积分不等式

血狼王

令$f\in C[0,1]$且$f(0)=f(1)=0.$
求证有$$4\int_0^1 f^2(x)\mathrm{d}x\leq \left( \int_0^1 |f'(x)|\mathrm{d}x \right)^2$$成立.

又：本题结论能否再加强？

hbghlyj

类似：Wirtinger's inequality for functions

Let y be a continuous and differentiable function on the interval $[0, L]$ with $y(0) = y(L) = 0$. Then
$ \int _{0}^{L}y(x)^{2}\,\mathrm {d} x\leq {\frac {L^{2}}{\pi ^{2}}}\int _{0}^{L}y'(x)^{2}\,\mathrm {d} x, $
and equality holds if and only if $y(x) = c \sin(⁠πx/L)$ ⁠for some number $c$.

血狼王

hbghlyj 发表于 2024-11-5 04:25
类似：Wirtinger's inequality for functions

Let y be a continuous and differentiable function on the ...

确实乍看起来象Wirtinger不等式，但是正弦显然不是正解
所以能证吗？

hbghlyj

血狼王 发表于 2024-11-5 16:58
所以能证吗？

令 $0<x_1<x_2<\dots<1$ 为 $f'(x)$ 改变符号的点，則 $\int|f'(x)|\rmd x = \sum \left|f(x_{j-1}) - f(x_j) \right| $.
其中一个点 $x_k$ 使 $|f|$ 达到最大值，即$|f(x_k)|=\max|f|$，則$\int|f'(x)|\rmd x\ge|0-f(x_k)|+|f(x_k)-0|$.
即$\int|f'(x)|\rmd x\ge2|f(x_k)|$.
而$|f(x_k)|\ge|f(x)|\;\forall x$,
即$\int|f'(x)|\rmd x\ge2|f(x)|$.
平方，$(\int|f'(x)|\rmd x)^2\ge4f(x)^2$.
积分，因左側是常数，$1\cdot(\int|f'(x)|\rmd x)^2\ge4\int_0^1f(x)^2$.

hbghlyj

血狼王 发表于 2024-11-4 16:18
又：本题结论能否再加强？

取$f_n(x)=\begin{cases}nx&0\le x\le\frac1n\\
1&\frac1n\le x\le1-\frac1n\\
n(1-x)&1-\frac1n\le x\le 1\end{cases}$
则$\int_0^1|f_n'(x)|\rmd x=2$,
$\int_0^1f_n(x)^2\rmd x\to\int_0^1\rmd x=1$,
$\frac{\left( \int_0^1 |f_n'(x)|\mathrm{d}x \right)^2}{\int_0^1 f_n(x)^2\mathrm{d}x}\to4$,
所以当$n\to\infty$时不等式趋于等式。所以常数$4$是最佳的。

hbghlyj

血狼王 发表于 2024-11-5 16:58
确实乍看起来象Wirtinger不等式

还有一个类似的（但不同的）：Friedrichs' Inequality
Proving Poincare in One Dimension

Let $G \subset \mathbb R^n$ be bounded domain.
Then for any $u \in W^{1, 2}_0(G)$:
$$\|u\|_{L^2(G)} \le \operatorname {diam}(G)\|\nabla u\|_{L^2(G)}$$
where $\operatorname {diam} (G) := \sup \limits_{x, y \in G} |x - y|$

取$G=[0,1]$，有$\operatorname {diam}([0,1])=1$，得到$$\forall u \in W^{1, 2}_0(G):\quad\|u\|_{L^2([0,1])} \le\|u'\|_{L^2([0,1])}$$
Poincarè inequality in dimension n=1这里也说$\bigg(\int_{[0,1]}u(x)dx\bigg)^2\leq C\int_{[0,1]}u'(x)^2dx$的最佳常数$C$是$1$.