Find the supremum of the following set (differential inequalities) Let $X=\Bbb{R}^{\Bbb{R}}\cap C^{2}$ that is the set of all functions $f:\Bbb{R}\to\Bbb{R}$ for which the second derivative exists on $\Bbb{R}$.
Let
$$
 A_f = \Vert f \Vert_\infty = \sup \{ |f(x)| : x \in \Bbb R \} \, ,\\
 B_f = \Vert f' \Vert_\infty = \sup \{ |f'(x)| : x \in \Bbb R \} \, , \\
 C_f = \Vert f'' \Vert_\infty = \sup \{ |f''(x)| : x \in \Bbb R \} \, .
$$
The task is to find
$$
 M = \sup \left\{ \frac{B_f^2}{A_f \, C_f} : f \in X; A_f, C_f < \infty \right\} \, .
$$
Very rough approximations that ignore the limits in definitions of derivatives indicate that the answer might be $\infty$ but that doesn't sit right with my intuition and I haven't been able to find an array of functions for which $\dfrac{B_f^2}{A_f\, C_f}$ could get arbitrarily large.
Any hints would be appreciated.
EDIT:
In Prove $\sup \left| f'\left( x\right) \right| ^{2}\leqslant 4\sup \left| f\left( x\right) \right| \sup \left| f''\left( x\right) \right| $ the upper bound $M \le 4$ is proven. An example of a function with $M=4$ would be sufficient :) 
 A: For bounded, twice differentiable functions $f: \Bbb R \to \Bbb R$ with bounded second derivative we have
$$
 \sup_{x \in \Bbb R} |f'(x)|^2 \leq 2 \sup_{x \in \Bbb R} |f(x)| \cdot \sup_{x \in \Bbb R} |f''(x)| 
$$
and the factor $2$ is best possible, i.e. $M=2$ is the sought supremum.
This is the case $k=1$, $n=2$ of the Landau–Kolmogorov inequality and was first proved by Edmund Landau in

Landau, E. (1913). "Ungleichungen für zweimal differenzierbare Funktionen". Proc. London Math. Soc. 13: 43–49.

First note that it suffices to prove the following (“Satz 4” and “Satz 5” in Landau's article):

Theorem: Let $f: \Bbb R \to \Bbb R$ be twice differentiable with $|f(x)| \le 1$ and $|f''(x)| \le 1$ for all $x \in \Bbb R$. Then $|f'(x)| \le \sqrt 2$ for all $x \in \Bbb R$. The constant $\sqrt 2$ cannot be replaced by a smaller value.

For the general case replace $f(x)$ by
$$
 \tilde f(x) = \frac{f(\sqrt{A_f/C_f}x)}{A_f} \, .
$$
Proof of the theorem: For $x \in \Bbb R$ we have, using Taylor's theorem,
$$
 f(x + \sqrt 2) = f(x) + \sqrt 2 f'(x) + f''(\xi_1) \, ,\\
 f(x - \sqrt 2) = f(x) - \sqrt 2 f'(x) + f''(\xi_2) \, .
$$
with some $\xi_1, \xi_2 \in \Bbb R$. By taking the difference of these two equations we get
$$
 2 \sqrt 2 |f'(x)| = |f(x + \sqrt 2) - f(x - \sqrt 2) - f''(\xi_1) + f''(\xi_2)| \le 4 \\
 \implies |f'(x)| \le \sqrt 2 \, .
 $$
It remains to show that the bound $\sqrt 2$ is best possible. Landau gives the following example: For arbitrary $\epsilon \in (0, \sqrt 2)$ define $h, f: [0, \sqrt 2] \to \Bbb R$ as
$$
 h(x) = \min(x / \epsilon, 1)
$$
and
$$
 f(x) = \int_0^x \int_t^{\sqrt 2} h(u) \, du dt \, .
$$
$f$ is then

*

*extended to a function on $[\sqrt 2, 2 \sqrt 2]$ by defining $f(2 \sqrt 2 - x) = f(x)$,

*extended to a function on $[- 2\sqrt 2, 2 \sqrt 2]$ by defining $f(x) = -f(x)$,

*and finally extended to all of $\Bbb R$ as a $4 \sqrt 2$-periodic function.

Elementary calculations show that
$$
 |f(x)| \le f(\sqrt 2) = \int_0^{\sqrt 2} \int_t^{\sqrt 2} h(u) \, du dt
 \le \int_0^{\sqrt 2} \int_t^{\sqrt 2} 1 \, du dt = 1 \, 
$$
$$
 |f''(x)| \le |h(x)| \le 1 \, ,
$$
and
$$
 |f'(x)| = \int_0^\sqrt 2 h(u) \, du \ge \sqrt 2 - \epsilon \, .
$$
for $x=0$ (and all integer multiples of $2 \sqrt 2$). This completes the proof.
