Problem on Rolle's Theorem I need to get hint/solution for the following problem:

Let $f(x)$ defined in $[0,1]$ be twice differentiable such that $$|f''(x)| \leq 1$$ for all $x$ belonging to $[0,1]$. If $$f(0) = f(1)$$ show that $$|f(x)| < 1$$ for all $x$ belonging to $[0,1]$.

I tried like this:
Integrating  $|f''(x)| \leq 1$
we get
$$|f'(x)| \leq x$$
and since $x \leqslant 1$
$$|f'(x)| \leq 1$$
Integrating again,
$$|f(x)| \leq x$$
and since $x \leqslant 1$
$$|f(x)| \leq 1$$
Is there any better approach? Am I doing anything wrong?
 A: I am also assuming you ment $f(0)=f(1)=0$.
Proof by contradiction:


*

*Suppose $f(x_0)=a \geq 1$ ($0<x_0<1$), then
$\frac{f(x_0)-f(0)}{x_0}=\frac a {x_0} > 1$.

*By Langrange there is $x_1$ such that $f'(x_1)=b > 1$.

*By Rolle's theorem there is a stationary point: $f'(x_s)=0$. Let's compute $|\frac{f'(x_1)-f'(x_s)}{x_1-x_s}|=|\frac {b} {x_1-x_s}| > 1$.

*By Lagrange there is $x_2$ with $|f''(x_2)| > 1$, contradiction.

A: This is the first half of the solution:
Since $f(0) =0= f(1)$ the function attains a maximum or minimum in $[0,1]$, wlog a maximum. In that place $x_0$, $f^\prime(x_0)=0$. So for any $t$ in $[0,1]$ 
$$|f^\prime(t)| = |\int_t^{x_0} f^{\prime\prime}(s) ds|\le |\int_t^{x_0}1ds|\le |x_0-t|$$
(note that for the inequlity in the middle actually two steps are needed).
Can you see how to proceed now?
A: I'm assuming you meant $f(0)=f(1)=0$. If $f$ is not $\equiv0$ then there is  a point  $\xi\in\>]0,1[\>$ where $|f|$ takes a positive maximum. One then has $f'(\xi)=0$, and by the MVT we can conclude that
$$|f'(t)|=|f'(t)-f'(\xi)|\leq\sup_\tau|f''(\tau)|\cdot |t-\xi|\leq|t-\xi|\qquad(0\leq t\leq1)\ .$$
It follows that
$$|f(x)-f(\xi)|=\left|\int_\xi^x f'(t)\>dt\right|\leq\int_\xi^x |t-\xi|\>|dt|={1\over2}|x-\xi|^2\ .$$
As $f(0)=f(1)=0$ this implies
$$|f(\xi)|\leq{1\over2}\min\{\xi^2,(1-\xi)^2\}\leq{1\over8}\ .$$
The function
$$f(x):={1\over8}-{1\over2}\left(x-{1\over2}\right)^2$$
shows that the obtained bound cannot be improved.
