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Let $\Omega$ be a sufficiently nice domain in $\mathbb{R}^n$.

If $ 1 \leq p < n $ and $ p^* = \frac{np}{n-p} $ then there exists a constant $C_1$ such that for all $ u \in W^{1,p}(\Omega) $ we have $$ (I)~~~~||u||_{L^{p^*}(\Omega)} \leq C_1||u||_{W^{1,p}(\Omega)}. $$ If $ p > n $ then there exists a constant $C_2$ such that for all $ u \in W^{1,p}(\Omega) $ we have $$ (II)~~~~||u||_{L^{\infty}(\Omega)} \leq C_2||u||_{W^{1,p}(\Omega)}. $$

In general, the constants $C_i$ depend on the domain $\Omega$. Can someone point me to some references that discuss the dependence between the embedding constants and the domain?

I'm interested in conditions under which, given some family of domains $ \Omega_\alpha $, I can get Sobolev embedding inequalities as above with a constant that doesn't depend on $\alpha$. To be even more specific, I'm interested in the case $ n = 2 $ and when the domains are families of balls or annuli.

For example, if I consider inequality (II) and a family of balls, then an obvious sufficient condition is to have both an upper and a lower bound on the radii of the balls. One can't get away without a lower bound (consider the function $u \equiv 1$) but can get away without an upper bound by using a translation argument. What about inequality (I)? What can I use to answer such questions?

Since one way to prove the inequalities above is to use an extension operator, and then "steal" the inequality from $\mathbb{R}^n$, this question is related to dependence of the minimal norm of an extension operator $W^{1,p}(\Omega) \rightarrow W^{1,p}(\mathbb{R}^n)$ on the domain

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    $\begingroup$ You may be interested in the items 30 and 37 on Villani's publication list, if you can get past the flashing download signs. $\endgroup$
    – user31373
    Commented Aug 1, 2012 at 18:55
  • $\begingroup$ Villani's list of publications now has a different URL, and the flashing signs are gone. (Yay!) The relevant papers are still numbered 30 and 37, they are both titles "Balls have the worst best Sobolev inequalities". $\endgroup$
    – user90090
    Commented Aug 16, 2013 at 20:10

1 Answer 1

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By a scaling argument, one does not need upper bound on the radii in (I) as well. The lower bound is needed because of the inhomogeneous term in the Sobolev norm in the right hand side. If you use the Sobolev seminorm $\|Du\|_{L^p}$ instead of the full norm you will have a scale independent inequality.

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  • $\begingroup$ I don't understand the sentences 2 and 3. Being on the right hand side, the inhomogeneous term cannot hurt the inequality by its presence. With or without this term, the constant $C_2$ blows up as the radius tends to zero. $\endgroup$
    – user31373
    Commented Aug 1, 2012 at 18:51
  • $\begingroup$ @Leonid: Sorry I was talking about (I), but mistakenly wrote (II). Now corrected. $\endgroup$
    – timur
    Commented Aug 1, 2012 at 20:20
  • $\begingroup$ Got it, but removing the inhomogeneous term from (I) makes $u\equiv 1$ a counterexample to the inequality. $\endgroup$
    – user31373
    Commented Aug 1, 2012 at 20:26
  • $\begingroup$ @Leonid: True.. So one has to require $u=0$ at the boundary, or the mean of $u$ equal zero. $\endgroup$
    – timur
    Commented Aug 1, 2012 at 20:33
  • $\begingroup$ Which is getting a bit further from the original question, but the point is that if the two sides scale differently there is no way the constant will be independent of the size. $\endgroup$
    – timur
    Commented Aug 1, 2012 at 20:35

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