Proving $\frac{1}{ x^2}$ is not uniformly continuous on $(0,2]$

Question

I am trying to prove the function in the title is not uniformly continuous on (0,2]. Different from the proof given in proof, I proceed to show the proof following lecture as follows: \begin{equation} |1/x^2 - 1/y^2|=\frac{|y-x||y+x|}{x^2y^2} \end{equation} (after simplification). Since both $x,y \in (0,2]$ above eq. reduces to \begin{equation} |1/x^2 - 1/y^2|= \frac{|y-x|}{4}<\frac{\delta}{4} (\text{by the definition of continuity}). \end{equation} Now if I pick $\delta=4\varepsilon$, I don't arrive at the contradiction. Any hint where I am wrong will help.

Answer 1

Let $\delta>0$, choose $x\in (0,2]$ such that $x<\frac{\delta}2$ and choose $y=\frac{x}2$. Then $$ |y-x| = \frac x2 < \frac \delta{2}<\delta $$ and since $x\leq 2$ $$ \left|\frac1 {x^2}-\frac 1{y^2}\right| = \left|\frac1 {x^2}-\frac 4{x^2}\right|=\frac{3}{x^2} \geq \frac 34. $$ This contradicts the definition of uniform continuity on $(0,2]$.

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If you are allowed to use some results on uniformly continuous functions there are several faster ways to solve the exercise. For instance, the extension theorem states that if $f:A\rightarrow \mathbb{R}$ is uniformly continuous then it is possible to extend $f$ to the closure of $A$ in such a way that the extension is uniformly continuous (and, in particular, continuous). Since there is no way to extend $f(x)=\frac 1{x^2}$ to $[0,2]$ to a continuous function, then $f$ is not uniformly continuous on $(0,2]$.