# continuity of convex functions, alternate proof

Let $f$ be convex and $y\in (a,b)$ be arbitrary but fixed. Then

$f(\lambda x+(1-\lambda )y)$ | $\le $ | $\lambda f(x)+(1-\lambda )f(y)$ | (1) | ||

$f(\lambda x+(1-\lambda )y)-f(y)$ | $\le $ | $\lambda (f(x)-f(y))\le \lambda |f(x)-f(y)|.$ | (2) |

Fix a number $c>sup\{|f(u)-f(v)|:u,v\in (a,b)\}$. Then

$$ | (3) |

Given $\u03f5>0$, let $\lambda $ range over $(0,\u03f5/c)$ if $$, or $\lambda =1$ otherwise. Then it is easy to see that $f(\lambda x+(1-\lambda )y)$ and $f(y)$ lie within $\u03f5$ distance of each other when $\lambda $ varies as specified.

Continuity of $f$ now follows–for $$, the left-hand limit equals $f(y)$ and for $$, the right-hand limit also equals $f(y)$, hence the limit is $f(y)$.

Title | continuity of convex functions, alternate proof |
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Canonical name | ContinuityOfConvexFunctionsAlternateProof |

Date of creation | 2013-03-22 18:25:28 |

Last modified on | 2013-03-22 18:25:28 |

Owner | yesitis (13730) |

Last modified by | yesitis (13730) |

Numerical id | 4 |

Author | yesitis (13730) |

Entry type | Proof |

Classification | msc 26B25 |

Classification | msc 26A51 |