Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ a open and connected set. Consider a function $z=2R^{-\alpha}v-v^2$ with $R$ that will be chosen suitably small, where $v$ is a harmonic function in $G$, and satisfies $$|x|^\alpha\leqslant v(x)\leqslant C_0|x|^\alpha. \ \ (*)$$ Then, $$\Delta z+f(z)=-2|\nabla v|^2+f(z)\leqslant-C|x|^{2\alpha-2}f(0)+Kz,$$ where $K$ is the Lipschitz constant of the function $f$. ($f$ is a Lipschitz-continuous function). I have no idea how to obtain this second inequality and I don't know if $(*)$ is really necessary.

share|cite|improve this question
what is C in the last inequality RHS term?? – dexter04 Nov 24 '12 at 19:09
$C$ is an arbitrary constant. – José Carlos Nov 25 '12 at 14:49
What do we know about $\alpha$? – Davide Giraudo Nov 29 '12 at 10:41
We just know that $\alpha>0$. I really dont know how to prove this second inequality. In the article H. Beresticky, L. A. Caffareli and L. Nirenberg, Monotonocity for elliptic equations in unbounded lipschitz domains, page: $1102$, he says that is a direct computation. – José Carlos Nov 29 '12 at 12:33
up vote 1 down vote accepted

This is an answer to a question raised in comment to another answer. Since it is of independent interest (perhaps of more interest than the original answer), I post it as a separate answer.

Question: How do we construct homogeneous harmonic functions that are positive on a cone?

Answer. Say, we want a positive homogeneous harmonic function $v$ on the cone $K=\{x\in\mathbb R^n:x_n>\kappa|x|\}$ where $\kappa\in (-1,1)$. Let $\alpha$ denote the degree of homogeneity of $v$: that is, $v(rx)=r^\alpha v(x)$ for all $x\in K$. The function $v$ is determined by the number $\alpha$ and by the restriction of $v$ to the unit spherical cap $S\cap K$. The Laplacian of $v$ can be decomposed into radial and tangential terms: $$\Delta v = v_{rr}+\frac{n-1}{r}v_r+\frac{1}{r^2}\Delta_S v$$ where $\Delta_S$ is the Laplace-Beltrami operator on the unit sphere $S$. Using the homogeneity of $v$, we find $v_r = \frac{\alpha }{r}v$ and $v_{rr}=\frac{\alpha(\alpha-1)}{r^2}v$. On the unit sphere $r=1$, and the radial term of $\Delta v$ simplifies to $\alpha(n+\alpha-2)v$.

Therefore, the restriction of $v$ to $S\cap K$ must be an eigenfunction of $\Delta_S$: $$\Delta_S v + \mu v =0, \ \ \ \ \mu = \alpha(n+\alpha-2) $$ Where do we get such a thing from?

Well, there is a well-developed theory of eigenfunctions and eigenvalues for $\Delta_S$ with the Dirichlet boundary conditions. In particular, it is known that the eigenfunction corresponding to the lowest eigenvalue $\mu_1$ has constant sign. So this is what we use for $v$. There are two drawbacks:

  • (a) we cannot choose $\alpha$ ourselves, because $\alpha(n+\alpha-2)=\mu_1$ and $\mu_1$ is determined by the domain $S\cap K$.
  • (b) the function $v$ vanishes on the boundary of spherical cap, and therefore is not bounded away from $0$.

Concerning (a), we know that $\mu_1$ is monotone with respect to domain: larger domains (in the sense of inclusion) have lower value of $\mu_1$. (Physical interpretation: a bigger drum emits lower frequencies.) Therefore, the value of $\alpha$ decreases when the domain is enlarged. Also, when $K$ is exactly half-space $\{x_n>0\}$, we know our positive harmonic function directly: it's $v(x)=x_n$, which is homogeneous of degree $\alpha=1$. Therefore, in the cones that are smaller than half-space we have $\alpha>1$, and in the cones that are larger than half-space we have $\alpha<1$.

