= Solution
Use $\widehat f(\xi)=\int e^{-ix\xi}f(x)dx$. For a <Schwartz function>, <Fourier inversion> is $f(x)=(2\pi)^{-1}\int e^{ix\xi}\widehat f(\xi)d\xi$. For a <tempered distribution> define $\langle\widehat T,\phi\rangle=\langle T,\widehat\phi\rangle$ for <Schwartz functions>; this agrees with the function transform in the stated convention. The half-indicator kernel has transform
$$
\widehat F_t(\xi)=\frac12\int_{-t}^t e^{-ix\xi}dx
=\frac{\sin(t\xi)}\xi,
$$
with value $t$ at zero. Transforming the <wave equation> gives $\widehat u_{tt}+\xi^2\widehat u=0$, zero initial value and initial derivative $\widehat g$. Hence $\widehat u=\widehat g\sin(t\xi)/\xi$, and convolution yields
$$
\boxed{u(t,x)=\frac12\int_{x-t}^{x+t}g(s)ds}.
$$
<Finite propagation speed> means that the solution at $(t,x)$ depends only on initial data in this interval, so disturbances cannot travel faster than one. The formula also makes sense for any smooth $g$, regardless of its growth at infinity, because the integration interval is compact. Direct differentiation verifies the equation and both initial conditions. Equivalently, locally replace $g$ by a compactly supported smooth function agreeing throughout the relevant domain of dependence.
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