Minty identification for a monotone Carathéodory flux

Fix arbitrary $\eta \in L^p(Q;{\mathbb{R}}^d)$. By [F1]–[F3], monotonicity gives the finite nonnegative integrals $0\le {\int }_Q(q_n-a(x,\eta ))\cdot ({\xi }_n-\eta )\mathrm{.}$ Expanding them and using the two weak convergences for the three terms linear in a weakly convergent factor, the limsup hypothesis yields $0\le {\int }_Q(q-a(x,\eta ))\cdot (\xi -\eta )\mathrm{.}$ Indeed the expanded nonnegative sequence has limsup at most the displayed right-hand side, so that right-hand side cannot be negative.

Let $\zeta \in L^p(Q;{\mathbb{R}}^d)$ and $0<\epsilon \le 1$. Substitute $\eta =\xi +\epsilon \zeta$ and $\eta =\xi -\epsilon \zeta$ into step 1.1. After dividing by $\epsilon$, the two inequalities give ${\int }_Q(q-a(x,\xi +\epsilon \zeta ))\cdot \zeta \le 0,{\int }_Q(q-a(x,\xi -\epsilon \zeta ))\cdot \zeta \ge 0.$ The fixed term $q\cdot \zeta$ is integrable by [F3]. For either sign, the flux term satisfies $\mathrm{\mid }a(x,\xi \pm \epsilon \zeta )\mathrm{\mid }\mathrm{\mid }\zeta \mathrm{\mid }\le C(g(x)+\mathrm{\mid }\xi {\mathrm{\mid }}^{p-1}+\mathrm{\mid }\zeta {\mathrm{\mid }}^{p-1})\mathrm{\mid }\zeta \mathrm{\mid },$ which is integrable by [F3]; here $g(x)$ is viewed as constant in time and $Q$ has finite time length. Carathéodory continuity gives pointwise $a(x,\xi \pm \epsilon \zeta )\to a(x,\xi )$ as $\epsilon \downarrow 0$. Dominated convergence [F4] on the flux term in both inequalities proves ${\int }_Q(q-a(x,\xi ))\cdot \zeta =0$ for every $\zeta \in L^p(Q;{\mathbb{R}}^d)$.

Put $w=q-a(x,\xi )\in L^{p^{\mathrm{\prime }}}(Q;{\mathbb{R}}^d)$ and choose $\zeta =\mathrm{\mid }w{\mathrm{\mid }}^{p^{\mathrm{\prime }}-2}w$, defined as zero where $w=0$. Since $(p^{\mathrm{\prime }}-1)p=p^{\mathrm{\prime }}$, this $\zeta$ belongs to $L^p$, and step 2.1 gives $0={\int }_{Q}w\cdot \zeta ={\int }_Q\mathrm{\mid }w{\mathrm{\mid }}^{p^{\mathrm{\prime }}}$. Thus $w=0$ almost everywhere, which is the claimed flux identification.