Prove that $\int_0^1|f''(x)|dx\ge4.$ Let $f$ be a $C^2$ function on $[0,1]$. $f(0)=f(1)=f'(0)=0,f'(1)=1.$ Prove that
$$\int_0^1|f''(x)| \, dx\ge4.$$
Also determine all possible $f$ when equality occurs.
 A: The inequality does not hold. Take
$$
f(x) = x^3 - x^2
$$
we have
$$
f'(x) = 3x^2 - 2x \\
f''(x) = 6x - 2 \\
f(0) = f(1) = f'(0) = 0 \\
f'(1) = 1
$$
But
$$
\int_0^1 \lvert f''(x)\rvert dx = \int_0^{1/3} (2 - 6x) dx + \int_{1/3}^1 (6x - 2)dx = 1/3 + 4/3 = 5/3 < 4
$$
Then, what can we say about the infimum of that integral?
$$
\int_0^1 \lvert f''(x) \rvert dx \geq \int_0^1 f''(x) dx = f'(1) - f'(0) = 1
$$
To show that $1$ is the best lower bound, choose a $C^2$ function $h$ defined on $[0, 1/2]$ such that
$$
h(0) = h'(0) = h'(1/2) = h''(1/2) = 0 \\
h(1/2) = -1
$$
Using the above function we can construct the map
$$
f(x) =
\begin{cases}
kh(x) & \text{if }0\leq x \leq 1/2 \\
-k & \text{if }1/2 < x \leq 1 - 3k \\
\frac {(x - 1 +3k)^3}{27k^2} - k & \text{if }1 - 3k < x \leq 1 
\end{cases}
$$
where $k$ is a positive constant lesser than $1/6$. $f$ satisfies all the constraints and
$$
\int_0^1 \lvert f''(x) \rvert dx = 1 + k\int_0^{1/2} \lvert h''(x) \rvert dx
$$
Choosing $k$ small enough we can make the integral as close to $1$ as we want.
