proof of Weierstrass’ criterion of uniform convergence

The assumption that $|f_{n}(x)|\leq M_{n}$ for every $x$ guarantees that each numerical series $\sum_{n}f_{n}(x)$ converges absolutely. We call the limit $f(x)$ .

To see that the convergence is uniform: let $\epsilon>0$ . Then there exists $K$ such that $n>K$ implies $\sum_{n>K}M_{n}<\epsilon$ . Now, if $k>K$ ,

|f(x)-\sum_{n=1}^{k}f_{n}(x)|=|\sum_{n>k}f_{n}(x)|\leq\sum_{n>k}|f_{n}(x)|\leq% \sum_{n>k}M_{n}<\epsilon

The $\epsilon$ does not depend on $x$ , so the convergence is uniform.

Title	proof of Weierstrass’ criterion of uniform convergence
Canonical name	ProofOfWeierstrassCriterionOfUniformConvergence
Date of creation	2013-03-22 16:26:28
Last modified on	2013-03-22 16:26:28
Owner	argerami (15454)
Last modified by	argerami (15454)
Numerical id	4
Author	argerami (15454)
Entry type	Proof
Classification	msc 40A30
Classification	msc 26A15