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Does there exist a constant $C >0$ such that $$\int_{\mathbb R^2} |\nabla w(x)|^2 dx + \int_{\mathbb R^2} w(x)^2\frac{1}{1+|x|^2} dx \geq C \left(\int_{\mathbb R^2} w(x)^6 dx\right)^{\frac 13},$$ for any function $w \in C_0^\infty(\mathbb R^2)$?

This inequality appears in my recent research. Does anyone know it or have references for it?

Thanks,

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  • $\begingroup$ These type of inequalities are called scale invariant family of Sobolev inequalities. Look at this paper (math.cornell.edu/~lsc/sprgsobc.pdf) for definitions and some more. I don't have a reference for the above result. $\endgroup$
    – T. Amdeberhan
    Commented Jan 4, 2017 at 3:39
  • $\begingroup$ @T.Amdeberhan: Thanks for your comment. However, this inequality is not scale invariant. $\endgroup$
    – nguyen0610
    Commented Jan 6, 2017 at 12:15
  • $\begingroup$ Take $w=1$ in the disk of radius $R$ centered at the origin and then descend to $0$ within the disk of radius $2R$ keeping the gradient of order $1/R$. Then it looks like the LHS is approximately $1+\log R$ while the RHS is $R^{2/3}$. Am I missing something? $\endgroup$
    – fedja
    Commented Jan 8, 2017 at 18:30
  • $\begingroup$ @fedja: Thanks for your comments. After posting this question, I realized that it does not hold in general. The argument is simple as follows. If there exists $C$, then for any $w$, applying the inequality for $w_a(x) = w(x-a)$, and letting $a\to \infty$ then the integral involving to $w_a^2$ tends to zero. Hence we get $$\int |\nabla w|^2 dx \geq C \left(\int w^6 dx\right)^{\frac 13},$$ for any $w\in C_0^\infty(\mathbb R^2)$. The last inequality does not hold because of lossing the scale invariant. $\endgroup$
    – nguyen0610
    Commented Jan 9, 2017 at 10:08

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