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I am looking at this exercise:

Let $f:A \to B, g: B \to \mathbb{R}$

If $f$ is uniformly continuous on $A$ and $g$ is uniformly continous on $B$,show that $g \circ f$ is uniformly continuous.

$f$ is uniformly continuous on $A$: $\forall \epsilon>0 \ \exists \ \delta=\delta(\epsilon)>0 \text{ such that } \forall x,y \in A \text{ with } |x-y|< \delta \text{ we have } |f(x)-f(y)|< \epsilon $ $$$$ $g$ is uniformly continuous on $B$:

$\forall \epsilon'>0 \ \exists \delta'=\delta'(\epsilon') \text{ such that } \forall u,v \in B \text{ with } |u-v|< \delta' \text{ we have } |g(u)-g(v)|< \epsilon'$

Is it right so far?

And to write a relation for $g \circ f$,do I have to pick $\epsilon=\delta'$ ?

But...in general don't we have to pick values for $\delta \text{ and not for } \epsilon$?? :/

evinda
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When you're trying to prove a limit of this type, you're granted $\epsilon$ and have to figure out what $\delta$ should be. But in this case, you're using the limits, which means you get to feed in any $\epsilon$ and you get the appropriate $\delta$ out. In this case, you're right about the approach. In order to prove (uniform) continuity of $g \circ f$, you are granted $\epsilon > 0$. Feed that into the proof of (uniform) continuity of $g$ to get $\delta'$, and then feed that in as $\epsilon'$ into the proof of (uniform) continuity of $f$ to get the final $\delta$ you need. The details are left as an exercise for the student!

Paul Z
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  • So,you mean that what I have written so far is right and to continue we pick $\epsilon=\delta'$ and then we have : $$|f(x)-f(y)|< \delta' \text{ and since } f(x),f(y) \in B \Rightarrow |g(f(x))-g(f(y))|< \epsilon'$$ ? Or have I understood it wrong? – evinda May 28 '14 at 14:07
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    @evinda: Since g is uniformly continuous, then the values $\epsilon', \delta'$ work for $g(f(x)$. – user99680 May 28 '14 at 18:14
  • @user99680 So,could I write it as I have done it above? – evinda May 28 '14 at 18:16
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    @evinda: Yes, since $g(f(x)$ is just a value of , a case of $g(y)$, with $y=f(x)$. So for any $\epsilon'$ ,the value $\delta'$ will work. – user99680 May 28 '14 at 18:22
  • Nice,thank you very much!!!! – evinda May 28 '14 at 18:25
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    Sure, glad it helped. – user99680 May 28 '14 at 18:45