# 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 $f(x_{0})=g(x_{0})$, $f^{\prime}(x_{0})=g^{\prime}(x_{0})$, $f^{\prime\prime}(x)\leq g^{\prime\prime}(x)$ for all $x$ in $[a,b]$, then $f(x)\leq g(x)$ 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,

 $\displaystyle h(x_{0})$ $\displaystyle=$ $\displaystyle g(x_{0})-f(x_{0})=0$ $\displaystyle h^{\prime}(x_{0})$ $\displaystyle=$ $\displaystyle g^{\prime}(x_{0})-f^{\prime}(x_{0})=0$ $\displaystyle h^{\prime\prime}(\xi)$ $\displaystyle=$ $\displaystyle g^{\prime\prime}(\xi)-f^{\prime\prime}(\xi)\geq 0$

so that

 $h(x)=\frac{1}{2}h^{\prime\prime}(\xi)(x-x_{0})^{2}\geq 0$

whence the thesis. ∎

Title a simple method for comparing real functions ASimpleMethodForComparingRealFunctions 2013-03-22 16:10:47 2013-03-22 16:10:47 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 10 Andrea Ambrosio (7332) Result msc 60E15