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In Function Theory of a Complex Variable I'm having trouble, proving the initial result in $(1.6)$ may I have a hint on how this could be achieved ?

$(0)$

If $f_{j}:U \rightarrow \mathbb{C}$ are holomorphic and $|f_{j}| \leq 2^{-j}$, then prove that:

$$\sum_{j=0}^{\infty}f_{j}(z)$$

converges to a holomorphic function on $U$.

$\text{Lemma (1)}$

The Open Set $U$ is within the subset of $\mathbb{C}$, since there exists an $r > 0$ such that $D(P,r) \subset A$,our holomorphic functions on $U$, as mentioned in $(0)$ are denoted by $\Psi(z)^{n}$ and $f_{j}$ can be defined by the following mappings $\Psi(z)^{n}: U \rightarrow \mathbb{C}$, and $f_{j}: U \rightarrow \mathbb{C}$.

$\text{Lemma (1.1)}$

To prove our original proposition in $(0)$, it suffices to show the conjecture as detailed in $(1.2)$

$(1.2)$

$$\sum_{j}^{\infty}f_{j}(z) \rightarrow \Psi(z)^{n}, \, \, \, |f_{j}| \leq 2^{-j}.$$

$\text{Lemma (1.3)}$

If $f_{j}$, $f$, $U$ are as in the theorem, then for any integer $ k \in \left\{ 0,1,2,3,5\right\}$ we have in $(1.4)$

$(1.4)$

$$(\partial_{z}f)^{k}f_{j}(z) \rightarrow (\partial_{z}f)^{k}f(z)$$

uniformly on compact sets

$\text{Lemma (1.4)}$

Applying $\text{Lemma (1.3)}$ to our conjecture in $\text{Lemma (1.2)}$, one achieves the following developments in $(1.5)$

$(1.5)$

$$(\partial_{z}f)^{k}\sum_{j}^{}f_{j}(z) \rightarrow (\partial_{z}f)^{k} \Psi(z)^{n}$$

The recent developments in $(1.5)$ become the following in $(1.6)$

$(1.6)$

$$(D_{z}f)^{k}\sum_{j}^{}f_{j}(z) \rightarrow (D_{z}f)^{k} \Psi(z)^{n}$$

$\text{Remark}$

The developments in $(1.5)-(1.6)$ were due to the fact that $\partial_{z}f = D_{z}f$.

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    $\begingroup$ It seems the best way to go about this is to use the M-Test $\endgroup$
    – Zophikel
    Commented Nov 11, 2017 at 21:58
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    $\begingroup$ M-test and Morera is a good way to go. $\endgroup$
    – zhw.
    Commented Nov 12, 2017 at 0:20
  • $\begingroup$ thanks for the hint zhw, sorry for posting an unsalvageable proof $\endgroup$
    – Zophikel
    Commented Nov 12, 2017 at 14:36

1 Answer 1

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The above attempt although accurate in it's developments doesn't exactly pan out, the best way to proceed would be to use the M-Test and Morrea's Theorem, the correct proof can be seen in the proceeding lemma's below

$\text{Lemma (1)}$

The Open Set $U$ is within the subset of $\mathbb{C}$, since there exists an $r > 0$ such that $D(P,r) \subset A$,our holomorphic functions on $U$, as mentioned in $(0)$ are denoted by $\Psi(z)^{n}$ and $f_{j}$ can be defined by the following mappings $\Psi(z): U \rightarrow \mathbb{C}$, and $f_{j}: U \rightarrow \mathbb{C}$.

$\text{Lemma (2)}$

Let $U \subset \mathbb{C}$ and $f_{n}$ be a sequence of functions on $E$. If for each $n \in \mathbb{N}$, there is $M_{n} \geq 0$ such that $|f_{n}(z)| \leq M_{n}$ for all $z \in E$ and $\sum_{n=1}^{\infty}f_{n}$ converges uniformly.

$\text{Lemma (3)}$

Morera's theorem states that a continuous, complex-valued function $f$ defined on an open set $U$ in the complex plane that satisfies

$$\oint_{\Gamma}f(z)dz = 0$$

$\text{Lemma (4)}$

Applying the M-Test to $\sum_{j=0}f_{j}(z)$ one can set $f_{n} = \frac{1}{2^{j}} \leq f_{j}(z)= M$ such that $\sum M_{n} = \sum \frac{1}{2^{j}}$ note that $\sum_{n}\frac{1}{2^{j}}$ converges by the p-test

$\text{Lemma (5)}$

Applying Morera's Theorem one let's $\partial\Delta$ such that it's a compact subset of $D$, so you know that $f_j\to \Psi(z)$ uniformly on $\partial\Delta$.So you get, for all $j$,

$$\left|\int_{\partial\Delta} \Psi(z)dz\right|=\left|\int_{\Delta} (\Psi(z)-f_n(z))dz\right|\leq\mathrm{length}({\partial\Delta})\sup_{z\in\partial\Delta}|\Psi(z)-f_n(z)|$$

By letting $n\rightarrow\infty$, you find that $\displaystyle\int_{\partial\Delta} \Psi(z)dz=0$.

By Morera's theorem, $f$ is holomorphic.

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  • $\begingroup$ looking back it at it seemed overkill to establish the estimate I could have just interchanged the limit and the integral due to uniform convergence but oh well $\endgroup$
    – Zophikel
    Commented Nov 12, 2017 at 16:14

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