一道积分不等式的证明

一道积分不等式的证明

题目：

设 \(f(x,y)\) 在 \(D=\{(x,y)|x^2+y^2 \le 1\}\) 上有连续的偏导数，且在 \(T=\{(x,y)|x^2+y^2=1\}\) 上恒为 \(0\)，证明：

\[\left| \iint\limits_D f(x, y)dx \right| \le \frac{\pi}{3}\max\limits_{(x,y)\in D}\left[ \left(\frac{\part f}{\part x}\right)^2+\left(\frac{\part f}{\part y}\right)^2 \right]^{\frac{1}{2}} \]

分析

考虑到不等式右边的形式类似于向量模长，尝试使用 \(Cauchy-Schwarz\) 不等式：

\[|(x, y)| \le \| x \| \cdot \| y \| \]

证明

首先对 \(\text{LHS}\) 中被积表达进行代换（ 约定 \(0\le r\le 1\) ）：

\[\begin{aligned} |f(x, y)| &= |f(r\cos \theta, r \sin \theta)| \\ &= \left|\int_r^1 \left( \frac{\part f}{\part x}\cos \theta+\frac{\part f}{\part y}\sin \theta \right) dt\right| \\ &\le \int_r^1 \left|\left( \frac{\part f}{\part x}\cos \theta+\frac{\part f}{\part y}\sin \theta \right)\right| dt \\ &\le \int_r^1 \left[ \left(\frac{\part f}{\part x}\right)^2+\left(\frac{\part f}{\part y}\right)^2 \right]^{\frac{1}{2}}dt \int_r^1[\cos^2\theta+\sin^2\theta]dt \\ &= (1-r) \int_r^1 \left[ \left(\frac{\part f}{\part x}\right)^2+\left(\frac{\part f}{\part y}\right)^2 \right]^{\frac{1}{2}}dt \\ &\le (1-r) \int_0^1 \left[ \left(\frac{\part f}{\part x}\right)^2+\left(\frac{\part f}{\part y}\right)^2 \right]^{\frac{1}{2}}dt \\ &=(1-r) \max\limits_{(x,y)\in D}\left[ \left(\frac{\part f}{\part x}\right)^2+\left(\frac{\part f}{\part y}\right)^2 \right]^{\frac{1}{2}}\\ \end{aligned} \]

将上述不等式代入 \(\text{LHS}\) 化简：

\[\begin{aligned} \text{LHS} &= \left| \int_0^{2\pi}\int_0^1 r\cdot f(r\cos\theta, r\sin\theta)drd\theta \right| \\ &\le \left| \int_0^{2\pi}\int_0^1 r\cdot (1-r) \max\limits_{(x,y)\in D}\left[ \left(\frac{\part f}{\part x}\right)^2+\left(\frac{\part f}{\part y}\right)^2 \right]^{\frac{1}{2}} drd\theta \right| \\ &= \left| \int_0^{2\pi}\int_0^1 (r-r^2)drd\theta \right| \cdot \max\limits_{(x,y)\in D}\left[ \left(\frac{\part f}{\part x}\right)^2+\left(\frac{\part f}{\part y}\right)^2 \right]^{\frac{1}{2}} \\ &= \frac{\pi}{3} \max\limits_{(x,y)\in D}\left[ \left(\frac{\part f}{\part x}\right)^2+\left(\frac{\part f}{\part y}\right)^2 \right]^{\frac{1}{2}} = \text{RHS}\\ \end{aligned} \]

QED.