- #1
Jonas Rist
- 7
- 0
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 don´t know how to go on :yuck:
I´d be glad to get some hints!
Thanks
Jonas
EDIT: Would it make sense to apply the Tayler series here?
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 don´t know how to go on :yuck:
I´d be glad to get some hints!
Thanks
Jonas
EDIT: Would it make sense to apply the Tayler series here?
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