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從二階導(dǎo)大于0證明下凸性的一種思路

2023-05-29 10:39 作者:~Sakuno醬  | 我要投稿

https://www.bilibili.com/read/cv23960056

在這篇文章中我們嘗試通過二階導(dǎo)函數(shù)大于零證明f(%5Cfrac%7Ba%2Bb%7D%7B2%7D)%20%5Cle%20%5Cfrac%7Bf(a)%2Bf(b)%7D%7B2%7D

這里因為%5Cfrac%7Ba%2Bb%7D%7B2%7D剛好到a的距離和到b的距離都是相等的所以我們通過加一項減一項的方式構(gòu)造出導(dǎo)數(shù)

f(a)%2Bf(b)-2f(%5Cfrac%7Ba%2Bb%7D%7B2%7D)%3Df(b)-f(%5Cfrac%7Ba%2Bb%7D%7B2%7D)%20-%20(f(%5Cfrac%7Ba%2Bb%7D%7B2%7D)%20-%20f(a))

%3Df'(%5Cbeta)(%5Cfrac%7Bb-a%7D%7B2%7D)%20-f'(%5Calpha)(%5Cfrac%7Bb-a%7D%7B2%7D)

假如我們換成更通用的形式,想證明

f(x%2B%5Calpha(y-x))%20%5Clt%20f(x)%20%2B%20%5Calpha%20(f(y)-f(x))? ? ? 其中?x%3Cy%20? ??0%3C%5Calpha%20%3C%201

首先我們還是先嘗試作差,因為?x%20%5Clt%20x%2B%5Calpha%20(y-x)%20%3C%20y 所以我們用最大項減去中間那項

在用中間那項減去第一項

f(y)%20-f(x%2B%5Calpha%20(y-x))%20-(f(x%2B%5Calpha%20(y-x))%20-%20f(x))

再運用拉格朗日中值定理

%3Df'(z_2)(y-x)(1-%5Calpha)%20-%20f'(z_1)(y-x)%5Calpha

我們發(fā)現(xiàn)減號左邊乘以一個?%5Calpha 減號右邊乘以一個?1-%5Calpha 就可以提取出f'(z_2)%20-%20f'(z1)

因此作出嘗試

%5Calpha(f(y)%20-f(x%2B%5Calpha%20(y-x)))%20-(1-%5Calpha)(f(x%2B%5Calpha%20(y-x))%20-%20f(x))

%3D%5Calpha%20f(y)%20-%20%5Calpha%20f(x%2B%5Calpha%20(y-x))%20-%20%20f(x%2B%5Calpha%20(y-x))%20%2B%20%5Calpha%20f(x%2B%5Calpha%20(y-x))%20%20%2B%20f(x)%20-%20%5Calpha%20f(x)

%3D%20f(x)%20%20%2B%20%5Calpha%20(f(y)%20-f(x))%20-%20f(x%2B%20%5Calpha(y-x))

巧合的是結(jié)果正是我們想證明的不等式

于是我們就有了

f(x)%20%20%2B%20%5Calpha%20(f(y)%20-f(x))%20-%20f(x%2B%20%5Calpha(y-x))

%3D%5Calpha(f(y)%20-f(x%2B%5Calpha%20(y-x)))%20-(1-%5Calpha)(f(x%2B%5Calpha%20(y-x))%20-%20f(x))

%3D%20%5Calpha%20(1-%5Calpha)%20(y-x)(f'(z_2)%20-%20f'(z_1))

因為?z_2%20%3E%20z_1而且一階導(dǎo)數(shù)單調(diào)增 所以必有?f(x)%20%20%2B%20%5Calpha%20(f(y)%20-f(x))%20-%20f(x%2B%20%5Calpha(y-x))%20%3E%200

也就是

f(x%2B%5Calpha(y-x))%20%5Clt%20f(x)%20%2B%20%5Calpha%20(f(y)-f(x))




從二階導(dǎo)大于0證明下凸性的一種思路的評論 (共 條)

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