# proof of criterion for convexity II

If $f$ was not convex, then there was a point $\xi \in (a,b)$ such
that $f(\xi )>h(x)=\frac{f(v)-f(u)}{v-u}(x-u)+f(u)$ for some $$ in
$(a,b)$. Since $f$ is continuous^{}, there would be a neighborhood^{}
$(\xi -\delta ,\xi +\delta ),\delta >0$, of $\xi $ such that $f(x)>h(x)$
for all $x$ in this neighborhood. (I.e., $f(x)$ was “above” the
line segment joining $f(u)$ and $f(v)$.) Let $s=\xi -\delta ,t=\xi +\delta $.

Using the two points $A=(s,f(s)),B=(t,f(t))$, we construct another line segment $\overline{AB}$ whose equation is given by $g(x)=\frac{f(s)-f(t)}{2\delta}(x-s)+f(s)$; we have $f(x)>g(x)$ for $x\in (s,t)$. In particular,

$$f(\xi )=f\left(\frac{s+t}{2}\right)>g(\xi )=\frac{f(s)+f(t)}{2}.$$ | (1) |

(One easily verifies $g(\xi )=(f(s)+f(t))/2$.) This contradicts
hypothesis^{}.

Note that we have tacitly used the fact that $h(x)=\lambda f(v)+(1-\lambda )f(u)$ for some $\lambda $ and $g(x)=\lambda f(s)+(1-\lambda )f(t)$ for some $\lambda $.

Title | proof of criterion for convexity II |
---|---|

Canonical name | ProofOfCriterionForConvexityII |

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

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

Owner | yesitis (13730) |

Last modified by | yesitis (13730) |

Numerical id | 5 |

Author | yesitis (13730) |

Entry type | Proof |

Classification | msc 52A41 |

Classification | msc 26A51 |

Classification | msc 26B25 |