Let $f: D\rightarrow \mathbb{R}$ be a continuously differentiable function defined on a domain $D$. Is it true that if $x$ is a non critical point of $f$, then there is a neighborhood of $x$ which contains no accumulation point of the set of zeros of $f$?

Thanks in advance!


If $f(x)\ne0$, then there is a neighborhood of $x_0$ with no zeros of $f$. Without loss of generality, asume $f\colon(-a,a)\to\Bbb R$, $f(0)=0$ and $f'(0)\ne0$. Then $$ f(x)=f'(0)\,x+h(x)\quad\text{with}\quad \lim_{x\to0}\frac{h(x)}{x}=0. $$ There exists $\delta>0$ such that $$ |x|<\delta\implies\Bigl|\frac{h(x)}{x}\Bigr|\le\frac{|f'(0)|}{2}. $$ Then, if $0<|x|<\delta$, $$ |f(x)|\ge|f'(0)|\,|x|-|h(x)|\ge|f'(0)|\,|x|-\frac{|f'(0)|}{2}\,|x|\ge\frac{|f'(0)|}{2}\,|x|>0. $$

  • $\begingroup$ Thanks for your answer. Is the statement true when $D$ is a domain in $R^n$? $\endgroup$ – Jiu Dec 5 '18 at 15:22
  • $\begingroup$ No. Consider $f(x,y)=x$. Then $(0,0)$ is not a critical point, but $f(0,y)=0$ for all $y$. $\endgroup$ – Julián Aguirre Dec 5 '18 at 15:26
  • $\begingroup$ you’re right. Thanks! $\endgroup$ – Jiu Dec 5 '18 at 15:28

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