Concerning (b), we do the following: carry out the above construction in a slightly larger cone $K'$ and then restrict $v$ to $K$. Since the closure of $K\cap S$ is a compact subset of $K'\cap S$, the function $v$ attains a positive minimum there. We can multiply $v$ by a constant and achieve $1\le v\le C$ on $K\cap S$, hence $|x|^\alpha\le v(x)\le C|x|^\alpha$ on all of $K$.

share|cite|improve this answer
one more time,thank you very much! You are helped me.I just didn't understand two steps in your answer. First: When you said that the function $v$ is determined by the number α and by the restriction of v to the unit spherical cap $S\cap K$. Why $v$ satisfies this? Because $v$ is homogeneous? Second: When you obtain the inequality $1\leq v\leq C$, the $v$ is restrict in the spherical cap $K\cap S$. How did you obtain the inequality $|x|^\alpha\leq v(x)\leq C|x|^\alpha$ in all $K$? We were in a small domain and $K$ can be as big as we want. I didn't understand this consequence. – José Carlos Dec 21 '12 at 22:06
The answer to both questions is: by homogeneity. For example: once you know that $u(0.6,0.8)=7$ and that the degree of homogeneity is $\alpha=2$, you can conclude that $u(3,4)=5^2\cdot 7 = 175$, and $u(30,40)=17500$. That is, the values on the entire half-line from $(0,0)$ through $(0.6,0.8)$ are determined. Similarly, if I only know that $1\le u(0.6,0.8)\le 7$, I can conclude that $25\le u(3,4)\le 175$, etc. – user53153 Dec 21 '12 at 23:30
I'm sorry for one more question but I'm trying to type this article and I really need understand this step. If $x\in K$, then $x=tz$, for some $t\geq0$ and $z\in S$. For each $z\in S$, the values $tz$ are determined, ok? But, in the end, we have $$|t|^\alpha\leq u(x)=u(tz)\leq C|t|^\alpha,$$ and I need $|x|^\alpha\leq u(x)\leq C|x|^\alpha,$ for all $x\in K$. – José Carlos Dec 26 '12 at 3:43
@JoséCarlos What is $|x|$, given that $x=tz$ where $z$ is a unit vector ($z\in S$)? – user53153 Dec 26 '12 at 4:10
Im sorry, Im getting retarded. LoL Thank you! – José Carlos Dec 26 '12 at 4:18

It's good that you gave a reference. In many questions of the kind "I don't follow the argument here" the underlying reason is that the OP did not notice the relevance of some information from the paper. Hence, the information included in the post is usually insufficient to give an explanation.

Here, the missing piece is that $v$ is homogeneous of degree $\alpha$. This means $v(tx)=t^\alpha v(x)$ for all $t\ge 0$. Differentiating with respect to $t$, we find $\frac{d}{dt}v(tx) = \alpha t^{-1} v(tx)$. Specializing this to $|x|=1$ and writing $y=tx$, we obtain $\frac{\partial}{\partial n}v(y)=\alpha |y|^{-1}v(y)$ where $\frac{\partial}{\partial n}$ is the derivative in radial direction. As a consequence, $|\nabla v(y)|\ge \alpha |y|^{-1}v(y)$ for all $y$. Notice that harmonicity is not used here; only homogeneity is needed.

Recalling that $|v(x)|\ge |x|^{\alpha}$, we get $|\nabla v(x)|\ge \alpha |x|^{\alpha-1}$. Therefore, $$-2|\nabla v|^2 \leqslant -2\alpha^2|x|^{2\alpha-2}$$ At the same time, the Lipschitz property of $f$ tells us that $$f(z)\le f(0)+Kz$$ Add the inequalities: $$-2|\nabla v|^2 + f(z) \leqslant -2\alpha^2|x|^{2\alpha-2}+f(0)+Kz$$ The plus sign before $f(0)$ is missing in the paper. Typos happen. If you look at the very next displayed formula in the paper, you will see the plus sign reappearing there.

share|cite|improve this answer
Thank you very much @Pavel M, you helped me a lot. As you have this article, could you explain something about this problem? In the page $1101$, he explains how to obtain the expression $(4.5)$ and talk about the constants $k'$ and $\alpha$. Do you understand this? Thank you!!! – José Carlos Dec 21 '12 at 12:18
@JoséCarlos Since this is a separate question, I wrote a separate answer. The proof of Lemma 2.1 is relevant, too. – user53153 Dec 21 '12 at 19:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.