# A funny exercise with bounded Lipschitz domain and Sobolev spaces

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 $$\nabla g = 0$$?

• 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 $\| f \|_{L^p} = 1$. Nov 20, 2019 at 16:36
• Need compactness based on extension theorems. See Gilbarg-Trudinger Section 7.12, as well as exercise 7.12. Nov 20, 2019 at 16:41
• @MichałMiśkiewicz It's quite long. I report the main ideas. Nov 24, 2019 at 18:56
• @YuDing see my additions Nov 24, 2019 at 19:15
• ouuuu maybe it's only the definition + Holder ineq. Nov 24, 2019 at 19:27

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.

• 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 $L^p \times L^p$). Dec 19, 2019 at 23:05
• Great idea, I didn’t think something like this Dec 20, 2019 at 19:39