TITLE: $f(0)=f'(0)=f'(1)=0$ and $f(1)=1$ implies $\max|f''|\geq 4$
QUESTION [12 upvotes]: Let $f\in C^2(\mathbb [0,1],\mathbb [0,1])$ such that
$f(0)=f'(0)=f'(1)=0$ and $f(1)=1$
Prove that  $\max_{[0,1]}|f''|\geq 4$

Progress
Applying Cauchy mean value theorem three times proves the existence of

$\xi\in (0,1)$ such that $f'(\xi)=1$
$\eta\in(\xi,1)$ such that $\displaystyle f''(\eta)=\frac{1}{\xi-1} <0$
$\beta\in(0,\xi)$ such that $\displaystyle f''(\beta)=\frac{1}{\xi}>0$

If $\displaystyle \xi\leq \frac{1}{4}$ or $\displaystyle \xi\geq \frac{3}{4}$, we're done.
What about other cases ?
I haven't used the continuity of $f''$ yet...

REPLY [8 votes]: Note that
$$f(1)=\int_0^1\left(\frac{1}{2}-t\right)f''(t)dt.$$
So
$$1\leq \left(\int_0^1\left\vert \frac{1}{2}-t\right\vert dt\right)\cdot \sup_{[0,1]}|f''|=\frac{1}{4} \sup_{[0,1]}|f''|.$$
and the desired conclusion follows.