prove that the sequence $A_n$ satisfies a particular formula 
Let $D = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2\leq 1\}.$ Let for each $n\in\mathbb{Z}, f_n(r,\theta) = r^{|n|} e^{in\theta}$. Consider $f(r,\theta) = \sum_{n\in\mathbb{Z}} A_n f_n(r,\theta)$, where $A_n \in \mathbb{C}$. Assume $\sum_{n\in\mathbb{Z}}|A_n| < \infty$ and that $\lim\limits_{r\to 1^-} f(r,\theta) =: g(\theta)$ uniformly in $\theta$. Prove that $f$ satisfies $f_{rr} + \frac{1}{r} f_r + \frac{1}{r^2} f_{\theta\theta} = 0$ and that $A_n = \frac{1}{2\pi} \int_{-\pi}^\pi g(\theta) e^{-in\theta}d\theta.$

One can show the first result by differentiating f. For instance, $f_r(\theta) = \sum_{n\in \mathbb{Z}} nr^{n-1}e^{in\theta}, f_{rr})\theta) = \sum_{n\in\mathbb{Z}}n(n-1)r^{n-2}e^{in\theta},$ and $f_{\theta\theta} = \sum_{n\in\mathbb{Z}} -n^2 r^ne^{in\theta}$ . From this, it's clear that $f_{rr} + \frac{1}r f_r + \frac{1}{r^2} f_{\theta\theta} = 0$.

Is there a more formal reason why I can just take the derivative in the infinite sum? If this doesn't always hold, then why might it not hold?

For the second result, I'm not sure how to show it. I'm not sure what $\lim\limits_{r\to 1} u(r,\theta) = g(\theta)$ uniformly in $\theta$ means. Perhaps there's a typo and it means $\lim\limits_{r\to 1} f_n(r,\theta) = g(\theta)$ uniformly in $\theta$? Or am I misunderstanding something?
I know that $\int_{-\pi}^\pi e^{in\theta} e^{-im\theta} = \begin{cases}0, &\text{ if $n\neq m$}\\
2\pi, &\text{if $n=m$}\end{cases}$. So the claim would follow easily if I could just show that $\begin{align}\int_{-\pi}^\pi g(\theta)e^{-in\theta} d\theta &=\int_{-\pi}^\pi\lim\limits_{r\to 1} f(r,\theta) e^{-in\theta} d\theta \\
&= \int_{-\pi}^\pi \lim\limits_{r\to 1} \sum_{m\in\mathbb{Z}}A_m f_m(r,\theta) e^{-in\theta}d\theta\\
& = \int_{-\pi}^\pi \sum_{m\in\mathbb{Z}}\lim\limits_{r\to 1} A_m f_m(r,\theta)e^{-in\theta}d\theta\\
& = \int_{-\pi}^\pi \sum_{m\in\mathbb{Z}} A_m e^{i(m-n)\theta}d\theta\\
&= \sum_{m\in\mathbb{Z}}\int_{-\pi}^\pi A_m e^{i(m-n)\theta}d\theta = 2\pi A_n.\end{align}$

My issue is a lot of these equalities seem unjustified, and it shouldn't be necessary to use theorems like the dominated convergence theorem to justify them. For instance, the ones I'd like to justify more formally are the third and fifth equalities.


For the third equality, I tried proving this using the Weierstrass M-test, but I can't seem to uniformly bound the $f_n(r,\theta)$'s, regardless of whether $r > 1$ or $r < 1$.

For the fifth equality, we prove that if $f_n$ and $f$ are continuous functions and $f_n\to f$ uniformly, then $g = \lim\limits_{n\to\infty} g_n$, where $g_n(x) = \int_a^x f_n(t)dt, g(x) = \int_a^x f(t)dt.$ Let $\epsilon > 0$. Since $f_n\to f$ uniformly, we may find $N$ so that $n\ge N\Rightarrow \lVert f_n - f\rVert_\infty < \epsilon/(b-a+1)$. Then $|g_n(x) - g(x)| = |\int_a^x f_n(t) - f(t)dt|\leq \int_a^x |f_n(t)-f(t)|dt \leq \int_a^x \lVert f_n - f\rVert_\infty dt \leq \int_a^b \lVert f_n - f\rVert_\infty dt < \epsilon,$ so $\lVert g_n - g\rVert_\infty \leq \epsilon$, proving that $g_n\to g$ uniformly.
And by the Weierstrass M-test, since for all $m\in\mathbb{Z}, |A_m e^{i(m-n)\theta}| \leq |A_m|$ and $\sum_{m\in\mathbb{Z}} |A_m| < \infty$, it follows that $\sum_{|m|\leq N} A_m e^{i(m-n)\theta}\to \sum_{m\in \mathbb{Z}} A_me^{i(m-n)\theta}$ uniformly as $N\to \infty$.
 A: With $z= r^{i \theta}$ we have
$$
 f(z) := f_n(r,\theta) = r^{|n|}e^{in\theta} = 
\begin{cases}
z^n & \text{if } n \ge 0 \, ,\\
\bar z^{-n} & \text{if } n < 0 \, .
\end{cases}
$$
and
$$
f(z) := f(r,\theta) = \sum_{n\in\mathbb{Z}} A_n f_n(r,\theta)
= \sum_{n\ge 0} A_n z^n + \sum_{n > 0} A_{-n} \bar z^n \, .
$$
If $\sum_{n\in\mathbb{Z}}|A_n| < \infty$ then both series
$$
 G(z) = \sum_{n\ge 0} A_n z^n \,, \, H(z) = \sum_{n > 0} A_{-n} z^n 
$$
converge absolutely and uniformly for $|z| \le 1$. It follows that
$$
 \lim_{r \to 1-} f(re^{i \theta}) = \sum_{n\ge 0} A_n e^{i n \theta} + \sum_{n > 0} A_{-n} e^{-i n \theta} =: g(\theta)
$$
uniformly in $\theta$,  and that both $G$ and $H$ are holomorphic in $|z| < 1$.
Application of the Laplace operator gives therefore zero:
$$
G_{rr} + \frac{1}{r} G_r + \frac{1}{r^2} G_{\theta\theta} = \Delta G 
= 4 \frac{\partial^2 G}{\partial z \partial \bar{z}} = 0 \, ,
$$
and the same applies to $H$ and to
$$
 f(z) = G(z) + H(\bar z) \, .
$$
For $0 < r < 1$ is
$$
 \frac{1}{2\pi}\int_{-\pi}^{\pi}f(re^{i\theta}) e^{-in \theta} \, d\theta
= \frac{1}{2\pi}\int_{-\pi}^{\pi} \sum_{k \in \Bbb Z} A_k r^{|k|} e^{ik\theta} e^{-in \theta} \, d\theta \\
= \sum_{k \in \Bbb Z}^\infty \frac{1}{2\pi}\int_{-\pi}^{\pi} A_k r^{|k|} e^{i(k-n)\theta} \, d\theta = r^{|n|} A_n \, ,
$$
exchanging the order of summation and integration is justified by the uniform convergence of the series. Finally,
$$
 A_n = \frac{r^{-|n|}}{2\pi}\int_{-\pi}^{\pi}f(re^{i\theta}) e^{-in \theta} \, d\theta \to \frac{1}{2\pi}\int_{-\pi}^{\pi}g(\theta) e^{-in \theta} \, d\theta
$$
for $r \to 1-$, because of the uniform convergence  of $f(re^{i\theta}) \to g(\theta)$.
