I'm in trouble with this exercise

Let $\Omega \subset \mathbb{R^n}$ be a bounded Lipschitz domain and let $p \in ]1,n[$. prove that there exists $C>0$ such that, for every $f \in W^{1,p}(\Omega)$ one has $\Vert f-f_{\Omega}\Vert_{L^p(\Omega)} \leq C \Vert Df\Vert_{L^p(\Omega)}$. And what about $p=1$ and $p \geq n$?

Here $f_{\Omega} = 1/$ L$^n({\Omega}) \int_{\Omega} f$ $d$L$^n$ and L is the Lebesgue measure.

I have an idea but I can't formalize it: I reason by contradiction and try to use the compactness theorem. Now I'm stuck. Some hints?

This is what I've found: For every $n > 0$ there exists $f_n$ such that:

$\Vert f_n-f_{n,\Omega}\Vert_{L^p(\Omega)} > n \Vert Df_n\Vert_{L^p(\Omega)}$

Now, I choose $g_n = \dfrac{f_n-f_{n,\Omega}}{\Vert f_n-f_{n,\Omega}\Vert_{L^p}}$. It's quite simple to see that $g_n \in W^{1,p}, g_{n,\Omega}=0$ and $ \Vert g_n \Vert_{L^p} =1 $ for every $n$.

By compactness theorem there exists a subsequnce $\{ g_{n_j} \}_j$ and $g \in W^{1,p}$ such that $g_{n_j} \to g$ in $L^p$. Now, $g_{\Omega} = 0$ and we may assume wlog $ \Vert g \Vert=1 $. From here issues start.

My idea is to show that $\nabla g = 0$ in $\Omega$.

If it holds then $g$ is constant in $\Omega$ and thus $g = 0$ necessarily, which is absurd since $ \Vert g \Vert=1 $.

How can I prove $\nabla g = 0$?

  • $\begingroup$ Show your reasoning by contradiction. I guess it's easier if you can restrict (without loss of generality) to functions $f$ having $f_\Omega = 0$ and $\| f \|_{L^p} = 1$. $\endgroup$ Nov 20, 2019 at 16:36
  • $\begingroup$ Need compactness based on extension theorems. See Gilbarg-Trudinger Section 7.12, as well as exercise 7.12. $\endgroup$
    – Yuval
    Nov 20, 2019 at 16:41
  • $\begingroup$ @MichałMiśkiewicz It's quite long. I report the main ideas. $\endgroup$ Nov 24, 2019 at 18:56
  • $\begingroup$ @YuDing see my additions $\endgroup$ Nov 24, 2019 at 19:15
  • $\begingroup$ ouuuu maybe it's only the definition + Holder ineq. $\endgroup$ Nov 24, 2019 at 19:27

1 Answer 1


Suppose that for every $m > 0$ there exists a function $f_m \in W^{1,p}(\Omega)$ such that

$\int_{\Omega} \vert f_m - (f_m)_{\Omega} \vert^p > m \int_{\Omega} \vert Df_m \vert^p$

In order to use compactness theorem we need a bounded sequnce of functions in $W^{1,p}$. We can choose $g_m := \dfrac{f_m - (f_m)_{\Omega}}{\Vert f_m - (f_m)_{\Omega} \Vert_{L^p}}$

Clearly $\Vert g_m\Vert_{L^p} = 1$ and $(g_m)_{\Omega} = 0$ for every $m$. Thus $\int_{\Omega} \vert D g_m \vert ^p < 1/m$ and so $\Vert g_m \Vert_{W^{1,p}}$ is bounded.

By the theorem there exists $\{ g_{m_j} \}_j \subset W^{1,p}(\Omega)$ and $g \in W^{1,p}(\Omega)$ such that $\Vert g_{m_j} - g \Vert_{L^p} \to 0$. Wlog we can suppose $\Vert g \Vert_{L^p} = 1$ and clearly $g_{\Omega} = 0$.

Now, our next goal is to prove that $g_{x_i} = \dfrac{\partial g}{\partial x_i} = 0$ for every $i=1,\dots ,n$. We use the definition of weak derivative and Holder's inequality.

For every $i =1, \dots, n$ and for every $ \phi \in C^1_c(\Omega)$ we have

$\vert \int_{\Omega} g \phi_{x_i} \vert = \lim_{j \to \infty} \vert \int_{\Omega} \phi (g_{n_j})_{x_i} \vert \leq \lim_{j \to \infty} (\int_{\Omega} \vert Dg_{n_j} \vert ^p)^{1/p} (\int_{\Omega} \vert \phi \vert^{p'})^{1/p'}$.

We have disovered $g \in W^{1,P}$ and $Dg = 0$ in $\Omega$. But $\Omega$ is a domain, in particular it is connected. Then, since $g_{\Omega} = 0$, we have $g = 0$ in the whole $\Omega$. On contrary $\Vert g \Vert_{L^p} = 1$. We have our absurd.

  • 1
    $\begingroup$ Right! The part with $\nabla g \equiv 0$ can be explained a bit differently - we have $g_{m_j} \to g$ in $L^p$ and $\nabla g_{m_j} \to 0$ in $L^p$, from which we can conclude (one can always view $W^{1,p}$ as isometrically embedded in $L^p \times L^p$). $\endgroup$ Dec 19, 2019 at 23:05
  • $\begingroup$ Great idea, I didn’t think something like this $\endgroup$ Dec 20, 2019 at 19:39

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