问题

分析 >> 数学分析 >> 定积分
Questions in category: 定积分 (Definite Integral).

Wirtinger 不等式(Wirtinger's inequality)

Posted by haifeng on 2017-03-14 08:38:32 last update 2022-04-20 10:42:51 | Answers (0) | 收藏


假设 $g\in C^1[a,b]$, 且 $g(a)=g(b)=0$. 则有

\[
\int_a^b g^2(t)dt\leqslant\Bigl(\frac{b-a}{\pi}\Bigr)^2\int_a^b|g'(t)|^2 dt.
\]

 


Remark:

高维的 Wirtinger 不等式又称为球面上的 Poincaré 不等式.

 

References:

https://arxiv.org/abs/1407.6871
https://en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions

Wirtinger's inequality is seen as the one-dimensional version of Friedrichs' inequality.