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Let $A_n=\{z\in \mathbb{C}: \epsilon_n\leq |z| \leq 1\}$ and $$ f_n\colon A_n\to \mathbb{C}, \quad n\in \mathbb{N}, $$ be a sequence of holomorphic functions such that $\lim_{n\to \infty} \epsilon_n=0$, and for any $r<1$, $\{f_n|_{\{z\in \mathbb{C}: r\leq |z| \leq 1\}}\}_{n\in \mathbb{N}}$ converges to $0$ in $C^\infty$-norm (Uniformly with all derivatives).

Is it true that either $\lim_{n\to \infty} ||f_n(x)||_{\infty} =0$ for all $x$ or (after passing to a subsequence) there are sequences of points $\{z_n\}_{n\in\mathbb{N}}$ and $\{w_n\}_{n\in \mathbb{N}}$ in the boundary components $\{|z|=\epsilon_n\}_{n\in \mathbb{N}}$ with $$ \Big(\lim_{n\to \infty} f_n(z_n) \neq \lim_{n\to \infty} f_n(w_n) \Big) \in \mathbb{C}\cup \{+\infty\} \cup \{-\infty\}. $$

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  • $\begingroup$ What is the meaning of $\lim_{a\to\infty}f_a\to 0$? $\endgroup$
    – Alexandre Eremenko
    Commented Sep 24, 2017 at 3:07
  • $\begingroup$ This remind me a problem in complex analysis of Ahlfors which states that the uniform limit of a sequence of injective holomorphic functions is either injective or constant. May be some strategies of the solution of that problem could be useful in this question. $\endgroup$
    – Ali Taghavi
    Commented Sep 24, 2017 at 6:37
  • $\begingroup$ I saw this problem in Ahlforse book about 20 years ago. $\endgroup$
    – Ali Taghavi
    Commented Sep 24, 2017 at 6:40
  • $\begingroup$ The injective case should be easy. You can argue geometrically then. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Sep 24, 2017 at 10:17
  • $\begingroup$ @MohammadF.Tehrani The injective case is an immediate consequence of the following lemma: Lemma:"If $f_n \to f $ uniformly on compact sets and $f$ is not a constant with f(a)=0 for some $a$ then there is a sequence $a_n \to a$ with $f_n(a_n)=0$" Proof: Choose a small circle $C_{\epsilon}$ around $a$ then $\int_{C_{\epsilon}} \frac{f_n'}{f_n} \to \int_{C_{\epsilon}} \frac{f'}{f} $ but these integrals count the number of zeros inside $C_{\epsilon}$. This lemma obviously proves the injectivity. Now I guess that (perhaps) the same strategy works in the $\endgroup$
    – Ali Taghavi
    Commented Sep 24, 2017 at 11:35

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I am not sure that I understand your question correctly. In particular, I am not sure what $+\infty$ and $-\infty$ mean.

As I first understood it, the answer is negative. Let $A(r,1)$ be the annulus of points with $r<z<1$.

Define $\newcommand{\eps}{\varepsilon}\eps_n := 1/n^2$, $A_n := A(\eps_n,1)$ and set $f_n\colon A_n \to \mathbb{C}; z\mapsto \frac{1}{nz}$.

Then clearly the functions converge uniformly to zero on every annulus $A(\eps,1)$. However, if $z_n$ is any sequence with $|z_n|=\eps_n$, then $$ |f_n(z_n)| = n \to \infty.$$

However, if by converging to $+\infty$ or $-\infty$, you mean that points should converge to infinity along different rays (e.g. asymptotically to the positive/negative real axis), I think the answer is positive.

Indeed, the maximum must be taken on the boundary, and because the function has no poles and takes values near zero, the image of the boundary cannot be contained in a small neighbourhood of some finite value. So, either you can find subsequences of values converging to finite values, or, for large $n$, the image of the boundary surrounds any fixed disc around the origin. In the latter case, it must contain arbitrarily large positive and negative real values.

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