Show that $f(x)=\frac{1}{x^{2}}$ is uniformly continuous on $[1,\infty $)

Question

Please check my proof.

Suppose a is arbitary number in $[1,\infty) $

Given $\epsilon ,\delta >0$

$$|x-a|\rightarrow |f(x)-f(a)<\epsilon $$

$$\Rightarrow |\frac{1}{x^{2}}-\frac{1}{a^{2}}|<\epsilon $$

$$\Rightarrow |\frac{a^{2}-x^{2}}{a^{2}x^{2}}|<\epsilon $$

$$\Rightarrow \frac{1}{a^{2}}|\frac{a^{2}-x^{2}}{x^{2}}|<\epsilon $$

$$\Rightarrow |\frac{a^{2}-x^{2}}{x^{2}}|<a^{2}\epsilon $$

Choose $\delta =a^{2}\epsilon $

$$\Rightarrow \frac{1}{a^{2}}|\frac{a^{2}-x^{2}}{x^{2}}|<\frac{a^{2}\epsilon}{a^{2}}=\epsilon $$

Then $f(x)=\frac{1}{x^{2}}$ is uniformly continuous on $[1,\infty $).

Answer 1

For a function $f:A \to B$ to be uniformly continuous we need that

$$ \forall \varepsilon > 0 \exists \delta >0 \forall x,y \in A: |x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon $$

We have in your case $f(x)=\frac{1}{x^2}$. We have $\forall x,y \in [1,\infty)$

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

Now, since $y+x > 0$ and $|y-x|=|x-y|$ we get

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

since $\frac{1}{xy^2} < \frac{1}{2}$ and $\frac{1}{x^2 y} < \frac{1}{2}$ for $x,y \in [1,\infty)$.

Therefore if you set $\delta = \varepsilon$ we get

$$ |f(x)-f(y)| < |x-y| < \delta =\varepsilon $$