Discrete Kruzhkov inequality for a monotone finite-volume scheme

If $s\ge k$, then $s\vee k=s$ and $s\wedge k=k$, so $G_k(s,s)=f(s)-f(k)=q_k(s)$. If $s<k$, then $s\vee k=k$ and $s\wedge k=s$, so $G_k(s,s)=f(k)-f(s)=q_k(s)$.

First suppose $k\in I$. Put $X=(U_{j-1}^n,U_j^n,U_{j+1}^n)$ and let $X^{+}$ and $X^{-}$ be its coordinatewise maximum and minimum with $(k,k,k)$. Both triples lie in $I^3$. By [F1], the two numbers $H(X)$ and $H(k,k,k)=k$ lie between $H(X^{-})$ and $H(X^{+})$; therefore $\mathrm{\mid }H(X)-k\mathrm{\mid }\le H(X^{+})-H(X^{-})\mathrm{.}$ Expanding the definition of $H$, the right side is $\mathrm{\mid }{U}_j^n-k\mathrm{\mid }-\lambda (G_k(U_j^n,U_{j+1}^n)-G_k(U_{j-1}^n,U_j^n)),$ which gives the claimed inequality after rearrangement.

If $k<a_{-}$, every state is above $k$ by the range assertion in [F1]. Then $\mathrm{\mid }{U}_j^n-k\mathrm{\mid }=U_j^n-k$ and $G_k(a,b)=F(a,b)-F(k,k)$ for state pairs; substituting the scheme recurrence makes the displayed left side zero. If $k>a_{+}$, every state is below $k$, so $\mathrm{\mid }{U}_j^n-k\mathrm{\mid }=k-U_j^n$ and $G_k(a,b)=F(k,k)-F(a,b)$; substitution again gives zero. The cases $k=a_{-}$ and $k=a_{+}$ are already covered by step 2.1.