Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 107 / 1

Fix $x\in \Omega$ and $R<\mathrm{dist}(x,\partial \Omega )$. The mean value property for harmonic functions isTo prove the spherical identity, translate $x$ to the origin and put $F(r)=|S^{n-1}{|}^{-1}{\int }_{S^{n-1}}u(r\theta )d{S}_{\theta }$. The divergence theorem givesHence $F(r)={\lim }_{s\downarrow 0}F(s)=u(0)$. Integrating the spherical identity in polar coordinates gives the ball identity.

Choose $0<r<\mathrm{dist}(x,\partial \Omega )$. Differentiating the ball mean-value formula with respect to its centre and then using the divergence theorem givesConsequently the interior derivative estimate for a harmonic function yieldsThe constant may depend on $x$ and its distance from the boundary, but it is independent of $u$.

Take a nonnegative radial mollifier ${\eta }_{\epsilon }$ supported in $B_{\epsilon }(0)$, with $\epsilon <\mathrm{dist}(x,\partial \Omega )$. Writing the convolution in polar coordinates and applying the spherical mean value property for harmonic functions at every radius shows directly thatThe convolution is smooth, so $u$ agrees locally with a smooth function. Since $x$ was arbitrary, $u\in C^{\infty }(\Omega )$.

Choose a cutoff function $\eta$ supported in $B_{\rho }(x)$, equal to one on $B_{{\rho }^{\prime }}(x)$, and satisfying $|D\eta |\le (\rho -{\rho }^{\prime }{)}^{-1}$. Use ${\eta }^{2}u$ as a test function in the weak formulation:The Cauchy-Schwarz inequality followed by Young inequality givesand henceThis is the Caccioppoli inequality. The numerical constant is convention-dependent and is normally absorbed into the displayed estimate; replacing the radii by fixed intermediate radii gives the stated form with one universal constant.

On a relatively compact ball, mollification commutes with the Laplacian, so $u_{\epsilon }=u\ast {\eta }_{\epsilon }$ is a smooth harmonic function. Repeated interior derivative estimates, together with the Caccioppoli inequality, bound every derivative of $u_{\epsilon }$ on a smaller ball by the local $L^2$ norm of $u$, uniformly as $\epsilon \downarrow 0$. The Arzela-Ascoli theorem and a diagonal argument give a smooth local limit, while mollification gives $u_{\epsilon }\to u$ in $L_{\mathrm{loc}}^2$. Thus the limit equals $u$ almost everywhere. After choosing this smooth representative, $u$ is harmonic pointwise. This is the Weyl lemma for an $H^1$ weak solution.

For $u\in L_{\mathrm{loc}}^2(\Omega )$, the correct formulation is the distributional identityThus $\Delta u=0$ as a distribution. The Weyl lemma applies already to locally integrable distributions, so $u$ agrees almost everywhere with a smooth harmonic function.