Timeline for A Sobolev type inequality
Current License: CC BY-SA 3.0
5 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jan 9, 2017 at 10:08 | comment | added | nguyen0610 | @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. | |
Jan 8, 2017 at 18:30 | comment | added | fedja | 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? | |
Jan 6, 2017 at 12:15 | comment | added | nguyen0610 | @T.Amdeberhan: Thanks for your comment. However, this inequality is not scale invariant. | |
Jan 4, 2017 at 3:39 | comment | added | T. Amdeberhan | 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. | |
Jan 4, 2017 at 0:38 | history | asked | nguyen0610 | CC BY-SA 3.0 |