Unifourm Continuity of f(x)=1/x on (0,+∞)

[ArixII]

Hi,
in this forum post, exactly at #4:
https://www.physicsforums.com/showthread.php?t=52795"

after a clarification for uniformly continuous function, it is written that:
"...For example, $$f(x)=\frac{1}{x}$$ is contiuous, but not uniformly continuous on the interval $$(0,+\infty)$$"

I failed to prove it :(

Could anyone please give me a clue, how to prove that this function is not uniformly continuous on the given interval?

 

[rasmhop]

Hi,
in this forum post, exactly at #4:
https://www.physicsforums.com/showthread.php?t=52795"

after a clarification for uniformly continuous function, it is written that:
"...For example, $$f(x)=\frac{1}{x}$$ is contiuous, but not uniformly continuous on the interval $$(0,+\infty)$$"

I failed to prove it :(

Could anyone please give me a clue, how to prove that this function is not uniformly continuous on the given interval?

Proof ideas/hints:
1. First suppose f is uniformly continuous in the interval. We now wish to obtain a contradiction.
2. Choose a fixed epsilon to deal with, say $$\epsilon = 1$$. This shouldn't matter because as we can see from the graph the difference in function values grows beyond all bounds as we approach 0 (which is what we are going to prove).
3. We then, by uniform continuity, have: "there exists some $$\delta > 0$$ such that if $$a,x \in (0,\infty)$$ and $$0 < |a-x| < \delta$$ then $$|f(x)-f(a)| < 1$$".
4. Considering the graph we see that as the gap between a and x becomes greater the difference in the function value becomes greater so we don't need to consider the case when they are close. Thus the proof will probably go through if we let a = x + delta/2. If it wouldn't go through then we could have been given delta/2 in the previous step and not being able to obtain a contradiction f would have been uniformly continuous. Since we know it isn't we can let $$a = x+\delta/2$$ and obtain a contradiction for some $$a \in (0,\infty)$$.
5. Now is it always true that $$0 < |a-x| < \delta$$ for positive x? If it is then it's always true that $$|f(x)-f(a)| < 1$$ for positive x.
6. Can you find some x expressed in terms of delta such that this leads to a contradiction? (You could for instance consider when equality occurs and then simply let x be that value).

 

[ArixII]

Proof ideas/hints:
1. First suppose f is uniformly continuous in the interval. We now wish to obtain a contradiction.
2. Choose a fixed epsilon to deal with, say $$\epsilon = 1$$. This shouldn't matter because as we can see from the graph the difference in function values grows beyond all bounds as we approach 0 (which is what we are going to prove).
3. We then, by uniform continuity, have: "there exists some $$\delta > 0$$ such that if $$a,x \in (0,\infty)$$ and $$0 < |a-x| < \delta$$ then $$|f(x)-f(a)| < 1$$".
4. Considering the graph we see that as the gap between a and x becomes greater the difference in the function value becomes greater so we don't need to consider the case when they are close. Thus the proof will probably go through if we let a = x + delta/2. If it wouldn't go through then we could have been given delta/2 in the previous step and not being able to obtain a contradiction f would have been uniformly continuous. Since we know it isn't we can let $$a = x+\delta/2$$ and obtain a contradiction for some $$a \in (0,\infty)$$.
5. Now is it always true that $$0 < |a-x| < \delta$$ for positive x? If it is then it's always true that $$|f(x)-f(a)| < 1$$ for positive x.
6. Can you find some x expressed in terms of delta such that this leads to a contradiction? (You could for instance consider when equality occurs and then simply let x be that value).

Thank you! Actually I didn't get a lot, but that the $$\epsilon = 1$$ helps as the function goes to infinity when x approaches zero. Puting $$a = x+\delta/2$$ I could not figure it out how this is useful.