# Ngô Quốc Anh

## August 9, 2015

### The third fundamental lemma in the method of moving spheres

Filed under: Uncategorized — Ngô Quốc Anh @ 4:03

In this post, we proved the following result (appeared in a paper by Y.Y. Li published in J. Eur. Math. Soc. (2004))

Lemma 1. For $n \geqslant 1$ and $\nu \in \mathbb R$, let $f$ be a function defined on $\mathbb R^n$ and valued in $[-\infty, +\infty]$ satisfying

$\displaystyle {\left( {\frac{\lambda }{{|y - x|}}} \right)^\nu }f\left( {x + {\lambda ^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) \leqslant f(y), \quad \forall |y - x| > \lambda > 0.$

Then $f$ is constant or $\pm \infty$.

Later, we considered the equality case in this post and proved the following result:

Lemma 2. Let $n\geqslant 1$, $\nu \in \mathbb R$ and $f \in C^0(\mathbb R^n)$. Suppose that for every $x \in \mathbb R^n$ there exists $\lambda(x)>0$ such that

$\displaystyle {\left( {\frac{\lambda(x) }{{|y - x|}}} \right)^\nu }f\left( {x + {\lambda(x) ^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) =f(y), \quad \forall |y - x| > 0.$

Then for some $a \geqslant 0$, $d>0$ and $\overline x \in \mathbb R^n$

$\displaystyle f(x) = \pm a{\left( {\frac{1}{{d + {{\left| {x - \overline x } \right|}^2}}}} \right)^{\frac{\nu }{2}}}.$

In this post, we consider the third result which can be stated as follows:

Lemma 3. For $n \geqslant 1$ and $\nu \in \mathbb R$, let $f$ be a function defined on $\mathbb R_+^n$ and valued in $[-\infty, +\infty]$ satisfying

$\displaystyle {\left( {\frac{\lambda }{{|y - x|}}} \right)^\nu }f\left( {x + {\lambda ^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) \leqslant f(y), \quad \forall |y - x| > \lambda > 0.$

Then $f$, restricted to $\partial\mathbb R_+^n$, is constant or $\pm \infty$.

In the statement of Lemma 3 above, by $\mathbb R_+^n$ we mean those points $z\in \mathbb R^n$ such that the last coordinate $z_n >0$. Then by $\partial\mathbb R_+^n$ we mean the boundary of $\mathbb R_+^n$. Hence, we may identify $\partial \mathbb R_+^n = \mathbb R^{n-1}$.

In addition, to make the difference $|y-x|$ possible for each $y \in \mathbb R_+^n$ and $x \in \partial\mathbb R_+^n$, we may identify $x$ by a new point $(x,0)$ sitting in $\overline{\mathbb R_+^n}$. Hence, by definition, we simply set

$\displaystyle |y-x|^2= (y'-x)^2 + y_n^2$

where, and from now on, we denote $y=(y', y_n)$ for each point $y \in \mathbb R_+^n$. We are now in a position to prove Lemma 3, see a paper by Dou and Zhu.

Proof of Lemma 3. For each $y' \ne z'$ we set $y^i \in \mathbb R_+^n$ with $y^i=(y',y_n^i)$ such that $y_n^i \to z_n$ as $i\to +\infty$. Then we choose $b_i>1$ in such a way that

$\displaystyle x^i := (x',0) = y^i + b_i(z-y^i) \in \partial\mathbb R_+^n.$

Note that this is possible since we can always choose $b^i$ such that $y_n + b_i(z_n - y_n)=0$. We also set

$\lambda_i := \sqrt {|z - x^i||y^i - x^i|} .$

Then we immediately obtain

$\displaystyle z={x^i + {\lambda ^2}\frac{y^i - x^i}{|y^i - x^i|^2}}.$

Therefore, by our assumption

$\displaystyle {\left( {\frac{\lambda_i }{{|y^i - x^i|}}} \right)^\nu }f(z) \leqslant f(y^i).$

Since

$\displaystyle\mathop {\lim }\limits_{i \to \infty } \frac{\lambda_i }{|y^i - x^i|} = \mathop {\lim }\limits_{i \to \infty } \sqrt {\frac{|z - x^i|}{|y^i - x^i|}} = 1$

and

$\displaystyle\mathop {\lim }\limits_{i \to \infty }y^i = (y',z_n),$

we have

$f(z',z_n)\leq f(y',z_n).$

Our lemma follows since $y' \ne z'$ are arbitrary.