Let $w \in \mathbb{C}$ be a complex number with $0 <|w|<1$. Show that the infinite product $$\theta(z) := \prod_{n=1}^\infty (1+w^{2n-1}e^z)(1+w^{2n-1}e^{-z})$$ converges locally uniformly and absolutely for $z \in \mathbb{C}$.

We have a theorem that states that for a sequence of holomorphic functions $f_n$, if the infinite series of $[f_n(z)-1]$ converges absolutely and locally uniformly, then the infinite product of $f_n(z)$ converges absolutely and locally uniformly. I believe we can use this and solve the question by proving the convergence of the corresponding infinite series using the Weierstrass M-test, but I cannot seem to bound the series from above. Or how should I do this?

  • $\begingroup$ Could you use the Jacobi triple product which relates it to an infinite sum? $\endgroup$ – Somos Apr 9 '20 at 18:31

Your series is $$ \sum_{n=1}^\infty\left(\big|w^{2n-1}e^z\big|+\big|w^{2n-1}e^{-z}\big|\right) $$ For $|w|<1$ a fixed number and $z$ in a bounded set, this should be easy to bound.

Suppose $z$ is in the set $\{z : |z| \le M\}$. Then $|e^z| = e^{\operatorname{Re} z} \le e^M$ and $|e^{-z}| = e^{-\operatorname{Re} z} \le e^M$, so

$$ \sum_{n=1}^\infty\left(\big|w^{2n-1}e^z\big|+\big|w^{2n-1}e^{-z}\big|\right) \le \sum_{n=1}^\infty\left(|w|^{2n-1}e^M+|w|^{2n-1}e^{M}\right) = \frac{2e^M |w|}{1-|w|^2} . $$ Therefore, by the Weierstrass M-test, the infinite series converges uniformly on the set $\{z : |z|\le M\}$. And so $\theta(z)$ converges uniformly and absolutely on $\{z : |z| \le M\}$.

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    $\begingroup$ Unless I am mistaken, a term $w^{4n-2}$ is missing. $\endgroup$ – Martin R Apr 9 '20 at 16:23
  • $\begingroup$ Corrected. Two terms in the sum correspond to two factors in the product. $\endgroup$ – GEdgar Apr 9 '20 at 21:26
  • $\begingroup$ I can't seem to see how it bounds? Let's say we look in $B(0,1)$, then $|w^{2n-1}e^{±z}| \leq |e^{±z}| \leq e$, won't each term be bounded by the constant $e$, so that the series doesn't converge? $\endgroup$ – JjL7 Apr 9 '20 at 22:43
  • $\begingroup$ Why isn't the last equality given by $\frac{2e^M}{ |w|} \cdot \frac{1}{1-|w|^2}$? $\endgroup$ – JjL7 Apr 10 '20 at 2:34
  • $\begingroup$ The sum is from $n=1$ to $\infty$, not $n=0$ to $\infty$. $\endgroup$ – GEdgar Apr 10 '20 at 12:58

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