Let $X$ be a compact subset of $\mathbb C$. Let $(f_n)$ be a sequence of functions on $X$ with continuous derivative such that $f_n$ converges uniformly to $f$ and $f_n '$ converges uniformly to $g$.
Suppose that, for each $z_0 \in X,\, \exists \, C = C_{z_0} > 0$ such that
$$ |f_n(z) - f_n(z_0)| \leq C |z - z_0| (|f_n|_X + |f_n'|_X), \quad \forall z \in X, \, \forall n \in \mathbb N \quad (I)$$
I'm trying to proof that $f'$ is continuous.
My attempt: Given $\epsilon > 0$. $\exists n_0$ s.t. $\forall z \in X , \, |g(z) - f_n'(z)| < \epsilon / 3$, since $f_n'$ converges uniformly to $g$. And, by hypothesis, $f_n'$ is continuous, then $|f_n' (z) - f_n ' (z_0)| < \epsilon / 3$, for $|z-z_0|< \delta$ with $\delta > 0$.
Hence,
$$|g(z) - g(z_0)| \leq |g(z) - f_{n_0} ' (z)| + |f_{n_0}' (z) - f_{n_0}' (z_0)| + |f_{n_0} ' (z_0) - g(z_0)| < \epsilon$$
for $|z-z_0| < \delta$. So, $g$ is continuous. Then, it's sufficient showing that $f' = g.$ It's true if the domain $X$ is a region, however in this case we don't have this hypothesis. I would like a hint to see that using the inequality in $(I)$.
Since
$$g(z_0) = \lim_{n\to \infty} f_n' (z_0) = \lim_{n\to \infty} \lim_{z \to z_0} \frac{f_n(z) - f_n(z_0)}{z-z_0}$$ and $$ f'(z_0) = \lim_{z\to z_0} \frac{f(z) - f(z_0)}{z-z_0} = \lim_{z\to z_0} \frac{\lim f_n(z) - \lim f_n(z_0)}{z-z_0} = \lim_{z\to z_0} \lim_{n\to \infty} \frac{f_n(z) - f_n(z_0)}{z-z_0}$$
So, why can I commute the limits in the expressios above?
Thank you