For a <harmonic function> $u$, let
$$
M(r)=\frac1{|\partial B_r|}\int_{\partial B_r(x)}u.
$$
The <divergence theorem> gives
$$
M'(r)=\frac1{|\partial B_r|}\int_{\partial B_r(x)}\partial_\nu u
=\frac1{|\partial B_r|}\int_{B_r(x)}\Delta u=0.
$$
Since $M(r)\to u(x)$ as $r\downarrow0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the <mean value property for harmonic functions>. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. <continuous function>[Continuity] makes $u$ constant on those spheres, and <connected space>[connectedness] propagates that value through the domain. Thus the <weak maximum principle for elliptic operators> gives
$$
\min_{\partial\Omega}u\leq u\leq\max_{\partial\Omega}u.
$$

For the derivative estimate, choose $r>0$ smaller than half the <distance to a set>[distance] from $\Omega'$ to $\partial\Omega$, and let $\rho_r$ be a smooth radial <mollifier> supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u*\rho_r=u$ on $\Omega'$. Hence, for every <multi-index> $\alpha$,
$$
D^\alpha u(x)=\int_\Omega D^\alpha\rho_r(x-y)u(y)\,dy,
$$
so <Holder inequality> gives
$$
\|D^\alpha u\|_{L^\infty(\Omega')}
\leq\|D^\alpha\rho_r\|_\infty\|u\|_{L^1(\Omega)}
\leq C(n,\alpha,\Omega,\Omega')\|u\|_{L^1(\Omega)}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 107 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-107/1/a/solution