For the classical Poincare's inequality, if $u \in H^1_0(\Omega)$, then $$\int_\Omega u^2 \,dx \le C \int_\Omega |\nabla u|^2 \,dx.$$ Do we have something similar with the difference quotient? That is, do we have that if $u \in L^2(\Omega)$ and $u=0$ on $\partial\Omega$, then $$\int_\Omega u^2 \,dx \le C \sum_{k=1}^n \int_\Omega |D^h_k u(x)|^2 \,dx,$$ where the difference quotient is defined by $$D^h_k u = \frac{u(x+he_k) - u(x)}{h}$$ with $e_k$ to be the standard normal coordinate unit vector. How can I show it, or where can I look for its proof?
I would like to imitate the proof of the original Poincare's inequality, but I don't know how to write the Fundamental Theorem of Calculus in difference quotient. I hope someone can show me.
Thank you.
It does indeed hold. Well, if $\Omega$ is bounded, $u=0$ on $\Omega^c$ not just $\partial \Omega$, and with a slightly bigger integral on the right. On the plus side you only need one $k$.
Your idea is already quite good. Now, you write $$u(x) = -h \sum_{l=0}^{N_h} D^h_k u(x+l h e_k),$$ where $N_h$ is such that $N_h h \geq diam(\Omega)$ and $N_h h \leq C$ (e.g., $C=diam(\Omega)+1$ if $h \leq 1$). Then you proceed as in the standard proof: $$\int_{\Omega} u^2 \leq h^2 N_h \int_{\mathbb{R}^n} \sum_{l=0}^{N_h} \lvert D^h_k u(x+l h e_k) \rvert^2 \,dx = h^2N_h^2 \int_{\mathbb{R}^n} \lvert D^h_k u(x) \rvert^2 \,dx $$