Values of local maxima are countable for a $C^1$ function

Question

Prove or disprove:

Let $f:[a,b]\rightarrow \mathbb{R}$ be a $C^1$ function, $C:=\{f(x) \mid f'(x)=0\}$. So $|C| \le \aleph_0$.

My try: I've shown that the set of the x-values (and the set of the points) of the strict local maxima is countable by selecting rational numbers and using their cardinality. However, I cannot use this fact for the set of the values...What can I do?

Thanks in advance.

Answer 1

This is in general not true. For example, consider the following function, constructerd as follows:

1. $f_0(x)=0$, $x\in[0,1]$
2. $f_1(x)=\frac{1}{3}(x-\frac{1}{3})(\frac{2}{3}-x)\chi_1(x)$, where $\chi_1$ is the characteristic funtion of the interval $A_1=[\frac{1}{3},\frac{2}{3}]$.
3. $f_2(x)=\frac{1}{9}\left((x-\frac{1}{9})(\frac{2}{9}-x)+(x-\frac{7}{9})(\frac{8}{9}-x)\right)\chi_2(x)$, where $\chi_2$ is the characteristic of the set $A_2[\frac{1}{9},\frac{2}{9}]\cup[\frac{7}{9},\frac{8}{9}]$.
4. $$f_n(x)=\frac{1}{3^n}\sum\limits_{k=0}^{3^n-1}\left(x-\frac{k}{3^n}\right)\left(\frac{k+1}{3^n}-x\right)\chi_n(x)$$ where $\chi_n$ is the chatacteristic function of the set $$A_n=\bigcup_{k=0}^{3^{n-1}-1}\left[\frac{3k+1}{3^n},\frac{3k+2}{3^n}\right]\cap A_{n-1}.$$

To explain this a little bit, we try to mimic the construction of the Cantor set $\mathcal{C}$ - for more about this, check out this link - and construct a function that its only roots are exactly the points of the Cantor set. So, we create, at first, the abovementioned sequence which takes the zero function and "lifts" it over the new intervals we cut at every step of the construction of the Cantor set. Now, we consider: $$g_n(x)=\sum_{k=0}^nf_k(x)$$ and note that a quadratic of the form $\left(x-\frac{k}{3^n}\right)\left(\frac{k+1}{3^n}-x\right)$, attains its maximum value at $x_0=\frac{2k+1}{2\cdot3^n}$ and, it is $\frac{1}{4\cdot3^{2n}}$, so, we note that: $$f_n(x)\leq\frac{1}{3^n}\sum_{k=0}^{3^n-1}\frac{1}{4}\cdot\frac{1}{9^n}\leq\frac{1}{3^n}\frac{1}{4}\sum_{k=0}^\infty\frac{1}{9^n}=\frac{1}{3^n}\frac{1}{4}\frac{1}{1-\frac{1}{9}}<\frac{1}{3^n}\tag{$\star$}$$ So, it follows that: $$g_n(x)=\sum_{k=0}^nf_k(x)<\sum_{k=0}^n\frac{1}{3^n}$$ Consider now the limit of the above sequence of functions: $$g:=\lim_{n\to\infty}g_n$$ Since the convergence of $g_n$ to $g$ is uniform - Weierstrass criterion using $(\star)$ - and $g_n$ are continuous, we have that $g$ is also continuous.

Also, note that, $g(x)=0$ iff $x\in\mathcal{C}$ - this is easy to prove - so the set of zeros of $g$ is exactly the Cantor set. Lastly, note that $g$ is not differentiable on $[0,1]$ since it is not on $\mathcal{C}$.

Now, let us consider the following function $f[0,1]\to\mathbb{R}$: $$f(x)=\int_0^xg(t)dt$$ It is clear that $f\in C^1([0,1])$, since $g$ is continuous and, moreover it is exactly $C^1$ - by that, I mean that it has no higher order derivatives over the entire interval $[0,1]$. For this function, we have that:

1. $f$ is strictly increasing, since $g(x)>0$ over $[0,1]\setminus\mathcal{C}$ and $\mathcal{C}$ contains no interval. To prove this formally, let $x,y\in[0,1]$ with $x<y$ and, since $(x,y)$ is not a subset of $\mathcal{C}$, there exists a $x_0\in(x,y)$ such that $g(x_0)>0$, so, there also exists an $\epsilon=\epsilon(x_0)>0$ such that, for every $t\in(x_0-\epsilon,x_0+\epsilon)$ - let us assume that $\epsilon<\min\{x_0-x,y-x_0\}$: $$g(t)>0$$ Now, note that: $$f(y)=\int_0^yg(t)dt\geq\int_0^xg(t)dt+\int_{x_0-\epsilon}^{x_0+\epsilon}g(t)dt>\int_0^xg(t)dt=f(x)$$ So $f(x)<f(y)$.
2. As a result, $f$ is $1-1$, so the cardinality of: $$C=\{f(x)|f'(x)=0\}$$ is equal that of the set: $$D=\{x|f'(x)=0\}$$

But we have that: $$f'(x)=g(x)$$ so: $$D=\{x|g(x)=0\}=\mathcal{C}$$ But, $\mathcal{C}$ is uncountable, so the statement is not true.