Algotssleeddynf

2022-01-05

I cannot prove that if $f\left(x\right)$ is convex on $\left[a,b\right]$ then
$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx\le \frac{f\left(a\right)+f\left(b\right)}{2}$

### Answer & Explanation

otoplilp1

The inequality
$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx$
is a special case of Jensens

sukljama2

For convenience, we can take
For the first inequality, use
$f\left(0\right)\le \frac{f\left(x\right)+f\left(-x\right)}{2}$
and integrate both sides from $x=-1$ to 1
For the second, use
$f\left(x\right)=f\left(\frac{1-x}{2}\left(-1\right)+\frac{1+x}{2}\left(1\right)\right)\le \frac{1-x}{2}f\left(-1\right)+\frac{1+x}{2}f\left(1\right)$
and integrate both sides from $x=-1$ to 1.

star233

Since f is convex ob [a,b], one has:

Therefore, one gets:
${\int }_{a}^{b}f\left(x\right)dx\le \frac{f\left(b\right)-f\left(a\right)}{b-a}\frac{\left(b-a{\right)}^{2}}{2}+f\left(a\right)\left(b-a\right)$
Finally, one has:
$\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx\le \frac{f\left(b\right)+f\left(a\right)}{2}$

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