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If I have an inequality: $\lVert u\rVert_{L^p(R^n)} \le C\lVert\nabla u\rVert_{L^q(R^n)}$ , where $C \in (0,\infty)$ and $u \in C_c^1(R)$, is there a relation between $p, q, n$ such that the inequality holds ?

Thank you for your help.

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Usually such questions can be answered quickly by means of a dimensional analysis. –  Giuseppe Negro May 30 '12 at 8:08
    
@GiuseppeNegro: is it true that the follwoing condition should hold $q=(np)/(n-p)$ ? Is it sufficient ? –  Theorem May 30 '12 at 9:03
    
    
Here is the 'dimensional analysis' I was referring to: Remark 10. The text refers to it as a 'scaling argument'. –  Giuseppe Negro Jun 3 '12 at 10:53

1 Answer 1

This result is a special case of what is known as Friedrich's Inequality, also sometimes known as Poincare's inequality. It is always true with $n$ arbitrary and $p = q$, although $C$ will depend on $n$ and the size and shape of the support of $u$. However, if you fix the supports of all your $u$'s to lie inside of some fixed set $\Omega$ (i.e. $ u \in C_c(\Omega)$), then you can choose $C$ depending only on $n$ and $\Omega$. Of course, since the support is compact and hence of finite Lebesgue measure, the inequality will also be true with $q \geq p$ simply because in this case $\|f\|_p \le C\|f\|_q$ by Holder's inequality. Any reasonable book on PDEs/Sobolev Spaces should have a proof of Friedrich's inequality, although its not too hard to cook one up yourself. Hint: integration by parts.

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I think the OP's inequality is supposed to hold with $C$ not depending on $u$. If you allow for dependency on the size and shape of the support, then you could also say "for each $u$ the inequality holds for some $C$", which would be trivial. (Of course the version you propose is not trivial.) –  Hendrik Vogt May 30 '12 at 12:13
    
@HendrikVogt Yes, the problem is that the OP only specified $u \in C_c(\mathbb{R}^n)$,rather than say $u \in C_c(\Omega)$ for some fixed bounded $Omega$ in which case $C$ could be chosen independently of $u$. –  user12014 May 30 '12 at 18:34
    
Yep, for bounded $\Omega$ that's Friedrich's inequality. I'm pretty sure, however, that the OP wants the Gagliardo–Nirenberg–Sobolev inequality after all. –  Hendrik Vogt May 30 '12 at 18:38
    
@HendrikVogt Hm I just checked Wikipedia, and yes that looks more like what the OP wants. I never realized one applied to more cases than the other... –  user12014 May 30 '12 at 18:40

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