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How to show that by replacing $\psi(x)$ with $\psi(\lambda x)$, an inequality of the form $$ \int|\nabla\psi|^2dx\geq C\left(\int|\psi|^qdx\right)^\frac{2}{q} $$ can only hold for $q=6$ in $N=3$?

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Let $u_{\lambda}(x):=\psi(\lambda x)$. If $\psi$ is infinitely many times differentiable, then $\frac{\partial }{\partial x_j}u_{\lambda}(\lambda x)=\lambda \frac{\partial u_{\lambda}}{\partial x_j}$, and when we integrate (assuming it's over $\Bbb R$), we make the substitution $t=\lambda x$, so the Jacobian of the inverse is $\frac 1{\lambda^3}$ and we would have $$\int_{\Bbb R^3}|\nabla \psi(x)|^2\lambda^2dx\lambda^{-3}=\frac 1{\lambda}\int_{\Bbb R^3}|\nabla \psi(x)|^2dx\geq C\frac 1{\lambda^{6/q}}\left(\int_{\Bbb R^3}|\psi(x)|^qdx\right)^{2/q}$$ hence $$\lambda^{6/q-1}\int_{\Bbb R^3}|\nabla \psi(x)|^2dx\geq C\left(\int_{\Bbb R^3}|\psi(x)|^qdx\right)^{2/q}.$$ If $q>6$, then $6/q-1<0$, and you get a contradiction when $\lambda\to 0^+$. If $q<6$, make $\lambda$ go to $+\infty$ to get what we want.


This argument helps us when we forget the critical exponent in Sobolev embeddings.

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@ David , i didn't understand few things : Jacobian of the inverse and how u introduced the inequality in the first line . Could u explain for me? :) –  Theorem May 13 '12 at 18:38
    
It's the term which appear when you do a substitution in a multiple integral. In the first line, I applied chain rule. –  Davide Giraudo May 13 '12 at 18:39
    
@DavideGiraudo Thanks! –  Ziqian Xie May 13 '12 at 18:42
    
Thanks David. I was just wondering what this inequality may mean ? any suggestions ? –  Theorem May 13 '12 at 18:44

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