For $\{f_n\}$ being holomorphic functions, we know that $\sum_{n=1}^\infty |f_n(z)|$ converges uniformly implies that the infinite product $\prod_{n=1}^\infty (1+f_n(z))$ converges uniformly as well, and so the product is holomorphic on $\mathbb{C}$. Does $\prod_{n=1}^\infty (1+f_n(z))$ being entire also imply that $\sum_{n=1}^\infty |f_n(z)|$ converges uniformly? I do not think my proof for the first statement works for this one, so I'm a bit curious as to whether or not this holds.


No. For a somewhat contrived example, take $f_n(z)$ to be the constant function $a_n$ where $a_n = (-1)^n/n$ or some other slowly decreasing alternating sequence. Then $\prod_{n=1}^\infty (1+f_n(z))$ is convergent (hence entire) but $\sum_{n=1}^\infty |f_n(z)|$ does not converge anywhere.

  • $\begingroup$ I see! In general, would we need uniform convergence of the infinite product for the infinite product to define an entire function? $\endgroup$ – mathstudent12 Oct 11 '12 at 7:18
  • $\begingroup$ @mathstudent12 Do you really mean uniform convergence or normal convergence? It is rare for entire functions to converge uniformly on all of $\mathbb C$ (see Liouville's theorem), but the typical scenario is that one proves uniform convergence on compact sets, which is called normal convergence. $\endgroup$ – Erick Wong Oct 11 '12 at 18:13

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