Poincaré's inequality proof for $u \in W_0^{1,p}(\Omega)$. I am trying to prove Poincare's inequality for $u \in W_0^{1,p}(\Omega)$, where $\Omega \subset \mathbb{R}^n$ is an open bounded set and $1 \leq p < \infty$. This is Poincare's inequality: 

$||u||_{L^p(\Omega)} \leq C ||\nabla u ||_{L^p(\Omega)}$.

For $p < N$ the inequality follows applying Sobolev-Gagliardo-Nirenberg inequality. Indeed, $$||u||_{p^*} \leq C||\nabla u||_{p}, $$ and from the fact that $\Omega$ is bounded we have $||u||_p \leq \tilde{C} ||u||_{p^*} \leq C||\nabla u||_{p}$.
How can I prove the inequality for $ N \leq p < \infty$ ?
 A: Take any $u \in C_c^\infty(\Omega)$. Let $x = (y,x_n)$ where $x_n \in \mathbb R, y \in \mathbb R^{n-1}$.Then due $u$ having compact support, we have some $z \in \mathbb R$ such that $(y,z) \in \Omega$ and $u(y,z) = 0$. Then $$ |u(x)| = |u(y,x_n) - u(y,z)| = \left| \int_z^{x_n} \frac{du}{dx_n}(y,t)dt \right| \le \int_z^{x_n} \left| \frac{du}{dx_n}(y,t)\right| dt $$
Applying holder inequality to the latter, and noting that $|x_n-y| \le diam(\Omega)$ we get $$ |u(x)| \le diam(\Omega)^{\frac{p-1}{p}} \Big( \int_{z}^x \left| \frac{du}{dx_n}(y,t) \right|^p dt \Big)^{\frac{1}{p}} \le diam(\Omega)^{\frac{p-1}{p}} \Big( \int_{-\infty}^\infty \left| \frac{du}{dx_n}(y,t) \right|^p dt \Big)^{\frac{1}{p}}$$  Hence $$ |u(y,x_n)|^p \le diam(\Omega)^{p-1} \int_{-\infty}^{\infty} \left| \frac{du}{dx_n}(y,t)\right|^p dt $$  Integrating over $x_n \in \mathbb R$ (in fact over $x_n \in \pi_n(\Omega)$ where $\pi_n:\mathbb R^n \to \mathbb R$ is a projection on last coordinate) we get $$ \int_{\mathbb R}|u(y,x_n)|^p dx_n \le diam(\Omega)^{p-1} \int_{\pi_1(\Omega)}\int_{-\infty}^{\infty} \left| \frac{du}{dx_n}(y,t)\right|^pdt dx_n \le diam(\Omega)^p \int_{-\infty}^{\infty} \left|\frac{du}{dx_n}(y,x_n)\right|^p dx_n$$
Integrating over $y \in \mathbb R^{n-1}$ we finally get ( by Fubinii - everything is non-negative) $$ \int_{\mathbb R^n} |u(x)|^p d\lambda_n(x) \le diam(\Omega)^p \int_{\mathbb R^n} \left|\frac{du}{dx_n}(x)\right|^p d\lambda_n(x)$$ where $\lambda_n$ is Lebesgue measure on $\mathbb R^n$. Please note that we could do the same with any of the derivatives $\frac{du}{dx_j}$ (but for $j \not \in \{1,n\}$ it's harder to write since we would need something like $x=(y_1,x_j,y_2)$ where $y_1 \in \mathbb R^{j-1}, y_2 \in \mathbb R^{n-j+1}$ and proceed analogously on $x_j$ as we did on $x_n$. Hence adding all those inequalities $$ \int_{\mathbb R^n}|u(x)|^pd\lambda_n(x) \le diam(\Omega)^p \int_{\mathbb R^n}|\frac{du}{dx_j}(x)|^p d\lambda_n(x) \qquad j \in \{1,...,n\}$$ we get $$ \int_{\mathbb R^n} |u(x)|^p d\lambda_n(x) \le \frac{diam(\Omega)^p}{n} \int_{\mathbb R^n} \sum_{j=1}^n |\frac{du}{dx_j}(x)|^p d\lambda_n(x) $$ Lastly, note that all norms on $\mathbb R^n$ are equivalent, hence there is some constant $A(n,p)$ such that $$\sum_{j=1}^n |\frac{du}{dx_j}(x)|^p \le A(n,p) \Big(\sum_{j=1}^n | \frac{du}{dx_j}(x)|^2\Big)^{\frac{p}{2}} = A(n,p)\|\nabla u(x)\|^p  $$ (the constant $A(n,p)$ can be calculated). Hence if $C(n,p,\Omega) = \frac{diam(\Omega)}{n}A(n,p)$ then $$ \int_{\mathbb R^n} |u(x)|^p d\lambda_n(x) \le C(n,p,\Omega) \int_{\mathbb R^n}\|\nabla u(x)\|^p d\lambda_n(x) $$ Or equivalently for some constant $\widehat{C} := \widehat{C}(n,p,\Omega)$ we get $\|u\|_{L_p(\Omega)} \le \widehat{C}\|\nabla u\|_{L_p(\Omega)}$. Since the result is true for any $u \in C_c^\infty(\Omega)$ (with the constant that does not depends on the function!), the result will remain true for any $u \in W_0^{1,p}(\Omega)$, since $C_c^\infty(\Omega)$ is dense in $W_0^{1,p}(\Omega)$ (so we can pass to the limit on both sides with our approximating sequence).
