I have searched all around for an answer to this, what I would imagine to be a rather natural question, but to no avail. How does one extend the Lindelöf maximum principle to sets of measure 0? Here's a precise formulation, which has a more restrictive condition on the domain of the harmonic function:

Let $f: \frac{1}{2}\mathbb{D} \to \mathbb{R}$ be harmonic and bounded. If $\limsup_{r\nearrow 1/2} f(re^{i\theta}) \leq 0$ for almost all $\theta$, then show $f \leq 0$ on $\frac{1}{2}\mathbb{D}$.

My attempt is as follows; I have tried to follow the usual technique used to prove the theorem when the exceptional set is a single point: let $E$ be the exceptional subset of the boundary, and let $S_{1, \epsilon}, S_{2, \epsilon}, \ldots, S_{N_\epsilon, \epsilon}$ be a collection of connected subsets of $\partial \frac{1}{2}\mathbb{D}$ whose lengths sum to $\epsilon$ and such that $E\subseteq \bigcup_{n=1}^{N_\epsilon} S_{n, \epsilon}$ (from the definition of Lebesgue measure). These naturally give rise to points $z_{i, \epsilon}$ and radii $r_{i, \epsilon}$ such that $S_{i, \epsilon} \subseteq \{|z-z_{i, \epsilon}| < r_{i, \epsilon}\}$ and such that $\sum_i r_{i, \epsilon} \leq \epsilon/2$. Then the function $$f_\epsilon :=f(z) + \sum_{i=1}^{N_\epsilon} \frac{1}{-\log(r_{i, \epsilon})} \log|z-z_{i, \epsilon}|$$ is a harmonic function which is less than 0 as one approaches any boundary point, and so is 0 throughout the disk. The problem is though that our modified function does not in general converge pointwise to $f$ as $\epsilon \searrow 0$ (since there is a priori no bound on $N_\epsilon$), which is what makes the usual proof of the Lindelöf maximum principle for a single point work.

Any pointers on how to adjust this proof or go about it differently would be greatly appreciated.

  • $\begingroup$ You assume $f$ is defined on $\frac{1}{2}\mathbb D$ and then talk about $f$ on $\mathbb D?$ $\endgroup$ – zhw. Aug 22 '18 at 3:03
  • $\begingroup$ Sorry, corrected. Everything is on the disk of radius 1/2. $\endgroup$ – Van Latimer Aug 22 '18 at 3:52
  • $\begingroup$ Why $\frac{1}{2}\mathbb D$ rather than $\mathbb D?$ Seems weird. Also, is $\limsup_{r\nearrow 1/2} f(re^{i\theta}) \leq 0$ for all $\theta?$ $\endgroup$ – zhw. Aug 22 '18 at 4:49
  • $\begingroup$ The question does not make sense. What are the exceptional points you are talking about? $\endgroup$ – Kavi Rama Murthy Aug 22 '18 at 6:01
  • $\begingroup$ So sorry, I forgot the crucial “for almost all theta”. The exceptional set refers to this. I set the scene in the disk of half radius so the log functions would strictly negative on the disk. $\endgroup$ – Van Latimer Aug 22 '18 at 8:24

The other answer is probably better than this one; potential theory, polar sets, etc get to the heart of the matter. But for the disk specifically one can give a very elementary argument. Here "elementary" means "just using things in for example Rudin Real and Complex Analysis".

If $u$ is a bounded harmonic function in $\mathbb D$ then $f(e^{it})=\lim_{r\to1^-}u(re^{it})$ exists for almost every $t$. It's clear that $f$ is measurable, so $f\in L^\infty(\Bbb T)$. And in fact $u=P[f]$, the Poisson integral of $f$. Since the Poisson kernel is positive it's clear that $f\le0$ almost everywhere implies $u\le 0$.

(That's a proof for bounded harmonic functions in $\Bbb D$; I'm totally stumped on how to transfer it to a proof for bounded harmonic functions in $\frac12\Bbb D$, sorry.)

  • $\begingroup$ Ah yes. The proof of Lindelöf for a single point and an arbitrary requires this trick with the logarithms, but when we have the Poisson kernel it becomes much easier; a waited average of a measurable function which is almost everywhere less than 0 will be less than 0. $\endgroup$ – Van Latimer Aug 22 '18 at 19:17

Adding logarithms is not going to be enough. You need to integrate them, i.e., to consider the logarithmic potential of a measure (which is a subharmonic function on $\mathbb{C}$ and harmonic outside of the support of the measure).

A set is called polar if there exists a subharmonic function on $\mathbb{C}$ (not identically $-\infty$) that is equal to $-\infty$ on that set. The maximum principle is properly generalized to

If a (sub)harmonic function $u$ is bounded above on $\Omega$, and $u\le M$ on $\partial \Omega\setminus P$, where $P$ is a polar set, then $u\le M$ in $\Omega$.

That a measure zero subset of the unit circle is a polar set is either immediate or takes some work with potentials of measures around that set. Depends on where you stand in potential theory. And since this box is too short for a book on potential theory, I'll recommend one:

Ransford, Thomas. Potential theory in the complex plane. London Mathematical Society Student Texts. 28. Cambridge: Cambridge University Press. (1995)

  • $\begingroup$ Thank you for this comprehensive answer. I hope to learn some potential theory soon but right now I do just have a year of graduate level complex analysis. $\endgroup$ – Van Latimer Aug 22 '18 at 19:14
  • $\begingroup$ I accepted the other answer mainly because this question came from an exam intended for students with just this level of knowledge, so I think the other answer is more the expected proof. $\endgroup$ – Van Latimer Aug 22 '18 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.