a simple method for comparing real functions

Theorem:

Let $f(x)$ and $g(x)$ be real-valued, twice differentiablefunctions 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 differentiablefunction 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.∎