Does d'Alembert's solution $u(x,t)=f(x-ct)+g(x+ct)$ of the wave equation remain valid when $f, g$ are not twice differentiable? If given $u(x,t)=f(x-ct)+g(x+ct)$ and $f, g$ are not twice differentiable, does the wave equation $u_{t,t} = c^2u_{x,x}$ still hold true?
If not, what's a counterexample?
 A: As cnick said, the equation still makes sense if we interpret the derivatives in a weak sense. Introduce the wave operator $\Box u = u_{tt}-c^2 u_{xx}$ and observe that for smooth compactly supported functions of $x,t$ we have $\int u(\Box v) = \int(\Box u)v$, by virtue of integration by parts. This leads to the definition: $\Box u=0$ in the weak sense if for every smooth compactly supported function $\phi$ 
$$\int u (\Box \phi) = 0 \tag1$$ 
With this definition, the equation makes sense provided that $u$ is  locally integrable. 
For two locally integrable functions $f,g$, the function $u(x,t)=f(x-ct)+g(x+ct)$ is a solution in the sense of (1). One way to prove this is to use a mollifier $\chi_\epsilon$: the functions $f_\epsilon = f*\chi_\epsilon$ and $g_\epsilon=g*\chi_\epsilon$ converge to $f,g$ locally in $L^1$, hence $$u_\epsilon(x,t):=f_\epsilon(x-ct)+g_\epsilon(x+ct)$$
converges to $u$ locally in $L^1$. Since $u_\epsilon$ satisfies the wave equation in the classical sense, it also satisfies (1). Pass to the limit $\epsilon\to 0$ in (1) to conclude that $u$ also satisfies it.
