Hi,


I need an inequality similar to this one for bounded domain [0,L].
<http://img94.imageshack.us/img94/3166/screenshot1qy.png>


My u(x) is not 0 on the boundary.


I will appreciate if you can help me about this question,


Edit: More precisely, is the following the statement true?


For $F\in C^1[0,L]$ with $F(0)=0$ or $\int\_0^L
F(x)dx=0$, one has the estimate
$$
\int\_0^L|F(x)|^px^{\beta}dx\leq
c\int\_0^L|F'(x)|^px^{\beta+p}dx
$$



Weighted Hardy Inequality for bounded domains
I don't think what you want to prove is true just assuming $f(0) = 0$. Let $f\_\sigma(x) = x^{\sigma}$. Then $x f\_\sigma'(x) = \sigma f\_\sigma(x)$, and $f(0) = 0$. (Okay, so $f$ is not strictly in $C^1$, but you can chomp off a bit to smooth it out and take limits.) Now the left hand side is equal to $\int\_0^1 f\_\sigma^2 dx$, and theright hand side is $c\*\sigma^2 \int\_0^1 f\_\sigma^2 dx$. Now take limit $\sigma \searrow 0$ and you derive a contradiction. 


But in the case where $f$ has mean zero your inequality is true. I'll prove it (per your comments) for the $L^2$ case with $x^2$ weight, but it works in general. 


Let $f\in C^1[0,1]$ be a function with mean $0$. Consider the function $g(x) = x f(x)^2$. By the fundamental theorem of calculus $\int\_0^1 g'(x) dx = f(1)^2$. On the other hand, $g'(x) = f(x)^2 + 2 x f(x) f'(x)$, so we have the following 
$$ \int\_0^1 f(x)^2 dx = f(1)^2 - 2 \int\_0^2 x f(x) f'(x) dx \leq f(1)^2 + \epsilon \int\_0^1 f(x)^2 dx + \epsilon^{-1} \int\_0^1 (x f'(x))^2 dx $$
using Young's inequality. 


Now $f(1) = \int\_0^1 (x f(x))' dx = \int\_0^1 f(x) dx + \int\_0^1 x f'(x) dx$ again by the fundamental theorem. Using that $f$ has mean 0, the first term drops out. By Holder's inequality
$$ \left|\int\_0^1 xf'(x) dx\right|^2 \leq \int\_0^1 1^2 dx \int\_0^1 (x f'(x))^2 dx $$
So we have that, collecting all the terms
$$ \int\_0^1 f(x)^2 dx \leq \frac{1+\epsilon^{-1}}{1-\epsilon} \int\_0^1 x^2 f'(x)^2 dx $$
for any $\epsilon \in (0,1)$. You can try to optimize what the minimal one should be.


Just remember that Hardy's inequality is a weighted version of integration by parts. 


