Can the harmonic function $u$ in $U$ be extended continuously to $\overline{U}$?

Question

Suppose that \begin{equation} u(z)=\arg \frac{1+z}{1-z} \end{equation} is harmonic in the unit disk $U$. Can this function be extended to a continuous real function on $\overline{U}$? I guess it can be, but I don't know how to prove it. Please help!

Answer 1

The answer is no because at $z= \pm 1$ there are $\pi$ jumps. Assume we take the principal branch of the argument with values in $(-\pi, \pi]$ (if we take the branch with values in $[0, 2\pi)$ same applies except that the values in cause will be $\pi/2, 3\pi/2$).

An easy way to see this is to note that if $z=e^{it}, -\pi \le t \le \pi$ we have $\frac{1+z}{1-z}=\frac{1+e^{it}}{1-e^{it}}=\frac{2 \cos t/2 e^{it/2}}{-2i \sin t/2 e^{it/2}}=i \cot t/2$ so the argument switches from $-$ to $+$ precisely at $t =0$ and since the function is purely imaginary, this means that $u(e^{it})=-\pi/2, -\pi \le t < 0$ and $u(e^{it})= \pi /2, 0< t \le \pi$ (where the extension at $-\pi= \pi$ (mod circle) is by continuity of course so there are left and right hand limits there too)

As an aside the above shows that $u(z)$ (interpreted using the principal branch with values in $(-\pi, \pi]$) is the solution to the Dirichlet problem on the circle with data that is $-\pi/2$ on the lower half-circle and $\pi/2$ on the upper one