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I'm currently trying to get an approximation to work:

Denote the set of all entire functions as $H(\mathbb C)$. Now I want to show the following:

Let $f, g \in H(\mathbb C)$, $\delta > 0$ and $A \subset \mathbb C$ compact so that $\mathbb C \setminus A$ is connected. Then there exist a $\varphi \in H(\mathbb C)$ and a $n \in \mathbb N$ with

$$ \Vert f - \varphi\Vert_A < \delta \qquad \text{and} \qquad \Vert g - \varphi^{(n)}\Vert_A < \delta,$$

where $\Vert f \Vert_A = \max_{z \in A} \vert f(z) \vert$ and $\varphi^{(n)}$ denotes the $n$-th derivative of $\varphi$.

What I can use is Runge's theorem. That gives me a polynomial $p \in H(\mathbb C)$, such that $\Vert f - p\Vert_A < \delta$ and $q \in H(\mathbb C)$ such that $\Vert g - q\Vert_A < \delta$. But I don't see how I can construct the $\varphi \in H(\mathbb C)$ from that. Sometimes one can get such a function by summation of polynomials, but I don't see how that could work in this context.

I also got the following lemma to work with: I know that $\varphi^{(-n)} \to 0$ locally uniformly for $n \to \infty$. So there exists a $n \in \mathbb N$ with $\Vert \varphi^{(-n)}\Vert_A < \delta \quad$ (or any other constant I might use. $\varphi^{(-n)}$ denotes the $n$-th antiderivative is this statement). Note that this lemma solves the case where $f \equiv 0$, but I fail to generalize it. I dont see how to facilitate this lemma in the general case.

I would appreciate some hints or good ideas on the topic. Thanks in advance :)

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    The fact you can choose $n$ as large as you want is quite important. And as usual for holomorphic functions, it reduces to the case $f,g$ are polynomials (the holomorphic functions being the closure of polynomials for the [uniform convergence on every compact](https://en.wikipedia.org/wiki/Compact_convergence)) – 2017-01-03

1 Answers 1

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You got all the ingredients, you just need to put them together.

$A$ is compact, so $A \subset \overline{D_R(0)}$ for some $R > 0$. Now let

$$q(z) = \sum_{k = 0}^d \frac{c_k}{k!}z^k\quad \text{and}\quad M := \sum_{k = 0}^d \frac{\lvert c_k\rvert}{k!}R^k.$$

Then $n$-fold integration of $q$ gives

$$q^{(-n)}(z) = \sum_{k = 0}^d \frac{c_k}{(k+n)!}z^{k+n}\quad\text{and}\quad \lvert q^{(-n)}(z)\rvert \leqslant \sum_{k = 0}^d\frac{\lvert c_k\rvert}{(k+n)!}R^{k+n} \leqslant M\frac{R^n}{n!}$$

on $A$. Now choose $n > \deg p$ so large that

$$M\frac{R^n}{n!} < \delta - \lVert f - p\rVert_A$$

and set

$$\varphi = p + q^{(-n)}.$$