5

A technical step in proving Hardy's inequality $$ \int_{B(0,r)}\frac{\mu^{2}}{|x|^{2}}dx\le C\int_{B(0,r)}(|D\mu|^{2}+\frac{\mu^{2}}{r^{2}})dx $$ where $n>3, r>0, \mu\in H^{1}(B(0,r))$ is to show that $$ \int_{B(0,r)}\mu D\mu\cdot \frac{x}{|x|^{2}}dx\le C\int _{B(0,r)}|D\mu|^{2}dx+\frac{C}{r}\int_{\partial B(0,r)}\mu^{2}dS $$ (see Evans, Partial Differential Equations, page 297). This seems to be assumed implicitly and should be easy to prove, but I do not know how to show it. My guess is the second term on the right hand side may not even be needed, I added at here for completeness.

One strategy is to regard $\mu D\mu=\frac{1}{2}D(\mu^{2})$, and then use integration by parts. But this create a difficulty that since $udv=uv-vdu$, we have $D\frac{x}{|x|^{2}}=\frac{Dx*|x|^{2}-x*D|x|}{|x|^{4}}=\frac{n}{|x|^{2}}-\frac{1}{2|x|^{3}}$. This term obviously did not show up in the right hand side, and I also do not know how to bound it. Therefore this seems a wrong-headed strategy.

Another strategy, also based on integration by parts is to consider $A=\mu, B=D\mu, C=\frac{x}{|x|^{2}}$. Then this give rise to the same problem since at some point we have to differentiate $DC$. I am sure integration by parts would work, but for the moment I am stuck how to apply it properly in this case. I am looking for a hint, not a solution.

Bombyx mori
  • 19,638
  • 6
  • 52
  • 112

1 Answers1

6

For context, I "quoted" the beginning of proof below. The line "and consequently" does not involve integration by parts: that was already done. Instead, it involves the Peter-Paul inequality $$2ab\le \epsilon a^2+\epsilon^{-1}b^2$$ which allows us to absorb one term into the left hand side, which we are estimating. Common thing in PDE, by the way.

Before "consequently" we have $$ \int_B \frac{u^2}{|x|^2}\lesssim -2\int_B u Du \frac{x}{|x|^2} + \frac1r \int_{\partial B} u^2 $$ and want to do something to the first term on the right. Triangle and Peter-Paul:

$$ \left|2\int_B u Du \frac{x}{|x|^2} \right|\le 2\int_B |Du| \frac{|u|}{|x|} \le \epsilon \int_B \frac{u^2}{|x|^2} + \epsilon^{-1}\int_B |Du|^2 $$

Evans

  • @900situpsaday Based on your referral here from my recent question, I used your answer here and applied this Peter-Paul inequality to the line before "and consequently" in the textbook in order to obtain this: $$(2-n-\epsilon) \int_{B(0,r)} \frac{u^2}{|x|^2} dx \le \epsilon^{-1} \int_{B(0,r)} |Du|^2 dx - \frac 1r \int_{\partial B(0,r)} u^2 dS$$ and the desired result of $\text{(18)}$ follows immediately from this. The $(2-n-\epsilon)$ part of my equation: Is this what you mean by "absorbing" one term to the left side? I wanted to be sure on that, and also if my above estimate is correct. – Cookie Jul 29 '14 at 06:09
  • 2
    @legâteauaufromage I somehow missed your comment. Yes, the estimate looks right, except I think the sign in front of $1/r$ should be $+$. The absorbing trick is: if you want to estimate $X$ from above, it is enough to get $X\le c X + Y$ with some $c<1$, because then $cX$ is absorbed on the left and we get $X\le (1-c)^{-1}Y$. When using this trick, it is important to know that $X$ is finite. Otherwise subtracting $cX$ from both sides is not legitimate. –  Aug 07 '14 at 19:58
  • Sorry, but how can I obtain $\int_{B(0,r)}\frac{u^2}{|x|^2}dx\le C\int_{B(0,r)} |Du|^2dx+\frac{C}{r}\int_{\partial B(0,r)}u^2dS$ of $(2-n-\varepsilon)\int_{B(0,r)}\frac{u^2}{|x|^2}dx\le \varepsilon^{-1}\int_{B(0,r)} |Du|^2dx+\frac{1}{r}\int_{\partial B(0,r)}u^2dS$? Cause, $(2-n-\varepsilon)<0.$ – Irddo Jun 03 '15 at 04:08
  • @Irddo Yeah, I guess dragon should have divided by $(2-n)$ before estimating. Let's not get distracted by the old comment exchange; What I wrote in the answer should be enough. You begin with an equality; divide it by $(2-n)$, and proceed from there. In my post, $\lesssim$ means "less than some positive constant times". –  Jun 03 '15 at 04:18
  • Fine, thanks you again @HomegrownTomato. – Irddo Jun 03 '15 at 04:22