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In the paper I am reading the authors state that $|\nabla u|_\infty$ can be replaced by $|u|_3$ using the Sobolev Lemma. I am trying to find this lemma but its turned out to be very difficult.

The context is the following:

  • a smooth bounded domain $\Omega \subset \mathbb{R}^3$
  • $|\cdot|_s$ denotes the Sobolev norm of the space $W^{s,2}(\Omega)=H^2(\Omega)$ and $|\cdot|_\infty$ the norm in $L^{\infty}(\Omega)$
  • $u$ is a vector valued function (the velocity of a fluid)

This has to be one of the many imbedding theorems which should give $$|\nabla u|_\infty \leq C \: |u|_3$$ where $C$ is a constant depending on $\Omega$ alone I suppose.

I'd appreciate if you can give me a reference as well. ThX!

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  • $\begingroup$ Your function $u$ belongs to what space? $\endgroup$ – Tomás May 17 '13 at 14:30
  • $\begingroup$ Should be in $W^{1,\infty}$ I suppose. $\endgroup$ – chango May 17 '13 at 14:35
  • $\begingroup$ I think I might have found the relevant theorem after all. It's Corollary 9.13 in Brezis's book FA, SS and PDEs (page 284). It's strange that I could not find it in Adams being that its a more thorough account of Sobolev's Spaces. $\endgroup$ – chango May 17 '13 at 14:37
  • $\begingroup$ Well, not really I just realized that this result if for the whole of $\mathbb{R}^N$. $\endgroup$ – chango May 17 '13 at 14:40
  • $\begingroup$ I do not understand you notation. What is $s$? and if $u$ is only in $W^{1,\infty}$ how can you estimare the norm $|u|_3$? $\endgroup$ – Tomás May 17 '13 at 14:45
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I'm assuming that your function $u$ belongs to $W^{3,2}(\Omega)$. We have the following result that can be found in any good book of Sobolev spaces (for example in Leoni's book or even in Brezi's book, but in the later you have to iterate the estimates that he find only to $W^{1,p}(\Omega)$).

If $\Omega$ is a bounded regular domain, $p\geq 1$, $k>\frac{n}{p}$ then $$W^{k,p}(\Omega)\hookrightarrow C^m(\overline{\Omega}),\ \forall\ 0\leq m<k-\frac{n}{p}$$

In your case: $p=2$, $n=3$ and $k=3$, then $m\in[0,3/2]$, which implies your result.

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  • $\begingroup$ Thank you! So how does this imply my result exactly. I know that with $m=1$, $\nabla u$ is continuous and therefore bounded in the compact set $\bar{\Omega}$. What norm do you define on $C^m(\bar{\Omega})$? Something like $||u|| = \max_{x \in \bar{\Omega}}{|u|} + \max_{x \in \bar{\Omega}}{|\nabla u|}$ $\endgroup$ – chango May 19 '13 at 10:49
  • $\begingroup$ Yes, thi norm is good. $\endgroup$ – Tomás May 19 '13 at 11:17

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