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).


\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


\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.


Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Blog at

%d bloggers like this: