1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ The proposition is true only in dimension 1. In higher dimensions, a constant function can be approximated in the Sobolev norm by continuous functions vanishing at finitely many points $\endgroup$ – almaz Feb 19 '17 at 5:29
  • $\begingroup$ Does $|u|_{W^{1,p}}$ here denote $\left(\int |\nabla u|^p\right)^{1/p}$? $\endgroup$ – Nate Eldredge Feb 19 '17 at 5:54
  • $\begingroup$ To build a bit upon what @almaz said: In one dimension, we have the embedding $W^{1,p}(\Omega) \subset C(\Omega)$. Point evaluations (i.e., integration w.r.t. a Dirac measure) are continuous functionals on $C(\Omega)$, regardless of the dimension. With the aforementioned embedding, in one dimension, they are also (well-defined) continuous functionals on $W^{1,p}(\Omega)$, so that their respective kernel is closed, and we can factor with respect to those kernels. The functional $f \mapsto \int_U f$ in well-defined and continuous on $W^{1,p}(\Omega)$ regardless of the dimension. $\endgroup$ – anonymous Feb 19 '17 at 13:52
  • $\begingroup$ @NateEldredge Yes. $\endgroup$ – Shuhao Cao Feb 19 '17 at 15:29
  • $\begingroup$ @almaz Any references? Thanks. $\endgroup$ – Shuhao Cao Feb 19 '17 at 15:38
4
$\begingroup$

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 $v(0)=0$.

$\endgroup$
  • $\begingroup$ Somehow on a second thought, Proposition 1 seems okay for $W^{k,p}$ that can be embedded in $C^{0,1}$. The biggest problem is in the proof but not the proposition itself. $\endgroup$ – Shuhao Cao Feb 20 '17 at 2:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.