# Rolle's theorem? [duplicate]

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 $$f'(c)=0$$?

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

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, $$f'(c)=0$$.

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 $$x_{n_k}$$.

• We are not given that $f'$ is continuous.
– M W
Dec 2, 2023 at 9:51
• @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 $x_{n_l}$ Dec 2, 2023 at 9:54
• You are trying to show $f'(c)=0$, right? How does that follow from $z_{n_k}$ having the same limit as $x_{n_k}$?
– M W
Dec 2, 2023 at 9:56

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.