Poisson Integral for half plane approaches argument

Question

I'm looking at an exercise in Ahlfors' Complex Analysis. It states the following:

Assume $U(\zeta)$ is piecewise continuous and bounded for all real $\zeta$, and assume $U$ has a jump at $0$ (i.e. $U(0^-)=1$ and $U(0^+)=0$). Define $P_U(z)=\frac{1}{\pi}\int^\infty_{-\infty} \frac{y}{(x-\zeta)^2+y^2}U(\zeta)d\zeta$ (the Poisson Integral on the half plane). Prove that $P_U(z)-\frac{1}{\pi}arg(z)\rightarrow0$ as $z\rightarrow 0$.

From what I can tell, the problem seems to boil down to showing that the integral portion of $P_U(z)$ approaches $arg(z)$ as $z\rightarrow 0$. I've shown in a previous problem that $P_U(z)$ is harmonic and its boundary values are equal to $U(\zeta)$, but don't quite know how to use it here. Any help is appreaciated!

Answer 1

Define $V: \Bbb R \to \Bbb R$ as $V(\zeta) = 1$ for $\zeta < 0$, $V(0)=U(0)$, and $V(\zeta) = 0$ for $\zeta > 0$. Show that

• $U-V$ is continuous at $\zeta=0$.
• $P_V(z) = \frac 1\pi \arg(z)$.

It follows that $$ P_U(z)-\frac{1}{\pi}arg(z) = P_U(z)-P_V(z) = P_{U-V}(z) \to (U-V)(0) = 0 $$ for $z \to 0$.