Let $u(z)$ be a bounded harmonic function in $D$ such that the limit $$\lim_{r→1^-}u(re^{iφ})$$ is equal to 1 when for $0 < φ < π$ and to 0 for $π < φ < 2π$. Find $u(1/2)$.
To solve this problem, I want to show by approximation that $$u(\frac{1}{2})=\frac{1}{2\pi} \int_0^{2\pi} \lim_{r→1^-}u(re^{iφ}) \frac{\frac{3}{4}}{1- \cos (\theta) +\frac{1}{4}} \, d\theta.$$ And by the hypothesis, the equation above is equal to $\frac{1}{2\pi} \int_0^{\pi} \frac{\frac{3}{4}}{1- \cos (\theta) +\frac{1}{4}} \, d\theta$.
I have trouble in the rigorous proof. I have tried the dominated convergence theorem but I don't know what is that dominating function(since this is a problem in complex analysis, I think there should be a way to avoid using results in real analysis). Thanks in advance!
The Poisson integral formula giving the value $u\bigl(r,\phi)$ when $u$ is given on a boundary circle of radius $R$ reads as follows (see 3.1-4 on page 2 of this document): $$u(r e^{i\phi})={1\over2\pi}\int_0^{2\pi}u(Re^{i\psi}){R^2-r^2\over r^2-2Rr\cos(\phi-\psi)+R^2}\>d\psi\qquad(0\leq r\leq R)\ .$$ You have referred to this formula in the special case $R=1$. In particular you have $$ u\left({1\over2}\right)={1\over2\pi}\int_0^{2\pi}u(Re^{i\psi}){R^2-{1\over4}\over {1\over4}-R\cos\psi+R^2}\>d\psi\qquad\left(R>{1\over2}\right)\ ,\tag{1}$$ whereby $u$ is a smooth function of $\psi$ as long as $R<1$, and the kernel part is bounded as long as $R$ stays away from ${1\over2}$. Now let $R\to1-$ in $(1)$, and use the dominated convergence theorem. The end result will of course be what you have found out already.