# Uniform convergent series

Let $\Omega$ be domain in $\mathbb C^2$. For each compact set $K_j$ define the holomorphic function $f_j$ on $\Omega$, such that $$\sup_{k_j}|f_j|<2^{-j}.$$ Define $$f= \prod_{j=1}^\infty(1-f_j)^j.$$

I need to show that this product converges uniformly on each compact set $K_l$.

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There is a theorem which gives a link between the convergence of a product and the sum. See for instance Rudin Real and complex analysis. – Davide Giraudo May 6 '12 at 12:59
In fact, after having looked at the proof, it seems it works even if $f\colon\Omega\subset\mathbb C^2\to \mathbb C$. – Davide Giraudo May 6 '12 at 20:10

We use the following result:

Let $S$ a set and $u_n\colon S\to \Bbb C$ functions such that $\sum_{n\geq 0}|u_n(s)|$ is uniformly convergent on $S$. Then the product $\sum_{n=1}^{+\infty}(1+u_n(s))$ is uniformly convergent on $S$.

We need a lemma:

If $N$ is an integer and $c_1,\ldots,c_N$ are complex numbers, and writing $p_N:=\prod_{j=1}^N(1+c_j)$, $p_N^*=\prod_{j=1}^N(1+|c_j|)$, we have $$p_N^*\leq \exp\left(\sum_{j=1}^N|c_j|\right)\mbox{ and }|p_N-1|\leq p_N^*-1.$$ The first inequality follows from $1+x\leq e^x$ if $x\geq 0$, and the second con be shown by induction.

Let $P_n(s):=\prod_{j=1}^n(1+u_j(s))$; we have \begin{align}|P_{n+m}(x)-P_n(s)|&=|P_n(s)|\left|\prod_{j=n+1}^{n+m}(1+u_j)-1\right|\\\ &\leq \exp\left(\sum_{j=1}^n|u_j(s)|\right)\left(\prod_{j=n+1}^{n+m}(1+|u_j|)-1\right)\\\ &\leq \exp\left(\sum_{j=1}^{+\infty}|u_j(s)|\right)\left(\exp\left(\sum_{j\geq n+1}|u_j(s)|\right)-1\right). \end{align} We choose, for a fixed $\varepsilon>0$ a $n_0$ such that $\sup_{s\in S}\sum_{j\geq n_0+1}|u_j(s)|\leq \varepsilon$, and we get the wanted result.

Now, we deal with this particular case. Let $u_j:=(1-f_j)^j-1$. we have to prove that $\sum_{j\geq 1}|u_j(s)|$ is uniformly convergent on $K_l$.

We have for $n\geq l$ that \begin{align*} |u_n(s)|&=\left|\sum_{k=0}^n\binom nk(-f_n(s))^{n-k}-1\right|\\\ &=\left|\sum_{k=1}^n\binom nk(-f_n(s))^{n-k}\right|\\\ &\leq \sum_{k=1}^n\binom nk|f_n(s)|^{n-k}\\\ &= \sum_{k=1}^n\binom nk|f_n(s)|^k\\\ &\leq \sum_{k=1}^n\binom nk 2^{-nk}\\\ &=2^{-n}\sum_{k=0}^{n-1}2^{-kn}\\\ &=2^{—n}+\sum_{k=0}^{n-1}2^{-n(k+1)}\\\ &\leq (n+1)2^{-n}, \end{align*} which is enough to conclude since $\sum_{n\geq 1}(n+1)2^{-n}$ is converging.

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thnx for the answer.. i will look upon the result stated and get back to you if need. – zapkm May 8 '12 at 17:41