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Let $f \in C^1([0,1])$ be a non-decrease function such that $f(0)=0, f(1)=1$.

Does there exist $[a,b] \subset [0,1]$ such that $\inf\limits_{x \in [a,b]}|f^{\prime} (x)| \geq \frac{1}{b-a}$?

rtybase
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user503375
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    Since function is non-decreasing then $f'(x)\geq0$ and if we assume such a segment $[a,b]$ exists, then from MVT $$0\leq \inf\limits_{x\in [a,b]}f'(x) \leq f'(c)=\frac{f(b)-f(a)}{b-a}\leq \sup\limits_{x\in [a,b]}f'(x)$$ but $$\inf\limits_{x\in [a,b]}f'(x) \leq f'(c)=\frac{f(b)-f(a)}{b-a}\leq\frac{1-0}{b-a}=\frac{1}{b-a}$$ Equality still may hold, though. – rtybase Nov 15 '17 at 19:01
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    @rtybase why are you posting this as a comment and not as an answer? – user190080 Nov 15 '17 at 19:11
  • @user190080 it may happen that OP committed a typo with the inequality. Just giving a chance to rectify ... – rtybase Nov 15 '17 at 19:32
  • @rtybase I think you found the key! And I was typing my counter-example in the mean time. I incorporated a reference to your comment if you don’t mind – user334639 Nov 15 '17 at 19:37

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Depends on $f$.

$f(x)=x^2$ is a counter example.

If $a>1/2$, the infimum is at most 2 and $b-a<1/2$ so the inequality is false.

If $a=0$, the infimum is zero so the inequality is definitely false.

If $0<a\le 1/2$, the infimum is at most 1 whereas $b-a<1$, so inequality is again false.

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Edit: in the mean time rtybase found the key: this is true only if $f$ stays constant equal zero, then increases straight to one, and then remains constant at 1. In this case $[a,b]$ will be exactly the interval in which the function grows linearly from 0 to 1. I let you work the details from rtybase’s comment until you prove this claim.

user334639
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