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Given:- $u(r,\theta)=\sum _{n \in Z} A_n r^{|n|} e^{i n \theta},\\$

Disk $D= \{(r,\theta) : 0 \leq r < 1, -\pi \leq \theta \leq \pi\}\\$ and $A_n$ is bounded.$\\$ Prove that :all $\frac{\partial^{j+k}u(r, \theta)}{\partial ^ jr \partial^k \theta }$ exist and are continuous

Can following be a somewhat correct approach at all?

$\frac{\partial^{j+k}u_n(r, \theta)}{\partial ^ jr \partial^k \theta }=A_n \frac{n!}{(n-j) !} r^{|n|} e^{in \theta} (in)^k, n \geq 0,$

and similarly calculable for $r < 0$.

Since $A_n$ is bounded, and $r < 1$, the series $\sum _{n \in Z} A_n r^{|n|} e^{i n \theta}$ converges. Therefore, series of it's partial derivatives also converges uniformly. Which means sum of partial derivatives and partial derivatives of sum are equal. Therefore, $\frac{\partial^{j+k}u(r, \theta)}{\partial ^ jr \partial^k \theta }=\sum_{n \in Z}\frac{\partial^{j+k}u_n(r, \theta)}{\partial ^ jr \partial^k \theta }$ each of which is calculable and continuous since each individual $u_n(r, \theta )$ is.

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    maybe if you see this first like a power serie? i mean, $r^{n}e^{in\theta}=z^{n}$, where $z=re^{i\theta}$. then you could conclude with theorems about holomorphic functions. – 2017-01-26
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    But is my approach totally Wrong? – 2017-01-27
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    Your approach is correct. I would think it is the preferred approach. You're using the Weierstrass M test, and facts about convergence. – 2017-01-27

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