\frac{1}{R_{2}-R_{1}}
    \sum_{x,y\in B_{R_{2}+1}}
    \big(
    {u}(x,\tau)^{p+q-2}
    +
    {u}(y,\tau)^{p-1}
    {u}(x,\tau)^{q-1}
    \big)
    {\omega}(x,y)
    \le
    \frac{C_{u}}{R_{2}-R_{1}}
    \,,
  \end{gathered}$$ where $C_{u}$ is independent of $R_{i}$. Owing to $p+q-2>q$ and to Remark \[r:lp\_scale\], to this end it is only left to observe that $$\begin{gathered}
    \sum_{x,y\in B_{R_{2}+1}}
    {u}(y,\tau)^{p-1}
    {u}(x,\tau)^{q-1}
    {\omega}(x,y)
    \\
    \le
    \Big(
    \sum_{y\in V}
    {u}(y,\tau)^{(p-1)q}
    {d_{{\omega}}}(y)
    \Big)^{\frac{1}{q}}
    \Big(
    \sum_{x\in V}
    {u}(x,\tau)^{q}
    {d_{{\omega}}}(x)
    \Big)^{\frac{q-1}{q}}
    \,,
  \end{gathered}$$ and to use once more Remark \[r:lp\_scale\], since $(p-1)q>q$.

The sought after estimates follows immediately upon applying Lemma \[l:monot\] with ${v}=0$ and then letting first $R_{2}\to\infty$ and then $R_{1}\to\infty$.

\[r:prelim\_diff\] Lemma \[l:cacc2\] is still in force if ${u}$ is the difference of two solutions to . The proof is the same, when we start from the difference of the two equations and recall Lemma \[l:monot\].

Existence and comparison {#s:prelim_ex}
------------------------

\[p:ex\] Let ${u}_{0}\in\ell^{q}(V)$, $q>1$. Then – has a solution in $L^{\infty}(0,+\infty;\ell^{q}(V))$. If ${u}_{0}\ge 0$ then ${u}\ge 0$.

Let ${u}_{0}\in\ell^{q}(V)$, $q>1$. Define for $n\ge 1$ ${u}_{n}$ as the solution to $$\begin{aligned}
{2}
    \label{eq:exist_pde_n}
    {\frac{\partial {u}_{n}}{\partial t}}(x,t)
    &=
    \operatorname{\Delta}_{p}{u}_{n}(x,t)
    \,,
    &\qquad&
    x\in B_{n}
    \,,
    t>0
    \,,
    \\
    \label{eq:exist_init_n}
    {u}_{n}(x,0)
    &=
    {u}_{0}(x)
    \,,
    &\qquad&
    x\in B_{n}
    \,,
    \\
    \label{eq:exist_dir_n}
    {u}_{n}(x,t)
    &=
    0
    \,,
    &\qquad&
    x\not\in B_{n}
    \,,
    t\ge 0
    \,.
  \end{aligned}$$ In practice this 