# Show that $|f'(x)| \le \frac{2M_0}{h} +\frac{hM_2}{2}$ and $M_1 \le 2\sqrt{M_0M_2}$

Let $f$ be a twice derivable function and $M_i =\sup_{x \in \mathbb{R}} |f^{(i)}(x)|$ and $|M_0|, |M_2|<\infty$. Preferably using the Taylor series on the interval $[x,x+h]$ show the following inequalities:

(a) $$|f'(x)| \le \frac{2M_0}{h} +\frac{hM_2}{2}$$

and

(b) $$M_1 \le 2\sqrt{M_0M_2}$$

Any hints and/or a solution would be appreciated.

P.S. I know that $(b) => (a)$ using the AM-GM inequality.

$$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(a_h)$$ and $$f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(b_h)$$
Then, for all $h\ne 0$, we have $$f'(x)=\frac{f(x+h)-f(x-h)}{2h}+\frac{h}{2}(f''(a_h)+f''(b_h))$$
Now you can use the definition of $M_i$ for (a) and variation for (b).
• Hey, thanks Krokop, after looking at your hint (and using it to solve the problem) i just realised that there is even no need to expand $f(x-h)$ but only $f(x+h)$ because the neccesary upper bound of $\frac{f(x+h)-f(x-h)}{2h}$ is the same as the one of $\frac{f(x+h)-f(x)}{2h}$. Similar argument applies to the second derivative part. fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-ash4/t1.0-9/… P.S.I also just realised that all i had to do in the end was look at the equality case in the said AM-GM and set h accordingly. – Marko Karbevski May 25 '14 at 19:16