Prove that $\left| f'(x)\right| \leq \sqrt{2AC}$ using integration Suppose that $f(x)$ is a $C^2$ function on $\mathbb{R}$ such that $\left| f(x) \right| \leq A$ and $\left| f''(x) \right| \leq C $ for $x \in \mathbb{R}$.
Prove that $\left| f'(x)\right| \leq \sqrt{2AC}$.
 A: Let $h > 0$ and $x\in \Bbb R$. Then $$f(x + h) = f(x) + \int_0^{h} f'(x + t)\, dt$$ and by integration by parts, $$\int_0^h f'(x + t)\,dt = f'(x)h + \int_0^h (h - t)f''(x + t)\, dt.$$ By the second mean value theorem for integrals, $$\int_0^h (h - t)f''(x + t)\, dt = f''(c)\int_0^h (h - t)\, dt = \frac{f''(c)}{2}h^2$$ for some $c\in (x, x+h)$. Thus $$f(x + h) = f(x) + hf'(x) + \frac{f''(c)}{2}h^2$$ which implies $$f'(x) = \frac{f(x+h) - f(x)}{h} - \frac{f''(c)}{2}h.$$ Similarly, 
$$f(x - h) = f(x) - hf'(x) + \frac{f''(d)}{2}h^2$$
for some $d\in (x - h,x)$. Hence
$$\frac{f(x + h) - f(x - h)}{2h} = f'(x) + \frac{f''(c) - f''(d)}{4}h.$$
By hypothesis and the triangle inequality,
$$\lvert f'(x)\rvert \le \frac{A}{h} + \frac{Ch}{2}.$$ Since this inequality holds for all $h > 0$, $$\lvert f'(x)\rvert \le \inf_{h > 0} \left\{\frac{A}{h} + \frac{Ch}{2}\right\} = 2\sqrt{\frac{AC}{2}} = \sqrt{2AC}.$$ Since $x$ was arbitrary, the result follows.
A: Since $f''(x)$ is assumed to be continuous we have
$$f'(x+x')-f'(x)=\int_{x}^{x+x'} f''(t)\,dt \tag 1$$
If we integrate $(1)$ with respect to $x'$, then we obtain
$$\begin{align}
\int_{0}^{h} \left(f'(x+x')-f'(x)\right)\,dx'&=f(x+h)-f(x)-f'(x)h\\\\
&=\int_{0}^{h}\int_{x}^{x+x'}f''(t)\,dt\,dx'\\\\
&=\int_{x}^{x+h} f''(t)(x+h-t)\,dt \tag 2
\end{align}$$
Rearranging $(2)$ reveals that
$$f'(x)=\frac{f(x+h)-f(x)}{h}-\frac1h\int_{x}^{x+h} f''(t)(x+h-t)\,dt \tag 3$$
We can replace $h$ with $-h$ in the preceding analysis and obtain
$$f'(x)=\frac{f(x-h)-f(x)}{-h}+\frac1h\int_{x}^{x-h} f''(t)(x-h-t)\,dt \tag 4$$
whereupon adding $(3)$ and $(4)$ yields
$$2f'(x)=\frac{f(x+h)-f(x-h)}{h}+\frac1h\left(\int_{x}^{x-h} f''(t)(x-h-t)\,dt-\int_{x}^{x+h} f''(t)(x+h-t)\,dt\right) \tag 5$$
Now, we are given that $|f''(x)|\le C$ and $|f(x)|\le A$.  Therefore, using these bounds in $(5)$ yields
$$|f'(x)|\le \frac{A}{h}+\frac{Ch}{2} \tag 6$$
for all $h>0$.  The maximum value of the right-hand side of $(6)$ occurs at $h=\sqrt{\frac{2A}{C}}$ and is equal to 
$$\frac{A}{\sqrt{\frac{2A}{C}}}+\frac{C\sqrt{\frac{2A}{C}}}{2}=\sqrt{2AC}$$ 
Therefore, we have
$$|f'(x)|\le \sqrt{2AC}$$
as was to be shown!
