twice continuously differentiable functionby Jonas Rist Tags: continuously, differentiable, function 

#1
May2404, 04:25 PM

P: 7

Hello again,
another problem: given: a function [tex] f:[0,\infty)\rightarrow\mathbb{R},f\in C^2(\mathbb{R}^+,\mathbb{R})\\ [/tex] The Derivatives [tex] f,f''\\ [/tex] are bounded. It is to proof that [tex] \rvert f'(x)\rvert\le\frac{2}{h}\rvert\rvert f\rvert\rvert_{\infty}+\frac{2}{h}\rvert\lvert f''\rvert\rvert_{\infty}\\ [/tex] [tex]\forall x\ge 0,h>0\\ [/tex] and: [tex] \rvert\rvert f'\rvert\rvert_{\infty}\le 2(\rvert\rvert f\rvert\rvert_{\infty})^{\frac{1}{2}}(\rvert\rvert f''\rvert\rvert_{\infty})^{\frac{1}{2}}\\ [/tex] I began like this: [tex] f'(x)=\int_{0}^{x}f''(x)dx\Rightarrow [/tex] [tex] \rvert f'(x)\rvert\le\rvert\int_{0}^{x}f''(x)dx\rvert\le\int_{0}^{x}\rvert f''(x)\rvert dx [/tex] But then already I donīt know how to go on Iīd be glad to get some hints! Thanks Jonas EDIT: Would it make sense to apply the Tayler series here? 



#2
May2604, 11:04 AM

P: 117

What is h ?




#3
May2604, 11:25 AM

Sci Advisor
HW Helper
P: 9,398

Don't include x as the variable in your integral and as a limit, it will only confuse you unnecessarily.



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