calculus - Local extrema of continous, non-differentiable function

Ok, so looking through some questions I've found this answer: https://math.stackexchange.com/a/1667/102636 containing proof, that there are either inifinitely or no local extrema of continous and nowhere differentiable function. I would love for someone to explain it to me in greater detail (I can't comment on this answer unfortunatelly...)

For recall, the proof goes like this:

Let's take $f:[a,b] -> R$ - continous and nowhere differentiable. Let's assume, $f$ has finite number of extrema, namly $c_1 < c_2 < .... < c_n$.

Now let's take some $c_i$ and $c_{i+1}$. We take $c$ - global maximum of $f$ on $[c_i, c_{i+1}]$ which occurs at an interior point. Therefore $c$ is local extrema of $f$.

I don't see why this global extremum on $[c_i, c_{i+1}]$ needs to be an interior point.

If someone could explain this to me I'd be really gratefull.

Answer

I think the answer was hastily written, and in particular "$c_1 < c_2<\dots f(c_2)$" needs to be fixed. Hopefully someone with commenting privilege will draw Akhil's attention to this.

Here is another way to phrase it. It is a theorem of Lebesgue that a function that is monotone on $[c,D]$ is differentiable at almost every point of $(c,d)$. Without getting into the technical meaning of "almost every", let's just say that the theorem guarantees $f$ to be differentiable at some point of $(c,d)$. Therefore, a nowhere differentiable function cannot be monotone on any subinterval of its domain.

If such a function had finitely many local extrema, there would be an interval $(c,d)\subset [a,b]$ on which it has no local extrema (you can order all extrema and pick an interval between two consecutive ones). On such an interval $f$ is monotone (this takes a proof), which contradicts the above.