The title might be misleading, but whether such a function exists is what boggles me about the following problem:

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function such that for all $a<b $ satisfying $f(a)=f(b)$, there exists $c$ in $(a,b)$ such that $f(a)=f(c)=f(b)$.

Prove that $f$ is monotonous on $\mathbb{R}$.

What I'm intuiting about this problem is that for every such pair $a, b$ the function f is constant on $[a,b]$. Therefore, $f$ could have no local extremum. However, I'm not sure how to go about proving this. Is the set $X=\{x\in [a,b]| f(x)=f(a)\}$ be dense in $[a,b]$? If it is, how could I prove it? What also troubles me, though, is the existence of nowhere-monotone functions such as the Weierstrass function. Does the Weierstrass function satisfy the problem condition?

Furthermore, I'd like to be able to prove that an arbitrary horizontal line $g(x)=u, u \in \mathbb{R}$ either intersects f at a single point, or at a compact interval $[a_{1},b_{1}]$ $(a_{1}<b_{1})$.

I'm not sure if these two conditions are enough to prove that the function is monotonous.

What is the best approach towards proving this problem? Am I on the right track?


Preimage $f^{-1}(v)$ of any value $v$ is a closed set, hence its complement $U(v)$ is open. This open set $U(v)$ is a disjoint union of intervals. If some interval is finite, say $(a,b)$, then $f(a)=f(b)=v$, but $f(c)\ne v$ for $a<c<b$. So, all intervals in $U(v)$ are infinite. Hence preimage of $v$ is connected: it is either a segment (possibly a single point), or a ray. Now we prove that $f$ is monotone. It suffices to prove that $f$ is monotone on any three points $\{a<b<c\}$. Assume the contrary, for example, $f(b)>v:=\max(f(a),f(c))$. We see that $f^{-1}(v)$ is not connected: it contains points on $[a,b)$ and $(b,c]$, but does not contain $b$.

  • $\begingroup$ Not sure about the formal definition of a ray, but the preimage of $v$ could also be the whole real line. Anyway, brilliant solution! $\endgroup$ – Matija Sreckovic Feb 12 '16 at 10:37
  • $\begingroup$ open ray is $(-\infty,a)$ or $(a,+\infty)$. Yes, preimage may be also empty or the whole line. $\endgroup$ – Fedor Petrov Feb 12 '16 at 11:02

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.