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通過定義證明下凸函數(shù)的積分不等式

2023-05-27 17:03 作者:~Sakuno醬 0人讀過 | 我要投稿

已知 $f(x)$ 是 $%5Ba%2Cb%5D$ 上的下凸函數(shù), 即對(duì)任意的 $x_1%2Cx_2%20%5Cin%20%5Ba%2Cb%5D%20$ 有? $f(%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D)%20%5Cle%20%5Cfrac%7Bf(x_1)%2Bf(x_2)%7D%7B2%7D%20$

證明1:

$(b-a)f(%5Cfrac%7Ba%2Bb%7D%7B2%7D)%20%5Cle%20%5Cint_%7Ba%7D%5E%7Bb%7D%7Bf(x)%7D%5Ctext%7Bd%7Dx$

網(wǎng)上的證明通常是通過泰勒公式證明的我這里嘗試通過積分的定義證明

首先用極限表示定積分

$%5Cint_%7Ba%7D%5E%7Bb%7D%7Bf(x)%7D%5Ctext%7Bd%7Dx%20%3D$ $%5Cint_%7Ba%7D%5E%7Bb%7D%7Bf(x)%7D%5Ctext%7Bd%7Dx%20%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7Bb-a%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20f(a%2B%5Cfrac%7Bb-a%7D%7Bn%7D%5Ccdot%20i)$

在這篇文章中我證明了通過中值定義的下凸函數(shù)可以推廣到更一般的形式，這里直接試使用結(jié)論了?https://www.bilibili.com/read/cv23934919

把? $%5Cfrac%7Bi%7D%7Bn%7D$ 看成? $t$ ?使用不等式? $f(a%2Bt(b-a))%20%5Cle%20f(a)%2Bt(f(b)-f(a))$

$%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7Bb-a%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20f(a%2B%5Cfrac%7Bb-a%7D%7Bn%7D%5Ccdot%20i)%20%5Cle%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7Bb-a%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20f(a)%2B%20%5Cfrac%7Bi%7D%7Bn%7D%20%5Ccdot%20(f(b)-f(a))$

$%5Cle%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7Bb-a%7D%7Bn%7D%20f(a)%5Ccdot%20n%2B%20%5Cfrac%7Bb-a%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cfrac%7Bi%7D%7Bn%7D%20%5Ccdot%20(f(b)-f(a))$

套用求和公式

$%5Cle%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D(b-a)%20%5Ccdot%20f(a)%2B%20%5Cfrac%7Bb-a%7D%7Bn%5E2%7D%20%5Ccdot%20(%5Cfrac%7Bn(n%2B1)%7D%7B2%7D)%20%5Ccdot%20(f(b)-f(a))$

$%5Cle%20(b-a)%20%5Ccdot%20f(a)%2B%20(%5Cfrac%7Bb-a%7D%7B2%7D)%20%5Ccdot%20(f(b)-f(a))$

$%5Cle%20(b-a)%20%5Cfrac%7Bf(a)%2Bf(b)%7D%7B2%7D$

另一個(gè)不等式也是同理

證明2:

$%5Cint_%7Ba%7D%5E%7Bb%7D%7Bf(x)%7D%5Ctext%7Bd%7Dx%20%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7Bb-a%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20f(a%2B%5Cfrac%7Bb-a%7D%7Bn%7D%5Ccdot%20i)$

這里把 $n$ 變成 $2n$ 然后首尾相加湊不等式

$%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7Bb-a%7D%7B2n%7D%5Csum_%7Bi%3D1%7D%5E%7B2n%7D%20f(a%2B%5Cfrac%7Bb-a%7D%7B2n%7D%5Ccdot%20i)%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7Bb-a%7D%7B2n%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20f(a%2B%5Cfrac%7Bb-a%7D%7B2n%7D%5Ccdot%20i)%20%2B%20f(a%2B%5Cfrac%7Bb-a%7D%7B2n%7D(2n%2B1-i))$