Let $D$ be the unit disk and define $\phi:\partial D \rightarrow \mathbb{C}$ by $\phi(\zeta)=\bar{\zeta}$. Show that there is no holomorphic $f$ on $D$ such that $\lim_{z\rightarrow \zeta} f(z)=\phi(\zeta)$ for all $\zeta\in \partial D$.
So, this is proving that no holomorphic solution to Dirichlet problem. I was gonna prove by contradiction but was stuck. If we assume there is such $f$ then $f$ can be extended continuously to $\bar{D}$ and $\lim_{z\rightarrow \zeta} f(z)=\phi(\zeta)=1/\zeta$. Now I do not see a way to proceed. Can somebody please help me?
You can use that your extended function is uniformly continuous on $\overline{D}$. Then it follows that if $\gamma$ is the unit circle, and $t\in [0,1[$, then $f(tz)\to f(z)=1/z$ uniformly on $\gamma$ if $t\to 1$ (as $|z-tz|=1-t$). Hence $\int_{\gamma}f(tz)dz=0\to \int_{\gamma}\frac{dz}{z}=2i\pi$.