Let $E_\delta =[0,1]^{N-1}\times [0,\delta]$, $p\in [1,\infty)$ and $1/p+1/p'=1$. Let $\varphi\in C^1(E_\delta)$ such that $$\varphi(x)=0,\ \forall \ x\in [0,1]^{N-1}\times \{0\}.$$

By the fundamental theorem of calculus and Holder inequality, if $x'=(x_1,\cdots,x_{N-1})$ and $x=(x',t)$ then $$|\varphi(x)-\varphi(x',0)|=\left|\int_0^t \frac{\partial \varphi(x',s)}{\partial s}ds\right|\le t^{1/p'}\left(\int_0^\delta\left|\frac{\partial \varphi(x',s)}{\partial s}\right|^pds\right)^{1/p},$$

which implies that $$|\varphi(x)|^p\le t^{p/p'}\int_0^\delta\left|\frac{\partial \varphi(x',s)}{\partial s}\right|^pds,$$

After integration and an application of Fubini's theorem in the above inequality, we may conclude that $$\int_{E_\delta} |\varphi(x)|^p\le \frac{\delta^p}{p}\int_{E_\delta}|\nabla \varphi(x)|^p.$$

Now let $\Omega\subset\mathbb{R}^N$ be a bounded regular domain.

So, we can think in the boundary of $\Omega$ as a finite union of sets of the above form, therefore, there is a constant $C>0$, such that if $\Omega_\delta=\{x\in \Omega:\ \operatorname{dist}(x,\partial\Omega)<\delta\}$ then $$\int_{\Omega_\delta} |\varphi(x)|^p\le C\delta^p\int_{\Omega_\delta}|\nabla \varphi(x)|^p.$$

By density, this is also true for functions in $W_0^{1,p}(\Omega)$. The last inequality implies that $$\lim_{\delta\to 0}\frac{1}{\delta^p}\int_{\Omega_\delta}|u|^p=0, \forall\ u\in W_0^{1,p}(\Omega). \tag{1}$$

My question is: can we conclude $(1)$ for function in $W_0^{1,p}(\Omega)$, without regularity on the boundary of $\Omega$. I mean, is it true for any open set $\Omega$?

  • $\begingroup$ So I think you are stating the result for $W_0^{1,p}$ Poincare inequality for domain bounded in one directions. This result is true as $u\in W_0^{1,p}(\Omega)$ not $W_0^{1,p}(\Omega_\delta)$. That is, for a fixed $u\in W_0^{1,p}(\Omega)$, as you narrow your domain, $u$ may not be in $W_0^{1,p}(\Omega_\delta)$ and hence the poincare is broken and you can not have $(1)$. Or am I missing something? $\endgroup$ – spatially Jan 9 '15 at 0:58
  • $\begingroup$ Sorry @wisher, but I can't understand your comment. I just proved that the result is true, so why are you saying it is not true? $\endgroup$ – Tomás Jan 9 '15 at 10:40
  • $\begingroup$ I'll write a better comment later $\endgroup$ – spatially Jan 9 '15 at 10:54

I would say it doesn't hold, See if this is true :

Take as $\Omega:= B(0,3) - \{0\}$ and take p=1.

$f:= \eta log(|x|)sin(|x|)$ where $\eta$ is a smooth function that is $1$ on $B(0;1)$ and $0$ outside $B(0,2)$.

First $f$ is in $W^{1,1}(\Omega)$:

$\int_{B(0,1)} |f(x)|dx = \int_0^{2\pi}\int_0^1 |log(r)sin(r)r|dr \leq C\int_0^1 |log(r)r|dr$ and an integration by part gives $\int_0^1 |rlogr|dr = [|\frac{r^2}{2}log(r)|]_0^1 - \int_0^1 1dr < \infty$ so that f is in $L^1$. Next, we have $\frac{d (log(r)sin(r))}{dr} = log(r)cos(r) + \frac{1}{r}sin(r)$ so that integrating as above with polar coordinates gives you that $\nabla f$ is $L^1$.

Next we shall prove that we have a sequence of $C^{\infty}_0(\Omega)$ functions that converges to $f$ in $W^{1,1}(\Omega)$ so that shows that $f \in W^{1,1}_0(\Omega)$:

