Let $I ⊂ \mathbb{R}$ be a compact interval. Show that for every $f ∈ C^2(I , \mathbb{R})$ it exists a $C > 0$ such that:

$|f(a)+f(b)-2f(\frac{a+b}{2})|≤C(b-a)^2 $

$∀a, b ∈ I .$

Let $a,b ∈I$ and $a<b$, it exists by definition a $ξ_1∈(a, (a+b)/2)$ with

$f(a)=f(\frac{a+b}{2})+f'(\frac{a+b}{2})(\frac{a-b}{2})+\frac{1}{2}f''(ξ_1)\frac{(a-b)^2}{4}$

and it exists a $ξ_2∈((a+b)/2, b)$ with

$f(b)=f(\frac{a+b}{2})+f'(\frac{a+b}{2})(\frac{b-a}{2})+\frac{1}{2}f''(ξ_2)\frac{(b-a)^2}{4}$

Summing the two identities we get

$f(a)+f(b)-2f(\frac{a+b}{2})=\frac{f''(ξ_1)+f''(ξ_2)}{8}(b-a)^2$

How can I say that the first part of the right side is equal to $C$ and introduce the absolute value with ≤?

Hint : Any continuous function $f $ from a compact space to $\Bbb R $ is bounded and attains its bounds.

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