# Blow up of a solution of the p-Laplace equation at a boundary point

My doubt is about a proof from the lemma 2.8 of the article 18 of this homepage: http://www.math.kth.se/~henriksh/Henriks_page/publications.html

The result that I said is:

Lemma: Let $D_1$ and $D_2$ two open , bounded , convex sets with $D_2 \supset \overline{D_1}$. Define $\Omega = D_2 \setminus \overline{D_1}.$ Suppose that the boundary of $\partial D_1$ is $C^1.$ Consider $u$ the p capacitary potential of $\Omega$ (for the definition of the capacitary potential see http://arxiv.org/pdf/1204.6578v1.pdf in the page 5).

Suppose that $|Du| \leq C$ uniformily in $\Omega.$ Fix $x_0 = 0 \in \partial D_1$ (suppose this ) and suppose that the exterior normal vector to $\partial \Omega$ is directed by the first coordinate vector. For a decreasing sequence $r_j$ of positive number converging to zero define the blow up sequence:

$$u_{r_j} (x)= \displaystyle\frac{1 - u(r_j x)}{r_j} \quad \text{(in the article there is a typo)}$$

Then exists a positive number $\alpha$ and a subsequence, also denoted by $u_{r_j}$, which uniformly converges to $u_0(x) = \alpha x_1$

Proof (I will write until the part that I have the doubt)

Thanks to the assumption, both $u_{r_j}$ and $|\nabla u_{r_j}|$ are bounded uniformly by the Lipschitz norm of $u$. Hence there is a subsequence converging to a limit function $u_0$, which is p-harmonic in the upper half plane (after suitable rotation and translation), and has concave level sets seen from the plane $\{ x_1 = 0\}$. This implies that the sets $\{ u_0 > t \} (t\geq 0)$ (this is the levels sets mencioned in the paper) are hyperplanes and $u_0$ dependsonly in $x_1$ ...

I am not understanding the part : "This implies that the sets $\{ u_0 > t \} (t\geq 0)$ are hyperplanes and $u_0$ depends only in $x_1$ ..."

someone can give me a help please?

any help will be appreciated.

The original function $u$ was zero on $\partial D_1$. After the blow up process, the limit $u_0$ is zero on the tangent plane to $\partial D_1$ at the point of blow up. This plane is $\{x_1=0\}$.
Therefore, the level surfaces $\{u_0=t\}$ for $t>0$ cannot intersect the plane $x_1=0$. On the other hand, they are concave, as the article mentions in the previous sentence. The only way for a concave surface to lie within a half-space without touching the boundary is that the surface is a plane of the form $x_1=s$.
Once we have that every surface $\{u_0=t\}$ is of the form $x_1=s$, it follows that $u_0$ depends only on $x_1$.