I am reading Conway's book about complex analysis. One question in it bothered me a lot recently. If given a piecewise continuous function with jump on the boundary of unit disk and it is bounded, we can define a function in the disk by using Poisson kernel. Is this function still harmonic as in the case of continuous boundary data? If it is, how does it behave around the jump on the boundary? Can I say it converges to the boundary data in weak sense, i.e., in the sense of distribution?


1 Answer 1


The function is still harmonic. At all points where the boundary function is continuous, the harmonic function continuously extends to the boundary. The behavior at the points of discontinuity is a bit trickier. For ease of stating the result, assume we're working on the upper half plane (since the two regions are conformally equivalent, there's no real different). Suppose the boundary function is $0$ on the positive reals and $1$ on the negative reals. Then the harmonic extension limits to $\frac{1}{\pi}\arg z$ as $z$ tends to zero. This is a problem in Ahlfors' text on complex analysis, and I think you can solve it by just directly computing the integral in question.

Your boundary function can get pretty crazy and the resulting extension will still be harmonic. You can even integrate against a finite Borel measure and get a harmonic function.

You can in fact say things about weak convergence and even $L^p$ convergence of the harmonic extension to the boundary. The best treatment I have seen of this is the first chapter of Garnett's Bounded Analytic Functions, but I believe it is fairly standard material and I'm sure you can find it elsewhere.

This is all true in higher dimensions too, and you can find proofs in chapter 2 of Stein and Weiss's Introduction to Fourier Analysis on Euclidean Spaces.


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