Does there exist bounded function with a bounded first derivative but unbounded second derivative? I know that a bounded function with a bounded second derivative also has a bounded first derivative.
But consider a bounded function $f(x)$ such that $f'(x)$ is also bounded.
Can we claim that $f''(x)$ will also be bounded. Intuition says NO !
But I am unable to find an example. Please help.
 A: The bounded analytic function $g: \mathbb{R} \rightarrow \mathbb{R}$ defined by $g(x)=\dfrac{\text{sin}(x^3)}{x^2+1}$, for all $x\in\mathbb{R},$ have the first derivative bounded. The derivatives are:
$$
\begin{align}
g'(x) = & \dfrac{x (3 (x + x^3) \text{cos}(x^3) - 2 \text{sin}(x^3)))}{(1 + x^2)^2} \\
g''(x)= & - \dfrac{6 x (x^4 - 1) \text{cos}(x^3) + (9 x^8 + 18 x^6 + 9 x^4 - 6 x^2 + 2) \text{sin}(x^3) }{(1 + x^2)^3},
\end{align}
$$ for all $x \in \mathbb{R}.$
The first derivative is bounded because the largest power multiplying the trigonometric functions in the numerator is 4, which is the same of the degree of the denominator. This means that the limit of these polynomial terms exists, implying that $g'(x)$ is bounded, since sine and cosine are bounded functions.
The second derivative is unbounded because the denominator have degree $6$ while the polynomial term following the sine have degree $8$, two degrees higher. Choosing a sequence of real numbers $\{x^k\}_{k\in\mathbb{N}}$ such that $\sin(x^{k})=1$ and $\cos(x^{k})=0$ diverging to infinity is enough for proving the unboundedness of the second derivative.
