# Taylor polynomials on a compact interval

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 $$a, it exists by definition a $$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.