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!

3 thoughts on “The Poisson Integral”

1. wardjm says:

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. Perpetua says:

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

3. Dylan says:

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.