I'm reading Chen and Wu's book, Second order elliptic equations and elliptic systems. In the chapter on Krylov and Safonov theorem for nondivergence form elliptic equations, the concept of normal mapping is introduced. On page 81, definition 1.4 it is said that if $\Omega \subset \mathbf{R}^n$ is a domain, $x_0 \in \Omega$, for $\lambda \in \mathbf{R}$ they set $w$ the function such that its graph is a cone surface with vertex at $(x_0, \lambda)$ and base $\Omega$.
I tried to construct such a function, but failed. For example if one takes $\Omega \subset \mathbf{R}^2$ to be the domain which boundary is: $B\ =\ A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6$ with:
- $A_1\ =\ \{(x,0): 1 \le x \le 2\}$
- $A_2\ =\ \{(1, y): 0 \le y \le 1\}$
- $A_3\ =\ \{(2,y): -1 \le y \le 0\}$
- $A_4\ =\ \{(x,-1): -1 \le x \le 2\}$
- $A_5\ =\ \{(-1, y): -1 \le y \le 1\}$
- $A_6\ =\ \{(x, 1): -1 \le x \le 1\}$
Then if we take $x_0\ =\ (0,0)$ and $\lambda =1$ I don't understand how to construct the function whose graph is a cone with vertex in $(x_0, 1)$ and base $\Omega$. In fact every point $(x,0)$ with $x > 0$ has "lots of lines" above it. Shall I assume $\Omega$ convex or am I doing something wrong?
I describe the geometric intuition for ABP-estimate.
Hessian could be looked as a Jacobi metrix of $(u_1,...,u_n)\to (e_1,...,e_n)$. Elliptic property is just said the graph of the function could not have very narrow cone. By area formula and a rescaling we will derive ABP estimate.