Let $D$ be the open unit disk.

I already know the Poisson integral formula:

"Let the complex-valued function $g(e^{i \theta})$ be piecewise continuous and bounded for $\theta$ in $[0,2\pi]$. Then the function $f(z)$ defined by

$$f(z)=\frac{1}{2\pi} \int_{0}^{2\pi} \frac{(1-|z|^2)}{(|e^{it}-z|^2)}g(e^{it})dt$$ is the harmonic extension of $g$ in the unit disk"

I also already proved:

"Let $\Omega$ be a subset of the complex plane that is a bounded convex domain whose boundary is a Jordan curve $\Gamma$. Let $g$ map the unit circle continuously onto $\Gamma$ and suppose that $g(e^{it})$ runs once around $\Gamma$ monotonically as $e^{it}$ runs around the unit circle. Then the harmonic extension given in the poisson integral formula is univalent in $D$ and defines a harmonic mapping of $D$ onto $\Omega$."

But now I have to prove the following statement:

"Let $f(z)$ be the harmonic extension (i.e. harmonic extension from the poisson integral formula, see above) of a stepwise constant function $g$ on the unit circle, i.e., if $\{e^{it_k}\}$ is a partition of the unit circle with $t_0<\cdots<t_k$ then $g(e^{it})=v_k$ for $t_{k-1}<t<t_k$. Suppose the vertices $v1; v2; ... ; vn$, when traversed in order, define a convex polygon whose interior is denoted by $\Omega$. Then $f(z)$ is univalent in $D$ and defines a harmonic mapping from $D$ onto $\Omega$."

Anyone an idea how to prove this last statement? You can use the previous two statements (I already proved them so that's okay).

A big thank you if you can help me!

Kind regards.

  • $\begingroup$ Latex for "f(z)=1/(2pi) integral_{0}^{2pi} (1-|z|^2)/(|e^{it}-z|^2)g(e^{it})dt" is $f(z)=\frac{1}{2\pi} \int_{0}^{2\pi} \frac{(1-|z|^2)}{(|e^{it}-z|^2)}g(e^{it})dt$. I have replaced it into your question. $\endgroup$
    – Jean Marie
    Mar 18 at 10:15
  • $\begingroup$ Besides, I think that "harmonic extension of f " should be "harmonic extension of g". $\endgroup$
    – Jean Marie
    Mar 18 at 10:18
  • $\begingroup$ I think you need a first mapping from the unit disk onto the upper half plane using $Z=i \frac{z+1}{z-1}$, then use the so-called Schwarz-Christoffel transformation $\endgroup$
    – Jean Marie
    Mar 18 at 10:23
  • 1
    $\begingroup$ Yeah that indeed works. But I am now looking for another method than Schwarz christoffel . $\endgroup$
    – anoniem
    Mar 18 at 10:35
  • 1
    $\begingroup$ Thank you, Jean Marie, but I think that's not an explanation why the poisson integral delivers a mapping from te unit disk to the polygon? My purpose was to find, besides schwarz christoffel, another method to map onto a polygon. And that should be the poisson integral. So I need to prove the statement in my question. $\endgroup$
    – anoniem
    Mar 18 at 10:37


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