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, $u(x)> 0$ is a smooth function defined on $\Omega$. I want to know if the following is true. $$\int_\Omega |\ln u(x)-\ln \bar{u}|^2dx\le C\int_\Omega |\nabla \ln u(x)|^2dx,$$ with $C$ 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 $M$.

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 $M$.

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, $u(x)> 0$ is a smooth function defined on $\Omega$. I want to know if the following is true. $$\int_\Omega |\ln u(x)-\ln \bar{u}|^2dx\le C\int_\Omega |\nabla \ln u(x)|^2dx,$$ with $C$ 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: Michael Renardy give an good counterexample. But I missed a condition that $u$ have an upper bound $M$, $C$ could depend on $M$.

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, $u(x)> 0$ is a smooth function defined on $\Omega$. I want to know if the following is true. $$\int_\Omega |\ln u(x)-\ln \bar{u}|^2dx\le C\int_\Omega |\nabla \ln u(x)|^2dx,$$ with $C$ 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 $M$.

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, $u(x)> 0$ is a smooth function defined on $\Omega$. I want to know if the following is true. $$\int_\Omega |\ln u(x)-\ln \bar{u}|^2dx\le C\int_\Omega |\nabla \ln u(x)|^2dx,$$ with $C$ 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.

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, $u(x)> 0$ is a smooth function defined on $\Omega$. I want to know if the following is true. $$\int_\Omega |\ln u(x)-\ln \bar{u}|^2dx\le C\int_\Omega |\nabla \ln u(x)|^2dx,$$ with $C$ 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: Michael Renardy give an good counterexample. But I missed a condition that $u$ have an upper bound $M$, $C$ could depend on $M$.