0

Suppose $f''(x)$ exists ($f(x)$ can be differentiated two times)

And the function and the second derivative is bounded : $\left|f(x)\right|\le P$, $\left|f''(x)\right|\le Q$

Then, how can I prove $\left|f'(x)\right|\le \sqrt{2PQ}$ ?

user127.0.0.1
  • 7,435
Chanhee Jeong
  • 1,048

1 Answers1

1

Note that by Taylor theorem, for each $h>0$ and for each $x$ there exist $c$ in $(x,x+2h)$ such that $$ f(x+2h)= f(x)+ 2hf'(x)+2h^2f''(c) $$ Note that this implies that $$ \vert f'(x)\vert \leq\frac{1}{2h} \lvert f(x+2h)-f(x)\vert + h\vert f'(c)\vert \leq \frac{P}{h}+Qh $$ Since this is valid for every $h>0$, it's valid for the minimum of the application $h\mapsto \frac{P}{h}+Qh$ and this minimum happens at $h=\sqrt{PQ^{-1}}$. Just evaluate at this point to obtain your result.

Brandon
  • 3,245
  • 2
  • 15
  • 23
  • Indeed, but i'm afraid it only claims that |f'(x)| <= 2*sqrt(PQ), not <= sqrt(2PQ) – Chanhee Jeong Jan 16 '14 at 06:04
  • Well, actually I think there's a mistake, because there are functions for which the equality $M= 2\sqrt{PQ}$ holds where $M$ is the least upper bound of the values of $\vert f'\vert$(I made a reference to the exercise 15 of chapter 5 of Rudin PMA) – Brandon Jan 16 '14 at 06:15