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Is it true that "an infinite product of entire functions $\{ f_{n}(z) \}_{n=1}^{\infty}$ is again an entire function"?

$$\prod_{n=1}^{\infty}f_{n}(z)$$ (entire function is a function which is analytic in the complex plane).

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If it converges locally uniformly, then yes. – Lukas Geyer Oct 10 '12 at 22:36
@Lukas: I don't have any further information, I just have an infinite sequence of entire functions and want to see if the product is entire or not. So I understand that it is not true in general! – math st. Oct 10 '12 at 22:38
Maybe you should post the explicit product you are interested in? In general, like for series there is no reason that these products converge in the whole plane. – Lukas Geyer Oct 10 '12 at 22:40
I already did.. – math st. Oct 10 '12 at 22:54
What if $f_n(z)=z$ for all $n$? – Berci Oct 10 '12 at 22:56

Start with a sequence rather than a product. There is a sequence of polynomials $p_n(z)$ such that $p_n(z) \to 0$ as $n \to \infty$ for every $z \ne 0$ but $p_n(0) = 1$. These can be obtained using Runge's Theorem so that $|p_n(z)| < 1/n$ on $K_n = \{r e^{i\theta}: 1/n \le r \le n, 0 \le \theta \le 2 \pi - 1/n\}$. Then if $f_n(z) = \exp(p_n(z) - p_{n-1}(z))$ for $n > 1$ with $f_1(z) = \exp(p_1(z))$, we have $\prod_{n=1}^\infty f_n(z) = \lim_{n \to \infty} \exp(p_n(z)) = 1$ for all $z \ne 0$ but $e$ for $z = 0$. So this infinite product is not analytic, in fact not continuous, at $0$.

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