# one-sided continuity by series

Theorem. If the function series

$\sum _{n=1}^{\mathrm{\infty}}}{f}_{n}(x)$ | (1) |

is uniformly convergent on the interval $[a,b]$, on which the ${f}_{n}(x)$ are continuous^{} from the right or from the left, then the sum function $S(x)$ of the series has the same property.

Proof. Suppose that the terms ${f}_{n}(x)$ are continuous from the right. Let $\epsilon $ be any positive number and

$$S(x):={S}_{n}(x)+{R}_{n+1}(x),$$ |

where ${S}_{n}(x)$ is the ${n}^{\mathrm{th}}$ partial sum of (1) ($n=\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots}$). The uniform convergence implies the existence of a number ${n}_{\epsilon}$ such that on the whole interval we have

$$ |

Let now $n>{n}_{\epsilon}$ and ${x}_{0},{x}_{0}+h\in [a,b]$ with $h>0$. Since every ${f}_{n}(x)$ is continuous from the right in ${x}_{0}$, the same is true for the finite sum ${S}_{n}(x)$, and therefore there exists a number ${\delta}_{\epsilon}$ such that

$$ |

Thus we obtain that

$|S({x}_{0}+h)-S({x}_{0})|$ | $\mathrm{\hspace{0.33em}}=|[{S}_{n}({x}_{0}+h)-{S}_{n}({x}_{0})]+{R}_{n+1}({x}_{0}+h)-{R}_{n+1}({x}_{0}|$ | ||

$\mathrm{\hspace{0.33em}}\leqq |{S}_{n}({x}_{0}+h)-{S}_{n}({x}_{0})|+|{R}_{n+1}({x}_{0}+h)|+|{R}_{n+1}({x}_{0})|$ | |||

$$ |

as soon as

$$ |

This means that $S$ is continuous from the right in an arbitrary point ${x}_{0}$ of $[a,b]$.

Analogously, one can prove the assertion concerning the continuity from the left.

Title | one-sided continuity by series |
---|---|

Canonical name | OnesidedContinuityBySeries |

Date of creation | 2013-03-22 18:34:03 |

Last modified on | 2013-03-22 18:34:03 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 10 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 40A30 |

Classification | msc 26A03 |

Synonym | one-sided continuity of series with terms one-sidedly continuous |

Related topic | OneSidedContinuity |

Related topic | SumFunctionOfSeries |