Show converges uniformly on compact subsets of C

1. Feb 22, 2015

Shackleford

1. The problem statement, all variables and given/known data

If α > 1, show: $∏ (1 - \frac{z}{n^α})$ 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

$|∏ f_j(n) - 1| < ε$, ∀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. Feb 23, 2015

RUber

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

3. Feb 26, 2015

Svein

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

4. Mar 1, 2015

Shackleford

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