# convex functions lie above their supporting lines

Let $f:\mathbf{R}\to \mathbf{R}$ be a convex, twice differentiable function on $[a,b]$. Then $f(x)$ lies above its supporting lines, i.e. it’s greater than any tangent line in $[a,b]$.

###### Proof.

:

Let $r(x)=f\left({x}_{0}\right)+{f}^{\prime}\left({x}_{0}\right)\left(x-{x}_{0}\right)$ be the tangent of $f(x)$ in $x={x}_{0}\in [a,b].$

By Taylor theorem, with remainder in Lagrange form, one has, for any $x\in [a,b]$:

$$f\left(x\right)=f\left({x}_{0}\right)+{f}^{\prime}\left({x}_{0}\right)\left(x-{x}_{0}\right)+\frac{1}{2}{f}^{{}^{\prime \prime}}\left(\xi \left(x\right)\right){\left(x-{x}_{0}\right)}^{2}$$ |

with $\xi \left(x\right)\in [a,b]$. Then

$$f\left(x\right)-r(x)=\frac{1}{2}{f}^{{}^{\prime \prime}}\left(\xi \left(x\right)\right){\left(x-{x}_{0}\right)}^{2}\ge 0$$ |

since ${f}^{{}^{\prime \prime}}\left(\xi \left(x\right)\right)\ge 0$ by convexity. ∎

Title | convex functions lie above their supporting lines |
---|---|

Canonical name | ConvexFunctionsLieAboveTheirSupportingLines |

Date of creation | 2013-03-22 16:59:20 |

Last modified on | 2013-03-22 16:59:20 |

Owner | Andrea Ambrosio (7332) |

Last modified by | Andrea Ambrosio (7332) |

Numerical id | 5 |

Author | Andrea Ambrosio (7332) |

Entry type | Result |

Classification | msc 52A41 |

Classification | msc 26A51 |

Classification | msc 26B25 |