# Question on harmonic functions

By the maximum principle, every harmonic function on a bounded domain is uniquely determined by its boundary values. However, for unbounded domains, we can have infinitely many harmonic functions with prescribed boundary values. My question, roughly, is this: since we may use conformal maps between bounded and unbounded domains to construct harmonic functions, how do we still maintain uniqueness?

More concretely, suppose we have a harmonic function $F$ on the upper half plane with prescribed boundary values. Consider the conformal map $u(z)=i(1-z)/(1+z)$ that takes the disc to the upper half plane. Then $F \circ u + cy$ for any constant $c$ gives an infnitely family of harmonic functions on the disc with given boundary values. But this contradicts the remark above that harmonic functions on bounded domains are unique. Where is the flaw in the reasoning?

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"its boundary values" presumes the existence of boundary values, usually in the form of continuous extension to the boundary. Which $(F+cy)\circ u$ lacks. – user31373 May 18 '12 at 21:04
I don't know what you mean by boundary values for the upper half plane, which, well, lacks a boundary. Given a set of boundary values on the unit disc, the single point $z=-1$ is mapped to $\infty$ in the Riemann sphere. All other points on the unit circle are mapped to the real line. So all you get to specify for the upper half plane is a single value, at $\infty.$ Furthermore, this single value had better be the limit of the data on the real line, separately as one goes to $\infty$ or to $-\infty.$ Otherwise the pullback is not continuous on the unit circle. – Will Jagy May 18 '12 at 21:12
@WillJagy I would not go as far as saying that $\{z\colon \Im z>0\}$ lacks a boundary. In the topology of $\mathbb C$ its boundary is $\mathbb R$; in the topology of $\mathbb C\cup \{\infty\}$ its boundary is $\mathbb R\cup \{\infty\}$. The issue underlying this question is that the maximum principle requires the consideration of $\mathbb R\cup \{\infty\}$ and not just $\mathbb R$. – user31373 May 18 '12 at 21:18
@LeonidKovalev, sounds good. – Will Jagy May 18 '12 at 21:40
Ok, so this is how I am interpreting your answer. Since $-1$ gets mapped to $\infty$, and if $F(\infty) = \infty$, for example, then there is no continuous extension to the boundary of the disc, since $(F + cy) \circ u$ blows up at -1. So do we just get a harmonic function on the disc with values in the boundary except the point -1? Thus, not specifying the value at even a single point on the boundary means we may no longer have uniqueness? – user94038190 May 18 '12 at 22:01