Suppose $f$ is two times differentiable in $(0,\infty)$ and that:
$|f(x)| \leq M_{0}, \forall x>0$;
$|f''(x)| \leq M_{2}, \forall x>0$.
a) Show that $$|f'(x)| \leq \frac{2}{h} M_{0} + \frac{h}{2}M_{2}, \forall h>0$$ b) Demonstrate that, for all $x>0$, $$ |f'(x)| \leq 2\sqrt{M_{0}M_{2}}$$
What I tried to do:
- For the first item, supposing $0<a<t<x$: $$\left |f(x) - [f(a) + f'(a)(x-a)] \right | = \left | \frac{f''(t)}{2}(x-a)^2 \right |$$
(That's merely the error comitted in such an approximation, in the Lagrange's form)
If we denote $h = (x-a) > 0$ $$\left |f(x) - [f(a) + f'(a)h] \right | = \left | \frac{f''(t)}{2}h^2 \right |$$
This would probably lead to the answer, but I cannot follow from here. I have tried many times and I won't get the expected inequality. For example, I tried to use the triangle inequality (without success):
$$\left |f(x) - [f(a) + f'(a)h] \right | \leq |f(x)| + |f(a)| + h|f'(a)| \leq 2M_{0} + h|f'(a)|$$
- For the second item: I don't have a clue from were this result might come. Does someone have at least a hint for some identity/inequality that could be useful?
You are very close to the solution for part (a), indeed $$f(x) = f(x_0) + f'(x_0)(x - x_0) + \frac{f''(\xi)}{2}(x-x_0)^2.$$ Rearranging and defining h as $x-x_0$ we have $$f'(x_0) = \frac{f(x) - f(x_0)}{h} - \frac{f''(\xi)}{2}h$$ which implies $$|f'(x_0)| \le \frac{2M_0}{h} + \frac{M_2}{2}h.$$
For part (b) define $g(h) = \frac{2M_0}{h} + \frac{M_2}{2}h$ and minimize with respect to $h$.