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Twice continuously differentiable function

by Jonas Rist
Tags: continuously, differentiable, function
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Jonas Rist
#1
May24-04, 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?
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kuengb
#2
May26-04, 11:04 AM
P: 117
What is h ?
matt grime
#3
May26-04, 11:25 AM
Sci Advisor
HW Helper
P: 9,396
Don't include x as the variable in your integral and as a limit, it will only confuse you unnecessarily.


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