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I want to use the contradiction argument and compact argument to prove the inequality below

$\forall\epsilon>0$,there exists $C_\epsilon>0$,$\forall u\in W^{1,p}(U)$,we have

$||u||_{L^p(U)}\leqslant\epsilon||Du||_{L^p(U)}+C_\epsilon|\bar u|$

where $\bar u:=\frac{1}{|U|}\int_U udx$.

PS:By the contradiction argument and compact argument,I can prove that there exist limit $u\in W^{1,p}(U)$ with $\bar u=0,||u||_{L^p(U)}=1$,but I failed when I want to find out a contradiction.

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I guess $U$ is an open set of finite measure. Any assumption(s) about $p$? – Davide Giraudo Jan 2 '13 at 12:52
@ Davide Giraudo: We can assume that $1<p<+\infty$. However, if $p$ could take value 1 will be nice! – Darry Jan 2 '13 at 13:04
The compactness argument should produce a function with zero norm of the gradient, hence constant. This is where the contradiction comes from. – user53153 Jan 2 '13 at 16:23
@ Pavel M:In fact,I know how to find the contradicion in standard procedure,e.g. the proof of poincare inequality stated in Evans'PDE.However,for the above one I failed.I can't prove $Du=0$,but only $Du\in L^p$.So wierd! – Darry Jan 2 '13 at 23:58

I think the inequality is false. Think of any function with zero average, for example $u=x-\frac{1}{2}$ in $[0,1]$, so you get that for any $\epsilon >0$

$$ ||u||_{L^2} \le \epsilon||Du||_{L^p} \\ \frac{1}{2\sqrt{3}} \le \epsilon $$

and this is false for any $\epsilon >0$

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I agree with above, this inequality is false in general. But what you may look at, or instead that you could use the equivalent norm in $W^{m,p}$, in which we have $$\|u\|_{L^p(\Omega)}\leq \epsilon\|D^\alpha u\|_{L^p(\Omega)}+C_\epsilon \|u\|_{L^p(\Omega)} $$ where $|\alpha|=m$.

If $\Omega$ is finite and $m=1$, you could instead have $$\|u\|_{L^p(\Omega)}\leq \epsilon\|\nabla u\|_{L^p(\Omega)}+C_\epsilon \|u\|_{L^q(\Omega)} $$ where $1\leq q\leq p$. Take $q=1$, I think this is the most closed answer I can think of. But generally, you can not use $\bar{u}$ to control the norm of $u$.

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