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Let $f:B_1(0)\subset \mathbb{C} \to \mathbb{C}$ be a holomorphic function with its taylor representation $$f(w)=\sum_{n=1}^{\infty} A_n w^n, A_n\in \mathbb{C}, \lvert w\rvert<1$$ $$f'(w)=\sum_{n=1}^{\infty} A_n\cdot n\cdot w^{n-1}$$ I want to verify the following: $$\int_{\lvert w\lvert <r} \lvert f'\rvert^2 dudv=\int_0^r \int_0^{2\pi} \lvert f'(te^{i\theta})\rvert^2 t\, d\theta dt \overset{!!!}{=} \pi \sum_{n=1}^{\infty} n \lvert A_n\rvert^2 r^{2n}$$

I already know how to calculate $$\int_{0}^{2\pi}\lvert f(re^{i\theta})\rvert^2\,d\theta$$ which is equal to $$2\pi \sum_{n=1}^{\infty} \lvert A_n\rvert^2 r^{2n} $$ (cf. here: Prove that $\int_0^{2\pi} \lvert \sum_{n=0}^{\infty} A_n (re^{i\theta})^n \rvert^2=2\pi \sum_{n=0}^{\infty} \lvert A_n\rvert ^2 r^{2n}$) It immediatly follows that $$\int_0^{2\pi} \lvert f'(te^{i\theta})\rvert^2\cdot t\,d\theta =2\pi \sum_{n=1}^{\infty} n^2 \lvert A_n\rvert^2 t^{2n-2}\cdot t=2\pi \sum_{n=1}^{\infty} n^2 \lvert A_n\rvert^2 t^{2n-1},$$ since you can basically just put a $2\pi$ in front of the sum and square the series inside the sum. Now if you could swap the series with the last integral, you would get $$2\pi \sum_{n=1}^{\infty} \int_0^r n^2 \lvert A_n\rvert^2 t^{2n-1}\,dt=2\pi \sum_{n=1}^{\infty} \left[\frac{n^2}{2n} \lvert A_n\rvert^2 t^{2n} \right]_0^r=\pi \sum_{n=1}^{\infty} n \lvert A_n\rvert^2 r^{2n},$$ which gives my desired result. However, I was not able to show that the series $$\sum_{n=1}^{\infty} n^2 \lvert A_n\rvert^2 t^{2n-1}$$ converges uniformly on $[0,r]$ for all $r\in [0,1)$. I have tried to calculate the radius of convergence, then move out the $\sqrt[n]{n^2}$ factor, which in the limit will be equal to one, but I wasn't able to continue.

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  • $\begingroup$ All terms are nonnegative, so exchanging the order of summation and integration can for example be justified with the monotone convergent theorem, compare math.stackexchange.com/a/1765051/42969. $\endgroup$
    – Martin R
    Commented Oct 30, 2020 at 8:19

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The radius of convergence of $\sum n^{2}|A_n|^{2}w^{n}$ is $\frac 1 {\lim \sup n^{2/n}|A_n|^{2/n}}$. Since $\sum A_nw^{n}$ has radius of convergence at least $1$ we get $\lim \sup |A_n|^{n} \leq 1$. Hence $\frac 1 {\lim \sup n^{2/n}|A_n|^{2/n}} \geq 1$. This implies uniform convergence of $\sum n^{2}|A_n|^{2}w^{n}$ for $|w| \leq r$ if $r<1$.

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