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Does there exists a function $f:[0,1]\to[0,1]$ such that:
$(1)$ $f$ is continuous everywhere.
$(2)$ $f$ is differentiable nowhere.
$(3)$ For every $c \in range(f)$, there is only a finite number of points $x_1,...,x_k$ in $[0,1]$ such that, $f(x_i) = c$ for every $i\in \{ 0,..,k \}$.

I saw a related post about A-continuous-nowhere-differentiable-but-invertible-function. Although I am not sure if we can use that result to prove the non-existence of the above function because I don't know any measure theory. I would be grateful if someone showed any old post that dealt with this condition.

Kripke Platek
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    Your condition (3) requires there to be no point in the range whose inverse image is an infinite set. It has been known for a long time (over 70 years) that every continuous nowhere differentiable function has the property that "most points" in the range are such that their inverse images have cardinality continuum -- see [12] in this mathoverflow answer. See also Banach's $(T1)$ condition on p. 277 in Theory of the Integral by Saks, and Theorem (6.2) on p. 278. – Dave L. Renfro Jun 01 '24 at 20:32
  • @DaveL.Renfro Ah ok. Thanks for the reference! – Kripke Platek Jun 04 '24 at 09:40

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