{\lvert{D}_{y}f_{i}(x,\tau)\rvert}^{p}
      {\omega}(x,y)
      \,.
  \end{aligned}$$ Since $b:=q/\nu<p$, it follows from Faber-Krahn inequality and Hölder’s and Young’s inequalities that $$\begin{gathered}
    \label{eq:fkb}
    \sum_{x\in V}
    f_{i+1}(x,\tau)^{b}
    {d_{{\omega}}}(x)
    \le
    {m}_{i+1}(\tau)^{1-\frac{b}{p}}
    {\Lambda_p}({m}_{i+1}(\tau))^{-\frac{b}{p}}
    {{D}}_{i+1}(\tau)^{\frac{b}{p}}
    \\
    \le
    {\varepsilon}^{\frac{p}{b}}
    {{D}}_{i+1}(\tau)
    +
    {\varepsilon}^{-\frac{p}{p-b}}
    {\Lambda_p}({m}_{i+1}(\tau))^{-\frac{b}{p-b}}
    {m}_{i+1}(\tau)
    \,.
  \end{gathered}$$ Here ${\varepsilon}>0$ is arbitrary and will be selected below.

We integrate over $(t_{i+1},t)$ to find $$\label{eq:linf_i}
    \begin{split}
      &\int_{t_{i+1}}^{t}
      \sum_{x\in V}
      f_{i+1}(x,\tau)^{b}
      {d_{{\omega}}}(x)
      {\,\textup{\textmd{d}}}\tau
      \le
      {\varepsilon}^{\frac{p}{b}}
      \int_{t_{i+1}}^{t}
      {{D}}_{i+1}(\tau)
      {\,\textup{\textmd{d}}}\tau
      \\
      &\qquad+
      {\varepsilon}^{-\frac{p}{p-b}}
      \int_{t_{i+1}}^{t}
      {\Lambda_p}({m}_{i+1}(\tau))^{-\frac{b}{p-b}}
      {m}_{i+1}(\tau)
      {\,\textup{\textmd{d}}}\tau
      \\
      &\quad\le
      {\varepsilon}^{\frac{p}{b}}
      \int_{t_{i+1}}^{t}
      {{D}}_{i+1}(\tau)
      {\,\textup{\textmd{d}}}\tau
      +
      {\varepsilon}^{-\frac{p}{p-b}}
      t
      {\Lambda_p}({M}_{i+1})^{-\frac{b}{p-b}}
      {M}_{i+1}
      \,.
    \end{split}$$ Next we invoke Lemma \[l:cacc2\] with $\tau_{1}=t_{i}$, $\tau_{2}=t_{i+1}$, $h=k_{i}$, to infer $$\label{eq:linf_ii}
    \begin{split}
      &
      L_{i}:=
      \sup_{t_{i}<\tau<t}
      \sum_{x\in V}
      f_{i}(x,\tau)^{b}
      {d_{{\omega}}}(x)
      +
      \int_{t_{i}}^{t}
      {{D}}_{i}(\tau)
      {\,\textup{\textmd{d}}}\tau
      \\
      &\quad
      \le
      \frac{\gamma 2^{i}}{t(\sigma_{2}-\sigma_{1})}
      \int_{t_{i+1}}^{t}
      \sum_{x\in V}
      f_{i+1}(x,\tau)^{b}
      {d_{{\omega}}}(x)
      {\,\textup{\textmd{d}}}\tau
      \\
      &\quad\le
      \frac{\gamma 2^{i}}{t(\sigma_{2}-\sigma_{1})}
      {\varepsilon}^{\frac{p}{b}}
      \int_{t_{i+1}}^{t}
      {{D}}_{i+1}(\tau)
      {\,\textup{\textmd{d}}}\tau
      \\
      &\qquad
      +
      \frac{\gamma 2^{i}}{\sigma_{2}-\sigma_{1}}
      {\varepsilon}^{-\frac{p}{p-b}}
      {\Lambda_p}({M}_{i+1})^{-\frac{b}{p-b}}
      {M}_{i+1}
      \,,
    \end{split}$$ where the second inequality follows of course from . For a $\delta>0$ to be chosen, select ($\gamma$ denotes here the constant in ) $$\frac{\gamma 2^{i}}{t(\sigma_{2}-\sigma_{1})}
    {\varepsilon}^{\frac{p}{b}}
    =
    \delta
    \qquad
    \text{i.e.,}
    \qquad
    {\varepsilon}=
    \gamma_{0}
    \delta^{\frac{b}{p}}
    t^{\frac{b}{p}}
    (\sigma_{2}-\sigma_{1})^{\frac{b}{p}}
    2^{-\frac{b}{p}i}
    \,.$$ On substituting this choice of ${\varepsilon}$ in we arrive at an estimate which can be successfully iterated, that is $$\label{eq:linf_iii}
    L_{i}
    \le
    \delta L_{i+1}
    