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

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...

• just a vague idea but tricks of the sort $f(1) = f(0) + \int_0^1 f'(x)dx = f(0) + \int_0^1 \big( f'(0) + \int_0^x f''(t) dt \big) dx$ possibly applied to an other function like $g(x) = f(x)(1-x)$ can be worth giving a try. – Myself Jun 15 '14 at 17:05

## 3 Answers

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.

• This is very elegant ! – Gabriel Romon Jun 15 '14 at 19:13

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.

• Very nice. I fixed some notations and a typo. – Gabriel Romon Jun 15 '14 at 17:23

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

• Could you please explain why we are allowed to assume this? Couldn't $|f''(x)| < 4$ be satisfied in patches throughout $[0,1]$. Thanks! – MathMan Jul 28 '18 at 5:33
• @MathMan Write down carefully what you mean. My assumptions do not eliminate anything with $|f’’|<4$. – Ted Shifrin Jul 28 '18 at 5:38
• I am sorry,I still don't get it. Could you please explain this? – MathMan Oct 11 at 14:32