Prove that there is $x \in [0,1]$ such that $|f''(x)| > 4$ 
Assume $f$ has a continuous second derivative with $f(0) = f'(0) = f'(1) = 0$ and $f(1) = 1$. Prove that there is $x \in [0,1]$ such that $|f''(x)| > 4$. 

We must also have that $f'$ is continuous by differentiability. Therefore, by Rolles theorem there exists a $c$ in $[0,1]$ such that $f''(c) = 0$. I am not sure how to use the fact that $f(1) = 1$ to show that $|f''(x)| > 4$.
 A: $$f(\frac{1}{2}) = f(0) + f'(0)(\frac{1}{2}) + \frac{f''(\theta_1)}{2}(\frac{1}{2})^2 = f''(\theta_1)\frac{1}{8}$$
$$f(\frac{1}{2}) = f(1) - f'(1)(\frac{1}{2}) + \frac{f''(\theta_2)}{2}(\frac{1}{2})^2 = 1 + f''(\theta_2)\frac{1}{8}$$
Here, $0\le \theta_1 \le 1/2$, and $1/2 \le \theta_2 \le 1$
Combining both, we have
$$ 1 = f''(\theta_1)\frac{1}{8} - f''(\theta_2)\frac{1}{8}$$
Take absolute value,
$$ 1 \le max \{|f''(\theta_1)|, |f''(\theta_2)|\} \frac{1}{4}$$
So c is one of $\theta_1, \theta_2$.
Let 
$$f(x) = \begin{cases} 2x^2\, \text{ for } 0 \le x \le 1/2\\ -2(x-1)^2 +1\, \text{ for } 1/2 < x \le 1 \end{cases}$$
$$f''(x) = 4$$, so the equality holds for this f.
A: We'll  prove strict inequality by contradiction.
Set $g=f'$. By hypothesis, $g$ is $C^1$  on $[0,1]$, $\;g(0)=g(1)=0$ and $\displaystyle \int_0^1g(x)\,\mathrm d\mkern1mu x=1$.
Suppose $\lvert g'(x)\rvert\le 4$. This means


*

*$g'\le 4$, which implies $g(x)\le 4x$ on $\bigl[0,\frac12\bigr]$ so that
$$\int_0^{1/2}\!\!g(x)\,\mathrm d\mkern1mu x\le4\int_0^{1/2}x\,\mathrm d\mkern1mu x=\frac12.$$ Furthermore, we have equality if and only if $g(x)=4x$ on  $\bigl[0,\frac12\bigr]$.

*$g'\ge -4$, which implies $g(x)\le 4(1-x)$ on $\bigl[\frac12,1\bigr]$ so that $$\int_{1/2}^1g(x)\,\mathrm d\mkern1mu x\le4\int_{1/2}^1(1-x)\,\mathrm d\mkern1mu x=\frac12,$$ and we have equality if and only if $ g(x)=4((1-x) $ on $\bigl[\frac12,1\bigr]$.


Now, as $\displaystyle \int_0^1g(x)\,\mathrm d\mkern1mu x=1$, we do have equality in both cases. Thus $g$ is defined by
$$g(x)=\begin{cases}4x &\text{if}\enspace x\in\bigl[0,\frac12\bigr],\\
4(1-x)&\text{if}\enspace x\in\bigl[\frac12,1\bigr].
\end{cases}$$
This contradicts derivability of $g$ at $x=\frac12$.
A: We have
$$
\eqalign{
1&=f(1)-f(0)=\int_0^1 f'(x)\,dx\cr
&=\int_0^{1/2}f'(x)\,dx+\int_{1/2}^1 f'(x)\,dx\cr
&=\int_0^{1/2}\int_0^xf''(t)\,dt\,dx-\int_{1/2}^1 \int_x^1f''(t)\,dt\,dx\cr
&=\int_0^{1/2}(1/2-t)f''(t)\,dt-\int_{1/2}^1 (t-1/2)f''(t)\,dt.\cr
}
$$
Consequently, if $|f''(t)|\le 4$ for all $t\in[0,1]$, then by the above and the triangle inequality,
$$
\eqalign{
1&\le 4
\int_0^{1/2}(1/2-t)\,dt+4\int_{1/2}^1(t-1/2)\,dt\cr
&=-2(1/2-t)^2\Big|_0^{1/2}+2(t-1/2)^2\Big|_{1/2}^1\cr
&=1/2+1/2=1.\cr
}
$$
It follows that
$$
\int_0^{1/2}(1/2-t)|f''(t)|\,dt=\int_{1/2}^1 (t-1/2)|f''(t)|\,dt=1/2,
$$
which togetther with the hypothesis $|f''(t)|\le 4$ for all $t$ implies that $|f''(t)|=4$ for all $t\in[0,1]$. As $f''$ is continuous, we must have either $f''(t)=4$ for all $t$ or $f''(t)=-4$ for all $t$. In the first case $f(x)=2x^2$ in violation of $f(1)=1$. Likewise  the second alternative leads to a contradiction.
