# proof of Weierstrass’ criterion of uniform convergence

The assumption that $|{f}_{n}(x)|\le {M}_{n}$ 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 $\u03f5>0$. Then there exists $K$ such that $n>K$ implies $$. Now, if $k>K$,

$$ |

The $\u03f5$ 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 |