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Hardy's inequality (for integrals, I think) presented in Evans' PDE book (pages 296-297) contains a formula whose notation is substantially different than the conventional estimate presentation of $\|F\|_p \leq \frac{p}{p-1} \|f\|_p$.

THEOREM 7 (Hardy's inequality). Assume $n \ge 3$ and $r > 0$. Suppose that $u \in H^1(B(0,r))$. $\quad$Then $\frac{u}{|x|} \in L^2(B(0,r))$, with the estimate $$\int_{B(0,r)} \frac{u^2}{|x|^2}\tag{17} dx \le C \int_{B(0,r)} |Du|^2 + \frac{u^2}{r^2} dx.$$ Proof. We may assume $u \in C^\infty(B(0,r))$. Note that $D \left(\frac 1{|x|} \right)=-\frac x{|x|^3}$. Thus \begin{align} \int_{B(0,r)} \frac{u^2}{|x|^2} dx&=-\int_{B(0,r)} u^2 D \left(\frac 1{|x|} \right) \cdot \frac x{|x|} dx \\ &=\int_{B(0,r)} 2uDu \cdot \frac x{|x|^2} + (n-1) \frac{u^2}{|x|^2} \, dx - \int_{\partial B(0,r)} u^2 \nu \cdot \frac x{|x|^2} \, dS. \end{align} Therefore, $$(2-n) \int_{B(0,r)} \frac{u^2}{|x|^2} dx= 2 \int_{B(0,r)} uDu \cdot \frac x{|x|^2} dx - \frac 1r \int_{\partial B(0,r)} u^2 dS.$$

The proof continues in the book, but my question is how is the last equality derived? I tried multiplying the second-to-last equality by $2-n$ on both sides, but it's not an easy derivation...

Cookie
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1 Answers1

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Observe that $$\int_{\partial B(0,r)} u^2 \nu \cdot \frac x{|x|^2} \, dS = \frac{1}{r}\int_{\partial B(0,r)} u^2 \, dS $$ because $\nu$ is pointing in the direction of $x$, which makes $\nu\cdot x = |x|$.

After you plug this into the equation before "therefore", all that's left to do is to rearrange the term. The integral with $u^2/|x|^2$ goes to the left where it's combined with another integral of that form. Since $1-(n-1) = 2-n$, you get the left hand side of the last line.

Also, you have a more difficult step waiting for you in this proof: see A technical step in proving Hardy's inequality

Community
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  • Your answer cleared everything up with my question, though now I have one more somewhat different question: when right after the first equality we have the integral $- \int_{B(0,r)} u^2 D\left(\frac 1{|x|} \right) \cdot \frac x{|x|} , dx$, the next line applied the integration by parts formula from Evans' Theorem 2 of appendix section C.3, and they distributed the minus sign to both terms as well. I think I understand where the second term of $\int_{\partial B(0,r)} u^2 \nu \cdot \frac x{|x|^2} , dS$ is coming from, but how did they get the first more-complicated term? – Cookie Jul 28 '14 at 21:49
  • ...the first more-complicated term of $$\int_{B(0,r)} 2u Du \cdot \frac x{|x|^2} + (n-1) \frac{u^2}{|x|^2} dx$$ -- I have a strong feeling the product rule was used. – Cookie Jul 28 '14 at 21:51
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    @glace yes, it's integration by parts followed by the product rule. Integration by parts throws the derivative from $1/|x|$ onto $u^2\cdot x/|x|$, which is a product. Then $D(u^2\cdot x/|x|)$ gets expanded by the product rule. For example, $D(u^2)=2uDu$, and so on. –  Jul 28 '14 at 21:56