${\varepsilon}B={\varepsilon}^{1-p}A$ is less than $1/16$; on distinguishing the cases ${\varepsilon}\le 1/16$, ${\varepsilon}>1/16$ we get the inequality in .

\[l:cacc2\] Let ${u}\in L^{\infty}(0,T;\ell^{q}(V))$, for a given $q> 1$, be a solution of in $V\times(0,T)$. Then for all $T>\tau_{1}>\tau_{2}>0$, $h\ge0$, we have for all $0<t<T$ $$\begin{gathered}
    \label{eq:cacc2_n}
    \sup_{\tau_{1}<\tau<t}
    \sum_{x\in V}
    {{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{q}
    {d_{{\omega}}}(x)
    +
    \int_{\tau_{1}}^{t}
    \sum_{x,y\in V}
    {\left|{D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{\frac{p+q-2}{p}}\right|}^{p}
    {\omega}(x,y)
    {\,\textup{\textmd{d}}}\tau
    \\
    \le
    \frac{\gamma}{\tau_{1}-\tau_{2}}
    \int_{\tau_{2}}^{t}
    \sum_{x\in V}
    {{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{q}
    {d_{{\omega}}}(x)
    {\,\textup{\textmd{d}}}\tau
    \,.
  \end{gathered}$$ We have also, if condition is satisfied, $$\begin{gathered}
    \label{eq:cacc2_nn}
    \sup_{0<\tau<t}
    \sum_{x\in V}
    {{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{q}
    {d_{{\omega}}}(x)
    +
    \int_{0}^{t}
    \sum_{x,y\in V}
    {\left|{D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{\frac{p+q-2}{p}}\right|}^{p}
    {\omega}(x,y)
    {\,\textup{\textmd{d}}}\tau
    \\
    \le
    \gamma
    \int_{0}^{t}
    \sum_{x\in V}
    {{\ifthenelse{\equal{*}{*}}{({u}_{0}(x)-h)_{+}}{\left({u}_{0}(x)-h\right)_{+}}}}^{q}
    {d_{{\omega}}}(x)
    {\,\textup{\textmd{d}}}\tau
    \,.
  \end{gathered}$$

Let us prove ; the inequality is proved similarly.

We multiply against $\zeta(x)\eta(t){{\ifthenelse{\equal{*}{*}}{({u}(x,t)-h)_{+}}{\left({u}(x,t)-h\right)_{+}}}}^{q-1}$; on integrating by parts as in the proof of Lemma \[l:cacc\] we obtain $$\label{eq:cacc2_i}
    \begin{split}
      &\frac{1}{q}
      \sum_{x\in V}
      \zeta(x)
      {{\ifthenelse{\equal{*}{*}}{({u}(x,t)-h)_{+}}{\left({u}(x,t)-h\right)_{+}}}}^{q}
      {d_{{\omega}}}(x)
      \eta(t)
      \\
      &\quad+
      \int_{0}^{t}
      \sum_{x,y\in V}
      {\lvert{D}_{y}{u}(x,\tau)\rvert}^{p-2}
      {D}_{y}{u}(x,\tau)
      \zeta(y)
      {D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{q-1}
      {\omega}(x,y)
      \eta(\tau)
      {\,\textup{\textmd{d}}}\tau
      \\
      &\quad+
      \int_{0}^{t}
      \sum_{x,y\in V}
      {\lvert{D}_{y}{u}(x,\tau)\rvert}^{p-2}
      {D}_{y}{u}(x,\tau)
      {D}_{y}\zeta(x)
      {{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{q-1}
      {\omega}(x,y)
      \eta(\tau)
      {\,\textup{\textmd{d}}}\tau
      \\
      &\qquad=
      \frac{1}{q}
      \int_{0}^{t}
      \sum_{x\in V}
      \zeta(x)
      {{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{q}
      {d_{{\omega}}}(x)
      \eta'(\tau)
      {\,\textup{\textmd{d}}}\tau
      \,.
    \end{split}$$ We estimate next the second integral in . The absolute value of the integrand is bounded from above by $$\begin{gathered}
    \frac{1}{R_{2}-R_{1}}
    \sum_{x,y\in B_{R_{2}+1}}
    {\lvert{D}_{y}{u}(x,\tau)\rvert}^{p-1}
    {{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{q-1}
    {\omega}(x,y)
    \\
    \le
    