Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 13 / 1

Write ${\omega }_n=|B_1|$. If $u$ is a $C^2$ harmonic function and $\overline{B_r(x)}\subset \Omega$, its two mean value properties for harmonic functions areTo prove the spherical property, express the normalized spherical mean on the fixed unit sphere:Differentiation and the divergence theorem giveSince continuity gives $M(r)\to u(x)$ as $r\downarrow 0$, the spherical mean equals $u(x)$ for every admissible radius. Integrate this identity in radial coordinates:This proves the ball mean property as well. For $n=1$, the sphere consists of the two endpoints and the same argument is the fundamental theorem of calculus.

The strong maximum principle for harmonic functions states that a $C^2$ harmonic function on a connected domain which attains its supremum at an interior point is constant. Applying it to $-u$ gives the corresponding assertion for an attained infimum.

Let $M=u(x_0)={\sup }_{\Omega }u$. The ball mean value property for harmonic functions gives, for every sufficiently small ball about $x_0$,The integrand is continuous and nonnegative. If it were positive at any point, it would be bounded below by a positive number on a smaller ball, contradicting the zero integral. Therefore $u=M$ on the whole ball. The set $A=\{x\in \Omega :u(x)=M\}$ is relatively closed by continuity and relatively open by this same argument at each of its points. It is nonempty. Connected space forces $A=\Omega$, proving the principle. If the open set is disconnected, constancy is asserted on the relevant connected component.

First suppose $u$ is a classical harmonic function. At a local maximum of $u-v$, its Hessian matrix is negative semidefinite, so $0=\Delta u(x_0)\le \Delta v(x_0)$. At a local minimum the Hessian inequality reverses and gives $\Delta v(x_0)\le 0$. Thus both test conditions hold.

Conversely, suppose only that $u$ is continuous and satisfies the two conditions. We prove the stronger conclusion that these conditions themselves imply $u\in C^{\infty }$ and $\Delta u=0$. Use the standard Dirichlet existence theorem on a ball: every continuous function on its boundary has a harmonic extension $h\in C^{\infty }(B_R(a))\cap C(\overline{B_R(a)})$ with exactly those boundary values. This follows, for example, from the Poisson integral. Apply it to $u$ on an arbitrary relatively compact ball, obtaining a harmonic replacement $h$.

If $u-h$ is positive somewhere, choose $\epsilon >0$ small enough that $u-h-\epsilon (R^2-|x-a{|}^2)$ is still positive there. It vanishes on the boundary and therefore has a positive maximum at an interior point. The upper testhas $\Delta v=-2n\epsilon <0$, contradicting the maximum test condition. Similarly, if $u-h$ is negative somewhere, $v=h-\epsilon (R^2-|x-a{|}^2)$ yields an interior negative minimum of $u-v$ for small $\epsilon$, but $\Delta v=2n\epsilon >0$, contradicting the minimum condition. Hence $u=h$ throughout the ball.

If the quantifier requires $v\in C^2(\Omega )$ rather than a local test, multiply the displayed smooth $v$ by a smooth cutoff supported within the ball and equal to one near the touching point, extending by zero outside. The resulting global test has the same germ and Laplacian there, so the argument remains valid. Since every point lies in such a ball, $\boxed{u\text{ is a smooth harmonic function throughout }\Omega }$. This is the continuous viscosity harmonic functions are classical characterization.