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Let $f:[0,1] \rightarrow \mathbb{R}$ be continuous in $[0,1]$ and differentiable in $(0,1)$, such that $f(0)=0.$ Prove that there is $x_0\in [0,1]$ such that $|f'(x_0)|\geq |f(x)|$ for all $x\in [0,1].$

Attempt. If we assume, by contradition, that for all $x\in [0,1]$ there is $z_x\in [0,1]$ such that: $$|f'(x)|<|f(z_x)|$$ then for all $n$ there is $z_n\in [0,1]$ such that $|f'(1/n)|<|f(z_n)|$. Even if we pass to a convergent subsequence, we do not get a contradiction, since $f'$ is not differentiable at $0$. I assume that this is not the right path to a solution.

Thanks in advance for the help.

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  • $\begingroup$ For some context, this is more than just an arbitrary exercise. It's a special case of $L^p$ monotonicity on $[0,1]$, since $\|f\|_{\infty} \le \|f'\|_1 \le \|f'\|_{\infty}$. $\endgroup$ – Shalop May 30 '18 at 22:07
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Since $|f|$ is continuous on $[0,1]$, it attains its maximum at some $c \in [0,1]$.

By MVT we have

$$|f(x)| \le |f(c)| = |f(c) - f(0)| = |f'(x_0)||c - 0| = |f'(x_0)||c| \le |f'(x_0)|, \forall x \in [0,1]$$

for some $x_0\in (0,c)$.

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