TITLE: How to prove this statement? (Real analysis)
QUESTION [6 upvotes]: This might be the basic question in real analysis.
A function $f$   is $ C^2 $ function on the closed interval$ [0,1]$
Also the function $ f $ is satisfying $ f(0) = f(1) =0 $ 
Plus, $\vert f''(x) \vert  \le \  A $ on the open interval $(0,1)$
Show $\vert f'(x) \vert  \le \frac  A2 $  on the interval $(0,1]$
I tried many times through the Rolle's thm, Mean value thm etc.
But failed. Please give me some hints.

REPLY [6 votes]: Two Taylor expansions at $x\in(0,1)$ are
\begin{eqnarray}
0&=&f(0)=f(x)+f'(x)(0-x)+\frac{f''(a)}{2}x^2,\\
0&=&f(1)=f(x)+f'(x)(1-x)+\frac{f''(b)}{2}(1-x)^2.
\end{eqnarray}
Here $a,b\in(0,1)$, hence $|f''(a)|\le A$ and $|f''(b)|\le A$. 
The second one minus the first one gives
$$
f'(x)=\frac{f''(a)}{2}x^2-\frac{f''(b)}{2}(1-x)^2.
$$
Now estimate
$$
|f'(x)|\le\frac{A}{2}(\underbrace{x^2+(1-x)^2}_{\le 1})\le\frac{A}{2}.
$$
The function $f'(x)$ is continuous, so the estimate can be extended to the closed interval.