# Thread: Uniform Continuity on an Interval

1. ## Uniform Continuity on an Interval

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

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 $\displaystyle \epsilon>0$, there is a fixed $\displaystyle \delta$ such that $\displaystyle |x-y|<\delta$ implies $\displaystyle |\frac{1}{x^2}-\frac{1}{y^2}|< \epsilon$.

To show that it's not uniformly continuous, choose any epsilon, and show that for any fixed delta, $\displaystyle |\frac{1}{x^2}-\frac{1}{(x+\delta)^2}|$ 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 $\displaystyle y^2$ with $\displaystyle (x+ \delta)^2$

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

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

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

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

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

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

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

So if we put :

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

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

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

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

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