How do I prove the existence of a limit of a harmonic function? Let $f:[0,2\pi]\rightarrow \mathbb{R}$ be a continuous function such that $f(0)=f(2\pi)$ and $R>0$.
Define $u(re^{i\theta})=\frac{1}{2\pi} \int_0^{2\pi} \frac{R^2 - r^2}{R^2 - 2rR \cos (\theta - \phi) + r^2} f(\phi) d\phi (r<R)$.
How do I prove that for each fixed $\theta\in [0,2\pi]$, $\lim_{r\to R} u(re^{i\theta})=f(\theta)$?
So far, I have proven that $u$ is harmonic on $B(0,R)$. (I don't know whether this is helpful for my question)
If $\theta$ is fixed, $u(re^{i\theta}) - f(\theta)= \frac{1}{2\pi} \int_0^{2\pi} \frac{R^2 - r^2}{R^2 - 2rR \cos (\theta - \phi) + r^2} (f(\phi)- f(\theta)) d\phi$. So it is sufficient prove that the right-hand side integration converges to $0$.
How do I prove this?
Thank you in advance.
 A: Here is a sketch of how I would approach the problem. It isn't complete, but hopefully it has the main ideas.
First suppose the argument of $u$ is real, i.e. $\theta=0$:
$$u(r)=\frac{1}{2\pi}\int_0^{2\pi}\frac{R^2-r^2}{R^2-2Rr\cos(\phi)+r^2}f(\phi)d\phi.$$
Using the triangle and reverse-triangle inequalities we can show that $|u(r)|$ is bounded, and so we can bring the limit $r\nearrow R$ inside the integral. Doing this we see that for $\theta\ne0,2\pi$ the integrand vanishes, and so for some small $\epsilon>0$
$$\lim_{r\nearrow R}u(r)\approx \lim_{r\nearrow R}\left(\frac{1}{2\pi}\int_0^\epsilon\frac{R^2-r^2}{R^2-2Rr\cos(\phi)+r^2}f(\phi)d\phi+\frac{1}{2\pi}\int_{2\pi-\epsilon}^{2\pi}\frac{R^2-r^2}{R^2-2Rr\cos(\phi)+r^2}f(\phi)d\phi\right).$$
Let's consider the integral from $0$ to $\epsilon$. For $\epsilon$ small enough, as $f$ is continuous $f(\phi)\approx f(0)$ up to some error of order $\epsilon$ (this can be made more rigorous with the $\epsilon-\delta$ definition of contuinuity). We want to play a similar trick with $\cos(\phi)$, but as it is on the denominator we have to be more careful. Expanding it as a Taylor series about $0$:
$$\approx\frac{1}{2\pi}\int_0^\epsilon\frac{R^2-r^2}{R^2+r^2 -2Rr\left(1-\frac{\phi^2}{2}+O(\phi^4)\right)}f(0)d\phi,$$
$$=\frac{1}{2\pi}\int_0^\epsilon\frac{(R+r)(R-r)}{(R-r)^2-Rr\phi^2+O(\phi^4)}f(0)d\phi.$$
Since $0\le\phi\le\epsilon$ and $\epsilon$ is small, we can discard the fourth order $o(\phi^4)$ terms as they will be negligible compared to the second order $Rr\phi^2$. We therefore get after re-arranging the integral
$$\frac{f(0)}{2\pi}\frac{(R+r)(R-r)}{Rr}\int_0^\epsilon \frac{d\phi}{\frac{(R-r)^2}{Rr}+\phi^2}.$$
Through trigonometric substitution we have the result $\int_0^b\frac{du}{a^2+u^2}=\frac{1}{a}\tan^{-1}(\frac{b}{a})$, and so this becomes
$$=\frac{f(0)}{2\pi}\frac{(R+r)(R-r)}{Rr}\frac{\sqrt{Rr}}{R-r}\tan^{-1}\left(\frac{\epsilon\sqrt{Rr}}{R-r}\right).$$
Consider now the argument of $\tan^{-1}$. Both $\epsilon$ and $R-r$ are small. However, $\epsilon$ is some small positive number, while $R-r$ approaches $0$, and so the argument of the arctan approaches $\infty$ as $R\rightarrow r$. Our calculation will then have some error of order $\epsilon$, from back when we set $f(\phi)\approx f(0)$. This is not a problem however as $\epsilon$ is arbitrarily small. Thus if we let $r\rightarrow R$, noting $\tan^{-1}(\infty)=\frac{\pi}{2}$, this becomes
$$\left(\frac{f(0)}{2\pi}\right)\left(\frac{2R^2}{R^2}\right)\left(\frac{\pi}{2}\right)=\frac{f(0)}{2}$$
The integral from $2\pi-\epsilon$ to $2\pi$ will give the same result, as $\cos(\phi)$ behaves similarly in the $0^+$ and $2\pi^-$ regions, and $f(0)=f(2\pi)$.
From this I believe we can generalise to the case where $\theta$ is arbitrary with the substitution $\phi'= \phi-\theta$, $d\phi'=d\phi$, $f(\phi)=f(\phi'-\theta)$.
