# Sequence of holomorphic functions converges to a holomorphic function (more details in post)

Given an open set $\Omega \subset \mathbb C$, and a sequence of holomorphic functions in $\Omega$ $\{f_n\}$, that converge uniformly in compact subsets of $\Omega$ to a function $f$, prove that $f$ is holomorphic in $\Omega$.

This is my work so far, theres probably an easier way to solve this, but I'd like to know if it's possible with this approach:

If a sequence $\{f_n\}$ of continuous functions converges uniformly to $f$ then $f$ is continuous.

Let $\alpha_n = f - f_n$ $\forall n$ then $\alpha_n$ is also a continuos function $\forall n$.

$f$ is holomorphic if the limit $\lim_{h \rightarrow 0} \frac{f(z+h)-f(z)}{h}$ exists.

Since all $f_n$ are holomorphic, their derivaties exists in all of $\Omega$.

Obviously $\alpha(z) = lim_{n \rightarrow \infty} \alpha_n (z) = 0$ $\forall z \in \Omega$ therefore and $\alpha \equiv 0$ and $\alpha ' \equiv 0$ in $\Omega$.

Now putting it all together:

$\frac{\alpha_n(z+h)-\alpha_n (z)}{h} = \frac{(f(z+h)-f_n(z+h))-(f(z)-f_n(z))}{h} = \frac{f(z+h)-f(z)}{h} - \frac{f_n(z+h)-(f_n(z)}{h}$

$\left| \frac{\alpha_n(z+h)-\alpha_n (z)}{h} \right|= \left| \frac{f(z+h)-(f(z)}{h} - \frac{f_n(z+h)-f_n(z)}{h} \right|$

In substets where $\{f_n\}$ is uniformly convergent for every $\epsilon \geq 0$ there is an $n_0$ such that $\forall n>n_0 \phantom{2} \left|\alpha_n(z)\right|<h \frac{\epsilon}{2} \phantom{2} \forall z \in K \subset \Omega \phantom{2} \forall h \in \mathbb{C}$, therefore:

$\epsilon \geq \left| \frac{\alpha_n(z+h)-\alpha_n (z)}{h} \right|= \left| \frac{f(z+h)-f(z)}{h} - \frac{f_n(z+h)-f_n(z)}{h} \right|$

And $f$ is holomorphic in $\Omega$ since every $z\in \Omega$ has a compact neighborhood in $\Omega$.

• Look at the integrals of $f$ on triangles inside $\Omega$. The integrals of $f_n$ on those triangles are all zero. Therefore, so are the integrals of $f$. Apply Morera's theorem to conclude that $f$ is analytic. – user525761 Feb 7 '18 at 2:08
• In the first line of the proof, you left out the word "uniformly." – saulspatz Feb 7 '18 at 2:11
• Thanks, I know it's easier using complex analysis tools but I'd like to know if it's can be proved withouth too. Edited in the uniformly, thanks. – potatosauce Feb 7 '18 at 2:12
• Morera's theorem's proof is so small that you can get rid of the name. Your argument is nowhere close to a proof, if that is what you are asking. – user525761 Feb 7 '18 at 2:19
• This cannot be correct because it would apply verbatim for a sequence of real-differentiable functions from ($0,1) \to \Bbb R,$ but $any$ continuous $f:(0,1)\to \Bbb R$ is the uniform limit,on all of $(0,1),$ of a sequence of real-differentiable functions from $(0,1)$ to $\Bbb R$.... The situation in $\Bbb C$ is quite different. – DanielWainfleet Feb 7 '18 at 4:05

A way to prove it without Morera is this. Let $\Gamma$ be the positively oriented boundary of a disk $D$ centred at $z$ contained in $\Omega$. For $h$ such that $z + h \in D$, $$f_n'(z) = \frac{1}{2\pi i} \oint_\Gamma \frac{f_n(\zeta)}{(\zeta - z)^2} d\zeta$$ and \eqalign{\frac{f_n(z+h) - f_n(z)}{h} - f_n'(z) &= \frac{1}{2\pi i} \oint_\Gamma f_n(\zeta) \left( \frac{1}{(\zeta-z-h) h} - \frac{1}{(\zeta - z) h} - \frac{1}{(\zeta - z)^2} \right)\cr&= \frac{1}{2\pi i} \oint_\Gamma f_n(\zeta) \frac{h}{(\zeta - z - h)(\zeta - z)^2}}
Thus $f_n'(z)$ converges to some limit $w$, and $$\left| \frac{f_n(z+h) - f_n(z)}{h} - f_n'(z) \right| \le C |h|$$ for some constant $C$ not depending on $h$ or $n$. Taking the limit of this inequality as $n \to \infty$, we get $$\left| \frac{f(z+h) - f(z)}{h} - w \right| \le C |h|$$ which implies $f'(z)$ exists and is $w$.