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I've been trying to solve this homework problem for a while but I can't seem to get any significant ideas about how to approach it, so I would really appreciate any hints that could help me solve it.

The problem is exercise 8.14 from Steven Krantz and Robert Greene's book Function Theory of One Complex Variable. It goes as follows:

Suppose that $$\sum |\alpha_n - \beta_n| < \infty$$ Then determine the largest open set of $z$ for which $$\prod_{n = 1}^{\infty} \frac{z - \alpha_n}{z - \beta_n}$$ converges normally.

What I've tried so far is writing the factors as

$$\frac{z - \alpha_n}{z - \beta_n} = 1 + \frac{z - \alpha_n}{z - \beta_n} - 1 = 1 + \frac{\beta_n - \alpha_n}{z - \beta_n}$$

so as to put the infinite product in the form $\displaystyle{\prod (1 + f_n(z))}$ to try to apply the basic convergence criteria I have available which says that this product would converge normally if the series

$$\sum |f_n(z)|$$ converges normally. Now I'm kind of stuck here because I think that maybe I would have to bound this sum with the sum $\sum |\alpha_n - \beta_n|$ but I'm not sure about how to proceed (assuming that this is the right way to follow).

So I would really appreciate some hints that would get me in the right track to solve this problem. Thank you very much.

GNUSupporter 8964民主女神 地下教會
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Adrián Barquero
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  • I think the product converges normally on any compact set $K\subset \mathbb{C}\setminus ((\beta_j)_j)$: With your notation, and using Hölder's inequality $\sum |f_n|_K \leq \sup_n |(\cdot-\beta_n)^{-1}|_K \sum |\alpha_n-\beta_n|$ and this implies the claim. – Jose27 Mar 05 '12 at 04:12
  • @Jose27 Can you please expand a little bit about what you wrote. I don't think I follow how to use Hölder's inequality here. – Adrián Barquero Mar 05 '12 at 04:36
  • Well, $$\sum |f_n(z)| = \sum |(z-\beta_n)^{-1}||\alpha_n-\beta_n| \leq \sup_n |(z-\beta_n)^{-1}|\left(\sum |\alpha_n-\beta_n|\right)$$. This shows that, on $K$ as above, the sum converges compactly. Now just use the same argument with $|\cdot|_K$ replacing $|\cdot|$. – Jose27 Mar 05 '12 at 04:49
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2 Answers2

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Hint:

Suppose that $z\in\mathbb{C}\setminus\overline{\{b_n\}}$, then there is an $\epsilon>0$ so that $|z-b_n|\ge\epsilon$ for all $n$. Then, $$ \sum_n\left|\frac{b_n-a_n}{z-b_n}\right|\le\frac{1}{\epsilon}\sum_n|b_n-a_n|<\infty\tag{1} $$ Inequality $(1)$ implies that $$ \prod_n\frac{z-a_n}{z-b_n}=\prod_n\left(1+\frac{b_n-a_n}{z-b_n}\right)\tag{2} $$ converges.

robjohn
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    I had Steven Krantz for both complex analysis and fourier analysis when he was at UCLA around 1980. He is an excellent teacher, one of the few math professors to get a Distinguished Teaching Award at UCLA. – robjohn Mar 07 '12 at 06:20
  • @robjohn how is this answer different than the one i provided. I guess other than the fact, that the largest set you found was $C- sup of b_{n}$ am i missing something here ? – Comic Book Guy Mar 07 '12 at 07:28
  • @robjohn should n't the open set be C- set of all values of $b_{n}$ ? – Comic Book Guy Mar 07 '12 at 07:35
  • @Hardy: $\mathbb{C}\setminus\overline{{b_n}}$ is the complement of the closure of the sequence. We want to exclude limit points of ${b_n}$ as well. – robjohn Mar 07 '12 at 07:44
  • @Hardy: your answer seems to assume that $\left|\frac{\beta_n - \alpha_n}{z - \beta_n}\right| < |\alpha_n - \beta_n|$, which is a difference. – robjohn Mar 07 '12 at 07:47
  • @robjohn In the interest of anyone else wanting to submit anything they have been working on i 'll keep this bounty open but no one does i shall grant thee the bounty in due time. – Comic Book Guy Mar 07 '12 at 09:53
  • @robjohn Suppose $\alpha_n=\beta_n$; then all of the factors in the product are identically $1$ and the largest open set where it converges is all of $\mathbb{C}$, regardless of what the sequence $(b_n)$ is. Is there any hypotheses over the sequences we may impose so that the largest domain is $\mathbb{C}\setminus \overline{{b_n}}$? – Fernando Martin Jul 05 '19 at 13:25
  • @robjohn After some thought, I think assuming $\alpha_n \neq \beta_n$ for all $n$ does the trick -- could you confirm? Thanks! – Fernando Martin Jul 05 '19 at 13:34
  • The product is not defined at any $z=b_n$; there are removable singularities at those points, so we can say that the function $1$ is equal everywhere but at $z=b_n$, but that is not the product. – robjohn Jul 05 '19 at 14:48
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I think you are very close.

The convergence criteria you are using is correct - consider this:

We observe that

$$|\beta_n - \alpha_n| = |\alpha_n - \beta_n| $$

so $$\left|\frac{\beta_n - \alpha_n}{z - \beta_n}\right| < |\alpha_n - \beta_n| $$ As you know that the absolute series converges as per the question's definition so $$\sum \left|\frac{\beta_n - \alpha_n}{z - \beta_n}\right| \leqslant \sum |\alpha_n - \beta_n| < \infty$$ so it converges by limit comparison theorem.

As per the largest open set, the condition needed for convergence is $\left|\frac{\beta_n - \alpha_n}{z - \beta_n} \right| \leqslant |\alpha_n - \beta_n| $ Solve for $z$.

Pedro
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Comic Book Guy
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  • Why is $\left| \frac{\beta_n-\alpha_n}{z-\beta_n}\right| < |\alpha_n-\beta_n|$ necessary for convergence? – Jose27 Mar 05 '12 at 04:35
  • @Jose27 As the author points out to get convergence we need $$\sum |f_n(z)|$$ u can also refer to http://en.wikipedia.org/wiki/Infinite_product if you doubt that. Under the information provided the only way to make this happen is the above condition although it shoudl be greater than equals i 'll update it. – Comic Book Guy Mar 05 '12 at 04:44
  • But a bound like $\left| \frac{\beta_n-\alpha_n}{z-\beta_n}\right| \leq C|\alpha_n-\beta_n|$ with $C$ not depending on $n$ would also ensure convergence (hence your condition is not necessary, only sufficient). – Jose27 Mar 05 '12 at 04:52
  • I guess i stand corrected u are right it should and the largest open set would be obtained when we solve for z and take the limit C goes to infinity – Comic Book Guy Mar 05 '12 at 05:10
  • I guess what i do n't like about this is that we end up with $|\beta_{n}| <= |z|$ which seems not an elegant answer. – Comic Book Guy Mar 05 '12 at 05:18