= Solution
Write $M=\|u_0\|_{L^1}$. Since $|U(0,y)|\le M$, choosing $y=x-ct$ gives $g((x-y)/t)=g(c)=0$ and $\max_yG_{t,x}(y)\ge-M$. At the maximizing foot $x_0$, the <inverse-flux quadratic bounds> give
$$
\max_yG_{t,x}(y)=U(0,x_0)+t g\left(\frac{x-x_0}{t}\right)\le M+\frac{k_+}{t}(x-x_0-ct)^2.
$$
Combining the two estimates and dividing by $-k_+>0$ gives
$$
\boxed{\left|\frac{x-x_0}{t}-c\right|\le\sqrt{\frac{2M}{-k_+t}},\qquad |u(t,x)|\le\frac{-2k_-}{\sqrt t}\sqrt{\frac{2M}{-k_+}}.}
$$
The last step uses $u=h((x-x_0)/t)$ and the sign-independent bound on $h$. This is <square-root decay before characteristic crossing>; it is proved only for $0<t<t^*$. No continuation past <characteristic crossing> or global-time smoothness is assumed. If $M=0$, the initial function and the solution vanish.
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