Let $f : \mathbb R \to \mathbb R$ be a differentiable function such that $f'(x) \geq 0, \forall x \in \mathbb R$. Suppose $f'(x) = 0$ at at-most countable points. Is $f$ strictly increasing? (maybe $\lambda$-almost everywhere?, where $\lambda$ is Lebesgue measure)

I define $U = \{(x,y) \in \mathbb R^2: x < y \text{ and } f(x) = f(y) \}$, then I want to prove that $U$ is at most countable (thus its Lebesgue measure is $0$). Is my approach correct?

  • 2
    $\begingroup$ You are trying too hard. Phone a calculus student friend who will tell you the function increasing. If it is not strictly increasing then it must be constant on some interval, etc. $\endgroup$ – B. S. Thomson Mar 10 '16 at 22:00
  • $\begingroup$ I can imagine a differentiable function which is strictly increasing with $f'(k) = 0$ for all $k\in\mathbb Z$. So, the answer to the question is: Yes, $f$ can be strictly increasing. $\endgroup$ – Friedrich Philipp Mar 10 '16 at 22:03

The function is strictly increasing.

Suppose $x<y$, but $f(x)=f(y)$. Pick $c$ such that $x<z<y$. Suppose that $f(x)>f(z)$. Then applying the Mean Value Theorem, there exists a point $a$ with $x<a<z$ with $f'(a)<0$. This is a contradiction, so $f(x)\leq f(c)$. Likewise, $f(z)\leq f(y)$, so $f(x)=f(z)=f(y)$. We conclude that $f$ is constant on $[x,y]$, a contradiction.

  • $\begingroup$ I just make it overcomplicated... $\endgroup$ – SiXUlm Mar 10 '16 at 22:05
  • $\begingroup$ Analysis teaches you to second guess things that seem obviously true because often they are actually false. But sometimes, they are still true. I think the second guessing is good, though. $\endgroup$ – Alex S Mar 10 '16 at 22:07
  • $\begingroup$ From $x < z < y$, you may conclude directly that $f(x) \leq f(z)$ (because I already assume that the function is increasing). $\endgroup$ – SiXUlm Mar 10 '16 at 22:12

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