How to show the poisson integral formula is harmonic? Suppose $$u(re^{i\theta}) =  \frac{1}{2\pi}\int_0^{2\pi}   \frac{(1-r^2)u(e^{it})}{1-2r\cos(\theta-t)+r^2} \,dt$$, how can I show that u is harmonic in the unit disc and continous on the boundary of unit disc?
 A: First, your integral is incorrect. We are defining $u$ in terms of some boundary data. In reality, we need $$u(re^{i\theta}) = \frac{1}{2\pi}\int_0^{2\pi}\frac{(1-r^2)\phi(\theta')}{1-2r\cos(\theta-\theta')+r^2}d\theta'$$ where $\phi$ is a continuous function on the circle. We want to check that $u$ is harmonic in the unit disc and extends to a continuous function to the closure of the unit disc.
Observe that $$P_r(\theta) = \frac{1-r^2}{1-2r\cos(\theta)} = \mathfrak{R}\left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\right)$$ Therefore the integrand is the real part of a holomorphic function, and therefore $u(re^{i\theta})$ is harmonic. To check continuity on the boundary of a disc, we must estimate $|u(re^{i\theta})-\phi(z_0)|$ where $0\le \theta_0<2\pi$. Since $\frac{1}{2\pi}\int_0^{2\pi}P_r(\theta)d\theta = 1$, 
\begin{align}
2\pi|u(re^{i\theta})-\phi(\theta_0)| &\le \int_0^{2\pi}P_r(\theta-\theta')|\phi(\theta)-\phi(\theta_0)|d\theta' \\
&\le \int_{|\theta-\theta_0|<\delta} P_r(\theta-\theta')|\phi(\theta)-\phi(\theta_0)|d\theta'+\int_{|\theta-\theta_0|\ge \delta}P_r(\theta-\theta')|\phi(\theta')-\phi(\theta_0)|d\theta' \\
&\le \epsilon\int_{|\theta-\theta_0|<\delta} P_r(\theta-\theta')d\theta'+2\sup(\phi)\int_{|\theta-\theta_0|\ge \delta}P_r(\theta-\theta')|d\theta'\\
&\le 2\pi \epsilon+2\sup(\phi)\int_{|\theta'|>\delta}P_r(\theta')d\theta'
\end{align}
where we used continuity of $\phi$ on the circle. Moreover, if $re^{i\theta}\to e^{i\theta_0}$ then $r\to 1$. This implies that the integral in the last inequality above goes to $0$. Hence, $|u(re^{i\theta})-\phi(\theta_0)| <\epsilon$. Therefore, $u$ extends to a continuous function on the closure of the unit disc and takes on the value $\phi$ on the boundary.
