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Prove $\frac{1}{x^2}$ is uniformly continuous on $[1,\infty)$ but not on $(0,1)$.

Proof

On $[1,\infty)$:

$$\left|f(x) - f(y)\right| = \left| \frac{1}{x^2} - \frac{1}{y^2} \right| = \frac{(x+y)\left|x -y\right|}{x^2y^2} = \left(\frac{1}{xy^2} + \frac{1}{x^2y}\right)\left|x-y\right|\leq 2 \left|x-y\right| $$

Therefore $f$ is a Lipschitz function on $[1,\infty)$ which implies uniform continuity.

On $(0,1)$:

Choose $\epsilon_0 = 1$ and consider the sequences defined by $x_n = \frac{1}{\sqrt{n}}$ and $y_n = \frac{1}{\sqrt{n+1}}$. We see that $\lim(x_n - y_n) = 0$ but

$$\left|f(x_n) - f(y_n)\right| = \left|n - (n+1)\right| = 1$$

therefore $f$ is not uniformly continuous on $(0,1)$.

Please comment on the validity, readability, and/or style. Thank you.

Zduff
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  • With a quick read it looks OK to me. This might help with intuition and lead to a somewhat briefer answer. https://math.stackexchange.com/questions/2540090/find-a-function-that-is-continuous-on-0-1-but-not-uniformly-continuous-on-t/2540103#2540103 – Ethan Bolker Feb 26 '18 at 18:36
  • A more roundabout proof of the first part: $f$ extends to a continuous function on the one-point compactification $[1, \infty]$, and therefore the extension is uniformly continuous; and then this passes to the subspace $[1, \infty)$. – Daniel Schepler Feb 26 '18 at 18:56

1 Answers1

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The only criticism I would have is that using sequences seems to complicate matters more than necessary.

For $0<x<1$ you could let $y = \dfrac x {\sqrt{1+x^2}}$ and then observe that

  • $0<y<x<1$ and
  • $|x-y| \to 0$ as $x\to0$ and
  • $|f(x)-f(y)|=1.$

That's essentially the same as what you did but without sequences; i.e. if $x=\text{your }x_n$ then $y$ as defined here coincides with your $y_n.$

Michael Hardy
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  • Maybe I should add that using sequences may be easier the first time through, but then one can rearrange the proof to read as in this answer. – Michael Hardy Feb 26 '18 at 18:50
  • So, let me understand, your way of thinking about it: you're saying that for any $\delta >0$ that your choice of $x$ and $y$ guarantees that the existence of $|x - y| < \delta$ (that's the limit part) and still we have $|f(x) - f(y)| = 1$. No sequence required. Does seem a lot cleaner. So when is it more convenient to use sequences? – Zduff Feb 26 '18 at 21:04
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    @Zduff : Off hand I don't know a general rule about when it's better to use sequences, except this: When the problem is about sequences. – Michael Hardy Feb 26 '18 at 22:57