In this paper here it is claimed in (1.3) that it is classical and immediate from the explicit solution of the wave equation in 3D
$$(\partial_t^2 -\Delta )u(t,x)=0$$ with $u(0,x)=0$ and $u_t(0,x)=g(x)$, that the solution satisfies for some $c>0$
$$\sup_x \vert u(t,x)\vert \le \frac{c}{t} \int \vert Dg(y) \vert dy.$$
Does anybody see how this estimate possibly follows?
The key is to estimate the integral (where $g\in C^\infty_c(\mathbb{R}^3)$ by assumption) $$\tag{A} \iint_{|x - y| = t} g(y)~ dS_y \lesssim \iiint_{|x-y| \geq t} |D g|~dy. $$ Once this estimate is found, the desired result follows from the representation formula (1.2) in the paper. This is an instance of a trace theorem, except in this case the argument is elementary.
Let $v^i = \frac{y^i - x^i}{|y-x|^3} \propto D^i \frac{1}{|y-x|}$, we see that outside of $|x-y| = t$ this vector field is divergence free (recall that the fundamental solution of the Laplacian in 3 dimensions is constant times $1/|x|$).
And so we have that by the Gauss-Green theorem (using $\nu$ for the appropriate normal vector field to the sphere; the RHS may be missing a minus sign which is inconsequential for the estimate) (recall that $g$ has finite support so there's no issue with boundary at infinity)
$$ \iint_{|x-y| = t} g(y) ~ dS_y = \iint_{|x-y| = t} t^2 (v\cdot \nu) g(y) ~dS_y = \iiint_{|x-y| \geq t} \mathrm{div}( t^2 v g) ~dy $$
which we evaluate to be
$$ = \iiint_{|x-y| \geq t} t^2 v \cdot Dg ~dy $$
Observe that as on our domain $|y-x| \geq t$, the vector field $t^2 v$ has norm $\leq 1$ exterior to the ball, so we have
$$ \leq \iiint_{|x-y| \geq t} |Dg| ~dy $$
