TITLE: Prove that $|f'(x)| \le \frac{A}2 \forall x \in [0,1] $
QUESTION [7 upvotes]: Let $f$ be twice differentiable in $[0,1]$
$f(0) = f(1) = 0$,
$|f''(x)|\le A$.
Prove that $|f'(x)| \le  \frac{A}2, \forall x \in [0,1] $.

Well this is what I came up with,
$f'(c_1) = \dfrac{f(x) -f(0)}{x-0} = \dfrac{f(x)}{x}$
$f'(c_2) = \dfrac{f(x) -f(1)}{x-1} = \dfrac{f(x)}{x-1}$
for $0\lt c_1,c_2 \lt 1$
from rolle's theorem we know that there is : $0\lt c_3 \lt 1$
such that
$f'(c_3) =0$

REPLY [3 votes]: I manage an answer using both time a taylor expansion center at a point $x \in \left [0, 1 \right ]$. I get 
$$
f(h) = f(x)+f'(x)(h-x) + \frac{1}{2} f''(\xi(h))(h-x)^2
$$
now for h = 0 and h = 1, I get 
$$0=f(x)-xf'(x)+\frac{1}{2}x^2f''(\xi(0))$$
and
$$0=f(x)+f'(x)-xf'(x)+\frac{1}{2}(1-x)^2f''(\xi(1)).$$
Subtracting one to the other to get $f(x)$ out and a little manipulation yield:
$$|f'(x)| = \frac{1}{2}|x^2f''(\xi(0))-(1-x)^2f''(\xi(1))|\leq\frac{1}{2}(|x^2f''(\xi(0))|+|(1-x)^2f''(\xi(1))|)\leq \frac{A}{2}(|x^2|+|1-x|^2) $$
since $x \in \left [0, 1 \right ], (|x^2|+|1-x|^2) \leq1 $ and we get the result.