This is exercise 12.25 from Leoni's book.

Update: @Lukas Geyer has provided a counterexample for the case that $\Omega$ has a very bad boundary, which suggests that original exercise might be wrong. However, I am still wondering that if we assume $\Omega$ has smooth boundary, could the result be hold? i.e., we assume that $\Omega$ has smooth boundary but still not bounded, only finite.

Given $\Omega\subset R^N$ is an open finite set, i.e., $|\Omega|<\infty$. But $\Omega$ may not have a nice boundary or being bounded, i.e., the usual Sobolev Embedding won't work on this set. Now I am trying to prove that given any $u\in W^{1,p}(\Omega)$, where $1<p<\infty$, I have for any $1< q<p$, that $$\left(\int_\Omega|u-u_\Omega|^q\right)^{\frac{1}{q}}\leq C\left(\int_\Omega|\nabla u|^p\right)^\frac{1}{p} \,\,\,\,\,\,\,\,\,\,\,\,(1)$$ where $C$ depends on $p$, $q$, and $\Omega$.

To do so, I think I need to use an exercise which I proved before: for $\Omega$ open finite, the space $W^{1,p}(\Omega)$ is compact embedded in $L^q(\Omega)$ for any $q<p$.

Next, I proceed as usual, suppose $(1)$ does not hold and I have $u_n\in W^{1,p}$ such that

$$\left(\int_\Omega|u_n-(u_n)_\Omega|^q\right)^{\frac{1}{q}}\geq n\left(\int_\Omega|\nabla u_n|^p\right)^\frac{1}{p} \,\,\,\,\,\,\,\,\,\,\,\,(2)$$

Next, by defining $$v_n:=\frac{u_n-(u_n)_\Omega}{\|u_n-(u_n)_\Omega\|_{L^p(\Omega)}} $$ we have $\|v_n\|_{L^p}\equiv 1$ and $\|\nabla v_n\|_{L^p}\to 0$.

Then, if we have $W^{1,p}$ is compact embedded in $L^p$ we would be done already and this is how usual Poincare proved. However, here we only have $W^{1,p}$ is compact embedded in $L^q$, thus, we only have $v_n\to v_0$ strongly in $L^q$ for some $v_0$ such that $\nabla v_0=0$. If we can prove $v_0$ is not $0$, then we done. However, since generally we have $\|v_n\|_{L^q}\leq \|v_n\|_{L^p}$ if $q<p$, I can not rule out the probability that $v_0=0$...

Any help would be really welcome!

  • $\begingroup$ I really doubt that this is true. In general Poincare inequalities hold because there are enough curves in your domain, and the constants get worse when domains have bottlenecks. Now if you don't require any regularity on the boundary, you can easily construct domains that on smaller and smaller scales have arbitrarily long and narrow bottlenecks, which should give counterexamples to your question from functions whose gradient is supported on the bottlenecks. $\endgroup$ Nov 2, 2014 at 17:43
  • $\begingroup$ I think this is exactly why we require $q<p$. The whole point is about compactness. This is an exercise from Leoni's book. Sorry I forgot to mention it in my OP. But could you explain to me in more details that " In general Poincare inequalities hold because there are enough curves in your domain, and the constants get worse when domains have bottlenecks." I never know this idea before... I only know how to prove Poincare... $\endgroup$
    – spatially
    Nov 2, 2014 at 17:58
  • $\begingroup$ I'm pretty sure the compact embedding you mention is also false in this generality. Are you sure the exercise you proved isn't that $W^{1,p}_0$ is compactly embedded in $L^q$? $\endgroup$
    – Jose27
    Nov 3, 2014 at 5:40
  • $\begingroup$ Hmm... The compact embedding I mentioned is exercise 11.17 on Leoni's... I think the key of proving is uniform integrability. Also exercise 11.18 states the same idea. When I prove 11.17, I mollify the boundary of $\Omega_n$ and use compact embedding inside, and kill the rest in $\Omega\setminus\Omega_n$ by using uniform integrability... $\endgroup$
    – spatially
    Nov 3, 2014 at 12:33

1 Answer 1


This is not true without some additional assumptions on the boundary. Just to make life simple, the following example uses $N=2$, but it is easy to modify for any $N \ge 2$.

Fix $p > q \ge 1$. Let $\Omega$ be a domain in the plane which consists of squares $(Q_k)_{k=0}^\infty$, of side length $2^{-k}$ centered at points on the $x$-axis, together with horizontal strips $(S_k)_{k=1}^\infty$, symmetric with respect to the $x$-axis, of width $2^{-k} t_k$ and height $2^{-k}t_k^{-1}$, joining $Q_{k-1}$ to $Q_k$, where $t_k \to \infty$ is determined below. By construction, the area of $\Omega$ is finite, since the areas of both $Q_k$ and $S_k$ are $4^{-k}$.

Now fix $n \ge 2$ and let $u$ be the function on $\Omega$ which is the constant $0$ on $Q_k$ and $S_k$ for $k<n$, the constant $1$ on $Q_k$ for $k \ge n$, and on $S_k$ for $k>n$, and is linear on $S_n$. Then $\nabla u$ vanishes everywhere, except on $S_n$, where it has constant (pointwise) norm $2^n t_n^{-1}$. The area of $S_n$ is $4^{-n}$, so $$ \int_\Omega \| \nabla u \|^p = 4^{-n} 2^{np} t_n^{-p} = 2^{n(p-2)} t_n^{-p} $$ It is easy to check that $u_\Omega \le 1/2$, so $|u-u_\Omega| \ge 1/2$ on $Q_n$, which also has area $4^{-n}$, implying $$ \int_\Omega |u-u_\Omega|^q \ge 4^{-n} 2^{-q} = 2^{-2n-q} $$ Combining both inequalities, we get $$ \frac{\| u-u_\Omega \|_q}{ \| \nabla u \|_p } \ge \frac{2^{-2n/q-1}}{2^{n(1-2/p)} t_n^{-1}} = 2^{-1+n(2/p-2/q-1)} t_n $$ Here it turns out that we did not even need to keep track of all the constants, we just need to choose any sequence $t_n$ which makes this last term go to infinity (e.g., $t_n = n^n$ works in any case), providing a counterexample.

  • $\begingroup$ I think the example you provide here is the famous "Rooms and Passengers"? I think the last term should be $2^{-1+n(2/p-2/q-1)}t_n$ but it does not really matter. So you are saying the exercise itself is wrong? omg i spend a lot of time on it... $\endgroup$
    – spatially
    Nov 2, 2014 at 22:00
  • $\begingroup$ But I still think this exercise makes some sense because usually the key step in proving the Poincare inequality is the compact imbedding from $W^{1,p}\subset\subset L^p$. In this case we do have the compact embedding from $W^{1,p}\subset\subset L^q$ for any $q<p$... $\endgroup$
    – spatially
    Nov 3, 2014 at 1:26
  • $\begingroup$ I don't have the book here myself, but if the exercise is stated as you said, then I would say it is wrong. And thanks for catching the sign error, fixed it. $\endgroup$ Nov 3, 2014 at 4:05
  • $\begingroup$ But, if in addition that I assume the boundary is smooth. Do you think this is true? $\endgroup$
    – spatially
    Nov 3, 2014 at 22:42
  • $\begingroup$ No, smoothness alone will not help, you can modify the example to have smooth boundary. $\endgroup$ Nov 4, 2014 at 4:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .