Prove or disprove: if $f:(0,1)\to\mathbb{R}$ is twice differentiable and $f$ and $f'$ are bounded then $f''$ is bounded. I believe this is not true and I wonder if this is a correct counterexample:
Define $F: [0,1] \to \mathbb{R}$ by $F(0)=0$ and $F(x)=\sin (1/x)$ elsewhere. Then define $f(x)=\int_{0}^{x} F$. Then $f$ is differentiable and bounded on $(0,1)$ and $f'(x)=\sin (1/x)$. On the other hand, $f''(x)=-(1/x^2) \cos (1/x)$.
Is this counterexample correct and or is there any flaw in my reasoning? 
Is there a function $f$ not defined in terms of integrals which does not satisfy the given property?

$f(x)=x^{3/2}$ is an example of a fucntion such that $f$ and $f'$ are bounded but $f''$ is not. 
