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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 an advice i got from another forum:

\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... :/

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    $\begingroup$ (1) What do you mean by $\sigma \in L^{\infty}_{+}(\Omega)$? Can $\sigma$ touch zero? In that case there is no chance for the a priori estimate you stated. (2) The idea to test with $u-g$ is right, but you wrote it in a confusing manner. Try obtaining something of the form $\| \nabla u \|_{L^p}^p \le C_1 \| u \|_{W^{1,p}}^\alpha + C_2 \| u \|_{W^{1,p}}^\beta$ with exponents $\alpha,\beta < p$, then use Young's inequality. $\endgroup$ – Michał Miśkiewicz Dec 11 '17 at 23:03
  • $\begingroup$ Also posted on mathoverflow: mathoverflow.net/questions/288181/… $\endgroup$ – Tommi Brander Dec 28 '17 at 15:58
  • $\begingroup$ @MichałMiśkiewicz $L_+^\infty$ is typically the functions in $L^\infty$ which are bounded from below by a positive constant (almost everywhere). $\endgroup$ – Tommi Brander Dec 28 '17 at 16:00

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