a simple method for comparing real functions

Theorem:

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