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From Ahlfors, exercise 5.2.3 #1:

Suppose that $a_n\to \infty$ and that the $A_n$ are arbitrary complex numbers. Show that there exists an entire function $f(z)$ which satisfies $f(a_n)=A_n$.

Hint: Let $g(z)$ be a function with simple zeros at the $a_n$. Show that $$\sum_{n=1}^\infty g(z)\frac{e^{\gamma_n(z-a_n)}}{z-a_n}\cdot \frac{A_n}{g'(a_n)}$$ converges for some choice of the numbers $\gamma_n$.

I see that $$\lim_{z\to a_i}g(z) \frac{e^{\gamma_n(z-a_n)}}{z-a_n}\cdot \frac{A_n}{g'(a_n)}=\begin{cases} A_n&n=i\\0&n\neq i \end{cases}$$

for any choice of $\gamma_n\in \mathbb{C}$, so the terms do what we want. In trying to choose $\gamma_n$ to ensure convergence, it doesn't seem like just a matter of making $\gamma_n$ be real numbers going to $-\infty$, for instance, since $e^{-z}$ takes every value in a neighborhood of infinity, not just values of small modulus. So we have to figure a way to use some symmetry to get the different $e^{\gamma_n(z-a_n)}$ to cancel, right?

Eric Auld
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  • I'm blank on what symmetries one might find because of the random nature of the $A_n$. Also, he didn't say g must be analytic or entire, but we must have g'($a_n$) exists for all the n's. So how do we know there is even such a g? Proving that might be equivalent to the original problem, unless Ahlfors has shown you how to construct it. If he did, that might throw some light on this. – Betty Mock Jan 04 '14 at 03:19
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    @BettyMock Weierstrass product theorem – Cocopuffs Jan 04 '14 at 03:49
  • @Cocopuffs Thank you for the tip. That theorem is very neatly done. I knew I could (trivially) construct a polynomial if the sequence is finite, and that extending it to be infinite would not work. Not being Weierstrass, I didn't think of his approach. – Betty Mock Jan 05 '14 at 02:41
  • Isn't it possible to write an $n$-dimensional analogue of the function in the above proof, so that we can have a holomorphic $f : \mathbb{C}^n \to \mathbb{C}$ having prescribed values at a discrete sequence? – SMS Dec 31 '16 at 12:50

1 Answers1

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get the different $e^{\gamma_n(z−a_n)}$ to cancel

No, that's not going to work.

$e^{-z}$ takes every value in a neighborhood of infinity

Yes, it does. But we are not looking to control the series at a neighborhood of infinity. What we need is its uniform convergence on every compact set.

So: think globally, act locally -- get the series to converge on every disk.

Let $$h_n(z)= \dfrac{A_n g(z)}{(z-a_n)g'(a_n)}$$ The goal is to have uniform convergence of $\sum e^{\gamma_n(z-a_n)}h_n$ on every disk $D_R=\{z:|z|\le R\}$. For this, it suffices to have $$ \max_{D_R}\left|e^{\gamma_n(z-a_n)}\right| \le \frac{2^{-n}}{ \max_{D_R} |h_n |} \tag{1} $$ for all $n\ge N_R$, where $N_R$ depends on $R$.

As you know, this means we want the real part of $\gamma_n(z-a_n)$ to be very negative on $D_R$. There is no way to do this until $a_n$ go out of $D_R$. Fortunately, they eventually do. If $|a_n|\ge 2R$, then by choosing $\gamma_n$ to have the same argument as $a_n$, we make sure that the argument of $\gamma_n(z-a_n)$ is between $\pi\pm \pi / 6$ for all $z\in D_R$. It remains to make the magnitude of $\gamma_n$ large enough. For example, make it large enough so that (1) holds for $R=n/2$.

Then for any fixed $R$ the inequality (1) holds when $n\ge 2R$.

Post No Bulls
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