Prove that $1/x^2$ is not uniformly continuous on $(0,\infty)$ using $\varepsilon$-$\delta$ arguments

Question

I am trying to prove that $f(x) = 1/x^2$ is not uniformly continuous on $(0,\infty)$ using $\varepsilon$-$\delta$ arguments.

This is what I got so far.

Proof: Let $\varepsilon = 1$. For every $\delta>0$ take $$c = \delta\, \text{ and }\, x = \frac{1}{\frac{1}{\delta} +\frac{\delta}{2}}.$$ Then $$|x-c| = \frac{\delta^3}{\delta^2+2}<\delta.$$ But $$|f(x) - f(c)| = \frac{1}{4}\left(\delta^2+4\right)\geq 1 = \varepsilon.$$

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

Is this a valid proof? Any advice would be appreciated. Thanks in advance. :)

Edit: I am trying to use part ii) from Bartle's Introduction to Real Analysis page 142:

Answer 1

Claim: $f:(0,\infty)\to \Bbb{R}$ defined by $f(x)=\frac{1}{x^2}$ is not uniformly continuous.

Preliminary:

$f$ is uniformly continuous if, for every $\epsilon>0$, there exists a $\delta>0$ such that for all $x,y\in (0,\infty)$ with $|x−y|<\delta$ implies $|f(x)−f(y)|<\epsilon$.

\begin{align}\forall\epsilon>0\space \exists \delta>0 \space \forall > x, y\in (0,\infty):|x-y|<\delta\implies |f(x) -f(y) |<\epsilon\end{align}

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$f$ is not uniformly continuous if, there exists $\epsilon>0$ such that for all $\delta>0$ there exists two points $x,y\in (0,\infty)$ with $|x−y|<\delta$ but $|f(x)−f(y)|\ge\epsilon$.

\begin{align}\exists\epsilon>0\space \forall \delta>0 \space \exists x, y\in (0,\infty):|x-y|<\delta\implies |f(x) -f(y)|\ge \epsilon\end{align}

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Proof : Let $\delta>0$ be given. Then by Archimedean property of $\Bbb{R}$ choose $N\in \Bbb{N}$ such that $\frac{1}{2N}<\delta$

Let $x=\frac{1}{N} $ and $y=\frac{1}{2N}$

Then $|x-y|=|\frac{1}{N} -\frac{1}{2N}|=\frac{1}{2N}<\delta$

$|f(x) -f(y) |=|\frac{1}{N^2} -\frac{1}{4N^2}|=\frac{3}{4N^2}$

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$\exists \epsilon (=\frac{3}{4N^2}) >0 $ such that $\forall \delta>0$ there exists two points $x=\frac{1}{N}, y=\frac{1}{2N}\in(0, \infty) $ with $|x-y|<\delta$ but $|f(x) -f(y) |\ge \epsilon$

•To get strict inequality , choose $0<\epsilon <\frac{3}{4N^2}$ ( You have plenty choices! )

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Alt:

$f: (X, d) \to (Y, d') $ is uniformly continuous iff $f$ preserve parallel sequences.

Two sequences $(x_n), (y_n) $ are called parallel sequence in $X$ if $d'(f(x_n), f(y_n)) \to 0$ whenever $d(x_n, y_n) \to 0$

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Now choose $x_n=\frac{1}{n} $ and $y_n=\frac{1}{2n}$

Then $|x_n-y_n|\to 0$ but $|f(x_n-f(y_n) |=3n^2\to\infty$

Hence $f$ is not uniformly continuous.

Comments: Your proof is also correct. $\forall\delta>0$ ,you need to find two points (doesn't matter whether those points depend on $\delta$ or not) which are atmost $\delta$ distance away but their images have distance at least $\epsilon$ for some fixed $\epsilon>0$.

So your proof is correct. But the above two proofs are somehow easier to execute. Isn't it?

Answer 2

Your reasoning is not valid since $x$ depends on $\delta$. In uniform continuity, the variables $x,y$ do not depend on anything, that is why it is uniform. An easy approach using sequences is as follows:

Take $x_n=\frac{1}{\sqrt{n+1}}$ and $y_n=\frac{1}{\sqrt n}$. Clearly $x_n-y_n\to 0$ as $n\to\infty$, but $$\vert f(x_n)-f(y_n)\vert=1\not\to 0$$ as $n\to\infty$.