A priori estimate of an inhomogeneous p-Laplace equation with Dirichlet boundary condition I'm currently working on this Dirichlet problem:
\begin{cases}
div(\sigma |\nabla u|^{p-2} \nabla u) = f &\quad  {in }~ \Omega\\
u = g &\quad in~\partial\Omega
\end{cases}
with $\sigma \in L^{\infty}_{+}(\Omega)$, boundary $g \in W^{1,p}(\Omega)$ and $f \in L^2(\Omega)$.
I think I've already proved that this problem has a unique solution, but I don't know how to prove the following a priori estimate
\begin{align}
||{u}||_{W^{1,p}} \leq C ({||g||}_{W^{1,p}} + {||f||}_{L^2}^{\frac{1}{p-1}})
\end{align}
This estimate is mentioned in this paper
https://pdfs.semanticscholar.org/7399/da07c625d51aa7ee72840789916b036019d2.pdf
(second page; equation (1.4))
but without giving proof. I guess it's seen "easy", but I just don't know how to get there. I know how to conclude an a priori estimate in the case $f=0$ or $g=0$, but since p-Laplace operator isn't linear, this doesn't help and I'm really clueless know. I hope someone can help me with with.
As far as i thought i can use the Poincaré inequality since $u - g \in W^{1,p}_0$:
\begin{align}
||u||_{W^{1,p}} &= ||u||_{L^p} + ||\nabla u||_{L^p}\\
& \leq ||g||_{L^p} + ||u - g||_{L^p} + ||\nabla u||_{L^p}\\
& \leq ||g||_{L^p} + C||\nabla(u - g)||_{L^p} + ||\nabla u||_{L^p}
\end{align}
The problem is that i can't find an upper bound for $||\nabla u||_{L^p}$ using $f$...
The inequality in the paper that i found implies that i should found something like
\begin{align}
||\nabla u||^p_{L^p} \leq ||f||_{L^2}~||\nabla u||_{L^p}
\end{align}
which only be valid, when the solution u is in $W^{1,p}_0$...
I hope someone can help me with with this...
Edit: Using fedjas advice i got
\begin{align}
\int_{\Omega} f~(u - g) &= \int_{\Omega} div(\sigma |\nabla u|^{p-2} \nabla u) (u - g)\\
&= \int_{\Omega} \sigma |\nabla u|^{p-2} \nabla u \nabla (u - g)\\
&\leq ||f||_{L^2} ||u - g||_{L^p} \\
&\leq ||f||_{L^2} ~C ||\nabla (u - g)||_{L^p}
\end{align}
Now if i would have $||\nabla (u - g)||_{L^p}^p \leq  \int_{\Omega} f~(u - g)$ it would be good... BUT I dont know how to get there...
$%\int_{\Omega} f~(u - g) \leq ||\nabla (u - g)||_{L^p}^p$
because the mapping $ x \mapsto |x|^p$ is convex we have always
$|x|^p \geq |y|^p + p |y|^{p-2}y (x-y)$ (characterization with first derivation)
and with $x = \nabla (u - g)$ and $y = \nabla u$ it follows
$|\nabla (u - g)|^p \geq |\nabla u|^p - p|\nabla u|^{p-2} \nabla u \nabla g$
don't know if it's useful here, but its valid for $p \geq 1$.
Furthermore i know that
$\frac{1}{p} (\int_{\Omega} |\nabla u|^p - |\nabla g|^p) \leq \int_{\Omega} f~(u - g)$
Because the solution u is the minimizer of the functional:
$J(v) = \frac{1}{p} \int_{\Omega} |\nabla v|^2 - \int_{\Omega} fv$ for all $v \in W^{1,p}_g$ (space with needed boundary values) and therefore $J(u)\leq J(g)$
Nevertheless i still don't see how to put all of this together to achieve the desired result... :/
 A: It looks like you are totally new to such estimates, so let me show you the old $\varepsilon$ vs $C_\varepsilon$ trick. If you need to estimate some product $xy$, you can write $|xy|\le |x|^2+|y|^2$, but sometimes you would prefer a smaller constant on $x$ (that you would later subtract from the other side) and do not really care how big one you get on $y$. In this case you can write 
$$
|xy|=|(\varepsilon^{1/2} x)(\varepsilon^{-1/2}y)|\le \varepsilon |x|^2+\varepsilon^{-1}|y|^2=\varepsilon |x|^2+C_\varepsilon |y|^2
$$
You can do it with any positive powers, so you can equally well write
$$
|x|^a|y|^b\le\varepsilon|x|^{a+b}+C_\varepsilon |y|^{a+b}
$$
(of course, $C_\varepsilon$ here depends on $a,b$ as well).
Now, you have your identity
$$
\int_\Omega f(u-g)=\int_\Omega\sigma|\nabla u|^{p-2}\langle\nabla u,\nabla u-\nabla g\rangle
$$
Rewrite it as 
$$
\int_\Omega\sigma|\nabla u|^{p}=\int_\Omega\sigma|\nabla u|^{p-2}\langle\nabla u,\nabla g\rangle+\int_\Omega f(u-g)\le
\int_\Omega\sigma|\nabla u|^{p-1}|\nabla g|+\int_\Omega |f||u-g|
$$
We certainly need $\sigma\ge \sigma_0>0$ in $\Omega$ (otherwise you can easily blow up everything). So, we have
$$
\sigma_0\int_\Omega|\nabla u|^{p}\le \int_\Omega\sigma|\nabla u|^{p-1}|\nabla g|+\int_\Omega |f||u-g|
$$
Now play $\varepsilon$ vs $C\varepsilon$ and recall that $\sigma\le\sigma_1<+\infty$:
$$
\sigma|\nabla u|^{p-1}|\nabla g|\le \sigma_1(\varepsilon|\nabla u|^{p}+C_\varepsilon|\nabla g|^p)\,.
$$
Assuming that $W^{1,p}\subset L^2$, we also have
$$
\int_\Omega |f||u-g|\le \|f\|_2\|u-g\|_2\le 
\|f\|_2\|u-g\|_{1,p}=(\|f\|^{1/(p-1)}_2)^{p-1}\|u-g\|_{1,p}
\\
\le 
\varepsilon \|u-g\|_{1,p}^p+C_\varepsilon \|f\|_2^{p/(p-1)}.
$$
Recalling the bound 
$$
\|u-g\|^p_{1,p}\le C[\|\nabla u|^p_p+\|g\|_{1,p}^p]\,,
$$
we get 
$$
\sigma_0\int_\Omega|\nabla u|^{p}\le
\sigma_1[\varepsilon\int_{\Omega}|\nabla u|^{p}+C_\varepsilon\int_{\Omega}|\nabla g|^p]+\varepsilon C[\int_{\Omega}|\nabla u|^{p}+\|g\|_{1,p}^p]+ CC_\varepsilon \|f\|_2^{p/(p-1)}
$$
Now choose $\varepsilon$ so small that after moving all the terms with $\int_\Omega|\nabla u|^p$ from the right to the left, you still have some positive coefficient on the LHS. 
That's all. I presented it with all details and spelled out what exactly is needed for what step since it looks like you are seeing this standard mumbo-jumbo for the first time. Try to to it yourself next time.
