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twice continuously differentiable function |
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| May24-04, 04:25 PM | #1 |
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twice continuously differentiable function
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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| May26-04, 11:04 AM | #2 |
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What is h ?
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| May26-04, 11:25 AM | #3 |
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Recognitions:
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Don't include x as the variable in your integral and as a limit, it will only confuse you unnecessarily.
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