[0,T]$, where we have made use of the equation again to obtain convergence for the time derivative. Finally owing to we have $$\label{eq:exist_energy_nj}
    \sum_{x\in V}
    {\lvert{u}(x,t)\rvert}^{q}
    {d_{{\omega}}}(x)
    \le
    {{\lVert{u}_{0}\rVert}_{\ell^{q}(V)}}^{q}
    \,,
    \qquad
    t>0
    \,.$$ It is easily seen that ${u}\in L^{\infty}(0,+\infty;\ell^{q}(V))$ is a solution to –. If ${u}_{0}\ge 0$, we appeal to our next result to prove that ${u}\ge 0$.

\[p:compare\] If ${u}_{1}$, ${u}_{2}\in L^{\infty}(0,T; \ell^{q}(V))$ solve – with ${u}_{01}$, ${u}_{02}\in\ell^{q}(V)$, ${u}_{01}\ge {u}_{02}$, then ${u}_{1}\ge{u}_{2}$.

According to Remark \[r:lp\_scale\] and to Definition \[d:sol\], we may assume $q>1$. Define ${w}={u}_{2}-{u}_{1}$. Then ${w}$ does not solve , but we may still apply (with $h=0$) to it, see Remark \[r:prelim\_diff\]. This proves ${{\ifthenelse{\equal{*}{*}}{({w})_{+}}{\left({w}\right)_{+}}}}=0$ and thus the statement.

Elementary inequalities {#s:prelim_elem}
-----------------------

We record for future use two immediate consequences of , : $$\begin{aligned}
2
  \label{eq:fkf_above}
  {\Lambda_p}(sa)^{-1}
  &\le
  s^{\omega}
  {\Lambda_p}(a)^{-1}
  \,,
  &\qquad&
  s\ge 1
  \,,
  a>0
  \,;
  \\
  \label{eq:fkf_below}
  {\Lambda_p}(\sigma a)^{-1}
  &\le
  \sigma^{\frac{p}{{N}}}
  {\Lambda_p}(a)^{-1}
  \,,
  &\qquad&
  0<\sigma\le 1
  \,,
  a>0
  \,.\end{aligned}$$ Also the following Lemma relies on and will be used in a context where it is important that $\nu<1/(p-1)$.

\[l:dcf\] Let $\nu={N}(p-2)/[({N}(p-2)+p)(p-1)]$ and $b>0$. Then the function $$\tau\mapsto \tau^{\nu}{\psi}_{1}^{(-1)}(\tau^{-1}b)^{\frac{p-2}{p-1}}
    \,,
    \qquad
    \tau>0
    \,,$$ is nondecreasing.

Equivalently we show that $$r\mapsto r^{-\alpha}{\psi}_{1}^{(-1)}(r)^{p-2}$$ is nonincreasing for 