Entire function with prescribed values

Question

I am trying to solve the following problem from Ahlfors' Complex Analysis Chapter 5, Section 2.3: Suppose that $\{a_n\}$ is a sequence of distinct complex numbers such that $a_n\to \infty$ and let $\{c_n\}$ be a sequence of arbitrary complex numbers. Show that there exists an entire function $f(z)$ satisfying $f(a_n)=c_n$.

The hint that is in Ahlfors' book is to let $g(z)$ be a function with simple zeros at each $a_n$. Such a function exists by Weierstrass' Theorem. Then, the hint says to look for appropriate $\gamma_n$ such that the following series converges $$ \sum_1^\infty g(z)\frac{e^{\gamma_n(z-a_n)}}{z-a_n}\frac{c_n}{g'(a_n)}. $$

I have been playing around with this for a while. I know that I need to find the $\gamma_n$ so that on any compact ball, $|z|\leq R$, the values

$$ \left|g(z)\frac{e^{\gamma_n(z-a_n)}}{z-a_n}\frac{c_n}{g'(a_n)}\right| $$ are bounded by some values $M_n(R)$ so that for each $R>0$, the sum $\sum M_n(R)$ converges. If I can do this then I know that the sum will converge uniformly on compact subsets, and so I know that the sum will be an analytic function, and then it will clearly have the right properties. However, I am having difficulty figuring out what I am supposed to choose for $\gamma_n$. I tried expressing $g$ as a Taylor series around each $a_n$ and then finding some upper bound of the terms in the sum which depended only on $R$, but have had no luck so far.

Can anyone provide a hint about how to go about finding such $\gamma_n$?

Answer 1

Unfortunately, Ahlfors' hint is very misleading, and there is in fact a simpler way to solve this problem, especially since at this point of the book Ahlfors has proven both Mittag-Leffler and Weierstrass Theorems.

Let $g$ be an entire function with simple zeros at $a_n$. Recall that Mittag-Leffler's Theorem not only asserts the existence of meromorphic functions with poles at $a_n$, but allows us to control the singular part of the function at each $a_n$. So let $h$ be a meromorphic function on $\mathbb{C}$ with simple poles at each $a_n$ with singular part $(c_n/g'(a_n))/(z-a_n)$. Then $f:=gh$ has the desired properties.

Answer 2

I think I might have a solution that Ahlfors intended.

Since $a_i$ are simple zeros we have $g(z) = (z-a_n)g_n(z)$ for some analytic function $g_n$ such that $g_n(a_n)\neq 0$ for all $n$. So we have: $$\frac{g(z)}{(z-a_n)g'(a_n)} = \frac{g_n(z)}{g'(a_n)} = h_n(z)$$ Where $h_n(a_n) = 1$.

Now consider a single summand: $$g(z)\frac{e^{\gamma_n(z-a_n)}}{z-a_n}\cdot\frac{c_n}{g'(a_n)} = h_n(z)e^{\gamma_n(z-a_n)}c_n$$ Let's just absorb $c_n$ in $h_n$ and consider: $$h_n(z)e^{\gamma_n(z-a_n)}$$

Now consider $z$ inside $|z|<R$ and remove all $a_n$ that are inside $|z|\leq 2R$. (The number of such $a_n$ is finite). So now we have only those $a_n$ that are outside of radius $2R$ and hence $|z-a_n|>R$.

Now consider any particular $n$ for which $a_n$ is outside $2R$. Now $|h_n(z)|$ must have a maximum inside $|z|\leq R$ because the disk is compact (which infact will occur at the boundary $|z| = R$). Let that maximum be $M_n$. Now let $\gamma_n = -\gamma_n'$. Then we have: $$|h_n(z)e^{\gamma_n(z-a_n)}|=\frac{|h_n(z)|}{|e^{\gamma_n'(z-a_n)}|}\leq \frac{M_n}{e^{\gamma_n'R}}$$ Now we choose $\gamma_n'$ large enough so that the RHS becomes less than $2^{-n}$.

Answer 3

If $b_n:=\frac{c_n}{g'(a_n)}=e^{O(n)}$, then you can simply take $\gamma_n= n |a_n|/a_n$ (if $a_n\neq 0$, otherwise set it to zero). For z in a compact set $K$, the $n$-th term in the series grows at most as

$$\frac{e^{-n\lvert{a_n}\rvert+O_K(n)}}{\lvert{a_n}\rvert+1}$$ and so the series converges since $\lvert{a_n}\rvert\to\infty$.

In general you can take for example $$\gamma_n= (n + |b_n|)\frac{\lvert{a_n}\rvert}{a_n}.$$