Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 9 / 5

For a real locally integrable function $f$, the distributional definition of a subharmonic function is $\Delta f\ge 0$: for every nonnegative $\phi \in C_c^{\infty }(\Omega )$,A superharmonic function has the opposite inequality, equivalently $-f$ is subharmonic. A harmonic function has $\Delta f=0$ distributionally, equivalently equality for every test function. These definitions concern almost-everywhere classes; the given upper semicontinuous representative supplies the pointwise version for subharmonic functions.

The statement as printed is false without an integrability condition. On any nonempty open set, the constant functions $f_n\equiv n$ are harmonic and therefore subharmonic, but their supremum is $+\infty$, which is not locally integrable.

A valid version assumes $g={\sup }_n{f}_n\in L_{\mathrm{loc}}^1(\Omega )$. First, the maximum of finitely many subharmonic functions is subharmonic. For a distributional proof, mollify two functions on smaller interior domains and replace their maximum byThis is smooth, convex, and nondecreasing in each argument. The chain rule shows that its Laplacian applied to two smooth subharmonic functions is nonnegative: the first-derivative terms multiply their nonnegative Laplacians, and the Hessian term is nonnegative by convexity. Mollifier convergence in $L_{\mathrm{loc}}^1$ and the uniform bound $|M_{\delta }-\max |\le \delta /2$ allow both regularizations to be removed. Continuity of distributional derivatives under this convergence proves the finite-maximum assertion.

Now $g_N={\max }_{n\le N}{f}_n$ increases to $g$ and satisfies $|g_N|\le |f_1|+|g|$ almost everywhere. The dominated convergence theorem gives convergence in $L_{\mathrm{loc}}^1$. For every nonnegative test function,Thus $g$ is subharmonic in the distributional sense. This is the locally integrable supremum theorem for subharmonic functions. For the pointwise upper semicontinuous convention, take its canonical representative; the raw supremum need not already have that regularity.

This is Weyl lemma, with the representative condition needed for pointwise smoothness. Choose a smooth radial nonnegative mollifier ${\rho }_{\epsilon }$ of mass one. On interior domains, $f_{\epsilon }=f\ast {\rho }_{\epsilon }$ is smooth and harmonic because $\Delta f_{\epsilon }=(\Delta f)\ast {\rho }_{\epsilon }=0$.

Fix a smaller interior neighborhood and a radial mollifier ${\rho }_r$ whose support stays inside the domain there. The mean value property for harmonic functions implies $f_{\epsilon }=f_{\epsilon }\ast {\rho }_r$ when $\epsilon$ is sufficiently small: a radial average is a weighted average of spherical means, all equal to the center value.

As $\epsilon \downarrow 0$, $f_{\epsilon }\to f$ in $L_{\mathrm{loc}}^1$, while $f_{\epsilon }\ast {\rho }_r\to f\ast {\rho }_r$ locally uniformly. The latter follows by bounding the difference by $\Vert {\rho }_r{\Vert }_{\infty }$ times the local $L^1$ error on a slightly larger compact set. Thus $f=f\ast {\rho }_r$ almost everywhere locally. The right side is smooth, and its Laplacian is zero, providing a smooth harmonic representative. These representatives agree on overlapping neighborhoods since continuous functions agreeing almost everywhere agree everywhere.

The given uniqueness of the canonical upper semicontinuous representative identifies this smooth function with $\tilde{f}$. Since the question assumes $f=\tilde{f}$, it follows that $\boxed{f\in C^{\infty }(\Omega )}$ pointwise. An arbitrary null-set modification of a harmonic distribution would not have that conclusion.

The strong maximum principle for subharmonic functions states that an upper semicontinuous subharmonic function on a connected open set which attains its finite global maximum at an interior point is constant. In particular this holds for a harmonic function; negating the function gives the corresponding minimum principle for superharmonic functions.

Let the maximum be $M$, attained at $x_0$. For every sufficiently small ball centered at $x_0$, the mean value inequality givesTherefore its average equals $M$. If some point in the ball had value less than $M$, upper semicontinuity would give an open neighborhood where $u\le M-\eta$ for some $\eta >0$. That positive-measure neighborhood would make the average strictly less than $M$, a contradiction. Thus $u=M$ throughout a neighborhood of $x_0$.

The same argument at each maximizing point shows that $\{u=M\}$ is open; it is also closed relative to the domain by upper semicontinuity. It is nonempty, so connectedness forces it to be the whole domain. Without connectedness, constancy is only guaranteed on the component containing the maximizing point.