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.

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