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Let $f: [0,1] \rightarrow \mathbb{R}$ be a continuous and weakly increasing function. Suppose $f$ is not a constant. Can we conclude that there exists $0 \leq \alpha < \beta \leq 1$ such that $f$ is strictly increasing over $(\alpha, \beta)$?

This result seems to me intuitive, but I do not know how to prove it formally. Thank you for any help.

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  • $\begingroup$ What's weakly increasing function? $\endgroup$ – asdf Dec 30 '17 at 18:07
  • $\begingroup$ No, you cannot conclude that there is such an interval. Look up devil's staircase. $\endgroup$ – Andrés E. Caicedo Dec 30 '17 at 18:18
  • $\begingroup$ See math.stackexchange.com/questions/1395960/… $\endgroup$ – Robert Z Dec 30 '17 at 18:18
  • $\begingroup$ @asdf It means $f(t) \geq f(s)$ if $t > s$. Strictly increasing function means $f(t) > f(s)$ if $t > s$. $\endgroup$ – user295959 Dec 30 '17 at 18:18

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