# Cut-off functions in Caccioppoli's inequality

Caccioppoli's inequality states that the solution $$u$$ of the equation $$-\nabla\cdot(A\nabla u)=0$$ in some bounded domain $$\Omega$$ satisfies $$\int_{B(0,\rho)}|\nabla u|^2dy\leq \frac{C}{(R-\rho)^2}\int_{B(0,R)}|u|^2~dy,$$ for $$0<\rho, $$2R$$ should be smaller than diameter of $$\Omega$$, etc. In the proof of Caccioppoli's inequality, a cut-off function $$\phi$$ is constructed on $$B(0,R)$$ satisfying $$|\nabla \phi|<\frac{C}{R-\rho}$$.

Is it possible to construct a cut-off function satisfying $$|\nabla \phi|<\frac{C}{(R-\rho)^2}$$ or with higher powers of $$R-\rho$$? What is the limit on this?

In general no, you can't expect to improve the exponent. Given such a $$\phi,$$ we have $$\phi(\rho e_1) = 1$$ and $$\phi(R e_1) = 0,$$ where $$e_1 = (1,0,\dots,0)$$ is the first of the standard basis vectors. Thus applying the mean value theorem to $$t \mapsto \phi(te_1)$$ we obtain $$\xi \in (\rho, R)$$ such that, $$\frac{\partial \phi}{\partial x_1}(\xi e_1) = \frac1{R-\rho}.$$ In particular, we see that, $$\sup_{B_R} |\nabla \phi| \geq \frac1{R-\rho}.$$