# Ngô Quốc Anh

## May 3, 2010

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

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 3:32

Today we continue to discuss the second fundamental lemma in the method of moving spheres. This lemma also comes from a paper due to Y.Y. Li published in J. Eur. Math. Soc. (2004).

Recall from the previous entry where the first lemma was considered if the following

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

holds then $f$ is constant or $\pm \infty$. We now consider the equality case. Precisely,

Lemma. 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 $\overline x \in \mathbb R^n$

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

Proof. It follows from the continuity of $f$ that

$\displaystyle\alpha : = \mathop {\lim }\limits_{|y| \to \infty } |y{|^\nu }f(y) = \lambda {(x)^\nu }f(x),\quad \forall x \in {\mathbb{R}^n}$.

If $\nu =0$ then $f \equiv \alpha$ and we are done. On the other hand, the case $\nu<0$ can easily be reduced to the case $\nu>0$ if we let

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

in the given identity. It is then sufficient to consider the only case $\nu>0$. If $\alpha=0$ then $f\equiv 0$ and we are also done. Otherwise, replacing $f$ by a nonzero multiple of $f$, we may assume that $\alpha=1$. Since $f(y)\to 0$ as $|y|\to \infty$ and since $f$ is continuous and positive, $f$ has a maximum point and we may assume that $f$ has a maximum point at the origin.

For $x \in \mathbb R^n$, we have for large $|y|$

$\displaystyle\begin{gathered} |y{|^\nu }f(y) = \lambda {(x)^\nu }{\left( {\frac{{|y|}}{{|y - x|}}} \right)^\nu }f\left( {x + \lambda {{(x)}^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) \hfill \\ \qquad= \lambda {(x)^\nu }\left( {1 + \frac{{\nu x \cdot y}}{{|y{|^2}}} + O(|y{|^{ - 2}})} \right)f\left( {x + \lambda {{(x)}^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) \hfill \\ \end{gathered}$

which yields

$\displaystyle |y|\left( {|y{|^\nu }f(y) - 1} \right) = |y|\lambda {(x)^\nu }\left( {f\left( {x + \lambda {{(x)}^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) - f(x)} \right) + \left( {\frac{{\nu x \cdot y}}{{|y|}} + O(|y{|^{ - 1}})} \right)f\left( {x + \lambda {{(x)}^2}\frac{{y - x}}{{|y - x{|^2}}}} \right)$.

Taking $x=0$ in the above and using the fact that $f$ has a maximum point at the origin, we obtain

$\displaystyle\mathop {\limsup }\limits_{|y| \to \infty } |y|\left( {|y{|^\nu }f(y) - 1} \right) \leqslant 0$

Claim. For any $\varepsilon>0$ there exists $M_\varepsilon$ such that for any $|y| \geqslant M_\varepsilon$ there exists $\widetilde x = \widetilde x(y)$ satisfying

$\displaystyle \widetilde x + \lambda {(\widetilde x)^2}\frac{{y - \widetilde x}}{{|y - \widetilde x{|^2}}} = 0 \quad \forall |\widetilde x| \leqslant \varepsilon$

Proof of claim. Remember that

$\displaystyle \lambda(x)=f(x)^{-\frac{1}{\nu}}$

for all $x$. For any $\varepsilon \in (0,1)$ pick $M_\varepsilon>1$ so that

$\displaystyle\frac{1}{{{M_\varepsilon } - \varepsilon }}\mathop {\max }\limits_{|x| \leqslant \varepsilon } f{(x)^{ - \frac{1}{\nu }}} < \frac{\varepsilon }{2}$.

Then for all $|y| \geqslant M_\varepsilon$

$\displaystyle\mathop {\max }\limits_{|x| \leqslant \varepsilon } \left| {\lambda {{(x)}^2}\frac{{y - x}}{{|y - x{|^2}}}} \right| = \mathop {\max }\limits_{|x| \leqslant \varepsilon } \frac{1}{{|y - x|}}f{(x)^{ - \frac{2}{\nu }}} \leqslant \frac{1}{{{M_\varepsilon } - \varepsilon }}\mathop {\max }\limits_{|x| \leqslant \varepsilon } f{(x)^{ - \frac{1}{\nu }}} < \frac{\varepsilon }{2}$.

Thus by a degree argument using the continuity of $f$, there exists the desired $\widetilde x$. The proof of claim follows.

With $x:=\widetilde x(y)$ in hand we obtain

$\displaystyle\mathop {\lim \inf }\limits_{|y| \to \infty } |y|\left( {|y{|^\nu }f(y) - 1} \right) \geqslant - \varepsilon \nu f(0)\mathop {\max }\limits_{|z| \leqslant \varepsilon } f{(z)^{ - \frac{1}{\nu }}}$.

Sending $\varepsilon$ to 0 we have

$\displaystyle\mathop {\lim \inf }\limits_{|y| \to \infty } |y|\left( {|y{|^\nu }f(y) - 1} \right) \geqslant 0$.

Thus

$\displaystyle\mathop {\lim \inf }\limits_{|y| \to \infty } |y|\left( {|y{|^\nu }f(y) - 1} \right) =0$.

Let

$\displaystyle\begin{gathered} {e_1} = \left( {1,0,0,...,0} \right), \hfill \\ {e_2} = \left( {0,1,0,...,0} \right), \hfill \\ \quad \;\vdots \hfill \\ {e_n} = \left( {0,0,0,...,1} \right). \hfill \\ \end{gathered}$

For any $x\in \mathbb R^n$, $1\leqslant i \leqslant n$ and $t \in \mathbb R$ let $y=y(x,t,i) \in \mathbb R^n$ defined by

$\displaystyle t{e_i} = \lambda {(x)^2}\frac{{y - x}}{{|y - x{|^2}}}$.

Taking this $y$ and sending $t$ to 0, we obtain

$\displaystyle\frac{{\partial f}}{{\partial {x_i}}}(x) = \mathop {\lim }\limits_{t \to 0} \frac{{f(x + t{e_i}) - f(x)}}{t} = - \frac{{\nu x \cdot {e_i}f(x)}}{{\lambda {{(x)}^2}}} = - \nu {x_i}f{(x)^{1 + \frac{2}{\nu }}}$.

By the continuity of $f$ we know that $f$ is in $C^1(\mathbb R^n)$ and the proof follows by writing the above system of PDEs as

$\displaystyle\frac{\partial }{{\partial {x_i}}}\left( {f{{(x)}^{ - \frac{2}{\nu }}} - |x{|^2}} \right) = 0$

and solving it.