Show that if $f \in C^3_c (\mathbb{R})$ is an odd function then for $|x|=r$ define \begin{equation} u(x,t) = \frac{f(r+t)+f(r-t)}{2r} \end{equation} then $u$ extends as a $C^2$ function that solves the wave equation on $\mathbb{R}^3 \times (-T,T)$.
It is only the part about it extending as a $C^2$ function I cannot come up with a satisfactory rigorous proof for.
It is easy to show continuity at $r=0$ since simply taking the limit at $r \rightarrow 0$ and using $u$ odd we have the definition of the first derivative of $f$ at $t$. I assume the first and second derivatives at $r=0$ follow similarly, but I can show this. I have tried looking at the limits of the partials as $r \rightarrow 0$ but I can't see that they clearly exist.
Use a bunch of Taylor expansions, starting with $$\begin{split} f(t+r)&=f(t)+rf'(t)+\frac{r^2}{2}f''(t)+\frac{r^3}{6}f'''(t)+o(r^3) \\ f'(t+r)&=f'(t)+rf''(t)+\frac{r^2}{2}f'''(t)+o(r^2) \\ f''(t+r)&=f''(t)+rf'''(t)+o(r) \end{split} $$ From here, $$\begin{split} f(t+r)-f(t-r) &= 2rf'(t)+\frac{r^3}{3}f'''(t)+o(r^3) \\ f'(t+r)-f'(t-r) &= 2rf''(t)+o(r^2) \\ f''(t+r)-f''(t-r) &= 2rf'''(t)+o(r) \end{split} $$
As $r\to 0$, $$\begin{split} u &= \frac{f(t+r)-f(t-r)}{2r} = f'(t)+\frac{r^2}{6}f'''(t)+o(r^2)\\ u_t& = \frac{f'(t+r)-f'(t-r)}{2r} = f''(t)+o(r) \\ u_{tt} &= \frac{f''(t+r)-f''(t-r)}{2r} = f'''(t)+o(1) \end{split}$$ so all of these have limits consistent with $u(0,t)=f'(t)$.
Next, derivatives in $r$. Note that by symmetry it suffices to consider $x=re_1$, which eliminates the distinction between $x$ and $r$. We have $$u_r = \frac{f'(t+r)+f'(t-r)}{2r} - \frac{f(t+r)-f(t-r)}{2r^2} \\= \frac{f'(t+r)-2f'(t)+f'(t-r)}{2r} - \frac{f(t+r)-f(t-r)-2rf'(t)}{2r^2} =O(r)$$ Also, $$u_{rr} = \frac{f''(t+r)-f''(t-r)}{2r} - \frac{f'(t+r)+f'(t-r)}{r^2} + \frac{f(t+r)-f(t-r)}{r^3} \\ = f'''(t)+o(1) - \frac{f'(t+r)-2f'(t)+f'(t-r)}{r^2} + \frac{f(t+r)-f(t-r) - 2rf'(t)}{r^3} \\ = \frac13 f'''(t) +o(1) $$ so these derivatives have limits too, consistent with the expansion $f'(t)+\frac{r^2}{6}f'''(t)+o(r^2)$ above.