TITLE: If $f(0)=f(1)=f(2)=0$, $\forall x, \exists c, f(x)=\frac{1}{6}x(x-1)(x-2)f'''(c)$
QUESTION [8 upvotes]: Let $f:[0,2]\to \mathbb R$ be a $C^3$ function such that $f(0)=f(1)=f(2)=0$
Prove that $\forall x\in[0,2], \exists c\in[0,2], f(x)=\frac{1}{6}x(x-1)(x-2)f'''(c)$

This problem got me stuck. I guess one has to use the mean value theorem at some point.
Using Rolle, there exists $0<\xi_1<\eta<\xi_2$ such that $f'(\xi_1)=f'(\xi_2)=0$ and $f''(\eta)=0$.

REPLY [7 votes]: Let $a \in [0, 2]$. We want to show that there exists a
$c\in[0,2]$ (depending on $a$) such that
$$ f(a)=a(a-1)(a-2) \, \frac{f'''(c)}{3!} \, .\tag 1$$
For $a = 0, 1, 2$ we can choose any $c$, so we assume $a \ne 0, 1, 2$
in the following. (The idea is to compare $f$ with a cubic polynomial
which interpolates $f$ at $0, 1, 2$ and at $a$.)
Define the function $g:[0,2]\to \mathbb R$ by
$$
g(x) = a(a-1)(a-2) \, f(x) - x(x-1)(x-2) \, f(a) \, .
$$ 
Then
$$
g'''(x) = a(a-1)(a-2) \, f'''(x) - 3! \, f(a) \tag 2
$$
and
$$
g(0) = g(1) = g(2) = g(a) = 0 \, .
$$
Repeated application of Rolle's theorem shows that there is
a $c\in[0,2]$ such that
$$
 g'''(c) = 0 \, .
$$
Now $(1)$ follows from setting  $x = c$ in $(2)$.