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 !