Suppose $A \subset \mathbb{C}$ is a compact set and $ f_n: A \rightarrow \mathbb{C}$ is a sequence of continuous functions uniformly converging to $f$.
Suppose $f(z_0) = 0$ and $f(z) \ne 0$ when $z \ne z_0$. Observe that $f$ is continuous but not necessarily holomorphic in any open subset of $A$.
We define a sequence of functions $g_n: A \rightarrow \mathbb{C} $ given by $g_n(z) = |f_n(z)|^{1/n}$. With a limiting function being $g$.
What can we say - if anything - about the indeterminate $g(z_0)$.
My initial thoughts:
I thought $g(z) = 1 {z \in A, z \ne z_0}$ should be constant and therefore holomorphic and bounded in open punctured disk around $z_0$ and therefore we should be able to define $g(z_0) =1$ by Riemann’s Removable Singularity Theorem.
Then I wondered what would be the difference if we have $g_n(z) = f_n(z)^{1/\log(n)}$, $g_n(z) = f_n(z)^{1/\sqrt{n}}$, $g_n(z) = f_n(z)^{1/n^2}$, etc.
Is there a theorem that addresses this already?
Thanks in advance.
Suppose $\{a_n\}, \{b_n\}$ are sequences of real numbers with $a_n \to a$ and $b_n \to 0$ as $n \to 0$. Further suppose that all $a_n > 0$ and $a > 0$. Then
$$\lim_n a_n^{b_n} = 1$$
Because $a > 0$, there exists $N$ such that for $n \ge N, |a_n - a| < a/2$. Let $$0 < m < \min\left\{\frac a2, \frac 2{3a}, a_n, \frac 1{a_n}\mid n < N\right\}$$ and $$M > \max\left\{\frac {3a}2, \frac 2a, a_n, \frac 1{a_n}\mid n < N\right\}$$
Then we have that $m < a_n < M$ and also $m < 1/a_n < M$ for all $n$. Therefore $$m^{|b_n|} < a_n^{b_n} < M^{|b_n|}$$
But because $b_n \to 0$, so does $|b_n|$ and therefore by continuity of the exponential function, $$\lim_n m^{|b_n|} = \lim_n M^{|b_n|} = 1$$
By the squeeze theorem, $a_n^{b_n} \to 1$ also.
The condition $a_n > 0$ can be dropped, provided $b_n \ge 0$ or one doesn't mind a finite number of undefined values in the sequence $a_n^{b_n}$ (finite because $a > 0$ means that eventually $a_n > 0$). However, the condition $a > 0$ is required.
So your conditions about $f_n$ converging uniformly and compactness of the domain - or even the domain being in $\Bbb C$ do not matter. By simple pointwise convergence, if $g_n = f_n^{b_n}$, then $g(z) = 1$ for all $z$ with $f(z) > 0$.