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一道积分不等式的证明

开发技术 开发技术 2021-11-14 175次浏览

一道积分不等式的证明

题目:

(f(x,y))(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| iintlimits_D f(x, y)dx right| le frac{pi}{3}maxlimits_{(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 (0le rle 1) ):

[begin{aligned} |f(x, y)| &= |f(rcos theta, r sin theta)| \ &= left|int_r^1 left( frac{part f}{part x}cos theta+frac{part f}{part y}sin theta right) dtright| \ &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^2theta+sin^2theta]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) maxlimits_{(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^{2pi}int_0^1 rcdot f(rcostheta, rsintheta)drdtheta right| \ &le left| int_0^{2pi}int_0^1 rcdot (1-r) maxlimits_{(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}} drdtheta right| \ &= left| int_0^{2pi}int_0^1 (r-r^2)drdtheta right| cdot maxlimits_{(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} maxlimits_{(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.

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