Suppose that $1 \leq p < \infty $ and $\Omega$ is a bounded open set. Then according to Poincare inequality, there exists a constant $C$ depending on $\Omega$ and $p$ such that
$$\|u\|_{L^p(\Omega)} \leq C\|\nabla u\|_{L^p(\Omega)}, \forall u\in W_{0}^{1,p}(\Omega)$$
Now, I want to apply Poincare inequality to functions that has non-zero boundaries, and I want the following inequality
$$\|u\|_{L^p(\Omega)} \leq C(\|\nabla u\|_{L^p(\Omega)}+\| u\|_{L^p(\partial \Omega)}), \forall u\in W^{1,p}(\Omega)$$
I have searched across the Sobolev space section in Brezis, Evans, and Maz'ya, but could not find something that exactly satisfy my need. I have found a paper titled "Best constant in Poincaré inequalities with traces", but the author displayed the equation without any proof, and I'm not sure how he get the inequality.
Best Answer
I will write a proof for the sake of future reference.
Proposition. Let $\Omega$ be a bounded open set of $\mathbb{R}^n$ with Lipschitz boundary, and let $p \in [1,\infty)$. Then there is a constant $C = C(\Omega, p)$ such that for all $u \in W^{1,p}(\Omega)$ we have $$||u ||_{L^p(\Omega)} \leq C\bigg(||\nabla u ||_{L^(\Omega)} + ||u ||_{L^p(\partial \Omega)}\bigg) $$
Proof. Suppose not, then we may find a sequence $u_k \in W^{1,p}(\Omega)$ such that $$||u_k ||_{L^p(\Omega)} > k\bigg(||\nabla u_k ||_{L^(\Omega)} + ||u_k ||_{L^p(\partial \Omega)}\bigg) $$ In particular, setting $v_k := \frac{u_k}{||u_k||_{L^p(\Omega)}}$ we find $$ ||\nabla v_k ||_{L^p(\Omega)} + ||v_k ||_{L^p(\partial \Omega)} < \frac{1}{k} $$ so that $\nabla v_k \to 0$ in $L^p(\Omega)$ and $v_k \to 0 $ in $L^p(\partial \Omega)$ in the sense of traces.
Now, since the sequence $||\nabla v_k ||_{L^p(\Omega)}$ converges, it follows that it is bounded, and using that $||v_k||_{L^p(\Omega)} = 1 $ we find $$|| v_k||_{W^{1,p}(\Omega)} = 1 + ||\nabla v_k ||_{L^p(\Omega)} < D$$
Thus the sequence $v_k$ is uniformly bounded in $W^{1,p}(\Omega)$ and hence there exists $v \in W^{1,p}(\Omega)$ such that a subsequence $v_k$ (which we don't relabel) converges weakly to $v$ in $W^{1,p}(\Omega)$. By the Sobolev embedding theorem, we also have that $v_k \to v$ strongly in $L^p$ and again passing to subsequence we may assume that $v_k \to v$ almost everywhere. Also notice that for any $\varphi \in C_c^\infty(\Omega)$ we have
$$ \int_\Omega v \varphi_{x_i} = \lim_{k \to \infty} \int_\Omega v_k \varphi_{x_i} = - \lim_{k \to \infty} \int_{\Omega} (v_k)_{x_i} \varphi = 0 $$
where the last equality follows from $$ \bigg | \int_{\Omega} (v_k)_{x_i} \varphi \bigg | \leq ||(v_k)_{x_i} ||_{L^p(\Omega)} ||\varphi||_{L^{p'}(\Omega)} \leq ||\nabla v_k ||_{L^p(\Omega)} ||\varphi||_{L^{p'}(\Omega)} \leq \frac{1}{k}||\varphi||_{L^{p'}(\Omega)} \to 0 $$ where $p'$ is the Holder expontent (in the case $p = 1$ set $p' = \infty$)
So this shows that the derivative of $v$ is $0$, therefore the function $v$ is constant on its connected components. Also notice that $v = 0$ on $\partial \Omega$ since $||v_k||_{L^p(\partial \Omega)} \to 0$. Now recall that if $C$ is a connected component of $\Omega$, then $\partial C \subset \partial \Omega$, hence $v = 0$ on $\partial C$ and since $v$ is constant in $C$ we have $v = 0$ in $C$ and consequently in all of $\Omega$ (because $v = 0$ in all connected components). So we conclude that $v \equiv 0$ but this is a contradiction because $||v||_{L^p(\Omega)} = \lim_{k \to \infty} ||v_k||_{L^p(\Omega)} = 1$