Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 12 / 5

Let $B_r(x_0)$ have closure contained in the domain of a harmonic function $u$. Put ${\omega }_n=|B_1(0)|$, so $|B_r|={\omega }_n{r}^n$ and $|\partial B_r|=n{\omega }_n{r}^{n-1}$. The sphere and ball mean value properties for harmonic functions areTo prove the sphere formula, define its average using a fixed unit sphere:Differentiating under the integral and applying the divergence theorem givesContinuity gives $M(r)\to u(x_0)$ as $r\downarrow 0$, proving the spherical average identity. Integrating that identity over radii with the polar-coordinate weight yieldswhich proves the ball identity. For $n=1$, the sphere average is the average of the two endpoints, and the same formulas hold with the counting surface measure.

A local form of the Harnack inequality for harmonic functions is: if $u\ge 0$ is harmonic on a neighbourhood of $\overline{B_{4r}(y)}$, thenFor $x,z\in B_r(y)$, the triangle inequality gives $B_r(x)\subset B_{3r}(z)\subset B_{4r}(y)$. The ball mean value property for harmonic functions and nonnegativity therefore implyTaking the supremum over $x$ and infimum over $z$ proves the inequality, including the case where the infimum is zero. No strict positivity assumption was used.

On a connected domain $\Omega$, the equivalent compact-set statement is that for each nonempty compact $K⋐\Omega$ there exists $C(K,\Omega ,n)$ withHere is the passage from balls to this statement. Cover $K$ by finitely many small balls $B_{r_j}(y_j)$ whose quadrupled closed balls lie in $\Omega$. Join their centres to a fixed centre by paths in $\Omega$. Each of the finitely many paths has compact image and positive distance from the boundary, so subdividing it gives a finite chain of overlapping small balls whose quadrupled closures also lie in $\Omega$. The local inequality compares any two points in each ball; using a point in each successive overlap propagates comparisons along the chain. The number of comparisons is bounded uniformly over this finite collection of chains and covering balls. Multiplying the factors $3^n$ gives one finite constant for all pairs $x,z\in K$, proving the compact-set inequality. In particular a nonnegative harmonic function vanishing at one interior point vanishes throughout a connected domain.

Let $m={\inf }_{{\mathbb{R}}^n}u$, which is finite because $u$ is bounded, and put $v=u-m$. Then $v\ge 0$ is harmonic. Choose points $y_j$ with $v(y_j)\to 0$. For any fixed $x$, choose $r_j$ large enough that $x,y_j\in B_{r_j}(0)$. Since $v$ is harmonic on all of ${\mathbb{R}}^n$, the Harnack inequality for harmonic functions on $B_{4r_j}(0)$ givesThe comparison factor is independent of the growing radius, which is essential. Thus $v(x)=0$ for every $x$ andThis proves the Liouville theorem for harmonic functions in every dimension. In fact the argument only needs an entire harmonic function to be bounded below; changing its sign gives the corresponding bounded-above result.

Set $X=Du/\sqrt{1+|Du{|}^2}$. Since $u\in C^2({\mathbb{R}}^n)$, the field is $C^1$, and $|X|\le 1$ everywhere, even when the gradient of $u$ is unbounded. The divergence theorem on $B_R(0)$ givesLetting $R\to \infty$ proves $\kappa =0$. This is the fact that an entire bounded vector field cannot have nonzero constant divergence; it holds in every dimension.

The equation is now exactly the minimal surface equation for a graph. Apply the supplied Bernstein theorem for minimal graphs in dimensions $1\le n\le 7$ to concludeThus no nonzero constant right-hand side is possible for an entire graph, and in the stated dimensions every possible graph is an affine plane.