If nonnegative $f: [0,1] \rightarrow \mathbb{R}$ has a continuous $f''$, then $\int_0^1 \Big| \frac{f''(x)}{f(x)} \Big| \,dx >4$

Since, $f$ is positive on the interval $(0,1)$ it suffices to prove that,

$\displaystyle \int_0^1 |f''(t)|\,dt \ge 4\max\limits_{x \in [0,1]} |f(x)|$

Since, $f(0) = f(1) = 0$ using integration by parts we have:

$\displaystyle f(x) = -(1-x)\int_0^x tf''(t)\,dt - x\int_x^1 (1-t)f''(t)\,dt$ for all $x \in [0,1]$.

Denoting $\displaystyle \phi(t) := \begin{cases} (x-1)t & \textrm{ for } t \in [0,x] \\ x(t-1) & \textrm{ for } t \in [x,1]\end{cases}$

We can write the above identity as, $\displaystyle f(x) = \int_0^1 \phi(t)f''(t)\,dt$

Thus, $\displaystyle |f(x)| \le \max\limits_{t \in [0,1]} |\phi(t)|\int_0^1 |f''(t)|\,dt = x(1-x)\int_0^1 |f''(t)|\,dt \le \frac{1}{4}\int_0^1 |f''(t)|\,dt$

for all $x \in [0,1]$.

Hence, $\displaystyle \int_0^1 |f''(t)|\,dt \ge 4\max\limits_{x \in [0,1]} |f(x)|$ as desired.

Integration by parts:

$\displaystyle f(x) - f(0) = \int_0^x f'(t)\,dt = xf'(x) - \int_0^x tf''(t)\,dt\tag{1}$

$\displaystyle \begin{align} f(1) - f(x) = \int_x^1 f'(t)\,dt &= f'(1) - xf'(x) - \int_x^1 tf''(t)\,dt \\ &= (1-x)f'(x) + \int_x^1 (1-t)f''(t)\,dt \tag{2} \end{align}$

Multiplying $(1)$ with $(1-x)$ and $(2)$ with $x$, and subtracting, we get:

$$\displaystyle f(x) = -(1-x)\int_0^x tf''(t)\,dt - x\int_x^1 (1-t)f''(t)\,dt$$