Intriguing simple question about Sobolev space $W^{1,p}(\Omega)$

Question

Let $w_1,w_2\in W^{1,p}(\Omega)$ be two functions with $w_1,w_2>0$ and $\dfrac{w_2}{w_1},\dfrac{w_1}{w_2}\in L^{\infty}(\Omega)$, where $\Omega\subset\mathbb{R}^N$ is a bounded domain (i.e. open and connected).

We know that $\dfrac{\nabla w_1}{w_1}=\dfrac{\nabla w_2}{w_2}$. How can we show that $\dfrac{w_2}{w_1}\equiv$ constant? I see it in many articles used in that hypotheses.

I tried to use the method indicated in this post A more general product rule for weak derivatives?, but I didn't succed in passing to the limit.

I also tried to use ACL, but it didn't work because the quotinent of two absolutely continuos function need not be absolutely continuous unless the domain is compact...

Answer 1

Assume that $\Omega$ is open and connected (if $\Omega$ is not connected, then the desired conclusion is clearly false). By convolution with a mollifier and approximation, we may assume that $w_1$ and $w_2$ are smooth.

The case $N=1$ is easy, and the case of any natural $N$ reduces to the case of $N=1$.

Indeed, if $v_j$ is the restriction of $w_j$ to any open interval $I$ in $\Omega$ (for $j=1,2$), then $\dfrac{dv_1}{v_1}=\dfrac{dv_2}{v_2}$, $d\ln v_1=d\ln v_2$, and hence $v_1=c_Iv_2$ for some real $c_I>0$, depending only on $I$. Since $\Omega$ is connected, it is linearly connected (in the sense that any two points of $\Omega$ can be connected by finitely many straight line segments). So, $c_I$ does not depend on $I$. $\quad\Box$

Answer 2

Another way to do this (again assuming that $\Omega$ is open and connected): $$\nabla\frac{w_1}{w_2}=\frac{w_1\nabla w_2-w_2\nabla w_1}{w_2^2} =\frac{w_1}{w_2}\Big(\frac{\nabla w_2}{w_2}-\frac{\nabla w_1}{w_1}\Big)=0.$$ So, $\dfrac{w_1}{w_2}$ is constant on $\Omega$.

Answer 3

@IosifPinelis This is too long for a comment so I post it as an answer.

My attempt was to consider the functions $w_{1,\epsilon}(x)=\begin{cases} \epsilon, w_1(x)\leq \epsilon\\ w_1(x), w_1(x)\in \left (\epsilon, \dfrac{1}{\epsilon} \right )\\ \dfrac{1}{\epsilon}, w_1(x)\geq \dfrac{1}{\epsilon}\end{cases}$ and $w_{2,\epsilon}(x)=\begin{cases} \epsilon, w_2(x)\leq \epsilon\\ w_2(x), w_2(x)\in \left (\epsilon, \dfrac{1}{\epsilon} \right )\\ \dfrac{1}{\epsilon}, w_2(x)\geq \dfrac{1}{\epsilon}\end{cases}$. We have that $w_{1,\epsilon}\to w_1$ and $w_{2,\epsilon}\to w_2$ pointwise. Moreover $w_{1,\epsilon},w_{2,\epsilon}\in W^{1,p}\cap L^{\infty}(\Omega)$. Also $\dfrac{w_{1,\epsilon}}{w_{2,\epsilon}},\dfrac{w_{2,\epsilon}}{w_{1,\epsilon}}\in L^{\infty}(\Omega)$ (with bounds independent of $\epsilon$.

So we have that $\dfrac{w_{1,\epsilon}}{w_{2,\epsilon}}\longrightarrow \dfrac{w_1}{w_2}$ pointwise. Being also bounded we deduce that $\dfrac{w_{1,\epsilon}}{w_{2,\epsilon}}\longrightarrow \dfrac{w_1}{w_2}$ in $L^p(\Omega)$ from Lebesgue dominated convergence theorem. Moreover we easily get that $\dfrac{w_{1,\epsilon}}{w_{2,\epsilon}}\in W^{1,p}(\Omega)$ (using product rule and chain rule for weak derivatives). Thus we have from the definition of the weak derivative:

$$\int_{\Omega} \dfrac{w_{1,\epsilon}}{w_{2,\epsilon}}\dfrac{\partial\phi}{\partial x_i}\ dx=-\int_{\Omega}\phi\left [\dfrac{\dfrac{\partial w_{1,\epsilon}}{\partial x_i}w_{2,\epsilon}-\dfrac{\partial w_{2,\epsilon}}{\partial x_i}w_{1,\epsilon}}{w_{2,\epsilon}^2}\right]\ dx$$

In the left hand side there is no problem when we take $\epsilon\to 0$. It converges to $\int_{\Omega} \dfrac{w_{1}}{w_2}\dfrac{\partial\phi}{\partial x_i}\ dx$. But in the right hand side although we have pointwise convergence it is hard for me to bound the function $\dfrac{\dfrac{\partial w_{1,\epsilon}}{\partial x_i}w_{2,\epsilon}-\dfrac{\partial w_{2,\epsilon}}{\partial x_i}w_{1,\epsilon}}{w_{2,\epsilon}^2}$ by a function from $L^1(\Omega)$ that is independent of $\epsilon$.

This is where I got stucked.