Twice continuously differentiable function

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SUMMARY

The discussion centers on proving bounds for the derivative of a twice continuously differentiable function \( f: [0, \infty) \rightarrow \mathbb{R} \) where both \( f \) and \( f'' \) are bounded. The key inequalities to demonstrate are \( |f'(x)| \leq \frac{2}{h} \|f\|_{\infty} + \frac{2}{h} \|f''\|_{\infty} \) for all \( x \geq 0 \) and \( h > 0 \), and \( \|f'\|_{\infty} \leq 2(\|f\|_{\infty})^{\frac{1}{2}}(\|f''\|_{\infty})^{\frac{1}{2}} \). The initial approach involves integrating \( f''(x) \) but the user seeks further guidance on progressing with the proof.

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Mathematicians, students studying real analysis, and anyone interested in understanding the behavior of derivatives of bounded functions.

Jonas Rist
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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
I´d be glad to get some hints!
Thanks
Jonas

EDIT: Would it make sense to apply the Tayler series here?
 
Last edited:
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What is h ?
 
Don't include x as the variable in your integral and as a limit, it will only confuse you unnecessarily.
 

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