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Let $f$ be a real-valued function (in my case, an orientation-preserving homeomorphims of $\mathbb{R}$) on the real line $\mathbb{R}$ which is not in any $L^p$ -space. Let us take the simplest example $f(t)=t$. Is there a $\textit{direct or indirect}$ way of computing the harmonic extension $H(f)$ to $\mathbb{H}$ of such a function. For $\textit{direct}$ way, if I try to use the standard Poisson formula with the Poison kernel $p(z,t)=\frac{y}{(x-t)^2+y^2}, z= x+iy$, then I am getting $H(f)(i)=\infty$ for $f(t)=t$. But all the sources [Evans, PDE, p. 38 or wikipidea http://en.wikipedia.org/wiki/Poisson_kernel] for Poison formula for $\mathbb{H}$ assumes that the boundary map $f$ is in some $L^p, 1\leq p \leq \infty $. But then how should I compute $H(f)$ for very nice functions like $f(t)=t$ and make that equal to $H(f)(z)=z $?

For $\textit{indirect}$ way, I know one solution to the problem could be to pass to the unit disk model $\mathbb{D}$ by a Möbius transformation $\phi$ that sends $\mathbb{H}$ to $\mathbb{D}$, and then solve the problem in $\mathbb{D}$, call the solution on disk $F$, and then take $\phi^{-1}\circ F \circ \phi :\mathbb{H}\to \mathbb{H}$. This does solve the problem for $f(t)=t$, but my concern is in general $\phi^{-1}\circ F \circ \phi $ may $\textit{not}$ be harmonic, because post-composition of harmonic with holomorphic is not harmonic in general. In that case, how would I solve this problem ? Answer or reference would be greatly appreciated !

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Is there necessarily always a harmonic extension? Why are you convinced there is? Note btw that homeomorphisms on $\mathbb R$ are not really the "nice" functions in this context... You're not doing topology here, you are trying to solve a PDE with given boundary conditions. So your well-behaved functions are those on which you know some bounds in whatever norm. –  Sam Oct 7 '12 at 1:42
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Having said this, integration against a kernel really does not seem suitable here. But maybe some variant of the Perron method might still show the existence of a solution. –  Sam Oct 7 '12 at 1:50
    
@ Sam L. : I agree that we are not doing topology here, but there are some good theorems connecting topology and harmonic extensions,for example, harmonic extension of the unit circle homeomorphim is homeomorphism of the closed unit disk(Rado-Kneser-Choquet theorem). In fact, there are techniques in low-dimensional topology which use harmonic extension of circle homeomorphisms. But for calculational simplifications, sometimes it is easier to use $\mathbb{H}$-model, of course, only if the harmonic extension exists ! Thanks though. –  Mathmath Oct 7 '12 at 2:41

1 Answer 1

First of all, the harmonic extension of $f(t)=t$ cannot be $F(z)=z$ because the natural harmonic extension of a real function is real. You have to add $+iy$ "manually", it does not come from the Poisson kernel. (Unless you interpret the boundary data as $f(t)+i\delta_\infty$, which makes some sense but is likely to be more confusing than helpful.)

Anyway, we have a legitimate question: given a orientation-increasing homeomorphism $f\colon \mathbb R\to\mathbb R$, how (and when) can we realize $f$ as boundary values of a real harmonic function $F(x,y)$ that is increasing with respect to $x$ for any fixed $y>0$? (The increasing property is what will make $F(x,y)+iy$ a homeomorphism.) It helps to look at the derivative $f'$, which is sure to exist at least as a positive Radon measure $\mu$. We want to get a positive harmonic function $u$ out of $\mu$; then $F$ will be a harmonic function such that $F_x=u$ (you can get $F$ by completing $u$ to holomorphic $u+i\tilde u$ and taking the antiderivative of that).

Thus, the problem is reduced to getting a positive harmonic function out of a positive measure $\mu$ on the boundary. This is possible if and only if the Poisson integral of $\mu$ converges. That is, if the integral converges at one point, then it converges at all points and gives what we want. If it diverges, there is no such positive function, hence no harmonic homeomorphic extension.

Two examples will illustrate the above. With $f(t)=t$ we have $f'=1$. The Poisson integral of $f'$ converges and gives $u=1$. Complete to holomorphic function (still $1$) and integrate: you get $z$.

But if $f(t)=t^3$, then $f'(t)=3t^2$ and the Poisson integral of $f'$ diverges. There is no harmonic homeomorphic extension of this $f$. It extends to a harmonic (indeed holomorphic) map $f(z)=z^3$ which is not injective.

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