Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 5 / 5

For a real locally integrable function, the definitions are distributional:Thus the Laplacian is respectively a positive distribution, a negative distribution, or zero. A superharmonic function is the negative of a subharmonic function; being both is equivalent to being harmonic. For a $C^2$ function these conditions are $\Delta f\ge 0$, $\Delta f\le 0$, and $\Delta f=0$ pointwise.

An $L_{\mathrm{loc}}^1$ equivalence class does not itself specify point values. For the maximum principle use the canonical representative of a subharmonic distributionThese normalized integrals are ball averages. To justify this construction, first mollify $f$: its smooth local regularizations satisfy $\Delta f_{\epsilon }\ge 0$. For any smooth subharmonic function, the spherical mean $m_x(r)$ obeysThis follows by differentiating the spherical integral and applying the divergence theorem. Its ball mean also increases with radius: its derivative is $n/r$ times the spherical mean minus the ball mean. Local $L^1$ convergence of the regularizations passes the monotonicity of the ball means to $f$. The decreasing-radius limit therefore exists, possibly as minus infinity at an exceptional point, and equals $f$ almost everywhere by the Lebesgue differentiation theorem. It is upper semicontinuous as the local infimum of continuous ball averages and satisfiesFor a superharmonic function use the corresponding lower semicontinuous representative and reverse the inequality.

For a distributionally harmonic function, the smooth regularizations are harmonic and have constant spherical means. A radial mollifier ${\rho }_{\delta }$ of integral one therefore reproduces them by averaging these means. At fixed interior radius $\delta$, pass to the local $L^1$ limit to obtain $f=f\ast {\rho }_{\delta }$ almost everywhere on a smaller region. The right side is smooth and has zero distributional Laplacian, giving the smooth harmonic representative. This proves the relevant Weyl lemma. Changing a value at a single point can spoil a pointwise maximum principle, so the representative convention is essential.

The strong maximum principle for subharmonic functions is: on a connected open domain, the canonical upper semicontinuous representative of a subharmonic function cannot attain a finite global maximum at an interior point unless it is constant. A local maximum instead forces constancy on an appropriate neighbourhood; it alone does not assert constancy on the whole domain unless that value is a global maximum.

Suppose $u\le M$ on $\Omega$ and $u(x_0)=M$. For every sufficiently small ball centred at $x_0$, the subharmonic mean inequality givesThe nonnegative function $M-u$ has integral zero, so $u=M$ almost everywhere on that ball. Its canonical representative is then $M$ at every point of the ball, by taking still smaller local averages.

The maximum set $E=\{x:u(x)=M\}$ is consequently open by the same argument at each of its points. It is closed relative to $\Omega$, because upper semicontinuity makes its complement $\{u<M\}$ open. It is nonempty, and connectedness gives $E=\Omega$. Therefore $u\equiv M$. This proves the principle, including its representative and connectedness hypotheses. Arbitrary representatives do not satisfy it: the function equal to zero almost everywhere but assigned value one at the origin has zero distributional Laplacian and an artificial pointwise maximum.

The local Harnack inequality for harmonic functions states that if $u\ge 0$ is harmonic on a neighbourhood of $\overline{B_{4r}(a)}$, thenMore generally, for any compact $K$ in a connected open domain $\Omega$, there is $C=C(K,\Omega ,n)$ such that ${\sup }_{K}u\le C{\inf }_{K}u$ for every nonnegative harmonic $u$ on $\Omega$. Positivity is essential; the assertion is not for arbitrary sign-changing functions.

First prove the mean value property for harmonic functions. For a smooth harmonic function the spherical mean has derivativeso it equals $u(x)$; integration in the radius gives the ball mean property. The preceding Weyl lemma supplies smoothness if harmonicity was initially distributional.

For $x,z\in B_r(a)$ the inclusion $B_r(x)\subset B_{3r}(z)\subset B_{4r}(a)$ and nonnegativity giveTaking the supremum over $x$ and the infimum over $z$ proves the local estimate directly. To obtain the compact-set form, connect centres of a finite ball cover of $K$ to a fixed interior point by paths in $\Omega$. The compact union of those paths stays a positive distance from the complement of $\Omega$. Subdivide the paths into finitely many steps small enough that the local estimate applies at consecutive points. Multiply the finitely many comparison constants, first from one point of $K$ to the reference point and then to any other point of $K$. This gives a constant independent of $u$. In particular a nonnegative harmonic function is either everywhere positive or identically zero on each connected component.

Fix a compact $K\subset \Omega$ and choose $\delta >0$ such that a closed $3\delta$-neighbourhood of $K$ is compactly contained in $\Omega$. The hypothesis gives a common bound $M$ on that neighbourhood. Choose a smooth radial mollifier ${\rho }_{\delta }$ supported in $B_{\delta }(0)$ with integral one. The mean value property for harmonic functions, integrated against its radial weights, givesin a neighbourhood of $K$. Differentiating this fixed-radius convolution gives the uniform interior estimateThus the sequence is equicontinuous there; taking its pointwise limit gives the same local modulus of continuity for $u$. A finite small-ball cover of $K$ now proves uniform convergence directly: choose the cover so both $u_j$ and $u$ vary by less than $\epsilon$ inside each ball, then use convergence at its finitely many centres to make $|u_j-u|<\epsilon$ there. The triangle inequality gives ${\sup }_K|u_j-u|<3\epsilon$ for large $j$.

Finally, pass to the limit in the reproducing convolution identity on smaller compact sets. It yields $u=u\ast {\rho }_{\delta }$, so $u$ is smooth locally. Also, for each compactly supported smooth test function $\varphi$, uniform convergence on its support givesHence its smooth Laplacian is zero. We have proved $u_j\to u$ uniformly on every compact subset and $u$ is harmonic. Using the continuous representatives in the pointwise hypothesis makes this conclusion pointwise, rather than only almost everywhere. This is compact convergence of locally bounded harmonic functions.