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.
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.
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?
Best Answer
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 $v(0)=0$.