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}.

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It’s a matter of principles (of convergence)

One thought on “It’s a matter of principles (of convergence)

  1. eborghini says:

    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.

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