Given that $f$ is $C^1$ with $f'$ absolutely continuous and $\vert f \vert \le 1$ and $\vert f'\vert, \vert f'' \vert \le 2$ show $$|f(b)-f(a)-(b-a)f'(a)|\le\frac{1}{2}(b-a)^2 \sup\vert f '' \vert.$$
From Taylor's theorem $f(b)=f(a)+(b-a)f'(a)+\frac{1}{2}(b-a)^2f''(a)+\cdots,$ so $$\vert f(b)-f(a)- (b-a)f'(a)\vert =\frac{1}{2}(b-a)f''(a)+\vert \cdots \vert$$
With no additional hypotheses on $\vert f^{(n)}\vert$, how can I show $$\frac{1}{2}(b-a)f''(a)+\vert \cdots \vert\le \frac{1}{2}(b-a)\sup \vert f''\vert ?$$
Note that, \begin{align*} f(b) -f(a) &= \int_a^{b} f'(v) dv \\ &= \int_a^{b} \big[f'(v) -f'(a) + f'(a) \big] dv \\ &= f'(a) (b-a) + \int_a^{b}\!\! \int_a^v f''(u)du dv\\ &= f'(a) (b-a) + \int_a^{b}\!\! \int_u^{b} f''(u)dv du\\ &= f'(a) (b-a) + \int_a^{b}f''(u)(b-u)du. \end{align*} Therefore, \begin{align*} |f(b) -f(a) - f'(a) (b-a)| &= \Big|\int_a^{b}f''(u)(b-u)du \Big|\\ &\le \int_a^b \sup|f''| \,(b-u)du\\ &=\frac{1}{2}(b-a)^2\sup|f''|. \end{align*}