# 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 $N \leq p < \infty$ ?

• It is proven in Evans, using a contradiction argument and Rellich's theorem.
– Ian
Jun 24, 2016 at 23:17
• @Ian Can it be proven without using Rellich's theorem?
– D1X
Jun 24, 2016 at 23:19
• I doubt it; the actual value of the constant is quite sensitive to specifics of the domain.
– Ian
Jun 24, 2016 at 23:30
• Okay, thank you very much.
– D1X
Jun 27, 2016 at 12:28

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).