$f(0)=f'(0)=f'(1)=0$ and $f(1)=1$ implies $\max|f''|\geq 4$

Solution 1:

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.

Solution 2:

Assume $|f''(x)|\le 4-\epsilon$ for all $x\in[0,1]$ with some $\epsilon>0$. Then for $x\in[0,1]$ $$ |f'(x)| \le \int_0^x |f''(t)|dt \le (4-\epsilon)x $$ and $$ |f'(x)| \le \int_x^1 |f''(t)|dt \le (4-\epsilon)(1-x). $$ This proves $|f'(x)| \le (4-\epsilon) \min(x,1-x)$. Then $$ |f(1)|\le \int_0^1 |f'(x)|dx \leq \int_0^{1/2} (4-\epsilon)x dx + \int_{1/2}^1 (4-\epsilon)(1-x)dx= \frac{4-\epsilon}4<1, $$ a contradiction.

Solution 3:

HINT: Suppose that $f''(x)<4$ for $0\le x\le 1/2$ and, symmetrically, that $f''(x)>-4$ for $1/2\le x\le 1$.