\psi) \cap \{ t \leq 0\} = \emptyset$. Then $\psi \equiv 0$.

*Proof*: It suffices to show that for every $f\in C^\infty_0(\mathbb{R}^{n+1})$ that $\langle \psi, f\rangle = 0$. Let $T$ be sufficiently large such that $f \equiv 0$ for $t \geq T$. Solve the Cauchy problem $\Box u = f$ with "initial" data $u(T,x) = \partial_t u(T,x) = 0$; this can be done using, e.g. the representation formula. Notice that the representation formula gives $u \equiv 0$ for $t \geq T$. Additionally, the representation formula (finite speed of propagation) shows that for any $S$, on $\{t \geq S\}$, there exists some $R$ such that $u(t,x) \equiv 0$ when $x \geq R$. Let $S < 0$ and choose a smooth cut-off $\chi$ such that $\chi(t) \equiv 1 $ when $t \geq 0$ and $\chi(t) \equiv 0$ when $t \leq S$.

The function $\chi(t) u(t,x) \in C^\infty_0(\mathbb{R}^{n+1})$, and hence

$$ 0 = \langle \Box \psi, \chi u\rangle = \langle \psi, \Box(\chi u) \rangle = \langle \psi, f \rangle + \langle \psi, - u \chi'' - 2 \chi' \partial_t u \rangle.$$

Both $\chi'$ and $\chi''$ are supported on $[S,0]$. So $-u \chi'' - 2 \chi' \partial_t u$ is $C^\infty_0(\mathbb{R}^{1+n})$ with support in $\{ t \leq 0\}$, and hence its pairing against $\psi$ vanishes by assumption.

The same argument fails for the heat equation due to the failure of finite speed of propagation. Solutions of $\partial_t u - \Delta u = f$ when $f$ has compact support may have spatial tails of size $\exp(-|x|^2)$. You can compensate this by not allowing arbitrary distributions but only distributions "growing no faster than $\exp(|x|^2)$" (interpreted suitably), and get a version of Tychonoff's uniqueness theorem.

notfollow from 1 if the wave equation only applies 'outside' the region, i.e. $t>0$ - as one would assume. $\endgroup$