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 $$ \int_0^L|F(x)|^px^{\beta}dx\leq c\int_0^L|F'(x)|^px^{\beta+p}dx $$
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.
Sorry, I can't access the image on my browser; it won't load for some reason. Could you please type out the inequality in Latex?
So I'll just have to guess what you might want; apologies if this is not what you're asking. The classical one-dimensional Hardy inequality
$$ \int_0^\infty \left| \frac{1}{t} \int_0^t f(s) ds \right|^2 dt \leq 4 \int_0^\infty |f(s)|^2 ds $$
has a generalisation for weights: for any functions $m,w \geq 0$, the best constant $M=M(m,w)$ in the inequality
$$ \int_0^\infty \left| \int_0^t f(s) ds \right|^2 m(t) dt \leq M \int_0^\infty |f(s)|^2 w(s)ds $$
is related to the quantity $$ S(m,w) = \sup_{R>0} \left( \int_R^\infty m(t) dt \right) \left( \int_0^R w(s)^{-1} ds \right) $$ by $S \leq M \leq 4S$. Also, if either $M$ or $S$ equals $+\infty$, then so does the other (i.e. we don't have any reasonable Hardy inequality).
There are generalisations to $L^p$, $L^q$ norms, and $d\mu(t)$ a measure instead of just $m(t)dt$. If you look up Hardy's Inequality with Weights by B. Muckenhoupt from about 1972, and newer related papers, you'll get loads of articles.
However, the multidimensional questions are more difficult and much less is known.
So, if you just wanted a one-dimensional version on $[0,L]$ instead of $[0,\infty)$, you can just set $m(t) = 0$ and $w(s) = +\infty$ for $t,s > L$.
Actually you need some boundary condition like "u(0)=0", or an integral condition like "u has zero mean", in order to get rid of the constant functions. Otherwise clearly you can't bound a norm of u with a norm of u'. Note that there is no problem with constant functions in the Hardy inequality in [0, ∞) for u is assumed to be in a weighted Sobolev space. Also note that the ratio LHS/RHS in your inequality does not depend on L, so (once you have chosen the constraint as I wrote) you may fix L=1.
