If $G_n(z)=(z-z_0)g_n(z)$ converges uniformly to $f(z)-f(z_0)$, what can I say about the convergence of $g_n(z)$?

Question

Let $G_n(z) = (z-z_0)g_n(z)$ be analytic on a region $\Omega$ for $n=1, 2,\dots$, with $z_0 \in \Omega$ and suppose $G_n(z)$ converges uniformly to $f(z)-f(z_0)$ on $\Omega$ for some analytic function $f$. Does $g_n(z)$ converge uniformly to $\frac{f(z)-f(z_0)}{z-z_0}$ on $\Omega$? I know by the uniform convergence of $G_n(z)$ that if $z\in \Omega$ with $|z-z_0|> \delta$, then there exists $N>0$ such that $|g_n(z)-\frac{f(z)-f(z_0)}{z-z_0}|< \epsilon$ for all $n \geq N$, but I am unsure how to show the uniform convergence inside $|z-z_0|<\delta$. I wanted to show $\{g_n(z)\}$ was uniformly bounded on $|z-z_0|<\delta$ but have gotten stuck. There may be something simple that I am not considering; any help is appreciated.

Answer 1

$\frac{f(z)-f(z_0)}{z-z_0}$ has a removable singularity at $z=z_0$, so that the maximum modulus principle can be applied to $$ h_n(z) = g_n(z)-\frac{f(z)-f(z_0)}{z-z_0} \, . $$

It follows that if $h_n \to 0$ uniformly on $|z-z_0| \ge \delta$ the same is true in the disk $|z-z_0| < \delta$.