Why Volterra's function's derivative is zero on Cantor Set?

Question

Why Volterra's function's derivative is zero on Cantor Set? Is it because Cantor Set is nowhere dense?

Answer 1

Volterra's function is usually constructed using a fat cantor set, with positive measure. That being said, the short, intuitive, and not entirely correct answer is that around each point in the Cantor set, the function looks like $x^2\sin(1/x)$ does around the origin.

There are weaknesses of the above argument: it only works for end points of intervals that are removed in the Cantor set construction (there are only countably many of those, so we haven't taken most of the Cantor set into account), and it is only true on one side of such points. But the basic idea for a general proof is the same: For any point $a$ in the Cantor set, the Volterra function lies between the parabolas $y = (x-a)^2$ and $y = -(x-a)^2$, and therefore has derivative $0$ at $a$.

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Edit: Some more details.

Let's take an arbitrary point $a$ in the Cantor set, and look at one specific step in constructing Volterra's function $f$, showing that the part of the function that we add at that stage stays between $y = \pm (x-a)^2$.

At one step in the construction, we take a single open interval $(r, t)$ as we remove it from the Cantor set, and we define $f(x)$ should be for $x$ in that interval. Let $s$ be the midpoint between $r$ and $t$. Then for some $s_0\in (r, s)$ $f(x) = (x-r)^2\sin(1/(x-r))$ on $(r, s_0)$, and then a constant on $[s_0, s]$. On the interval $(s, t)$ the function looks like an exact mirror image.

Here it is important to note that $\sin$ is always between $\pm 1$, so multiplying by a sine term will never increase the absolute value of a function. Specifically, $g(x)\cdot \sin(\theta)$ always lies between $\pm g(x)$ for any function $g(x)$, for any $\theta$ (even if $\theta$ is itself a function of $x$).

Thus on the interval $(r, s)$, the value of $f(x)$ is between $\pm (x-r)^2$ and on the interval $(s, t)$ it lies between $\pm (x-t)^2$. This means that on the entire interval $(r, t)$, $f(x)$ lies between both $\pm (x-r)^2$ and $\pm (x-t)^2$.

There are now two cases: $a\leq r$ or $a\geq t$ (we can't have $a\in (r, t)$ because that interval is removed from the Cantor set).

If $a\leq r\leq x$, then $(x-a)^2\geq (x-r)^2$, and since $-(x-r)^2\leq f(x)\leq (x-r)^2$ we get $$ -(x-a)^2\leq f(x)\leq (x-a)^2 $$ In the other case, swap out $r$ with $t$, set $a\geq t\geq x$, and the proof is completely analoguous.