Is a continuous real function with vanishing derivative in all but countably many points constant?

Question

The Cantor function is a standard example of a function $f:[0,1]\to\mathbb{R}$ that is continuous, has almost everywhere zero derivative, and nonetheless is not constant. More specifically, the number of points in which it is not differentiable is uncountable (with the cardinality of the continuum).

I'm trying to get a better understanding of which features allow it to not be constant while still satisfying $f'(x)=0$ almost everywhere in a compact set.

I've seen a comment in this answer hinting at the fact that if a continuous function is differentiable with zero derivative in all but a countable number of points, then it must be constant. This would indicate that the uncountable number of non-differentiable points is indeed a crucial feature of the Cantor function (certainly not surprising, but still not obvious). The post Is the set of non-differentiable points for a singular continuous function nowhere dense? also seems highly relevant, but I can't parse the answer enough to figure out whether it addresses my particular point.

However, that comment cites a link that seems to be broken, and I haven't found this statement mentioned in other places. Is this true? If so, how would I prove it?

Answer 1

Yes, this is true. I try to outline a proof. The main work is in the following Lemma:

Lemma: If $u:[0,1] \to \mathbb{R}$ is continuous and if $u'(t) \le 0$ except at a countable set $C\subseteq [0, 1]$, then $u$ is nonincreasing on $[0,1]$.

Proof. It suffices to show that $u(t) \le u(0)$ $(t \in [0, 1])$. Assume not, then there exists a $t_0 \in (0, 1)$ such that $u(t_0) - u(0) =: p > 0$. For $\varepsilon >0$ define $$ h_\varepsilon(t) := u(t) - u(0) - \varepsilon t. $$ We have $h_\varepsilon(0)=0$, $h_\varepsilon(t_0)=p- \varepsilon t_0$ and $h_\varepsilon$ is continuous on $[0, 1]$.

We fix some $q \in (0,p)$. Then we have: $$ \text{If} ~~ \varepsilon < \frac{p-q}{t_0} ~~\text{then} ~~ h_\varepsilon(t_0)= p- \varepsilon t_0> p-(p-q)=q. $$ Now, for such an $\varepsilon$ set $$ M:=\{t \in [0,t_0): h_\varepsilon(t)\le q\}. $$ Note that $0 \in M$ and that $t_0$ is an upper bound of $M$. Set $s:=s(\varepsilon):=\sup M$. Now $$ 0 < s < t_0 < 1, \quad h_\varepsilon(s)=q, $$ and $$ (\ast) \quad \quad \forall t \in (s,t_0]: ~ h_\varepsilon(t)>q. $$ If $s \notin C$, then $u'(s) \le 0$ and $(\ast)$ yields $h_\varepsilon'(s)\ge 0$. On the other hand $u'(s) \le 0$ implies $h_\varepsilon'(s) \le -\varepsilon$, a contradiction.

At this point we know that $s=s(\varepsilon)\in C$ for each $$ 0< \varepsilon < \frac{p-q}{t_0}, $$ that is, the function $\varepsilon \mapsto s(\varepsilon)$ maps the interval $I:=(0,(p-q)/t_0)$ into $C$. On the other hand this function is injective: $$ s_0:=s(\varepsilon_1)=s(\varepsilon_2) ~~ \Rightarrow ~~ h_{\varepsilon_1}(s(\varepsilon_1))=q= h_{\varepsilon_2}(s(\varepsilon_2)) ~~ \Rightarrow ~~ $$ $$ 0= h_{\varepsilon_1}(s_0)- h_{\varepsilon_2}(s_0) = (\varepsilon_2-\varepsilon_1)s_0 ~~ \Rightarrow ~~ \varepsilon_2=\varepsilon_1. $$ We have obtained an injective mapping from the interval $I$ into the countable set $C$, a contradiction. $\square$

So, if $u:[0,1] \to \mathbb{R}$ is continuous and if $u'(t) =0$ except at a countable set $C\subseteq [0, 1]$ then the Lemma can be applied to $\pm u$. Thus $u$ and $-u$ are nonincreasing, hence $u$ is constant.