11
$\begingroup$

Let $\mathbb C$ be the complex plane, $H(\mathbb C)$ the set of all entire functions, and $D(\mathbb C)$ the set of all non-negative divisors in $\mathbb C$.

Consider the map $Z:H(\mathbb C)\to D(\mathbb C)$ which to every entire function $f\in H(\mathbb C)$ puts into correspondence its divisor of zeros. There are natural topologies which make this map continuous: on $H(\mathbb C)$ this is uniform convergence on compact subsets, and on $D(\mathbb C)$ the induced topology of weak convergence of measures. (A divisor can be thought of as a discrete measure that takes integer values on all bounded sets).

The Weierstrass factoriation theorem says that this map is surjective, and he actually constructed a right inverse $W$, so that $Z\circ W=id_{D(\mathbb C)}$.

Question: Does there exist a CONTINUOUS right inverse?

Weierstrass map is evidently not continuous. (His map depends on the ordering of zeros by absolute value, and this ordering can change when the divisor varies continuously). I can prove that there is no analytic right inverse, with the natural analytic structures on $H(\mathbb C)$ and $D(\mathbb C)$. I can also prove that there is no multiplicative continuous right inverse (multiplicative means that the sum of divisors corresponds to the product of functions).

EDIT. Let me add for completeness a brief explanation why there is no analytic $W$. In MR2280501, Michigan Math. J. 54 (2006), no. 3, 687–696, for every compact $E\in U$ of zero log capacity in the unit disk $U$, I constructed an holomrphic function $F(z,w)$ of two variables $(z,w)\in\mathbb C\times U$, with the property that $F(z,w)\neq 0$ when $w\in E, z\in\mathbb C$ but $z\mapsto F(z,w)$ has zeros for $w\in U\backslash E$. Take some uncountable $E$. Let $D(w)$ be the divisor of the entire function $f_w=F(.,w)$ in the $z$-plane. Assuming that an analytic Weierstrass map exists, denote $g=W(\emptyset)$. Then the set $\{ w:W(D(w))=g\}$ must be analytic (that is either discrete of the whole $w$-disk, but this is not so because it contains $F$ and is not equal to the whole disk.

$\endgroup$

1 Answer 1

6
$\begingroup$

The Weierstrass product has the form $W(z)=\prod E_{N(a)}(z/a)$, where the product is over the set of the desired zeros $a$, and the integers $N(a)$ can be chosen freely; they only need to be large enough asymptotically to ensure convergence.

To avoid the problem you mentioned, we must make sure that $N(a)$ depends continuously on the divisor. In particular, we will also need $E_N(z)$ for non-integer $N$, but this can be done by just interpolating in a straightforward way: if $N=n+d$ with $n\in\mathbb N$, $0\le d<1$, then we set $$ E_N(z) = (1-z)\exp \left( z + \frac{z^2}{2} + \ldots + \frac{z^n}{n} + d\frac{z^{n+1}}{n+1}\right) . $$

Fix a continuous, compactly supported $F: \mathbb R\to [0,1]$ with $F(x)=1$ for $|x|\le 2$. I claim that taking $$ N(a) = 2+|a|+ \sum_b F(|b|-|a|) $$ will give us continuous dependence of $W$ on the zero set. (The sum is over the zeros, with multiple zeros contributing the corresponding number of summands.)

A divisor $D'$ is close to the given divisor if it has almost exactly the same zeros $|a'|\le R$, and $D'$ can do anything whatsoever on $|z|>R$. To show that $W, W'$ are close in the desired topology, we must compare them on a fixed compact set, say $|z|\le r$, and of course we will take $R\gg r$.

Obviously, the finitely many factors corresponding to the $|a'|\le R$ are almost the same as before. So we need to show that for any (discrete) set of zeros $|a'|>R$, the corresponding part of the Weierstrass product is almost $1$ on $|z|\le r$ (provided $R$ was chosen large enough to start with).

Let's look at the contribution from the $a'$ with $n\le |a'|<n+1$. If there are $K$ of these, then they all contribute to my modified counting function $F$, so $N(a')\ge K+2$. Recall that $|E_N(z)-1|\le |z|^{-N-1}$ for $|z|\le 1$. So the total contribution to $\sum\log |E_{N(a')}(z/a')-1|$ from these $a'$ is $\lesssim K n^{-K-3} \le n^{-4}$ (because this function is decreasing in $K\ge 1$). This is summable, so the claim follows.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.