Proof Verification/Doubt: Showing $f(x)=1/x^2$ is not uniformly continuous on $(0,\infty)$

Question

Show that $f(x)=1/x^2$ is not uniformly continuous on $(0,\infty)$

So I was trying to attempt this question and did the proof as follows:

$f(x)$ is not defined at the point $x=0$. Then, by the continuous extension theorem, $f(x)$ is not uniformly continuous on the interval $(0,\infty)$

I checked the answer to this question in the solution manual, and this was not the approach they used. Rather, they used the non uniform continuity sequential criteria.

My question is, is my proof correct? I am particularly concerned about the interval being $(0,\infty)$ rather than some interval $(a,b)$ where $b \in \mathbb{R}$, so I'm now unsure if the continuous extension theorem is even applicable at all in this case.

Thank you.

Answer 1

The issue is not whether the function is defined at $0$ but rather whether it can be extended by continuity there. The key lemma you want to use here is that if $f$ is uniformly continuous on a bounded open interval $(a,b)$ then it can be extended by continuity to the extremities $a$ and $b$, i.e., defined there in such a way that the extended function is continuous on the closed interval $[a,b]$.

Then show that $\frac{1}{x^2}$ cannot be extended by continuity at $0$ and you are done.

Answer 2

$f$ is uniformly continuous on $[a,\infty)$ for any $a>0$ so you can concentrate on what happens near $0$. The correct argument is the following: if $f$ is uniformly continuous and $x_n$ decreases to $0$ then $\{f(x_n)\}$ is a Cauchy sequence (as you can see from definition of uniform continuity) hence convergent; since $\frac 1 {x_n^{2}} \to \infty$ we have a contradiction. I believe this is what you had in mind but the way you expressed the idea is not correct.

Answer 3

In a comment, A.Asad has clarified:

Theorem 1 (The Continuous Extension Theorem): If $I=(a,b)$ is an interval, then $f:I\to\mathbb{R}$ is a uniformly continuous function on $I$ if and only if $f$ can be defined at the endpoints $a$ and $b$ such that $f$ is continuous on $[a,b]$.

Using this theorem, you need to show that $f(0)$ cannot be defined so that $f$ is continuous. Simply saying that "$f(0)$ is not defined" is not quite enough. Something like "$\lim\limits_{x\to0}f(x)$ does not exist, so there is no value of $f(0)$ that would make $f$ continuous at $0$" would be good.

Regarding your concern at the end of your question: if $f$ is not uniformly continuous on $(0,1)$ then it is not uniformly continuous on $(0,\infty)$.

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We can simply use the definition of uniform continuity. Note that $$ f\left(\frac1{\sqrt{n+2}}\right)-f\left(\frac1{\sqrt{n\vphantom{1}}}\right)=2 $$ yet $$ \begin{align} \left|\frac1{\sqrt{n\vphantom{1}}}-\frac1{\sqrt{n+2}}\right| &=\frac2{\sqrt{n\vphantom{1}}\sqrt{n+2}\left(\sqrt{n\vphantom{1}}+\sqrt{n+2}\right)}\\ &\le\frac1{n^{3/2}} \end{align} $$ So we can't find a $\delta\gt0$ so that $|x-y|\le\delta\implies|f(x)-f(y)|\le1$.

Answer 4

$$f(x) = \frac{1}{x^2}$$ is uniformly continuous on the interval $[3, \infty)$ and is not uniformly continuous on the interval $(0, 3)$.