By Douglas Arnold

Lecture notes.

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**Example text**

The difference between this function and u satisfies the mean value property and is zero on the boundary of the disk, so it vanishes on the disk. Corollary. If a sequence of harmonic functions converges uniformly on compact subsets, then the limit function is harmonic. From the expressions Pr (θ) = we have 1 − r2 1 − r2 = (1 + r)2 − 2r(1 + cos θ) (1 − r)2 + 2r(1 − cos θ) 1−r 1+r ≤ Pr (θ) ≤ . 1+r 1−r Now suppose u is harmonic on B(0, R) and non-negative, and reiθ is an arbitrary point of B(0, R).

Let G, G+ , G− , and G0 be as above. If f is a continuous complex-valued function on G+ ∪ G0 , which is analytic on G+ and real on G0 , then f admits a unique extension to a harmonic function on all of G. The extension is given by f (z) = f (¯ z ) for z ∈ G− . Proof. We could base a proof on the previous result applied to Im f and harmonic conjugates, but it is also easy to verify this directly using Morera’s theorem. 39 Of course the line of symmetry could be any line, not just the real axis. Thus if G is symmetric with respect to any line, defined and analytic on one side of the line, and real on the line, it can be extended to be analytic on the whole domain.

Proof. If such an extension exists it is certainly unique, so it suffices to show that the stated extension defines a harmonic function in G. It certainly defines a continuous extension which is harmonic in G+ ∪ G− , so it suffices to show that it is harmonic in a neighborhood of z0 ∈ G0 . By translating and extending we can assume that z0 = 0 and that G contains ¯ Define U on D ¯ as the solution of the Dirichlet problem with the closed unit disk, D. boundary data u|∂D (where u has been extended to G− by the formula above).