The following is the beginning of a proof of the Poincaré inequality:
Here are my questions:
- How is the limiting argument made so that "it is sufficient to prove the inequality for $u\in C_c^\infty(\Omega)$?
- What does "The inequality is invariant under rotations and translations" mean? Could anyone write it out explicitly?
1) Suppose you prove the inequality for all smooth $u$ with compact support in $\Omega$. Let $u \in H^1_0(\Omega)$. Then by density of $C_c^\infty(\Omega)$ in $H_0^1$, there exists a sequence $\phi_k \in C_c^\infty(\Omega)$ converging to $u$ in the $H^1_0(\Omega)$ norm. Therefore $$\lim_{k\to \infty} \int_\Omega \phi_k^2 \, dx = \int_\Omega u^2 \, dx \ \ \ \text{ and } \ \ \ \lim_{k\to \infty} \int_\Omega |D\phi_k|^2 \, dx = \int_\Omega |Du|^2 \, dx.$$ Since the estimate holds for each term $\phi_k$ in the sequence, with the same uniform constant $C$, it must hold for $u$ as well. This is an argument by density or approximation.
2) A rotation and translation is a transformation of the form $y=Ax + b$, where $b \in \mathbb{R}^n$ and $A$ is an orthogonal matrix ($A^TA=AA^T=I$, and $\text{det}(A)=1$). Suppose we change variables and set $v(y)=u(x)=u(A^T(y-b))$. Then $dy = |\text{det}(A)|dx = dx$ and $Dv(y) = ADu(x)$. Since $A$ is orthogonal, $|Dv(y)|=|Du(x)|$ and we have $$\int_\Omega u(x)^2 \,dx = \int_{\Omega'} v(y)^2 \, dy \ \ \ \text{ and } \ \ \ \int_\Omega |Du(x)|^2 \,dx = \int_{\Omega'} |Dv(y)|^2 \, dy,$$ where $\Omega' = A\Omega + b$. Since both sides of the estimate are invariant under rotations and translations, we may as well assume the bounded direction lies in the $x_n$ direction between $0 < x_n < a$.