We take $f_{\epsilon}:= \eta_{\epsilon}f$ where $\eta_{\epsilon}$ is $0$ on $B(0,\epsilon)$ $1$ outside $B(0,2\epsilon)$ and goes linearly from $0$ to $1$ as $|x|$ varies from $\epsilon$ to $2\epsilon$. That $f_{\epsilon}$ converges to $f$ in $L^1$ is clear. Furthermore, $\nabla \eta_{\epsilon}$ is $0$ on $B(0,\epsilon)$ outside $B(0,2\epsilon)$ and is $\frac{1}{\epsilon}$ on $B(0,2\epsilon)-B(0,\epsilon)$ so that $\int_{\Omega} |(\nabla \eta_{\epsilon}) f| = \frac{1}{\epsilon} \int_{B(0,2\epsilon)-B(0,\epsilon)}f\leq \frac{1}{\epsilon}|log(\epsilon)| \int_{B(0,2\epsilon)-B(0,\epsilon)}1 = C\frac{1}{\epsilon}|log(\epsilon)|\epsilon^2 \rightarrow 0$ as $\epsilon$ goes to $0$. And since $\eta_{\epsilon}\nabla f$ converges to $\nabla f$ in $L^1$ we have that $\nabla (\eta_{\epsilon}f)=\eta_{\epsilon}\nabla f + (\nabla \eta_{\epsilon}) f$ converges to $\nabla f$ in $L^1$. This ends to prove our claim.

Finally let's see the averages:

We have $|\frac{1}{\epsilon}\int_{B(0,\epsilon)}f|\leq \frac{1}{\epsilon}\int_{B(0,\epsilon)}|f|$, and therefore we will show that the left hand side does not converge to $0$ as $\epsilon$ goes to $0$ so that it will prove that the rhs cannot go to $0$.

$\int_{B(0,\epsilon)}f=\int_0^{2\pi}\int_0^{\epsilon}log(r)sin(r)rdr= C\int_0^{\epsilon}log(r)sin(r)rdr= C([-rlog(r)cos(r)]^{\epsilon}_0 + \int_0^{\epsilon}(1+log(r))cos(r)dr) =C( -\epsilon log(\epsilon)cos(\epsilon) + sin(\epsilon) + \int_0^{\epsilon} log(r)cos(r)dr)$

dividing by $\epsilon$ we get

$\frac{1}{\epsilon}\int_{B(0,\epsilon)}f=C(-log(\epsilon)cos(\epsilon) + \frac{sin(\epsilon)}{\epsilon} + \frac{1}{\epsilon}\int_0^{\epsilon} log(r)cos(r)dr)$

The first term in the rhs goes to $+\infty$ and the second term goes to $1$. For the last term we have:

$\frac{1}{\epsilon}\int_0^{\epsilon} log(r)cos(r)dr \geq - \frac{1}{\epsilon}\int_0^{\epsilon} log(r)dr=- \frac{1}{\epsilon}([rlog(r)]_0^{\epsilon} - \int_0^{\epsilon}1dr)= -log(\epsilon) + 1 \rightarrow +\infty$

This proves that $\frac{1}{\epsilon}\int_{B(0,\epsilon)}f \rightarrow +\infty$ and achieves the proof.

  • $\begingroup$ Ok so here is how I see it: $\nabla log(|x|)$ is the application $b \rightarrow \frac{<x;b>}{|x|}\frac{1}{|x|}$ so that $|\nabla log(|x|)|=\frac{1}{|x|}$. Similarly $(\nabla sin(|x|) ;b) = \frac{<x;b>}{|x|}cos(|x|)$ and therefore $|\nabla sin(|x|)|=|cos(|x|)|$. Now $|\nabla(log(|x|)sin(|x|)|=|sin(|x|)\nabla log(|x|) + log(|x|)\nabla (sin(|x|)| \leq |\frac{1}{|x|}| + |log(|x|)|$ And $\int_{B(0;1)}\frac{1}{|x|} < \infty$ and also $\int_{B(0,1)} |log(|x|)|=\int_0^{2\pi}\int_0^1 r|log(r)|dr=C([-\frac{r^2}{2}log(r)]_0^1 + \int_0^1 r dr)< \infty$ $\endgroup$ – incas Jan 10 '15 at 13:22
  • $\begingroup$ It seems right. Do you mean $|\nabla \eta_\epsilon|<1/\epsilon$ in $B(2\epsilon)\setminus B(\epsilon)$? $\endgroup$ – Tomás Jan 10 '15 at 14:44
  • $\begingroup$ Yes, I meant that $|\nabla \eta_{\epsilon}|=\frac{1}{\epsilon}$ in $B(2\epsilon)-B(\epsilon)$ $\endgroup$ – incas Jan 10 '15 at 15:09
  • $\begingroup$ So does the answer convince you ? I think the trick here is that the boundary is too small to make the space $W^{1,1}_0(B(0,1))$ different from $W^{1,1}_0(B(0,1)-\{0\})$, the boudary $\{0\}$ is in some sense invisible, and the functions at 0 can have any kind of behavior as long as they remain integrable as well as their derivatives. $\endgroup$ – incas Jan 11 '15 at 15:31
  • $\begingroup$ Your answer seems good. I am studying it more. What was your intention when your multiplied $\log$ by $\sin$? I mean, it seem that it will also work, if we cut off the function $\log{|x|})$ $\endgroup$ – Tomás Jan 11 '15 at 15:35

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