I'm continuing my studies about the space $\mathbb{T}$ and I reach the point in which are introduced the Harmonic functions. Well up to now I have a little trouble with understanding the Poisson's integral. I know that this particular integral is defined as:

DEF: If $\mu\in\ M(\mathbb{T})$ we define the Poisson integral of $\mu$ by the expression for $z= re^{i\theta}$: $P(\mu)(z)=\int_{-\pi}^{\pi}P(ze^{-it})d\mu (t)$.

Now to understand better how to use it, I am trying doing some exercise about this Poisson integral. In particular, I'm trying to calculate $P(\mu)$ for this two cases:

  1. For $\mu=\delta_{0}-\delta_{\pi/2}$ where $\delta_{x}$ is the Dirac measure for the points

  2. $d\mu=D_{N}(t)dm(t)$ where $D_{N}$ is the kernel of Dirichelet.

I've tried to apply directly my definition but something goes wrong!

Can someone help me?


1 Answer 1


Note that the easiest way to do this type of question is to use either the harmonic representation of $P$ (for term by term integration):

$P(z)=\sum_{n <0}r^{|n|}e^{in\theta}+\sum_{n \ge 0}r^{n}e^{in\theta}=1+\sum_{n \ge 1}(\bar z^n+z^n), z=re^{i\theta}$

or the closed-form of $P$ (for substitution)

$P(z)=\Re \frac{1+z}{1-z}=\frac{1-r^2}{1-2r\cos \theta +r^2}$

Then by definition the first integral is $P(z)-P(ze^{-i\frac{\pi}{2}})=\frac{1-r^2}{1-2r\cos \theta +r^2}-\frac{1-r^2}{1-2r\sin \theta +r^2}$

while the second is:

$\frac{1}{2\pi}\int_{-\pi}^{\pi}(\sum_{n <0}r^{|n|}e^{in\theta}e^{-int}+\sum_{n \ge 0}r^{n}e^{in\theta}e^{-int})(\sum_{-N}^{N}e^{ikt})dt=$

$=\sum_{-N \le n <0}r^{|n|}e^{in\theta}+\sum_{0 \le n \le N}r^{n}e^{in\theta}=1+\sum_{1}^{N}(\bar z^n + z^n)$,

which is the $N$ truncation of it

  • $\begingroup$ @ Conrad why the $sin(\theta)$ at the denominator? $\endgroup$
    – Adam
    May 9, 2020 at 15:54
  • $\begingroup$ $cos(\theta-\frac{\pi}{2})=\sin \theta$ $\endgroup$
    – Conrad
    May 13, 2020 at 14:11

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