Using the Rellich-Kondrachov theorem to prove Poincare inequality for a function vanishing at one point

Question

Suppose $\Omega\subset \mathbb{R}^n$ is bounded Lipschitz domain. When $1\leq p < n$, apparently $p< p^* = np/(n-p)$, hence $$ W^{1,p}(\Omega )\subset \subset L^{p}(\Omega ), $$ and for any $u\in W^{1,p}(\Omega)$ $$ \|u\|_{L^p}\leq C \|u \|_{W^{1,p}}\;. $$ To bound using the seminorm like the following Poincare inequality $$ \|u\|_{L^p}\leq C |u |_{W^{1,p}} := C \left(\int_{\Omega} |\nabla u|^p\right)^{1/p}\;,\tag{$\dagger$} $$ we need some extra conditions like $$ u = 0 \;\text{ on }\partial \Omega,\tag{1} $$ or $$ \int_{\Omega}u = 0,\tag{2} $$ or $$ \int_{U} u =0, \text{ where } U\subset \Omega, \text{ and }\operatorname{meas}_{n}(U) \neq 0. \tag{3} $$ The proof is pretty standard by using the sequential compactness to reach a contradiction.

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Normally the conditions like (1), (2), or (3) avoid using pointwise value because $W^{1,p}(\Omega)$ may not necessarily embedded in $C^{0,\alpha}$ spaces.

Here if we assume the extra continuous condition like $u$ being continuous, I could prove the following:

• Proposition 1: $u\in V:= W^{1,p}(\Omega)\cap C^0(\overline{\Omega})$, $u(z) = 0$ for some $z\in \overline{\Omega}$, then ($\dagger$) is true: $$ \|u\|_{L^p}\leq C |u |_{W^{1,p}}\;.\tag{$\dagger$} $$

Proof: Assume otherwise, then there exists a sequence $\{v_n\}\subset V$ such that $$ \|v_n\|_{L^p} = 1, \;\text{ and }\; |v_n |_{W^{1,p}} < \frac{1}{n}. $$ Now by Rellich-Kondrachov compactness theorem, there exists a subsequence $\{v_m\}\subset \{v_n\}$ such that $$ v_m\to v \; \text{ in } \|\cdot\|_{L^p}. $$ Moreover, the $L^p$ convergence further implies there exists a subsequence $\{v_j\}\subset \{v_m\}$ converging a.e., and by $V$ itself being $C^0$, we have $$ v_j\to v \; \text{ in } \|\cdot\|_{L^p}, \; \text{ and } v(z) = 0. $$ By triangle inequality it is straightforward to verify that $$\|v\|_{L^p} = 1.\tag{4}$$

Now, by integrating against a smooth test function $\phi\in C^{\infty}_c(\Omega)$, $|v_j|_{W^{1,p}}<1/j$, and bounded convergence theorem, we have $$ \int_{\Omega} v\,\partial_{x_i} \phi = \int_{\Omega} \lim_{j\to \infty}v_j\,\partial_{x_i} \phi \lim_{j\to \infty}\int_{\Omega} v_j\,\partial_{x_i} \phi =-\lim_{j\to \infty}\int_{\Omega} \partial_{x_i} v_j\,\phi = 0. $$ As a result, the limit $v$ is a constant, also $v$ now must be 0 due to $v(z) = 0$. This contradicts with (4). Q.E.D.

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The result first does not look quite right, since no point values are allowed in general (as aforementioned conditions (1), (2), and (3) do).

So now my question is:

• Is Proposition 1 correct?

• If not, is there any counterexample in simple domains like a ball?

Answer 1

A typical counterexample is like the following. For concreteness, let's take $p=1$ and $n=2$, and $\Omega = B(0,1) \subset \mathbb{R}^2$.

Let $$f_n(t) = \begin{cases} nt, & 0 \le t \le 1/n \\ 1, & t > 1/n.\end{cases}$$ Set $v_n(x) = f_n(|x|)$, so $v_n$ is continuous and $v_n(0)=0$. (If you like you may modify the function $f_n$ slightly to make it $C^\infty$.)

You can check that $|\nabla v_n(x)| = n$ for $0 < |x| < 1/n$ and $|\nabla v_n(x)| = 0$ for $1/n < x < 1$. Thus $|v_n|_{W^{1,1}} = n \cdot m(B(0,1/n)) = \pi/n \to 0$. But $v_n \to 1$ monotonically almost everywhere, so $v_n \to 1$ in $L^1(\Omega)$.

The flaw in your proof is in looking at the properties of $v$. You defined $v$ as the $L^p$ and a.e. limit of the $v_n$, so it's only well-defined up to null sets. In general you can't choose a continuous representative for it (you claimed you could but didn't justify it), so speaking of $v(0)$ doesn't really make sense. In this example you can choose a continuous representative (namely 1) but you don't have $v_n \to v$ pointwise everywhere, so using that representative you cannot conclude $v(0)=0$.