\(\mathrm {Exercise \ \oplus \ Problem } \ 20 \ \)
\(\mathrm {Exercise \ \oplus \ Problem } \ 20 \ \) |
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\( \qquad \)你好，这里是我的个人网站数学分析的每周一题栏目(数学分析每周一题，其中数学分析指的是数学中的分析学, 主要包括微积分，实分析，复分析) \(\qquad \ \)——————Alina Lagrange
设 \(\left\{K_{\varepsilon}\right\}_{\varepsilon>0}\) 是一类恒同逼近(核)，证明存在常数 \( C >0\) 对于所有可积函数 \( f\in L(\mathbb R ^n )\) 有 $$ \sup _{\varepsilon>0}\left|\left(f * K_{\varepsilon}\right)(x)\right| \leq C f^*(x). $$ 在这里 \( f^* \) 表示极大函数.
\(\mathcal{P}roof. \)
$$\begin{aligned} \left|\left(f * K_{\varepsilon}\right)(x)\right| & =\left|\int_{\mathbb{R}^n} f(x-y) K_{\varepsilon}(y) d y\right| \\ & \leq \int_{\mathbb{R}^n}|f(x-y)|\left|K_{\varepsilon}(y)\right| dy \\ & \leq \int_{|y| \leq \varepsilon}|f(x-y)|\left|K_{\varepsilon}(y)\right| dy +\sum_{k=0}^{\infty} \int_{2^k \varepsilon<|y| \leq 2^{k+1} \varepsilon} |f(x-y)|\left|K_{\varepsilon}(y)\right| dy \\& \leq \frac{A}{\varepsilon^n } \int_{|y| \leq \varepsilon}|f(x-y)| d y+ \sum_{k=0}^{\infty} \int_{|y| \leq 2^{k+1} \varepsilon} \frac{A \varepsilon}{\left(2^k \varepsilon\right)^{n+1}}|f(x-y)| d y \\ & = \frac{A}{\varepsilon^n} \int_{|y| \leq \varepsilon}|f(x-y)| d y+\frac{A}{\varepsilon^n } \sum_{k=0}^{\infty} \frac{1}{2^{k(n +1)}} \int_{|y| \leq 2^{k+1} \varepsilon}|f(x-y)| d y \\ &= \frac{A}{\varepsilon^n } I_1 + \frac{A}{\varepsilon^n } I_2 \end{aligned} $$ 考虑 \(I_1 \) $$ \int_{|y| \leq \varepsilon}|f(x-y)| d y=\int_B|f| \leq m(B) f^*= \frac {\pi ^{\frac n2 }}{\Gamma ( \frac n2 +1 )} \cdot \varepsilon ^n f^* .$$ 其中 \( B\) 是 \(\mathbb R^n\) 中半径 \(\varepsilon \) 的球. 考虑 \( I_2 \) $$ \frac{A}{\varepsilon^n} \sum_{k=0}^{\infty} \frac{1}{2^{k(n+1)}} \int_{|y| \leq 2^{k+1} \varepsilon}|f(x-y)| d y \leq \frac{A}{\varepsilon^n} \sum_{k=0}^{\infty} \frac{1}{2^{k(n+1)}}\cdot \frac {\pi ^{\frac n2 }}{\Gamma ( \frac n2 +1 )} \cdot\left(2^{k+1} \varepsilon\right)^n f^* $$ 所以 $$\begin{aligned} \sup _{\varepsilon>0 } \left|\left(f * K_{\varepsilon}\right)(x)\right| &\leq \sup _{\varepsilon>0 } \frac{A}{\varepsilon^n } ( I_1 + I_2 ) ) \\& \leq \left ( A\frac {\pi ^{\frac n2 }}{\Gamma ( \frac n2 +1 )} +A \frac {\pi ^{\frac n2 }}{\Gamma ( \frac n2 +1 )} \sum_{k=0}^\infty \frac{1}{2^{k-n } } \right) f^* \\&= A \frac {\pi ^{\frac n2 }}{\Gamma ( \frac n2 +1 )} (2^{n+1 }+1 ) f^*\\&=Cf ^* \end{aligned} $$\(\qquad \qquad \)