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Show converges uniformly on compact subsets of C

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data

    If α > 1, show: [itex] ∏ (1 - \frac{z}{n^α})[/itex] converges uniformly on compact subsets of ℂ.

    2. Relevant equations

    We say that ∏ fn converges uniformly on A if

    1. ∃n0 such that fn(z) ≠ 0, ∀n ≥ n0, ∀z ∈ A.

    2. {∏ fn} n=n0 to n0+0, converges uniformly on A to a non-vanishing function g such that ∃δ > 0 with |g| ≥ δ.

    Theorem: ∏ fn on A converges uniformly ⇔ ∀ε > 0, ∃n0 such that

    [itex] |∏ f_j(n) - 1| < ε[/itex], ∀n ≥ m ≥ n0, and z ∈ A. converges uniformly on compact subsets of ℂ.

    3. The attempt at a solution

    Of course, the limit of z/nα is 0, and there exists an n0 such that for all larger n fn ≠ 0, i.e. when |z/nα| < 1.

    Moreover, as n → ∞, each nth term in the term approaches 1, so the infinite product definitely converges, and I'd say that it satisfies the theorem given.
     
  2. jcsd
  3. Feb 23, 2015 #2

    RUber

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    Homework Helper

    If A is compact, does it have a least upper bound on |z|? That might be important to mention in your proof.
     
  4. Feb 26, 2015 #3

    Svein

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    This is correct with one caveat: The n0 you mention is dependent of z. Here you need the compactness: Since [itex] C\subset \bigcup_{N} \left\{n>N\Rightarrow \frac{z}{n^{\alpha}}<\epsilon\right\}[/itex], the compactness says that a finite number of these is sufficient. Take the maximum of these N, and you are done.
     
  5. Mar 1, 2015 #4
    Oh, yes. If it's compact, then it's close and bounded. Thanks for the refinement.
     
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