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Suppose $$ (*) \quad \|u\|_{L^q(\mathbb{R^N})} \le c \|\nabla u\|_{L^q(\mathbb{R^N;\mathbb{R^N}})}. $$ Assume $u\in C_c^1(\mathbb{R}^N)$ and for $r>0$ defined the rescaled function $$u_r(x) := u(rx), \quad x \in \mathbb{R}^N.$$

If $(*)$ holds for $u_r$ then \begin{align} \bigg(\int_{\mathbb{R}^N}|u(rx)|^q dx\bigg)^{1/q} & = \bigg(\int_{\mathbb{R}^N}|u_r(x)|^q dx\bigg)^{1/q} \\ & \le c\bigg(\int_{\mathbb{R}^N}|\nabla u_r(x)|^p dx\bigg)^{1/p} \\ & = c\bigg(r^p \int_{\mathbb{R}^N}|\nabla u(rx)|^p dx\bigg)^{1/p} \\ \end{align}

I don't see where the $r^p$ term is appearing from?

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    $\begingroup$ Keyword: Change of variable. $\endgroup$ – Did Sep 29 '17 at 11:32
  • $\begingroup$ I dont see how change of variables gets used. By definition $u_r(x) = u(rx)$ so my attempt is $$c\bigg(\int_{\mathbb{R}^N}|\nabla u_r(x)|^p dx\bigg)^{1/p} = c\bigg(\int_{\mathbb{R}^N}|\nabla u(rx)|^p dx\bigg)^{1/p}.$$ What did I do wrong? $\endgroup$ – Riggs Sep 29 '17 at 13:59
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    $\begingroup$ But $\nabla u_r(x)\ne\nabla u(rx)$... Why not come back to the basics when confused as you seem? In the present case, $u_r(x+h)-u_r(x)=u(rx+rh)-u(rx)\approx\nabla u(rx)\cdot(rh)=r\nabla u(rx)\cdot h$ for every $h$ hence... $\endgroup$ – Did Sep 29 '17 at 15:22
  • $\begingroup$ @Did I don't see how you can just say $\nabla u_r(x) \neq \nabla u(rx)$. By definition $u_r(x) = u(rx)$, so if we take the gradient of both side we get precisely $\nabla u_r(x) = \nabla u(rx)$. $\endgroup$ – Riggs Oct 8 '17 at 6:54
  • $\begingroup$ So... obviously you simply did not read my comment. Please note that the simple computation in it disproves your claim (except if $r=1$ or $\nabla u(rx)=0$, of course). $\endgroup$ – Did Oct 8 '17 at 8:33
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Hint: By Chain rule we have $\partial_i [u(rx) ]= r[\partial_i u](rx) .$ This yields

$$[\nabla u_r](x) =[\nabla u(\cdot r)](x)=r [\nabla u](rx)$$ Then integrates on both side.

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  • $\begingroup$ This notation is a bit confusing..is the gradient with respect to $rx$ in the last item? $\endgroup$ – eurocoder Sep 29 '17 at 13:52

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