For any function $f(x)$ we denote $\bar{f}:=\frac{1}{\Omega}\int_\Omega f(x)\,dx$. Let $\Omega\subset \mathbb{R}^n$ be a bounded smooth domain and $u(x)> 0$ be a smooth function defined on $\Omega$. I want to know whether the following is true: $$\int_\Omega |\ln u(x)-\ln \bar{u}|^2dx\le C\int_\Omega |\nabla \ln u(x)|^2dx,$$ for some $C$ independent of $u$.

I think this general result must be fault, is it true when $\Omega$ is convex?

If it is true for convex $\Omega$, I have a further question: use "$F$" instead of "$\ln$" which kinds $F$ support this type inequality.

Remark: I missed a condition that $u$ have an upper bound $M$, $C$ could depend on $M$.

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    $\begingroup$ I do not think an upper bound helps. Note that both sides of the inequality are unchanged if u is replaced by Ku. So if there is a constant that works if u is bounded above by M, that same constant will work for any u that is bounded above, and then a limit argument would extend it to unbounded u as in my counterexample. $\endgroup$ – Michael Renardy Jan 16 '14 at 23:21

It is false for a ball. In three dimensions, consider $u=r^a$, where $a>-3$ and let $\Omega$ be the unit ball. We find $\bar u=3/(3+a)$. As a approaches -3, the right hand side of your inequality remains bounded, the left hand side does not.

  • $\begingroup$ Your answer is correct and very nice. Thank you. But I very sorry, I had make a mistake that what I really concern, there is one more condition: $u$ has a uniform upper bound $M$. Now I edit the problem. $\endgroup$ – user44565 Jan 16 '14 at 20:38
  • $\begingroup$ The domain is bounded, why the left hand side of your example is unbounded? $\endgroup$ – user44565 Jan 16 '14 at 21:41
  • $\begingroup$ Because $\tilde u\to\infty$ as $a\to -3$ and $|\ln u|^2$ remains integrable. $\endgroup$ – Michael Renardy Jan 16 '14 at 22:28
  • $\begingroup$ Oh, I see. Sorry, I alway keep $u\le M$ in my mind! Yes, you are right, but what I really concern, need $u$ has an upper bound. $\endgroup$ – user44565 Jan 16 '14 at 22:32

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