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The function $f$ is differentiable in $[0,1]$ and $f$ has infinite roots in $[0,1]$. Prove that there exists $c\in [0,1]$ such that $f(c)=f'(c)=0.$

My attempt: Assume $x_1, x_2,...,x_n,...$ are roots of $f(x)=0$. Since $(x_n)$ is bounded, there exists $({x_n}_k)$ converging to $c\in [0,1]$. Since $f$ is continuous in $[0,1]$, we deduce that $f(c)=0$.
Can $f'(c)=0$?

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    $\begingroup$ Choose your subsequence to be increasing or decreasing to c. Then what do you get if you apply Rolle's Theorem to each pair $x_n,x_{n+1}$? $\endgroup$ Dec 1, 2023 at 16:12
  • $\begingroup$ @FrancisAdams But it's not the same $c$ such that $f(c)=0$. $\endgroup$
    – Harry
    Dec 1, 2023 at 16:22

3 Answers 3

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Yes, your approach works. You have$$f'(c)=\lim_{x\to c}\frac{f(x)}{x-c},$$and therefore, if $(a_n)_{n\in\Bbb N}$ is a sequence of elements of $[0,1]$ such that $\lim_{n\to\infty}a_n=c$, then$$f'(c)=\lim_{n\to\infty}\frac{f(a_n)}{a_n-c}.$$But$$\frac{f(x_{n_k})}{x_{n_k}-c}=0,$$for any $k\in\Bbb N$, and $\lim_{k\to\infty}x_{n_k}=c$. So, $f'(c)=0$.

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Another way to see that $f(c)=f’(c)=0$, with the same $c$, is as follows. Rolle’s theorem gives a sequence $(z_n)$ such that $z_n\in (x_n,x_{n+1})$ and $f’(z_n)=0$ for each $n\in \Bbb N$. It has a convergent subsequence $(z_{n_k})$. Now, given any $\epsilon\gt 0$, there is $N\in \Bbb N$ such that $|z_{n_k}-x_{n_k}|\le|x_{n_k}-x_{n_l}|\lt \epsilon$ whenever $k,l \ge N$. Thus, $z_{n_k}$ have the same limit as $x_{n_k}$.

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  • $\begingroup$ We are not given that $f'$ is continuous. $\endgroup$
    – M W
    Dec 2, 2023 at 9:51
  • $\begingroup$ @MW I’m not sure where you saw the continuity of $f’$ mentioned, the inequality follows from the fact every $z_{n_k}$ must lie between $x_{n_k}$ and some other $x_{n_l}$ $\endgroup$
    – Divide1918
    Dec 2, 2023 at 9:54
  • $\begingroup$ You are trying to show $f'(c)=0$, right? How does that follow from $z_{n_k}$ having the same limit as $x_{n_k}$? $\endgroup$
    – M W
    Dec 2, 2023 at 9:56
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I would like to suggest another approach by proving the contrapositive:

Let $f\colon [0,1] \to \Bbb R$ be differentiable and such that whenever $f(x) = 0$, then $f'(x) \neq 0$. Then $f^{-1}(0) \subset [0,1]$ is finite.

Indeed, assume that $x_0\in f^{-1}(0)$. In a neighbourhood of $x_0$, write $$ f(x) = f'(x_0)(x-x_0) + o(x-x_0). $$ In $\varepsilon-\delta$ language, this means $$ \forall \varepsilon >0, \exists \delta >0, \forall x \in (x_°-\delta,x_0+\delta), \quad |f(x) - f'(x_0)(x-x_0)| \leqslant \varepsilon|x-x_0|. $$ Choose $\varepsilon = \frac{|f'(x_0)|}{2}$, which is nonzero by assumption: $$ \exists \delta >0, \forall x \in (x_0-\delta,x_0+\delta), \quad |f(x) - f'(x_0)(x-x_0)| \leqslant \frac{|f'(x_0)|}{2}|x-x_0|. $$ The generalized triangle inequality now yields, for $x \in (x_0-\delta,x_0+\delta)$, $$ |f(x)| \geqslant \big|\,|f(x) - f'(x_0)(x-x_0)| - |f'(x_0)(x-x_0)|\,\big| \geqslant \frac{|f'(x_0)|}{2}|x-x_0|. $$ In particular, if $|x-x_0|< \delta$ and $x\neq x_0$, then $f(x) \neq 0$. It follows that $f^{-1}(0) \cap (x_0-\delta,x_0+\delta) = \{x_0\}$, that is to say, that $x_0$ is an isolated point of $f^{-1}(0)$.

Hence, any point of $f^{-1}(0)$ is an isolated point: $f^{-1}(0) \subset [0,1]$ is then a discrete subset. Since $[0,1]$ is compact, $f^{-1}(0)$ is finite.

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