欢迎光临散文网 会员登陆 & 注册

从二阶导大于0证明下凸性的一种思路

2023-05-29 10:39 作者:~Sakuno酱  | 我要投稿

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

在这篇文章中我们尝试通过二阶导函数大于零证明f(%5Cfrac%7Ba%2Bb%7D%7B2%7D)%20%5Cle%20%5Cfrac%7Bf(a)%2Bf(b)%7D%7B2%7D

这里因为%5Cfrac%7Ba%2Bb%7D%7B2%7D刚好到a的距离和到b的距离都是相等的所以我们通过加一项减一项的方式构造出导数

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)

假如我们换成更通用的形式,想证明

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

首先我们还是先尝试作差,因为 x%20%5Clt%20x%2B%5Calpha%20(y-x)%20%3C%20y 所以我们用最大项减去中间那项

在用中间那项减去第一项

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

再运用拉格朗日中值定理

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

我们发现减号左边乘以一个 %5Calpha 减号右边乘以一个 1-%5Calpha 就可以提取出f'(z_2)%20-%20f'(z1)

因此作出尝试

%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))

巧合的是结果正是我们想证明的不等式

于是我们就有了

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))

因为 z_2%20%3E%20z_1而且一阶导数单调增 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))




从二阶导大于0证明下凸性的一种思路的评论 (共 条)

分享到微博请遵守国家法律