# The Poisson Integral

So I’m sure I’m losing readers/upsetting some readers with my recent mumbo-jumbo, and I shall continue in that vein. I will answer a long standing question: What is the Poisson Integral? Before I begin, I will not be doing this in all its glory.

So let’s clear up what I consider to be a misnomer. This should not be called the Poisson integral, but rather the Poisson integral representation. Let U be the unit disc in the complex plane. Let T be the unit circle. Then if A is a vector space of continuous complex functions on the closed unit disc $\overline{U}$, A contains all polynomials and $\sup_{z\in U}|f(z)|=\sup_{z\in T}|f(z)|$ for every $f\in A$, then the Poisson integral representation

$\displaystyle f(z)=\frac{1}{2\pi}\int_{-\pi}^\pi \frac{1-r^2}{1-2r\cos(\theta -t) +r^2}f(e^{it})dt$ where $z=re^{it}$ is valid for every $f\in A$ and every $z\in U$.

OK. Let’s break this down. It isn’t so bad. (Why is this in my real analysis textbook?!?) So first note that A contains all polynomials. It may not contain anything extra than that, and that is OK. If we throw extra stuff in, then we need to make sure it is still a vector space. That next condition can be stated in a different way. It is sort of a “maximum modulus” type condition. We want $\|f\|_U=\|f\|_T$, where we use the typical norm for an operator on a Banach space ($\|f\|_K=\sup\{\|f(z)\| : z\in K, \|z\|\leq 1\}$).

Now to get the integral representation, fix a z in U. Now be the Riesz Representation Theorem we have that there is a Borel measure, $\mu_z$, such that $f(z)=\int_T fd\mu_z$. Since z is complex we can write it as $z=re^{i\theta}$ for some |r|<1. Let $u_n(w)=w^n$, then we have the that the collection $span\{u_n\}_n$ is dense in A by Stone-Weierstrass, so let’s see what we can get out of the integral representation for this collection.

$u_n(z)=z^n=r^ne^{in\theta}=\int_T u_nd\mu_z$ for n=0,1,2,… and since $u_{-n}=\overline{u_n}$ on T, we have $r^{|n|}e^{in\theta}=\int_T u_nd\mu_z$ now for $n=0,\pm 1, \pm 2, \ldots$.

Now define $\displaystyle P_r(\theta -t)=\sum_{-\infty}^\infty r^{|n|}e^{in(\theta-t)}$ where t is real. But by the Dominated Convergence Theorem we have that $\displaystyle r^{|n|}e^{in\theta}=\frac{1}{2\pi}\int_{-\pi}^\pi P_r(\theta-t)e^{int}dt$, but here we have constructed what we want $\int_T fd\mu_z=\frac{1}{2\pi}\int_{-\pi}^\pi f(e^{it})P_r(\theta - t)dt$.

But you say this doesn’t look the same as you said. Well, let’s deconstruct $P_r(\theta-t)$ which is called the Poisson kernel. Since the series is symmetrical about zero, we can break it into $\displaystyle \sum_{-\infty}^\infty r^{|n|}e^{in(\theta-t)}=1+2\sum_{n=1}^\infty (ze^{-it})^n=\frac{e^{it}+z}{e^{it}-z}=\frac{1-r^2+2ir\sin(\theta - t)}{|1-ze^{-it}|^2}$. But we only care about the real part (why?), so $\displaystyle P_r(\theta-t)=\frac{1-r^2}{1-2r\cos(\theta - t)+r^2}$, i.e. the Poisson integral in the form first stated!

Also, if anyone actually read all this and understood it, please leave a comment that you did.

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### Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

### 3 thoughts on “The Poisson Integral”

1. I can say that read it and understood it. There’s no way I would have been able to write it though. Some things I know but haven’t had a lot of practice with like measure and Riesz.

2. I read but did not understand. As the kids say, “my bad.”

3. BOOM. Read it AND understood it. Though this may have to do with the fact that we just recently did Riesz and are about to do Stone-Weierstrass in Reals… I quite like this development especially in comparison to the complex-analytic way I saw this proven last year.