TITLE: A function $f$ twice differentiable, such that $f$ and $f''$ are bounded imply $f'$ bounded?
QUESTION [5 upvotes]: Let $f:(0,\infty) \to \mathbb{R} $ twice differentiable. Suppose $A$, $C$ in $(0,\infty)$ such that for each $x>0$ we have $|f(x)|<A$ and $|f''(x)|<C$. Prove that for each $x>0$ and each $h>0$ we have $|f'(x)| \le (A/h)$+$Ch$
Using Mean Value Theorem
$f'(x)-f'(y) = f''(c) (x-y)$ for some $c$. Since $f''$ is bounded by $C$
$f'(x)-f'(y) \le  C (x-y)$. Letting $h=x-y$ we got
$f'(x) \le  Ch + f'(y)$
How can I bound $f'(y)$ by $A/h$ to obtain the desired bound??

REPLY [5 votes]: Since $f$ is twice differentiable, we can use the MVT to write
$$f(x+h)=f(x)+f'(x)h+\frac12 f''(\xi)h^2$$
for some $x<\xi<x+h$.  Rearraning we see that  
$$f'(x)=\left(\frac{f(x+h)-f(x)}{h}\right)-\frac12 f''(\xi)h$$
Then, 
$$\begin{align}
|f'(x)|&\le\left|\frac{f(x+h)-f(x)}{h}\right|+\left|\frac12 f''(\xi)\right|h \\
&\le \frac{1}{h}\left(|f(x+h)|+|f(x)|\right)+\frac12 h|f''(\xi)|\\
&\le 2(A/h)+\frac{1}{2}Ch
\end{align}$$