Suppose f is twice differentiable, $f'(0)=f'(1)=0$, $f(0)=0$ and $f(1)=1$. Prove that $\exists x \in [0;1]: |f''(x)| > 4$ This is exercise 39 in Chapter 11 of Calculus by Spivak.
Suppose f is twice differentiable, $f'(0)=f'(1)=0$, $f(0)=0$ and $f(1)=1$. Prove that $\exists x \in [0;1]: |f''(x)| > 4$

So part $(a)$ only requires you to prove $|f''(x)| \geq 4$, which can be handled.
Here's how i do it:
Let's assume that $f''(x) < 4$ on $[0;\frac{1}{2}]$ and $f''(x) > -4$ on $[\frac{1}{2};1]$.
Now:
$$\forall x\in (0;\frac{1}{2}]: \exists x_0\in(0;x): f''(x_0) = \frac{f'(x) -f'(0)}{x-0} < 4 $$. Or: $f'(x) <4x,\forall x\in(0;\frac{1}{2}]$
Consider $g(x) = f(x) - 2x^2$.
So, $g'(x) = f'(x) - 4x$, which is $\leq 0$ and takes zero only once on $[0;\frac{1}{2}]$. We conclude that $g$ is decreasing on $[0;\frac{1}{2}]$. Therefore, $g(x) < g(0) = 0$. Or: $f(x) < 2x^2$. Choosing $x=\frac{1}{2},$ we have: $f(\frac{1}{2}) < \frac{1}{2}$.
In addition to that, consider $g(x) = 1- f(1-x)$ on $[0;\frac{1}{2}]$, we have the same thing: $g(\frac{1}{2}) < \frac{1}{2}$, or: $f(\frac{1}{2}) > \frac{1}{2}$
So, $|f|$ must be $\geq 4$ at some point in one of the invervals
However, it goes tricky when it comes to prove $|f''(x)| > 4$.
Let's assume that $f′′(x)\leq 4$ on $[0;\frac{1}{2}]$ and $f′′(x)\geq −4$ on $[\frac{1}{2};1] (*)$. Three cases to be considered.
Case 1:
$f''$ constantly takes the value of $4$ and $-4$ respectively on $[0;\frac{1}{2}]$ and $[\frac{1}{2};1]$
This is absurd since $f''(\frac{1}{2})  = 4 = -4$
Case 2: There exists some $x$ in $[0;\frac{1}{2}]$ that $f''(x) < 4$
Case 3: There exists some $x$ in $[\frac{1}{2};1]$ that $f''(x) > -4$
We will try to exclude case 2 and case 3 will be trivial.
The idea is to show: $f(\frac{1}{2}) < \frac{1}{2}$ just like part $(a)$ and $f(\frac{1}{2}) \geq \frac{1}{2}$. It's the idea from the answer book and i'm trying to follow it.

$f'(x) \leq 4x,\forall x\in[0;\frac{1}{2}]$ by $(*)$, Mean value theorem and the fact that $f'(0) = 0$ $(**)$
If we had $f'(x) = 4x, \forall 0\leq x\leq \frac{1}{2}$ then clearly $f''(x) = 4, \forall 0\leq x\leq \frac{1}{2}$, which is absurd due the condition of case 2. So:
$$\exists x_0 \in (0;\frac{1}{2}]: f'(x_0) < 4x_0$$. And $\forall \frac{1}{2} \geq x>x_0$: Applying Mean-value theorem into $[x_0;x]$, we have: $$\frac{f'(x) - f'(x_0)}{x-x_0}\leq 4$$ or: $f'(x) - 4x \leq f'(x_0) - 4x_0 <0 $. Hence: "$f'(x) < 4x$ for larger x in $(0;\frac{1}{2}]$". But this time, when we consider $g(x) = f(x) - 2x^2$, we have nothing like $g(0)$ to compare with! We do have $(**)$ but the number of zeros might be infinite, so we can't know whether f is decreasing! I'm trying to extend the fact somehow to $[0;x_0)$. I'd like to hear your opinions on how to proceed.
 A: Assume that $|f''(x)| \le 4$ for all $x \in [0, 1]$.
Using the mean-value theorem we determine an upper bound for $f'$: On the interval $[0, 1/2]$ we have
$$
 f'(x) = \underbrace{f'(0)}_{=0} + (x-0) \underbrace{f''(c)}_{\le 4} \le 4x
$$
and on the interval $[1/2, 1]$ we have
$$
 f'(x) = \underbrace{f'(1)}_{=0} + (x-1) \underbrace{f''(d)}_{\ge -4} \le -4(x-1) = 4(1-x) \, .
$$
So we have
$$ \tag{*}
 f'(x) \le \begin{cases}
 4x & \text{ for } 0 \le x \le 1/2 \, ,\\
 4(1-x) & \text{ for } 1/2 \le x \le 1  \, .
\end{cases}
$$
Then
$$ 
1 = f(1) - f(0) = \int_0^1 f'(x) \, dx \underset{(**)}{\le} \int_0^{1/2} 4x \, dx + 
\int_{1/2}^1 4(1-x) \, dx = \frac 12  + \frac 12 = 1
$$
so that equality holds at $(**)$. Since $f'$ is continuous, it follows that equality holds in $(*)$ for every $x \in [0, 1]$.
But then $f''(1/2)$ does not exist, contrary to the assumption that $f$ is twice differentiable.
A: So, consider this set:
$$ A=  \{x\in (0;\frac{1}{2}]|f'(x) -4x<0\}$$
We know this set isn't empty since $\exists x_0 \in (0;\frac{1}{2}]: f'(x_0) < 4x_0$. Furthermore, $A$ bounded below by $0$. So we must have: $\exists \alpha \geq 0 = infA$.
This means: $\forall \epsilon>0: \exists x_0 \in A: 0<x-\alpha<\epsilon$.
Therefore:
If $\alpha = 0$ then for every such $x_0$, we can prove: $\forall x\in(x_0;\frac{1}{2}]: f'(x) <4x$. Or: $f'(x) -4x$ must $<$ $0$ on $(0;\frac{1}{2}]$, and consequently on $[0;\frac{1}{2}]$, we have: $g(0) > g(\frac{1}{2})$ since $g$ is decreasing.
Else, $\alpha >0$, similarly, we have: $g$ is decreasing on $[\alpha;\frac{1}{2}]$ but remains constant on $[0;\alpha)$.
