Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is $f(x)=\frac{1}{x}$ uniformly continuous then x $\in (0,1)$?

I think that it's not uniformly continuous so I am trying to prove that there exists an epsilon>0 for all deltas>0 and there exist x,y such that

$|f(x)−f(y)|≥ϵ$ if $ |x−y|<δ$

I started by choosin epsilon=1 . Then:

$|\frac{1}{x}−\frac{1}{y}|=\frac{|x−y|}{xy}$

Now, I think, I need to choose such values of x and y expressed through delta such that the equality above would be greater or equal than one, but I am having trouble thinking of such values. Is my approach any good?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

We can see that it's not uniformly continuous, because a function that is uniformly continuous on an interval must be bounded, whereas the function is not. This alone is actually enough for a proof. If we want to do it from first principles however, we can do the following.

let $x = \delta$, $y = \frac{\delta}{2}$. Then $|x-y|< \delta$, and

$$|\frac{1}{x} - \frac{1}{y}| = \frac{|x-y|}{xy} = \frac{\frac{\delta}{2}}{\frac{\delta^2}{2}} = \frac{1}{\delta} > 1$$

Since we must have $\delta < 1$.

share|improve this answer
    
Why delta must be less than 1? Shouldn't this proof work for every delta? –  user1242967 Dec 11 '12 at 19:10
    
@user1242967 It does, since we are working in $(0,1)$, $x, y\in (0,1) \implies |x-y| < 1$. Even if we weren't, since we are trying to prove $f$ is not uniformly continuous, we don't have to prove "...for every $x,y$ such that $|x-y|< \delta$..." but only "...there exists $x,y$ such that $|x-y| < \delta$..." So if it is true for some small $\delta$, it must also be true for any $\alpha > \delta$ since $|x-y| < \delta \implies |x-y| < \alpha$. –  Tom Oldfield Dec 11 '12 at 19:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.