Representing a function as a Poisson Integral.

Question

This is a question I came across in Ahlfors' book Complex Analysis. It is found on page 171 of the 3rd Edition, Exercise 2.

"Prove that a function $T(z)$ which is harmonic and bounded in the upper half plane [and] continuous on the real axis can be represented as a Poisson Integral."

How would I go about performing this proof?

From a previous exercise I can prove that if $U(\xi)$ is piecewise continuous and bounded for all $\xi \in \mathbb{R}$, then $$P_U(x,y) = \frac{1}{\pi}\int_{\mathbb{R}} \frac{y}{(x-\xi)^2 + y^2}U(\xi)~d\xi$$ is a Poisson integral (of $U$) in the upper half plane.

I also know that if $U$ is continuous at $\xi$ then $$\lim_{(x,y) \to (\xi,0)} P_U(x,y) = U(\xi).$$ This is a sort of extension of what Ahlfors calls the "Schwarz Theorem."

Ahlfors suggests that we use the fact that a harmonic function attains its maximum value on the boundary of the domain, then look at $G(z) = T-P_T-\varepsilon~\Im m(\sqrt{iz})$ for $\varepsilon >0$ and then let $\varepsilon \rightarrow 0$.

My main hang up is showing $G(z) \rightarrow -\infty$ as $y \rightarrow\infty$ and also that $G(z) \rightarrow 0$ as $|x| \rightarrow\infty$. If I can get this, I have it I believe.

Answer 1

Notice that $\sqrt{iz}$ maps the upper half plane into a wedge in the upper half plane bounded by $y = \pm x$. $G(z)$ is a harmonic function because all three components are.

For any half disc bounded by the real axis and $|z| = R$ with $\operatorname{Im}z > 0$, we have $\epsilon \operatorname{Im}(\sqrt{iz}) \geq 0$ on its boundary. On the half circle in particular, we have $\epsilon \operatorname{Im}(\sqrt{iz}) \geq \epsilon \sqrt{R/2}$.

This means if we take $R$ big enough, we can guarantee that $G(z) \leq 0$ on the boundary, because

• On the half circle, both $U(z)$ and $P_U(z)$ are bounded, while $\epsilon \operatorname{Im}(\sqrt{iz}) \rightarrow \infty$ as $R\rightarrow\infty$.
• On the segment $|x| \leq R$, $U(z) - P_U(z) = 0$ and $\epsilon \operatorname{Im}(\sqrt{iz}) \geq 0$.

Now we can apply the maximum principle for harmonic functions, we have $G(z) \leq 0$ in the half disc, since $R$ can be arbitrarily large, $G(z) \leq 0$ in the closed upper half plane. Therefore $U(z) - P_U(z) \leq \epsilon \operatorname{Im}(\sqrt{iz})$. Since $\epsilon$ can be made arbitrarily small, we have $U(z) \leq P_U(z)$.

Swap $U(z)$ and $P_U(z)$ in the definition of $G(z)$, we have $U(z) \geq P_U(z)$.

Conclusion: $U(z) = P_U(z)$.