An example of a function for which the equality $M_1 = 2 \sqrt{M_0M_2}$ holds. Let $f$ be twice differentiable on $(a,\infty),a\in \Bbb R$ and let 
$$M_k = \sup \{|f^k(x)|\mid x \in (a, \infty) \} < \infty, k=0,1,2.$$
$a)$ Prove that $M_1 \leq 2 \sqrt{M_0M_2}$.
$b)$ Give an example of a function for which the equality $M_1 = 2 \sqrt{M_0M_2}$ holds.
I have done the first part. But help needed for the example in the second part!!
 A: This example is given in Rudin's mathematical analysis text. Let $a = -1$ and set
$$f(x) = \begin{cases}2x^2 - 1 & (-1 < x < 0)\\ \\\dfrac{x^2 - 1}{x^2 + 1} & (0 \le x < \infty)\end{cases}$$
Then $$f'(x) = \begin{cases}4x & (-1 < x < 0)\\ \\ \dfrac{4x}{(x^2 + 1)^2} & (0 \le x < \infty)\end{cases}$$ and $$f''(x) = \begin{cases} 4 & (-1 < x \le 0)\\ \\ \dfrac{1 - 3x^2}{(x^2 + 1)^3} &  (0 < x < \infty)\end{cases}$$
Now, $|f| \le 1$, $|f'| \le 4$, and $|f''| \le 4$, with $|f(0)| = 1$, $\lim_{x\to -1} |f'(x)| = 4$, and $|f''(0)| = 4$. Therefore $M_0 = 1$, $M_1 = 4$, and $M_2 = 4$. Then $$2\sqrt{M_0M_2} = 2\sqrt{(1)(4)} = 2(2) = 4 = M_1$$
A: For $a,b,c>0$, consider the functions $f(x)=a\arctan(bx)+c$ where $x\in (0,\infty)$. 
Note that $M_0=\frac{a\pi}{2}+c$ and $M_1=ab$. To find $M_2$, calculate the points where $f'''(x)=0$. You will find two points, one for $x<0$ and the other one is $x=\frac{1}{\sqrt{3}b}$. 
Substitute it in $f''$ to find that $M_2=\frac{9ab^2}{8\sqrt{3}}$. Now you can play with the equation $$a^2b^2=4\left( \frac{a\pi}{2}+c\right)\frac{9ab^2}{8\sqrt{3}}.$$ 
