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Let $u\in \mathcal C^1_c(\mathbb R^n)$. I have that $$\int_{R_1}|u|^r\leq C\left(\int_{R_1}|\nabla u|^p\right)^{ar/p}\left(\int_{R_1}|u|^q\right)^{(1-a)r/q},$$ where $R_\rho=\{x\in \mathbb R^n\mid \rho\leq |x|<2\rho\}.$

Then it's written in my course that : Rescaling and multiplying by $\rho^{\gamma r}$ we get $$\int_{R_\rho}|x|^{\gamma r}|u|^r\leq C\left(\int_{R_\rho}|x|^{\alpha r}|\nabla u|^p\right)^{ar/p}\left(\int_{R_\rho}|x|^{\beta q}|u|^q\right)^{(1-a)r/q},$$ where $$\frac{1}{r}+\frac{\gamma }{n}=a\left(\frac{1}{p}-\frac{\alpha -1}{n}\right)+(1-a)\left(\frac{1}{q}+\frac{\beta }{n}\right),$$ $\gamma =a\varphi+(1-a)\beta $.

Question

Could someone explain me how to get this ? I tried a substitution as $y=\rho^{\gamma r}x$, but no $|x|$ appear.

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  • $\begingroup$ YOUR MISTAKE*: Don't Forget that $x\in R_1=\{x\in \mathbb R^n\mid 1\leq |x|<2 \} $ is set So putting $$y=\rho^{\gamma r}R_1$$ Does not make any sense as change of variables but rather you should put $$y=\rho^{\gamma r}x~~~\text{with}~~x\in R_1$$ $$R_\rho=\{x\in \mathbb R^n\mid \rho\leq |x|<2\rho\}.$$ $\endgroup$ – Guy Fsone Oct 15 '17 at 16:22
  • $\begingroup$ @GuyFsone: Thanks a lot for your answer. Unfortunately, after the substitution the $|x|^\gamma $ doesn't appear, no ? Because $dy=\rho^{\gamma r}dx.$ $\endgroup$ – user380364 Oct 15 '17 at 17:58
  • $\begingroup$ becarefull your jacobian should be raise to power n. check again $\endgroup$ – Guy Fsone Oct 15 '17 at 18:12
  • $\begingroup$ @GuyFsone: Yes indeed, thank :-) But I still no have $|x|$ that appear. $\endgroup$ – user380364 Oct 16 '17 at 8:52
  • $\begingroup$ you finally completly changed your post now the problem absolutely clear $\endgroup$ – Guy Fsone Oct 20 '17 at 18:25
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Let $u : R_\rho \to \mathbb{R}$ and then rescale by defining $v : R_1 \to \mathbb{R}$ via $v(x) = u(\rho x)$. We apply your inequality to $v$ to get $$ \int_{R_1}|v|^r\leq C\left(\int_{R_1}|\nabla v|^p\right)^{ar/p}\left(\int_{R_1}|v|^q\right)^{(1-a)r/q}. $$ Now we change variables: $$ \int_{R_1} |v(x)|^r dx = \int_{R_1} |u(\rho x)|^r dx = \rho^{-n} \int_{R_\rho} |u(y)|^r dy, $$ $$ \int_{R_1} |v(x)|^q dx = \int_{R_1} |u(\rho x)|^q dx = \rho^{-n} \int_{R_\rho} |u(y)|^q dy, $$ and $$ \int_{R_1} |\nabla v(x)|^p dx = \int_{R_1} \rho^p |\nabla u(\rho x)|^p dx = \rho^{p-n} \int_{R_\rho} |\nabla u(y)|^p dy. $$

Now, for $y \in R_\rho$ we have that $\rho \le |y| \le 2 \rho$, and so $\rho^\delta \le |y|^\delta \le 2^\delta \rho^\delta$ for any $\delta >0$. Thus $$ \rho^{-n} \int_{R_\rho} |u(y)|^r dy \ge \frac{\rho^{-\gamma r-n}}{2^{\gamma r}} \int_{R_\rho} |y|^{\gamma r}|u(y)|^r dy, $$ $$ \rho^{-n} \int_{R_\rho} |u(y)|^q dy \le \rho^{-\beta q-n} \int_{R_\rho} |y|^{\beta q} |u(y)|^q dy, $$ and $$ \rho^{p-n} \int_{R_\rho} |\nabla u(y)|^p dy \le \rho^{p-n-\alpha p} \int_{R_\rho} |y|^{\alpha p}|\nabla u(y)|^p dy. $$

Combining all these shows that $$ \frac{\rho^{-\gamma r-n}}{2^{\gamma r}} \int_{R_\rho} |y|^{\gamma r}|u(y)|^r dy \le C \left( \rho^{p-n-\alpha p} \int_{R_\rho} |y|^{\alpha p}|\nabla u(y)|^p dy \right)^{ar/p} \left( \rho^{-\beta q-n} \int_{R_\rho} |y|^{\beta q} |u(y)|^q dy \right)^{(1-a)r/q}. $$

Now, if $\alpha,\beta,\gamma,p,q,r,a$ are chosen so that $$ (1-a)\left( \frac{\beta}{n} + \frac{1}{q} \right) +a\left(\frac{1}{p} + \frac{\alpha-1}{n} \right) = \frac{\gamma}{n} + \frac{1}{r} $$ then the $\rho$ terms exactly cancel out and we're left with $$ \int_{R_\rho} |y|^{\gamma r}|u(y)|^r dy \le 2^{\gamma r} C \left( \int_{R_\rho} |y|^{\alpha p}|\nabla u(y)|^p dy \right)^{ar/p} \left( \int_{R_\rho} |y|^{\beta q} |u(y)|^q dy \right)^{(1-a)r/q}. $$ I realize this isn't exactly what you've asked for, but I suspect there are two typos in your post. First, I suspect you want the power to be $\alpha p$ in the gradient integral, and second I suspect there's a sign error on the $\alpha-1$ term. At least, this is what I get when I adjust the power in the gradient term.

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  • $\begingroup$ In fact it answer totally to my answer. Thank you very much :-) $\endgroup$ – user380364 Oct 21 '17 at 9:43

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