Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $K_l$.

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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