I need to show that $1 \over x^2$ is not uniformly continuous on the interval $(0,2]$ using the definition of uniform continuity.

Definition of Uniform Continuity on a set A:

Let $A \subseteq \Bbb R$, let $f:A \rightarrow \Bbb R$.

$f$ is uniformly continuous if $ \forall \epsilon > 0$ $\exists \delta > 0$ such that if for any $x \in A$, $x$ satisfies the inequality:

$|x - u| < \delta \implies |f(x) - f(u)| < \epsilon$

To prove this function is not uniformly continuous by the given definition, I negated the definition such that:

$f$ is not uniformly continuous on the interval $(0,2]$ whenever

$\exists \epsilon_0 > 0$ $ \forall \delta > 0$ $\exists$ $x_\delta \in A$ satisfies $|x_\delta-u_\delta| < \delta $ but $|f(x_\delta) - f(u_\delta)| > \epsilon_0$

I chose to let $\epsilon_0 = 2$

For any $ 0< \delta < 2$ choose $x_\delta = \delta$ and $u_\delta$ = $\delta \over 2$

Then $|x_\delta - u_\delta|$ = $|\delta -$ $\delta \over 2|$ = $\delta \over 2$

But $|f(x_\delta) - f(u_\delta)|$ = |$1 \over \delta^2$ - $\frac 1 1 \over \delta^2$ | = |$(\frac 1\delta)^2 - \frac 2\delta^2)|$ = $3 \over \delta^2$ $> 2 = \epsilon_0$

But on the interval $(0,2]$ this does not always hold true.

Can anyone help me with this? I am thinking that my choice of $\delta$ is wrong but I'm not sure.

  • $\begingroup$ Something's not right with the formatting in the third to last line. $\endgroup$ Feb 9, 2016 at 16:50
  • $\begingroup$ Your negation is incorrect, the 'whenever $x \in A$ should be 'there exists $x \in A$ satisfying'. $\endgroup$
    – copper.hat
    Feb 9, 2016 at 16:54
  • $\begingroup$ Thank you, I will correct that. $\endgroup$
    – Iff
    Feb 9, 2016 at 16:57
  • $\begingroup$ It doesn't make sense to define uniform continuity in a point $c$. That's only the definition of continuity in $c$. $\endgroup$
    – user302982
    Feb 9, 2016 at 16:59
  • $\begingroup$ You're right @sigmabe thank you, I corrected the definition. $\endgroup$
    – Iff
    Feb 9, 2016 at 17:02

1 Answer 1


Here is another way:

If $f$ was uniformly continuous, then for some $\delta>0$ we would have $f(x)-f(x+\delta) < 1$ for all $x$ such that $x,x+\delta \in (0,2]$.

Let $f(x) = {1 \over x^2}$, and let $\phi(x) = f(x)-f(x+\delta)$ for $x \in (0,2-\delta)$.

Since $f$ is bounded on any compact subinterval of $(0,2]$, we see that $\lim_{x \downarrow 0} \phi(x) = \infty$. Hence $f$ is not uniformly continuous.

  • $\begingroup$ Thank you. Unfortunately I must prove this using the definition. It's a study guide for an upcoming exam so I'm trying to understand using and negating the definition. $\endgroup$
    – Iff
    Feb 9, 2016 at 16:52
  • $\begingroup$ Well, this shows that there is $\epsilon_0 = 1$ such that for all $\delta>0$ we can find some $x \in (0,2]$ such that $|f(x)-f(x+\delta)| \ge \epsilon_0$. $\endgroup$
    – copper.hat
    Feb 9, 2016 at 16:56
  • $\begingroup$ Oh I see, thank you @copper.hat - so I can set $\epsilon_0$ = 1? I guess I was assuming I had to set it equal to the endpoint. If I set $\epsilon_0$ = 1 I think the rest of my proof works. $\endgroup$
    – Iff
    Feb 9, 2016 at 17:05
  • $\begingroup$ To show that it is not uniformly continuous, you need to show that there exists some $\epsilon_0>0$ such that blah, blah, blah. With this particular function, any $\epsilon_0>0$ will work, so you can choose it to suit your computations. $\endgroup$
    – copper.hat
    Feb 9, 2016 at 17:06
  • $\begingroup$ Thanks! I appreciate your help. $\endgroup$
    – Iff
    Feb 9, 2016 at 17:08

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