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

Question

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 $(0,1)$.

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

Answer 1

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 $y_n.$