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If $h$ is uniformly continuous on $\mathbb{R}$, then $\forall \epsilon > 0$, $\exists \delta > 0$ s.t. $\forall x,y \in dom(h)$, $|x - y| < \delta$ implies $|f(x) - f(y)| < \epsilon$.

My proof:

Let $\epsilon > 0$. Since $g$ is uniformly continuous, $\exists \delta > 0$ s.t. $\forall x,y \in \mathbb{R}$, $|x - y| < \delta$ implies $|g(x) - g(y)| < \epsilon$.

Now, $f$ uniformly continuous means that $\exists \delta' > 0$ s.t. $\forall x',y' \in \mathbb{R}$, $|x' - y'| < \delta'$ implies $|f(x') - f(y')|< \delta$.

So, for any $x', y' \in \mathbb{R}$ with $|x' - y'| < \delta'$, we have $|f(x') - f(y')|< \delta$. Hence $|g(f(x')) - g(f(y'))| < \epsilon$. This is exactly what it means for $g \circ f: \mathbb{R} \rightarrow \mathbb{R}$ to be uniformly continuous, so we are done.

Thank you for your feedback.

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  • $\begingroup$ Your proof is very incorrect, unfortunately. You've negated the definition of uniform continuity wrongly. There's also not a need to go by contradiction here. $\endgroup$ – user296602 Oct 28 '18 at 22:25
  • $\begingroup$ @T.Bongers What is the correct negation? $\endgroup$ – Jake Oct 28 '18 at 22:27
  • $\begingroup$ Give it a shot and edit your question to include it. Start by writing (in quantifiers, if you please) what it does mean for a function $h$ to be uniformly continuous. A giant red flag here is that if $x = y$ then $|x - y| < \delta$ but $|g \circ f(x) - g \circ f(y)| = 0$ is not $\ge \epsilon$. $\endgroup$ – user296602 Oct 28 '18 at 22:30
  • $\begingroup$ @T.Bongers I made these edits. How do they look? I am also working on a direct proof at the moment... $\endgroup$ – Jake Oct 28 '18 at 22:43
  • $\begingroup$ It looks better, but it's not a proof yet. You haven't yet used that $g$ or $f$ are uniformly continuous, for example. So why is it a problem that $f(x), f(y) \in \mathbb{R}$? $\endgroup$ – user296602 Oct 28 '18 at 22:44
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To prove that $g\circ f$ is uniformly continuous on $\mathbb{R},$ one needs to prove that for any $\varepsilon>0$, there exists $\delta>0$ such that for all $x,y\in \mathbb{R},$ if $|x-y|<\delta,$ then $$|g(f(x)) - g(f(y))|<\varepsilon.$$


Now, fix $\varepsilon>0.$ Since $g$ is uniformly continuous on $\mathbb{R},$ there exists $\eta>0$ such that for all $x,y\in \mathbb{R},$ if $|x-y|<\eta,$ then $$|g(x) - g(y)|<\varepsilon.$$ Since $f$ is uniformly continuous on $\mathbb{R},$ there exists $\delta>0$ such that for all $x,y\in\mathbb{R},$ if $|x-y|<\delta,$ then $$|f(x)-f(y)|<\eta.$$

We claim that such $\delta>0$ will work. Indeed, fix $x,y\in\mathbb{R}$ such that $|x-y|<\delta.$ By uniform continuity of $f$, we have
$$|f(x)-f(y)|<\eta.$$ By uniformly continuity of $g,$ we have $$|gf(x))-gf(y))|<\varepsilon.$$ This concludes that $g\circ f$ is uniformly continuous on $\mathbb{R}.$

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