2
$\begingroup$

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 $\Omega$?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.