Show converges uniformly on compact subsets of C

[Shackleford]

Homework Statement

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

Homework Equations

We say that ∏ f_n converges uniformly on A if

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

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

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

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

The Attempt at a Solution

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

Moreover, as n → ∞, each n^th term in the term approaches 1, so the infinite product definitely converges, and I'd say that it satisfies the theorem given.

 

[RUber]

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

 

[Svein]

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.

This is correct with one caveat: The n₀ you mention is dependent of z. Here you need the compactness: Since C\subset \bigcup_{N} \left\{n&gt;N\Rightarrow \frac{z}{n^{\alpha}}&lt;\epsilon\right\}, the compactness says that a finite number of these is sufficient. Take the maximum of these N, and you are done.

 

[Shackleford]

Oh, yes. If it's compact, then it's close and bounded. Thanks for the refinement.