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.




---


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?



Using the Rellich-Kondrachov theorem to prove Poincare inequality for a function vanishing at one point
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 $v(0)=0$. 


