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Suppose $f_{n}$ is a sequence of holomorphic functions in the ball $B\left(0,r\right)$ such that $\sum_{n=1}^{\infty}\left|f_{n}\left(z\right)\right|$ converges uniformly on $B\left(0,r\right)$. Show that $\sum_{n=1}^{\infty}\left|f_{n}^{'}\left(z\right)\right|$ converges uniformly on compact subsets of of $B\left(0,r\right)$.

My line of thought was to try using Cauchy's formula to represent the values of the derivatives through the values of the functions themselves and maybe use Weirstrass M-Test but I didn't quite manage to make it work. I'm also a bit confused why the requirement is convergence only on compact subsets and not on the entire ball.

Help would be appreciated!

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  • $\begingroup$ Yes, this proof will work. Every compact subset of the open disk is compained is a closed subdisk, so you can use Cauchy's formula using the boundary circle of the subdisk. This explains why convergence is only on compacts. $\endgroup$ – Moishe Kohan Jan 17 '14 at 23:13
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My line of thought was to try using Cauchy's formula to represent the values of the derivatives through the values of the functions themselves

That's the right idea.

I'm also a bit confused why the requirement is convergence only on compact subsets and not on the entire ball.

In general, the series/sequence of the derivatives will not converge uniformly on the entire disk, even when the series/sequence of functions does. Consider for a trivial example

$$\sum_{n=1}^\infty \frac{z^n}{n^2}.$$

The series converges uniformly on the closed unit disk, but the series of derivatives,

$$\sum_{n=1}^\infty \frac{z^{n-1}}{n}$$

does not, as the limit function is unbounded for $z\to 1$.

Cauchy's integral formula for the derivatives gives you an estimate

$$\begin{align} \lvert f'(z) \rvert &= \left\lvert \frac{1}{2\pi i}\int_{\lvert z\rvert = \rho} \frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta\right\rvert\\ &= \frac{1}{2\pi} \left\lvert \int_0^{2\pi} \frac{f(\rho e^{i\varphi})}{(\rho e^{i\varphi}-z)^2} \rho e^{i\varphi}\,d\varphi\right\rvert\\ &\leqslant \frac{\rho}{2\pi} \int_0^{2\pi} \frac{\lvert f(\rho e^{i\varphi})\rvert}{\lvert \rho e^{i\varphi}-z\rvert^2}\,d\varphi\\ &\leqslant \frac{\rho}{(\rho - \lvert z\rvert)^2}\cdot \max_{\lvert \zeta\rvert = \rho} \lvert f(\zeta)\rvert. \end{align}$$

That estimate grows to infinity as $\lvert z\rvert \to \rho$, but it gives a finite uniform bound on every disk of smaller radius than $\rho$. Thus on every disk $D_{r_1}(0)$ with $r_1 < r$, for every $r_1 < \rho < r$, we have a uniform bound

$$\lvert f'(z)\rvert \leqslant \frac{\rho}{(\rho - r_1)^2} \cdot \sup_{\lvert \zeta\rvert \leqslant \rho} \lvert f(\zeta)\rvert$$

valid for all holomorphic functions on $D_r(0)$. For fixed $r_1$ and $\rho$, the factor with which the supremum of the moduli of $f$ is multiplied is constant, and thus uniform convergence of a series/sequence $f_n$ of functions holomorphic on $D_r(0)$ implies the uniform convergence of the series/sequence of the $f_n'$ on the smaller disk $D_{r_1}(0)$. Since every compact subset of $D_r(0)$ is contained in such a smaller disk, that is the locally uniform or compact convergence of the series/sequence of the derivatives.

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    $\begingroup$ Hey Daniel, I'm a bit unclear on as to why the bound you presented combined with uniform convergence of the series of $f_{n}$ implies convergence of the series of $f^{'}_{n}$. $\endgroup$ – Serpahimz Jan 18 '14 at 6:56
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    $\begingroup$ Fix $r_1$. Choose $\rho\in(r_1,r)$ (if you have uniform convergence of $\sum f_n$ on the closed disk $\overline{D_r(0)}$, you can choose $\rho=r$). Given $\varepsilon>0$, there is an $N$ such that$$\left\lvert\sum_{n=k}^mf_n(z)\right\rvert\leqslant\frac{\varepsilon(\rho-r_1)^2}{\rho}$$for all $N\leqslant k\leqslant m$ on $\lvert z\rvert=\rho$. The estimate then yields$$\left\lvert\sum_{n=k}^mf_n'(z)\right\rvert\leqslant\varepsilon$$on $\overline{D_{r_1}(0)}$ for these $k,m$. $\endgroup$ – Daniel Fischer Jan 18 '14 at 9:55

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