proof of Weierstrass’ criterion of uniform convergence


The assumption that |fn(x)|Mn for every x guarantees that each numerical series nfn(x) converges absolutely. We call the limit f(x).

To see that the convergence is uniform: let ϵ>0. Then there exists K such that n>K implies n>KMn<ϵ. Now, if k>K,

|f(x)-n=1kfn(x)|=|n>kfn(x)|n>k|fn(x)|n>kMn<ϵ

The ϵ does not depend on x, so the convergence is uniform.

Title proof of Weierstrass’ criterion of uniform convergenceMathworldPlanetmath
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