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Please note that the set of entire functions is a complete metric space $(H(\mathbb C), d)$ for a suitable metric.

Let $(f_k)_{k \in \mathbb N}$ a sequence in $H(\mathbb C)$, such that $f_k$ has no zeros in $\{z: \vert z \vert \leq k\}$ and $f:= \lim_{k \to \infty} f_k \in H(\mathbb C)$ has no zeros in $\mathbb C$.

Now I want to show that $\sup_{n \in \mathbb N}(d(f_k^{(n)}, f^{(n)})) \to 0$ for $k \to \infty$.

I can use the following three statements that I have already proven, so you can take them as given:

  1. A sequence $(f_k)_{k \in \mathbb N}$ converges in $(H(\mathbb C), d)$ to $f$ $\Leftrightarrow$ $(f_k)_{k \in \mathbb N}$ converges on every compact set to $f$.

  2. For every $\epsilon > 0$ there exists a $\delta > 0$ and a compact set $A \subseteq \mathbb C$ with $$ \max_{z \in A} \vert f(z) - g(z) \vert < \delta \quad \Rightarrow \quad d(f,g) < \epsilon.$$

  3. For every $\delta > 0$ a compact set $A \subseteq \mathbb C$ there exists a $\epsilon > 0$ with $$ d(f,g) < \epsilon \quad \Rightarrow \quad \max_{z \in A} \vert f(z) - g(z) \vert < \delta.$$

I tried many things but I don't see how to achieve that result. I can easily get that $d(f_k^{(n)}, f^{(n)}) \to 0$ for $k \to \infty$ by using $f= \lim_{k \to \infty} f_k$, lemma 1, the Weierstrass theorem and again lemma 1. But I can't get the conclusion that $\sup_{n \in \mathbb N}(d(f_k^{(n)}, f^{(n)})) \to 0$ from that. I would really appreciate some hints :)

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    As the sequence $(z^k/k!)$ shows, the condition of having no zeros must play an important role (if the assertion is true, which I don't know yet, gotta think about it in bed). – 2017-01-05
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    I may be that the statement doesn't hold. But it is a statement that would complete a proof I am working on. – 2017-01-06
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    It doesn't hold, unfortunately. What are you trying to prove? It may follow from something weaker already. – 2017-01-06
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    @DanielFischer I used an estimate prior in my proof. I think it might be too strong then. I will try to work it out without using it. If I don't suceed, I might post it here :) – 2017-01-06
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    Good luck then. Although I'd be interested, I hope you don't need to post it here. – 2017-01-06

1 Answers 1

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The desired conclusion does not follow.

For concreteness, define $p_m(f) = \max \{ \lvert f(z)\rvert : \lvert z\rvert = m\}$ for $m \in \mathbb{N}$ and

$$d(f,g) = \max \biggl\{ 2^{-m}\frac{p_m(f-g)}{1 + p_m(f-g)} : m \in \mathbb{N}\biggr\}.$$

Let $f \equiv 1$ and $f_k = 1 + g_k$, where $g_k(z) = e^{k(z-k)}$. Then $f_k \to f$ locally uniformly, $f$ has no zeros at all, and $f_k$ has no zeros in the disk $\{ z : \lvert z\rvert \leqslant k\}$ (for $\operatorname{Re} z < k$, we have $\lvert g_k(z)\rvert < 1$, and $f_k(k) = 2$).

But for $n > 0$, we have $f_k^{(n)}(z) = g_k^{(n)}(z) = k^ng_k(z)$, and hence

$$\sup_{n\in \mathbb{N}} d\bigl(f^{(n)},f_k^{(n)}\bigr) \geqslant \sup_{n\in \mathbb{N}\setminus \{0\}} \frac{p_0\bigl(g_k^{(n)}\bigr)}{1 + p_0\bigl(g_k^{(n)}\bigr)} = \sup_{n\in \mathbb{N}\setminus \{0\}} \frac{k^n e^{-k^2}}{1 + k^ne^{-k^2}} = 1$$

for all $k \geqslant 2$.