Smooth weak equivalence and moving jump residual

By [A2] and [F2], $P$ and $P^{-1}$ preserve the Lebesgue sigma-algebra and measure. Pullback by $P$ bijects nonnegative simple minorants and preserves their level-set measures, so the nonnegative integral is preserved by its defining supremum; applying positive and negative parts gives the same conclusion for the integrable tested scalar products in [F1]. Thus the time-first coordinates and the spatial-first weak definition give the same test integrals. For a test $\varphi$ supported in $\mathcal{O}$, apply the product rule to the space-time current $(U^{\pm },F_1^{\pm },F_2^{\pm },F_3^{\pm })$ separately on ${\mathcal{O}}^{-}$ and ${\mathcal{O}}^{+}$. Coordinatewise integration by parts in $t,x_1,x_2,x_3$ gives ${\int }_{\mathcal{O}}(U\cdot {\mathrm{\partial }}_t\varphi +F:{\mathrm{\nabla }}_x\varphi +S\cdot \varphi )=-{\int }_{{\mathcal{O}}^{-}}{W}^{-}\cdot \varphi -{\int }_{{\mathcal{O}}^{+}}{W}^{+}\cdot \varphi -{\int }_{J\times D}{K}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}}\cdot \varphi (t,x^{\mathrm{\prime }},h(t,x^{\mathrm{\prime }}))d{x}^{\mathrm{\prime }}dt$, where $W^{\pm }={\mathrm{\partial }}_t{U}^{\pm }+{\mathrm{div}}_x{F}^{\pm }-S^{\pm }$ and $K_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}}=[F_3]-{\mathrm{\partial }}_{t}h[U]-{\mathrm{\nabla }}_{x^{\mathrm{\prime }}}h\cdot [F_{1,2}]$. Compact support removes all outer-boundary terms; the remaining graph term is the difference of the two side conormal fluxes.

The unnormalised space-time conormal from minus to plus is $(-{\mathrm{\partial }}_{t}h,-{\mathrm{\nabla }}_{x^{\mathrm{\prime }}}h,1)$. Its contraction with the jump current equals $K_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}}=[F_3]-{\mathrm{\partial }}_{t}h[U]-{\mathrm{\nabla }}_{x^{\mathrm{\prime }}}h\cdot [F_{1,2}]=\lambda ([F]n-V_n[U])$. Since $\lambda >0$, the graph coefficient vanishes exactly when $[F]n-V_n[U]=0$, and has the same sign as the scalar entropy jump coefficient.

If the weak identity holds, choose tests supported wholly inside either side in step 1.1. Each continuous component of $W^{\pm }$ must vanish: a nonzero value would have a fixed strict sign on a small ball and contradict the identity for a nonnegative $C^1$ bump there. With these bulk terms zero, step 1.1 gives zero integral of $K_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}}$ against every trace of a test. Any compactly supported $C^1$ graph test $\zeta (t,x^{\mathrm{\prime }})$ extends to the patch as $\varphi (t,x^{\mathrm{\prime }},x_3)=\zeta (t,x^{\mathrm{\prime }})\eta (x_3-h(t,x^{\mathrm{\prime }}))$, where $\eta$ is a $C^1$ cutoff with $\eta (0)=1$ and sufficiently small support. Such a uniform tube lies in $\mathcal{O}$ because the graph over $\mathrm{supp}\zeta$ is compact in this open patch. A nonzero component of continuous $K_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}}$ would have a fixed sign on a small graph patch and contradict the identity. Hence $W^{\pm }=0$ and $K_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}}=0$. Conversely, these equalities make the weak integral zero by step 1.1, and step 1.2 gives the stated jump condition.

For the scalar inequality, step 1.1 gives the weak integral as the negative of the bulk integrals of $g^{\pm }={\mathrm{\partial }}_t{H}^{\pm }+{\mathrm{div}}_x{G}^{\pm }-R^{\pm }$ and the negative graph integral with coefficient $K_H=[G_3]-{\mathrm{\partial }}_{t}h[H]-{\mathrm{\nabla }}_{x^{\mathrm{\prime }}}h\cdot [G_{1,2}]$. If both bulk residuals and $K_H$ are nonnegative, the weak integral is nonpositive for every nonnegative test. Conversely, a negative bulk residual at a point admits a nonnegative bump supported on that side and violates the weak inequality. If $K_H<0$ at a graph point, continuity gives a smaller graph patch where $K_H$ is bounded above by a negative constant. Choose a nonnegative test with fixed positive trace on that patch and support in a tube of width $\epsilon$ about the graph. The negative interface contribution to the weak integral is positive and independent of $\epsilon$, while the bounded bulk terms have magnitude $O(\epsilon )$; for small enough $\epsilon$ the weak integral is positive, a contradiction. Hence both the side inequalities and $K_H\ge 0$ are necessary. Step 1.2 shows $K_H=\lambda ([G]\cdot n-V_n[H])$, so the stated jump sign is equivalent.