Real Analysis – Continuous Function from [0,1] to ? with Uncountably Many Local Maxima

Question

Does there exist a continuous function from $[0,1]$ to $R$ that has uncountably many strict local maxima?

Answer 1

A strict local maximum has a punctured neighborhood in which the values are less that that value. Every neighborhood has a rational number, so each strict local maximum is associated with a rational number. The rationals are countable, thus the strict local maxima are as well.

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My previous argument is not rigorous and it does have some problems. Here is a more rigorous one.

For each strict local maximum location $a\in[0,1]$ there is a $\delta>0$ such that $0<|x-a|<\delta\implies f(x)<f(a)$. Let $\delta_a$ be the least upper bound (supremum) of such $\delta$'s for the given $a$. Then $0<\delta_a\le 1$.

If $a\ne b$ are both strict maxima, then $|a-b|$, the distance between $a$ and $b$, may be less than $\delta_a$ if $f(b)<f(a)$, but then $|a-b|$ must be at least $\delta_b$. So $|a-b|\ge\min(\delta_a,\delta_b)$. The $\delta$'s are a measure of "spread" of the strict local maxima.

So for any given $n\in\Bbb Z^+$, the number of $a$ such that $\delta_a\ge\frac 1n$ is at most $n+1$.

We can now count the strict local maxima: first list all $a$ such that $\delta_a\ge \frac 11$, then those such that $\delta_a\ge\frac 12$, and so on. Each stage gives us finitely many strict local maxima, there are countably many stages, and the union gives us all strict local maxima. Therefore there are at most countably many strict local maxima.