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If $f,g$ is uniformly continuous on some domain $S$,then $fg$ still uniformly continuous. I think i could find a counter example but when i try to explain why the statement fail by considering some property the uniform continuous functions have, it seems that i don't quite see(or don't quite get the big picture) why they fail. So my question is can it be explained by considering the property of the uniform continuity to explain the statement above

The definition of uniform continuity I am studying: $ \forall \epsilon>0, \exists \delta>0 s.t.\forall x,y \in S,d(x,y)<\delta \implies d(f(x),f(y))<\epsilon$

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  • $\begingroup$ Is $fg$ continuous when $f$ and $g$ are? What do you know about continuous functions on compact domains? $\endgroup$ – Quinn Culver May 28 '13 at 3:40
  • $\begingroup$ @QuinnCulver I better change it to some domains $\endgroup$ – johnny May 28 '13 at 3:41
  • $\begingroup$ johnny, change the title too, then. $\endgroup$ – Quinn Culver May 28 '13 at 3:45
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Well, look at the function $f(x) = x$ over the whole real line. It is uniformly continuous but its product with itself is not. Uniform continuity is about how the function changes on the domain with respect to the independent variable. If the change is too steep, like in $f(x) = x^{2}$, you are not likely to get uniform continuity. If the change is reasonably slow, like in $f(x) = x$ or $f(x) = \sqrt{x}$ for example, you are likely to get uniform continuity. Thus, while $f$ and $g$ themselves may change "reasonably" slowly, their product may change faster and hence may not be uniformly continuous.

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  • $\begingroup$ Or can we say the function that are not uniform continuous cannot be so 'slow' because there always exist some $\epsilon, $ such that no matter how small the change is in the domain, there is always exist $x,y\in S$ such that the function can't change less than $\epsilon$ amount of unit so the change is considered to be 'large' $\endgroup$ – johnny May 28 '13 at 4:07
  • $\begingroup$ Well yeah, to put it formally, there exists an $\epsilon > 0$ such that for every $\delta > 0$ there are $x,y \in S$ such that $|x-y| < \delta$ and $|f(x)-f(y)| \geq \epsilon$. But this does not help you explaining why product of two uniformly continuous functions may not be uniformly continuous. $\endgroup$ – Vishal Gupta May 28 '13 at 4:13
  • $\begingroup$ You seem to be unhappy about the explanation. Could you pour in some more thoughts? $\endgroup$ – Vishal Gupta May 28 '13 at 4:28
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    $\begingroup$ uniform continuity is related to not having rapid changes and not so much about growth. A function can be bounded yet not uniformly continuous (on the real line). The function $x\mapsto x^2$ exhibits rapid change when $x$ is large. It is caused by the rapid growth of the function, but that is a side issue. $\endgroup$ – Ittay Weiss May 28 '13 at 4:34
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    $\begingroup$ @IttayWeiss O now i see the point. The uniform continuity is talking about no any rapid changes in function on the domains. If it is not uniform continuous, then the function must have some region which is rapidly changing. Thx $\endgroup$ – johnny May 28 '13 at 8:33

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