If for each neighborhood $U$ of $x_0$ there exists $x\in U$ such that $f(x)=f(x_0)$, then $f'(x_0)=0$ or $f'(x_0)$ does not exist.

Question

Prove the following:

If for each neighborhood $U$ of $x_0$ there exists $x\in U$ such that $f(x)=f(x_0)$ with $x\neq x_0$, then $f'(x_0)=0$ or $f'(x_0)$ does not exist.

I am not sure how to start this problem. Any help is appreciated.

Answer 1

If $f'(x_{0})$ does not exist, then we are done.

Assume that $f'(x_{0})$ exists, then consider the ball $B_{1/n}(x_{0})$, then by assumption we have some $x_{n}\in B_{1/n}(x_{0})$ such that $x_{n}\ne x_{0}$ and $f(x_{n})=f(x_{0})$. Given $\epsilon>0$, there is some $\delta>0$ such that \begin{align*} \left|\dfrac{f(x_{0}+h)-f(x_{0})}{h}-f'(x_{0})\right|<\epsilon,~~~~0<|h|<\delta. \end{align*} Find $n_{0}$ so large such that $0<|x_{n_{0}}-x_{0}|<\delta$, then \begin{align*} |f'(x_{0})|=\left|\dfrac{f(x_{n_{0}})-f(x_{0})}{x_{n_{0}}-x_{0}}-f'(x_{0})\right|<\epsilon, \end{align*} since this is true for all $\epsilon>0$, we have $f'(x_{0})=0$.