term $J_{231}$ into $J_{22}$, for a suitable choice of ${\varepsilon}$. To this end we observe that by a change of variables we have $$\begin{split}
      J_{22}
      &=
      \frac{1}{4}
      \int_{0}^{t}
      \sum_{x,y\in V}
      {\lvert{D}_{x}{u}(y,\tau)\rvert}^{p-1}
      {\lvert{D}_{x}{{\ifthenelse{\equal{*}{*}}{({u}(y,\tau)-h)_{+}}{\left({u}(y,\tau)-h\right)_{+}}}}^{\theta}\rvert}
      \zeta(x)^{p}
      {\omega}(y,x)
      \eta(\tau)^{p}
      {\,\textup{\textmd{d}}}\tau
      \\
      &=
      \frac{1}{4}
      \int_{0}^{t}
      \sum_{x,y\in V}
      {\lvert{D}_{y}{u}(x,\tau)\rvert}^{p-1}
      {\lvert{D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{\theta}\rvert}
      \zeta(x)^{p}
      {\omega}(x,y)
      \eta(\tau)^{p}
      {\,\textup{\textmd{d}}}\tau
      \\
      &=
      \frac{1}{8}
      \int_{0}^{t}
      \sum_{x,y\in V}
      {\lvert{D}_{y}{u}(x,\tau)\rvert}^{p-1}
      {\lvert{D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{\theta}\rvert}
      (\zeta(x)^{p}+\zeta(y)^{p})
      {\omega}(x,y)
      \eta(\tau)^{p}
      {\,\textup{\textmd{d}}}\tau
      \,.
    \end{split}$$ Then by elementary calculus $$\begin{gathered}
    \label{eq:cacc_j}
    \chi_{x,y}
    {\lvert{D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}^{\theta}\rvert}
    \ge
    \chi_{x,y}
    \theta
    {\lvert{D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}\rvert}
    H(x,y;h)^{\theta-1}
    \\
    \ge
    \chi_{x,y}
    \theta
    {\lvert{D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}\rvert}
    H(x,y;k)^{\theta-1}
    \,.
  \end{gathered}$$ Next we discriminate three cases in , aggregating equivalent symmetric cases: i) ${u}(x,\tau)>h$, ${u}(y,\tau)>h$. In this case clearly $${\lvert{D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}\rvert}
    =
    {\lvert{D}_{y}{u}(x,\tau)\rvert}
    \,.$$ ii) ${u}(x,\tau)>2h$, $h\ge {u}(y,\tau)$. Then $${\lvert{D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x,\tau)-h)_{+}}{\left({u}(x,\tau)-h\right)_{+}}}}\rvert}
    \ge
    \frac{{u}(x,\tau)}{2}
    \ge
    \frac{1}{2}
    {\lvert{D}_{y}{u}(x,\tau)\rvert}
    \,.$$ iii) $2h \ge {u}(x,\tau)>h\ge {u}(y,\tau)$. In this case $J_{22}$ does not offer any help. We rather bound directly this part of $J_{231}$ as shown below.

Collecting the estimates above we see that, provided ${\varepsilon}\le 1/16$, $$\begin{gathered}
    J_{231}
    \le
    J_{22}
    +
    {\varepsilon}2^{p+2}
    h^{p}
    (h-k)^{\theta-1}
    \int_{\tau_{2}}^{t}
    {\mu_{{\omega}}}(B_{R_{2}}\cap\{2h\ge {u}(x,\tau)>h\})
    {\,\textup{\textmd{d}}}\tau
    \,.
  \end{gathered}$$ Hence we have transformed into $$\label{eq:cacc_jj}
    J_{1}+J_{21}
    \le
    J_{3}
    +
    \gamma {\varepsilon}B
    +
    \gamma {\varepsilon}^{1-p}
    A
    \,,$$ where $A$ and $B$ are as in the statement.

Finally we check whether the root ${\varepsilon}$ of 