Let $u$ be a positive solution of the elliptic equation $\mathcal Lu = 0$ on $B^+_1 \subset \mathbb{R}^n$ such that $u$ vanishes continuously on $\{x_n = 0\}$. To fix ideas, we may take $\mathcal L = - \Delta$, ie $u$ harmonic.

Then we can prove the comparison principle stated in STEP 1 below (Caffarelli, 1998 p. 40):

$$u\big|_{B^+_{1/2}} \le M.$$

I've some questions on the proof below (PROOF OF STEP 1).

  • In particular, what does the (c) observation mean? What does the author mean by $M>M_0$ large enough? What is the "absurde assumption" that will lead to a contradiction?

  • Why after proving that

"if $M \ge M_0$ large, we can construct a sequence of points, $X_k$ all contained in $B^+_{3/4}$, $X_k → \{x_n = 0\}$, and such that $u(X_k)$ goes to $+\infty$"

do we get that the proof of STEP 1 is complete?

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  • $\begingroup$ Iterate and consider a sequence of nested balls $B_{1/2^{n+1}}\subset B_{1/2^{n}}$, I presume? $\endgroup$ – leo monsaingeon Jan 28 at 9:39
  • $\begingroup$ @leomonsaingeon My mistake: I meant the $(c)$ step. $\endgroup$ – user123456 Jan 28 at 9:57
  • $\begingroup$ By Item (b), if $u(X_0)=M$, for $X_0\in B_{1/2}^+$ is large then one must have $x_n(X_0)$ small. This should yield the contradiction, I'm not sure why the points $X_k$ are said to lie in $B_{3/4}^+$ and not $B_{1/2}^+$. Is this this is because the balls are open? $\endgroup$ – RBega2 Feb 16 at 21:23
  • $\begingroup$ Something must be wrong in the formulation. If $u$ is harmonic and positive in $B_1$, and if it vanishes at some interior point (here for $x_n=0$), then $u\equiv0$. $\endgroup$ – Denis Serre Feb 18 at 12:14
  • $\begingroup$ @DenisSerre You're right. There is a typo in the first line: the function is harmonic on $B^+_1$, i.e. the part of the ball above $\{x_n=0\}$. $\endgroup$ – user123456 Feb 26 at 18:38

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