Dirichlet Problem on the Disk - Question about this proof of solution using Poisson transform I'm looking at a proof for the Dirichlet problem on the disk. The problem is as follows.

Let $D=U(w,\rho)$ and$\phi : \partial B(w,\rho) \to \mathbb{R}$
  continuous. Then $$g = \begin{cases} P_D \phi & \text{in }U(w,\rho),\\
 \phi & \text{in }\partial B(w,\rho)  \end{cases}$$ is the unique
  solution to the Dirichlet problem, i.e. $g$ is harmonic in $U(w,\rho)$
  and $g$ is continuous in $B(w,\rho)$ and $g = \phi$ on $\partial
> B(w,\rho)$. $P_D$ is the Poisson kernel given by the rule $\phi
 \mapsto P_D \phi$, $$P_D \phi(z) = \frac{1}{2\pi}\int_0^{2\pi} P
 \left( \frac{z-w}{\rho}, \mathrm{e}^{i\theta}\right) \phi(w +
 \rho\mathrm{e}^{i\theta}) \, \mathrm{d}\theta.$$

So to prove that $g$ actually solves the problem we need to show that $P_D \phi$ is harmonic in $U(w,\rho)$ and that $P_D \phi \to \phi$ as $z$ approaches the boundary from the interior. WLOG, we assume that $D=U(0,1)$. When proving harmonicity, one ends up with $$P_D \phi = \frac{1}{2\pi} \mathrm{Re}\left( \int_0^{2\pi} \frac{z + \mathrm{e}^{i\theta}}{\mathrm{e}^{i\theta}  -z}  \phi(\mathrm{e}^{i\theta}) \, \mathrm{d}\theta \right).$$ Knowing that a harmonic function satisfies $u=\mathrm{Re}f$ where $f$ is analytic, we need to show that the parenthesis is analytic. So I would do this by using Fubini's theorem and Morera's theorem. By Morera's theorem, the parenthesis is analytic if $$\int_T \int_0^{2\pi} \frac{z + \mathrm{e}^{i\theta}}{\mathrm{e}^{i\theta}  -z}  \phi(\mathrm{e}^{i\theta}) \, \mathrm{d}\theta  \mathrm{d} z=0.$$ By Fubini's theorem, we can switch the integrals if $$\int_T \int_0^{2\pi}  \left| \frac{z + \mathrm{e}^{i\theta}}{\mathrm{e}^{i\theta}  -z}  \phi(\mathrm{e}^{i\theta}) \right| \, \mathrm{d}\theta  \mathrm{d} z < \infty.$$ Hence we want to show that what is in $| \, |$ is bounded. $\phi$ is bounded because it is continuous. However, with the fraction, I get $$ \left|  \frac{z + \mathrm{e}^{i\theta}}{\mathrm{e}^{i\theta} -z} \right| = \frac{| z + \mathrm{e}^{i\theta} |}{|\mathrm{e}^{i\theta} - z|} \leq \frac{2}{|\mathrm{e}^{i\theta} - z|}.$$ How can I bound this?
Now that the boundedness is taken care of, how do I know that $$\int_T \frac{z + \mathrm{e}^{i\theta}}{\mathrm{e}^{i\theta}  -z}  \phi(\mathrm{e}^{i\theta}) \, \mathrm{d}z =0?$$ How do I see that this is analytic?
And finally, this is a question on the Poisson transform. Is it true that I can write $$\phi = \frac{1}{2\pi} \int_0^{2\pi} P(z,\mathrm{e}^{it}) \phi(\zeta_0) \, \mathrm{d}t$$ for some complex $\zeta_0$ by the nature of the Poisson transform?
Wow this got huge. If you don't need all the background information on the proof of the Dirichlet problem on a disk, feel free to edit it out. Or tell me to do so.
 A: Analyticity  and harmonicity are local properties. When proving them, you focus on a small neighborhood of a  generic point $z_0$ in the disk. It helps to make this neighborhood small enough so that it stays away from the boundary. For example, let it be the disk of radius $(1-|z_0|)/2$ with center $z$, denoted $N$ below. By the reverse triangle inequality, 
$$|z-e^{i\theta}|\ge \frac{1}{2}(1-|z_0|)$$
for all $z\in N$ and all real $\theta$. This takes care of the boundedness of $2/|z-e^{i\theta}|$. 

Concerning $$\int_T \frac{z + \mathrm{e}^{i\theta}}{\mathrm{e}^{i\theta}  -z}  \phi(\mathrm{e}^{i\theta}) \, \mathrm{d}z $$
note that $\theta$ is a constant here, and the integrand is a holomorphic function of $z$ in $D$. Indeed, the denominator does not vanish, because $z\in D$  and $e^{i\theta}\in\partial D$.

I am not sure I understand your last question.  You wrote 
$$\phi = \frac{1}{2\pi} \int_0^{2\pi} P(z,\mathrm{e}^{it}) \phi(\zeta_0) \, \mathrm{d}t \tag{1}$$
where on the left the argument is suspiciously absent.  For any fixed $\zeta_0$ the expression on the right is 
$$\frac{\phi(\zeta_0)}{2\pi} \int_0^{2\pi} P(z,\mathrm{e}^{it})  \, \mathrm{d}t = \phi(\zeta_0)$$
So, (1) is true if you put $\phi(\zeta_0)$ on the left.
