# Thread: Uniform Continuity on an Interval

1. ## Uniform Continuity on an Interval

Show $f(x)=\frac{1}{x^2}$ is not uniformly continuous on $I=(0,\infty)$ .

2. Basically, it can't be uniformly continuous in that interval because the slope goes to (negative) infinity as x goes to zero. To prove this rigorously, remember the definition of uniform continuity: for any $\epsilon>0$, there is a fixed $\delta$ such that $|x-y|<\delta$ implies $|\frac{1}{x^2}-\frac{1}{y^2}|< \epsilon$.

To show that it's not uniformly continuous, choose any epsilon, and show that for any fixed delta, $|\frac{1}{x^2}-\frac{1}{(x+\delta)^2}|$ goes to infinity as x goes to 0.

3. so to rigorously prove it can I just choose a delta like 0?

4. oh ok never mind that was a stupid question lol

the only thing I dont see is how to replace $y^2$ with $(x+ \delta)^2$

5. You can also use the fact that you can find a Cauchy sequence $(s_{n})\in (0,\infty )$ where $(f(s_{n}))$ is NOT Cauchy.

6. Originally Posted by frenchguy87
Show $f(x)=\frac{1}{x^2}$ is not uniformly continuous on $I=(0,\infty)$ .
There is a theorem concerning uniform continuity that says:

f is uniform continuous on its domain ,and in our case (0, $\infty$) iff for any pair of sequences on (0, $\infty$ { $x_{n}$},{ $y_{n}$} we have:

$lim_{n\to\infty} |x_{n}-y_{n}| = 0\Longrightarrow$ $lim_{n\to\infty}|f(x_{n})-f(y_{n})| =0$

Hence to prove that : $\frac{1}{x^2}$ is not uniform continuous on (0, $\infty$) we must find a pair of sequences that:

$lim_{n\to\infty} |x_{n}-y_{n}| = 0$ ,and

$lim_{n\to\infty}|f(x_{n})-f(y_{n})|\neq 0$

So if we put :

$x_{n}=\frac{1}{\sqrt{n}}$ and

$y_{n} =\frac{2}{\sqrt{n}}$,then

$lim_{n\to\infty} |x_{n}-y_{n}| = 0$ and

$lim_{n\to\infty}|f(x_{n})-f(y_{n})|\neq 0$

7. Originally Posted by frenchguy87
Show $f(x)=\frac{1}{x^2}$ is not uniformly continuous on $I=(0,\infty)$ .
Notice that uniformly continuous functions preserve Cauchy sequences. Is $\frac{1}{\sqrt{n}}$ Cauchy? Is $n$?