# Ngô Quốc Anh

## January 5, 2010

### On a property of log-function appeared in conformally invariant equations

Suppose $f \in L^1(\mathbb R^n) \cap L_{loc}^\infty (\mathbb R^n)$ with $f \geq 0$. Define $\displaystyle Sf\left( x \right) = \int_{\mathbb{R}^n } {\log \frac{{\left| y \right|}}{{\left| {x - y} \right|}}f\left( y \right)dy}$.

Show that $Sf(x)$ is finite for all $x \in \mathbb R^n$ and $Sf \in L_{loc}^1(\mathbb R^n)$.

Proof. We consider the equivalent form of $Sf(x)$ as follows (this is nothing but $-Sf(x)$) $\displaystyle S'f\left( x \right) = \int_{\mathbb{R}^n } {\left( {\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}$.

We firstly consider the case when $|x| \geq 4$. From the identity $\displaystyle \mathbb{R}^4= \underbrace {\left\{ {y:\left| {x - y} \right| \geq\frac{{\left| x \right|}}{2}} \right\}}_A \cup \underbrace {\left\{ {y:\left| {x - y} \right| \leq\frac{{\left| x \right|}}{2}} \right\}}_B$

if $y \in B$, then $\left| y \right| \geq \left| x \right| - \left| {x - y} \right| \geq\frac{{\left| x \right|}}{2} \geq\left| {x - y} \right|$ which implies $\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}} \leq0$. Therefore $\displaystyle S'f\left( x \right) \leq\int_A {\left( {\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}$.

Since $|x - y|\leq |x|+|y| \leq |x| |y|$ provided $|x|, |y| \geq 2$ and $\log |x-y| \leq \log |x| + C$ for $|x| \geq 4$ and $|y| \leq 2$, then we have $\displaystyle \begin{gathered}\int\limits_A {\left( {\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}= \int\limits_{A \cap \left\{ {y:\left| y \right| \geqslant 2} \right\}} {\left( {\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}+ \int\limits_{A \cap \left\{ {y:\left| y \right| < 2} \right\}} {\left( {\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}\hfill \\ \qquad\qquad\qquad\qquad\leq\log \left| x \right|\left( {\int\limits_{A \cap \left\{ {y:\left| y \right| \geqslant 2} \right\}} {f\left( y \right)dy} } \right) + {\rm const}. \hfill \\ \end{gathered}$

Thus, $S'f(x)$ is bounded from above for all $|x| \geq 4$. For $0< |x| \leq 4$, consider $S'f\left(\frac{4^2}{x} \right)$. More precisely, $\displaystyle\begin{gathered} + \infty> S'f\left( {\frac{4}{{\left| x \right|}}x} \right) = \int_{{\mathbb{R}^n}} {\left( {\log \frac{{\left| {\frac{4}{{\left| x \right|}}x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}\hfill \\ \qquad= \frac{4}{{\left| x \right|}}\int_{{\mathbb{R}^n}} {\left( {\log \frac{{\left| {x - \frac{{\left| x \right|}}{4}y} \right|}}{{\left| {\frac{{\left| x \right|}}{4}y} \right|}}} \right)\widetilde f\left( {\frac{{\left| x \right|}}{4}y} \right)d\left( {\frac{{\left| x \right|}}{4}y} \right)}\hfill \\ \qquad= \frac{4}{{\left| x \right|}}S'f\left( x \right) \hfill \\ \end{gathered}$

where $\widetilde f\left( y \right): = f\left( {\frac{4}{{\left| x \right|}}y} \right)$. Note that $\displaystyle 0 \leq\int_{\mathbb{R}^n } {\widetilde f\left( y \right)dy}= \frac{{\left| x \right|}}{4}\int_{\mathbb{R}^n } {f\left( {\frac{4}{{\left| x \right|}}y} \right)d\left( {\frac{4}{{\left| x \right|}}y} \right)}<+ \infty$.

If $|x| = 0$, then $Sf'(0) = 0$. Thus, $S'f(x)$ is bounded from above in $\mathbb R^n$ which implies that $Sf(x)$ is bounded from below in $\mathbb R^n$. Similar, we can prove that $Sf(x)$ is bounded from above in $\mathbb R^n$.

Finally, from the above estimates, clearly $S'f$ is of class $S'f \in L_{loc}^1(\mathbb R^n)$, and thus, so is $Sf$.