散文網(wǎng) » 科技 »學習 » 二階可導(dǎo)的下凸函數(shù)，導(dǎo)數(shù)一定非負嗎

二階可導(dǎo)的下凸函數(shù)，導(dǎo)數(shù)一定非負嗎

2023-05-27 17:40 作者:~Sakuno醬 0人讀過 | 我要投稿

是的下面給出證明

假設(shè)函數(shù) $f(x)$ 在 $%5Ba%2Cb%5D$ 上下凸，即對任意 $x_1%2Cx_2%20%5Cin%20%5Ba%2Cb%5D%20$ 有 $f(%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D)%20%5Cle%20%5Cfrac%7Bf(x_1)%2Bf(x_2)%7D%7B2%7D%20$

并且 $f(x)$ 在 $(a%2Cb)$ 上二階可導(dǎo)

通過定義寫出 $f(x_0)$ 二階導(dǎo)數(shù)

$f''(x_0)%20%3D%20%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7Bf'(x_0%2Bx)-f'(x_0)%7D%7Bx%7D$

$f''(x_0)%20%3D%20%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7Bf'(x_0-x)-f'(x_0)%7D%7B-x%7D$

把上面兩個相加得到

$2f'(x_0)%20%3D%20%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7Bf'(x_0%2Bx)-f'(x_0-x)%7D%7Bx%7D$

因為 $y%3Dx%5E2%20$ 和? $y%3Df(x)$ 都是一階可導(dǎo)并且導(dǎo)函數(shù)連續(xù) 所以可以反向用一下洛必達法則

$%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7Bf'(x_0%2Bx)-f'(x_0-x)%7D%7Bx%7D%3D%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7Bf(x_0%2Bx)%2Bf(x_0-x)-2f(x_0)%7D%7Bx%5E2%7D%5Ccdot2$

套用不等式? $f(%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D)%20%5Cle%20%5Cfrac%7Bf(x_1)%2Bf(x_2)%7D%7B2%7D%20$

得到

$%5Cfrac%7Bf(x_0%2Bx)%2Bf(x_0-x)-2f(x_0)%7D%7Bx%5E2%7D%5Ccdot2%20%5Cge%20%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7B2f(x_0)-2f(x_0)%7D%7Bx%5E2%7D%5Ccdot2%20%5Cge%200$

嚴謹?shù)脑捒梢詷?gòu)造數(shù)列? $a_n%20%3D%5Cfrac%7B1%7D%7Bn%7D$ ?把? $x$ ?換成 $a_n$ ?

構(gòu)造出? $g(a_n)%3D%5Cfrac%7Bf(x_0%2Ba_n)%2Bf(x_0-a_n)-2f(x_0)%7D%7Ba_n%5E2%7D$ ?結(jié)果就是 $g(a_n)$ 每一項都是非負的

因此? $%5Clim_%7Bn%20%5Cto%20%5Cinfty%7Dg(a_n)%20%5Cge%200$

所以 $f''(x_0)%20%5Cge%200$

同樣二階導(dǎo)非負也可以證出下凸性

$f(x_1)-f(%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D)%20%2Bf(x_2)-%20f(%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D)%20%0A$

$%3Df(x_2)-%20f(%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D)%20-(%20f(%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D)%20-%20f(x_1))$ ?

運用拉格朗日中值定理

$%3Df'(%5Calpha)%20-f'(%5Cbeta)%20$ 其中? $%5Cbeta%20%5Cle%20%5Calpha$

根據(jù)一階導(dǎo)數(shù)的單調(diào)性

$f(x_1)%2Bf(x_2)%20-%202f(%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D)%3D%20f(x_1)-f(%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D)%20%2Bf(x_2)-%20f(%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D)%20%0A%20%5Cge%200$