Fix $x\in\Omega'$ and $0<\rho<R(\Omega')$. Differentiate the ball mean-value formula and apply the <divergence theorem>:
$$
\partial_{x_j}u(x)
=\frac1{\omega_d\rho^d}\int_{B(x,\rho)}\partial_j u
=\frac1{\omega_d\rho^d}\int_{\partial B(x,\rho)}u\nu_j\,dS.
$$
Since $|\nu_j|\leq1$ and $|\partial B(x,\rho)|=d\omega_d\rho^{d-1}$,
$$
|\partial_{x_j}u(x)|\leq\frac d\rho\max_\Omega|u|.
$$
Letting $\rho\uparrow R(\Omega')$ and taking both maxima proves
$$
\boxed{\max_{1\leq j\leq d}\max_{\Omega'}|\partial_{x_j}u|\leq\frac d{R(\Omega')}\max_\Omega|u|.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 107 / 1 / v / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-107/1/v/solution