2017-2018-2偏微分方程复习题解析9

Problem: Let $K,f,g$ be in $\calD(\bbR^d)$, and $K$ is radial (for definition, see Problem 2). Show that $$\bex \int (K*f)(x)g(x)\rd x=\int f(x)(K*g)(x)\rd x. \eex$$ 

 

Proof: $$\beex \bea \int (K*f)(x)g(x)\rd x &=\int \sez{\int K(x-y)f(y)\rd y} g(x)\rd x\\ &=\int \sez{\int K(y-x)g(x)\rd x} f(y)\rd y\\ &=\int (K*g)(y)f(y)\rd y. \eea \eeex$$

posted @ 2018-05-14 12:04 张祖锦 阅读(...) 评论(...) 编辑 收藏