TITLE: Proving bounded first derivative
QUESTION [5 upvotes]: I am trying to prove that if a function $f:\mathbb{R} \to \mathbb{R}$ is bounded and the second derivative $f''$ is bounded, then the first derivative $f'$ is also bounded. My hint is to use Taylor's theorem.
I know for any $x$ and $y$, $f(y) = f(x) + f'(x)(y-x) + \frac{1}{2} f''(c)(y-x)^2,$ is  the Taylor approximation where $c$ is some point in between $x$ and $y$.  From this I get
$$f'(x)(y-x) = f(y) - f(x) - \frac{1}{2}f''(c)(y-x)^2.$$
If $|f(x)| < C$ and $|f''(x)| < D$ are bounds, then
$$|f'(x)| < \frac{|f(x)| + |f(y)|}{|y-x|} + \frac{1}{2} |f''(c)| |y-x| < \frac{2C}{|x-y|}+ \frac{D}{2}|x-y|.$$
I seem to be stuck at this point because both terms can be unbounded depending on $x$ and $y$.
Thank you for any help.

REPLY [2 votes]: Here is a solution with Tyler's expansion:
$$
\forall x\in\mathbb{R},\forall h>0,\\\
f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(\eta_1)\text{ ,where $\eta_1\in(x,x+h)$}\\
f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(\eta_2)\text{ ,where $\eta_2\in(x-h,x)$}\\
\Rightarrow f(x+h)-f(x-h)=2hf'(x)+\frac{h^2}{2}[f''(\eta_1)+f''(\eta_2)]\\
\text{Hence,  }|2hf'(x)|\leq|f(x+h)-f(x-h)|+|\frac{h^2}{2}[f''(\eta_1)+f''(\eta_2)]|\leq\\ 2\sup|f|+h^2\sup |f''|
$$
We deduce that: $$2\sup|f'|\leq\frac{2\sup|f|}{h}+h\sup|f''|$$
Since $\sup|f^{(k)}|$ is a constant. We can let 
$$h=\sqrt{\dfrac{2\sup|f|}{\sup|f''|}}$$
We deduce that $$\boxed{\sup |f'| \le  \sqrt{2}\cdot \sqrt{\sup |f| \cdot \sup|f''|}}$$
The first derivative is bounded since the function and its second derivative are bounded.