# It’s a matter of principles (of convergence)

Let $(f_n)_{n\in \mathbb{N}}$ be a sequence of entire functions such that, for every $z \in \mathbb{C}$, there exists $F(z)=sup_{n\in \mathbb{N}}|f_n(z)| \in \mathbb{R}$. Suppose that $F:\mathbb{C} \to \mathbb{R}$ is continuous and $F(0)=0$. Let $g$ be an entire function such that $g'(0) \neq 0$. Prove that $g \circ f_n \to g$ over compact sets if and only if $f_n(z) \to z$, for every $z \in \mathbb{C}$.

Recall that convergence over compact sets is given by the following metric on the set ${\mathcal H}(\mathbb{C})$ of entire functions:
$d(f,h) = \sum_{n \geq 1} \frac{1}{2^n} \frac{ d_{\infty}(f|_{n\mathbb{D}},h|_{n\mathbb{D}}) }{ 1 + d_{\infty}(f|_{n\mathbb{D}},h|_{n\mathbb{D}}) },$
where $\mathbb{D}$ stands for the unit disk and $f,h$ are entire functions.
Suppose now that $f_n(z) \to z$ for every $z \in \mathbb{C}$. Let $K \subseteq \mathbb{C}$ be a compact set. Since by hypothesis the set ${\mathcal F} = \{ f_n|_{K} \}_{n \in \mathbb{N} }$ is pointwise bounded, Montel’s theorem implies ${\mathcal F}$ is precompact. Thus, every subsequence of $(f_n|_{K})_{n \in \mathbb{N} }$ admits a uniformly convergent subsequence. Since $(f_n)_{n \in \mathbb{N} }$ converges pointwise to the identity function, it converges uniformly over compact sets to the identity function. It follows easily that $g \circ f_n \to g$ uniformly over compact sets.
For the other implication, fix $k \in \mathbb{N}$ and note that again by Montel every subsequence of $(f_n)_{n \in \mathbb{N} }$ admits a convergent subsequence $(f_{n_j})_{j}$ to a certain function $f$ in ${\mathcal H}(k\mathbb{D})$. Hence, $g \circ f_{n_j} \to g \circ f$ uniformly over $k\mathbb{D}$. Since $g'(0) \neq 0$, there are open sets $U \ni 0$, $V \ni g(0)$ such that $g(U) = V$ and $g$ admits an inverse $V \to U$ which is holomorphic. By continuity of the function $F$, for a sufficiently small $\varepsilon > 0$ all the sets $f_n(\varepsilon \mathbb{D})$ are contained in $U$. Therefore, by precomposing with the holomorphic inverse of $g$ we see that $f$ coincides with the identity over $\varepsilon \mathbb{D}$. By the identity principle, it is $f(z) = z$ for every $z \in k\mathbb{D}$. Since $k \in \mathbb{N}$ was arbitrary, the conclusion follows.