I've also posted this question on physics SE in case it is more appropiate there.
Consider the Klein-Gordon equation: $$(\square + m^2)\phi = (\partial_t^2 - \Delta + m^2)\phi = 0 \tag{1}$$ The solution to this equation is typically obtained by writing $\phi$ as an inverse Fourier transform: $$\phi(x,t) = \frac{1}{(2\pi)^{3/2}} \int_{\mathbb{R}^3} e^{i (k \cdot x)} \hat{\phi}(t,k) d^3k$$ and then substituting this into (1) to get $$(\partial_t^2 - k^2 + m^2)\hat{\phi} = 0. \tag{2}$$
From here it is often stated that the general solution to (2) must be a distribution, but why is this?
The answer is a very physical one: All wave equations $$f(t+dt,x)-2f(t,x)+f(t-dt,x) = \text{mean}(f(t,x+\xi), |\xi| < \epsilon \ )$$ are local maps of amplitude and rapidity $$(f(t,x),\partial_t f(t,x)) \to (f(t+dt,x),\partial_t f(t+dt,x)). $$ In the easiest case of free linear, time- and space independent coefficients the map is a simple shift in space $$(f(t,x),\partial_t f(t,x)) \to (f(t,x-c dt),\partial_t f(t,x-c dt)) $$ So the translation map is working on functions f without any smoothness requirements. Since smoothness means summability of the energy of Fourier modes, the wave equation has applications far wider than states in Hilbert spaces.