Theorem: Let $u\in C^2(\mathbb{R}\times[0,\infty))$ be a solution of the problem $$u_{tt}-u_{xx}=0\text{ in }\mathbb{R}\times[0,\infty).$$ Given $x_0\in\mathbb{R}$ and $t_0>0$ consider the cone $$C_{x_0,t_0}=\{(x,t)\in\mathbb{R}\times[0,t_0];\;0\leq t\leq t_0,\,|x-x_0|\leq t_0-t\}.$$ If $u(x,0)=u_t(x,0)=0$ for all $x\in(x_0-t_0,x_0 t_0)$, then $u=0$ in $C$.
Problem: let $u\in C^2(\mathbb{R}\times[0,\infty))$ be a solution of the problem $$\left\{\begin{matrix} u_{tt}-u_{xx}=0 &\text{in}&\mathbb{R}\times(0,\infty) \\ u=g,\,u_t=h &\text{on}&\mathbb{R}\times\{0\} \end{matrix}\right.$$ where $h,g$ have compact support in $\mathbb{R}$. Prove that, for each $t>0$, the function $u(\cdot,t)$ has compact support in $\mathbb{R}$.
Hint: theorem above.
My (not successful) approach: it's enough to prove that there exists a bounded set $K\subset\mathbb{R}$ such that $u(x,t)=0$ for all $(x,t)\in(\mathbb{R}\backslash K)\times(0,\infty)$. We know that there is a bounded set $K$ such that $$\text{supp}(g)\cup\text{supp}(h)\subset K.$$
Given $(y,s)\in(\mathbb{R}\backslash K)\times(0,\infty)$, there are two cases: $s<d(K,y)$ or $s\geq d(K,y)$. In the first case we can consider the cone $C_{y,s}$ and conclude (by theorem above) that $u(y,s)=0$ (because $u(y,0)=g(y)=0=h(y)=u_t(y,0)$ for all $y\in\mathbb{R}\backslash K\supset(y-s,y+s)$). This argument doesn't work for the second case. So, I need help.
Thanks.
The direction which you are heading towards in your approach is false. Your approach would prove:
That statement is false for the wave equation.
In general, you need to be careful with the order of quantifiers!
The statement you should be proving is that
And in this light the answer should be much more easily found. Hint: let $K_0 = \overline{ \supp g \cup \supp h}$. You can define $K_t$ as a set of points "sufficiently far" (depending on $t$) from $K_0$.