# a simple method for comparing real functions

Let $f(x)$ and $g(x)$ be real-valued, twice differentiable functions on $[a,b]$, and let ${x}_{0}$ $\in [a,b]$.

If $f({x}_{0})=g({x}_{0})$, ${f}^{\prime}({x}_{0})={g}^{\prime}({x}_{0})$, ${f}^{\prime \prime}(x)\le {g}^{\prime \prime}(x)$ for all $x$ in $[a,b]$, then $f(x)\le g(x)$ for all $x$ in $[a,b]$.

###### Proof.

Let $h(x)=g(x)-f(x)$; by our hypotheses, $h(x)$ is a twice differentiable function on $[a,b]$, and by the Taylor formula with Lagrange form remainder (http://planetmath.org/RemainderVariousFormulas) one has for any $x\in [a,b]$:

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

where $\xi =\xi (x)\in [x,{x}_{0}]$.

Then by hypotheses,

$h({x}_{0})$ | $=$ | $g({x}_{0})-f({x}_{0})=0$ | ||

${h}^{\prime}({x}_{0})$ | $=$ | ${g}^{\prime}({x}_{0})-{f}^{\prime}({x}_{0})=0$ | ||

${h}^{\prime \prime}(\xi )$ | $=$ | ${g}^{\prime \prime}(\xi )-{f}^{\prime \prime}(\xi )\ge 0$ |

so that

$$h(x)=\frac{1}{2}{h}^{\prime \prime}(\xi ){(x-{x}_{0})}^{2}\ge 0$$ |

whence the thesis. ∎

Title | a simple method for comparing real functions |
---|---|

Canonical name | ASimpleMethodForComparingRealFunctions |

Date of creation | 2013-03-22 16:10:47 |

Last modified on | 2013-03-22 16:10:47 |

Owner | Andrea Ambrosio (7332) |

Last modified by | Andrea Ambrosio (7332) |

Numerical id | 10 |

Author | Andrea Ambrosio (7332) |

Entry type | Result |

Classification | msc 60E15 |