$|f''(x)|\leq 1$, $\exists\ x_1\neq x_2, \in [0,1]$ such that $f(x_1)=f(x_2)=0$. Show that $|f(x)|\leq 1, \forall\ x\in [0,1].$ 
$|f''(x)|\leq 1$, $\exists\ x_1\neq x_2, \in [0,1]$ such that $f(x_1)=f(x_2)=0$. Show that $|f(x)|\leq 1, \forall\ x\in [0,1].$

If $x_1=0, x_2=1$, then Taylor expansion would help. But here we only know that $x_1\neq x_2$! What this helps?
 A: $f(x_1)=f(x_2)=0$ implies that $f'(x_0) = 0$ for some $x_0$ between $x_1$ and $x_2$. Now use the mean-value theorem twice: First,
$$
 f'(x) = f'(x_0) + (x-x_0) f''(\xi)
$$
shows that $|f'(x)| \le 1$ for  all $x \in [0, 1]$. Then
$$
 f(x) = f(x_1) + (x-x_1) f'(\eta)
$$ shows that $|f(x)| \le 1$ for  all $x \in [0, 1]$.

But a better estimate is possible, one can actually show that $|f(x)| < 1/2$:
For $c \in [0, 1]$, $c \ne x_1, x_2$, consider the function
$$
 g(x) = (c-x_1)(c-x_2)f(x) - (x-x_1)(x-x_2)f(c) \, .
$$
Then $g(x_1) = g(x_2) = g(c) = 0$, and repeated application of Rolle's theorem shows that $g''(\xi) = 0$ for some $\xi$. It follows that
$$
 0 = g''(\xi) = (c-x_1)(c-x_2)f''(\xi)- 2 f(c)
$$
and therefore
$$
 |f(c)| = \left| \frac 12 (c-x_1)(c-x_2)f''(\xi)\right| < \frac 12 \, .
$$
This is the best possible bound, as can be seen by considering the functions $f_\epsilon(x) = \frac 12 x(x-\epsilon)$ with $\epsilon$ close to zero.
A: Since $f(x_1)=f(x_2)$ then by Rolle's theorem $f'(c) = 0$ for some $c$ between $x_1$ and $x_2$.
If $f''$ is continuous then we have $ \left| f'(x) \right|=\left| \int_{c}^{x} f''(t)dt \right|\leq \left|x-c\right|$ for all $x \in [0,1]$.
Similarly ,  $ \left| f(x) \right|=\left| \int_{x_1}^{x} f'(t)dt \right|\leq | \int_{x_1}^{x}  \left| t-c \right| dt |\leq \frac{1}{2}(c-x_1)^2+\frac{1}{2}(x-c)^2\leq 1 $.
