I encountered this question on my textbook but I think that the function $f(x)$ is not uniformly continuous for all $\alpha$. For example let us say $\alpha=10^{-10}$ and fix $\epsilon=1$. Then $\exists \delta>0$ s.t. $\forall x,y\in\mathbb{R}$ with $|x-y|<0$, $|\frac{1}{x}-\frac{1}{y}|<1$. Now, pick $x\in[10^{-9},10^{-8}]$ with $x<\delta$ and set $y=x/10$. Then $$|x-y|=|\frac{9x}{10}|<\delta, \text{but}$$ $$\lvert \frac{1}{x}-\frac{1}{y}\rvert=|-\frac{9}{x}|>1\quad \text{since $x\in[10^{-9},10^{-8}]$}$$ However, this contradicts with $f$ being uniformly continuous on its domain.

If there is a mistake, please help me to find out and to solve the exercise.

  • 2
    $\begingroup$ You can consider this like the order below: For fix $\epsilon=1$ then "there exists $\delta>0$". This $\delta$ is decided by $|1/x-1/y|<\epsilon=1$. When you calculate that $|1/x-1/y|=|x-y|/xy \leq |x-y|/\alpha^2$, now you can decide that $\delta=\epsilon\alpha^2$. So your "$x<\delta$" is wrong. More generally, you don't know what $\delta$ exactly is. it is just a abstract number. $\endgroup$
    – Houa
    Commented Jul 27, 2020 at 15:04
  • $\begingroup$ @Houa, thanks a lot, now I get it. $\endgroup$
    – confused
    Commented Jul 28, 2020 at 8:20

1 Answer 1


Choose $\epsilon\gt0$. Then for any $x,y\in[\alpha,\infty)$ we have that, for $\delta=\alpha^2\epsilon\gt0$, \begin{align} |x-y|\lt\delta &\implies|x-y|\lt xy\epsilon\\ &\implies\frac{|x-y|}{xy}\lt\epsilon\\ &\implies\left|\frac{x-y}{xy}\right|\lt\epsilon\\ &\implies\left|\frac1y-\frac1x\right|\lt\epsilon\\ &\implies\left|\frac1x-\frac1y\right|\lt\epsilon\\ \end{align} as required.


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