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Definition A.1.4. The infinite product $\prod_{n=1}^{\infty}(1+a_{n}(x))$, where $x$ is a real or complex variable in a domain, is uniformly convergent if $p_{n}(x)=\prod_{m=k}^{n}(1+a_{n}(x))$ converges uniformly in that domain, for each $k$.

Theorem A.1.5. If the series $\sum_{n=1}^{\infty}|a_{n}(x)|$ converges uniformly in some region, then the product $\prod_{n=1}^{\infty}(1+a_{n}(x))$ also converges uniformly in that region.

Corollary A.1.6. If $a_{n}(x)$ is analytic in some region of the complex plane and $\prod(1+a_{n}(x))$ converges uniformly in that region, then the infinite product represents an analytic function in that region.

They are taken from Special Functions page 596-597.

I want to prove Theorem A.1.5 and Corollary A.1.6.

Proof: There exists a convergent series $\sum_{n=1}^{\infty}M_{n}$ such that $|a_{n}(x)|\leq M_{n}$, for all $x\in A\subseteq \mathbb{C}$. Note that $|1+a_{m}(x)|\leq 1+|a_{m}(x)|<2M_{m}$, so $$|p_{n}(x)|<2\prod_{m=k}^{n}M_{m}\leq 2\prod_{m=1}^{n}M_{m}=:2C_{n}.$$ We know that $M_{m}$ is bounded for all $1\leq m\leq n$, so the partial products $C_{n}$ must be bounded too. Therefore $2\sum_{n=1}^{\infty}C_{n}<\infty$, and this shows that the infinite product is uniformly convergent on $A$.

Is this correct? Is there another way to prove it? I am not sure how to provethe corollary. Any hints?

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  • $\begingroup$ Have another look at the details of the proof(s) that for a non-negative real sequence $(r_n)_n$ we have $\sum_n r_n<\infty \iff \prod_n (1+r_n)<\infty.$ $\endgroup$ Mar 21, 2016 at 22:16

2 Answers 2

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Thm A.1.5 is false. On $(0,1)$ define $a_1(x) = 1/x,$ $a_n \equiv 1/2^n, n > 1.$ Then $\sum_n |a_n(x)| = \sum_n a_n(x)$ is uniformly convergent on $(0,1)$ (to $1/x+ 1$). But if we set $p_N(x) = \prod_{n=1}^N (1+a_n(x)),$ then for $1<M<N,$ we have

$$p_N(x) - p_M(x) = p_M(x)\left (\prod_{n=M+1}^N (1+a_n(x)) - 1)\right ) \ge (1/x)2^{-N},$$

which is unbounded on $(0,1).$ Thus there is no chance for the partial products to be uniformly Cauchy on $(0,1),$ hence $P_N$ is not uniformly convergent.

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No, that's totally wrong. For the series to be convergent, you certainly need $M_n \to 0$. But $1 + |a_n| > 1$, which is certainly not bounded by $2 M_n$.

Hint: What you need to do is use logarithms: $$\prod_{n=1}^N (1 + a_n(x)) = \exp\left(\sum_{n=1}^N \log(1+a_n(x))\right)$$

As for the corollary, that's just the fact that a uniform limit of analytic functions in a region is analytic there.

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