2017-2018-2偏微分方程复习题解析6

Problem: If $a$ is a smooth homogeneous function of degree $m$, show that $$\bex |\dot \lap_ju(x)|\leq C2^{jm}(Mu)(x), \eex$$ where $$\bex (Mf)(x)=\sup_{r>0}\f{1}{|B(x,r)|} \int_{B(x,r)}|u(y)|\rd y \eex$$ is the Hardy-Littlewood maximal function.

Proof: $$\beex \bea \dot \lap_ja(D)u &=\calF^{-1}\sex{\phi(2^{-j}\xi)a(\xi)\hat u(\xi)}\\ &=c_d\calF^{-1} \sex{\phi(2^{-j}\xi)a(\xi)}*u\\ &=c_d\calF^{-1}\sex{\phi(2^{-j}\xi)a(2^{-j}\cdot 2^j\xi)}*u\\ &=c_d 2^{jd} \calF^{-1}\sex{\phi(\xi)a(2^j\xi)}(2^j\cdot)*u\\ &=c_d 2^{jd} \calF^{-1}(\phi(\xi)2^{jm}a(\xi))(2^j\cdot)*u\\ &=c_d2^{jm}2^{jd} h_a(2^j\cdot)*u\qx{ h_a=\calF^{-1}(\phi a)\in \calS }. \eea \eeex$$ And consequently by 1.17 in the book, $$\beex \bea &\quad|\dot\lap_ja(D)u(x)| \leq 2^{jm} \int_{\bbR^d} \sez{ \sup_{|y'|\geq |y|} 2^{jd} |h_a(2^jy')|}\rd y\cdot Mu(x)\\ &\leq 2^{jm} \int_{\bbR^d} \sup_{|z'|\geq |z|} |h_a(z')|\rd z\cdot Mu(x)\\ &=2^{jm} \int_{\bbR^d} \sup_{|z'|\geq |z|} \f{1}{(1+|z'|)^{d+1}} \cdot (1+|z'|)^{d+1}|h_a(z')|\rd z\cdot Mu(x)\\ &\leq C2^{jm} \int_{\bbR^d} \f{1}{(1+|z|)^{d+1}}\rd z\cdot Mu(x) \leq C2^{jm}Mu(x). \eea \eeex$$

posted @ 2018-05-11 19:36 张祖锦 阅读(...) 评论(...) 编辑 收藏