 Solution

Since the derivative of $|z-k|$ is $\mathrm{sgn}(z-k)$ away from $k$, choose the Kruzhkov entropy fluxOn either side of $k$ its derivative is $f^{\prime }(z){\eta }_k^{\prime }(z)$. The factor $f(z)-f(k)$ vanishes at the change of sign, so the function is continuous and there is no extra delta contribution to its weak derivative. Adding a constant to the flux changes none of the test integrals. Substitution into the entropy inequality pairs $|u-k|$ with this flux, not with an arbitrary antiderivative unrelated to the chosen entropy.