# If $f$, $g$ are uniformly continuous on $\mathbb{R}$, then $g \circ f$ is uniformly continuous on $\mathbb{R}$.

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.

• 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. – user296602 Oct 28 '18 at 22:25
• @T.Bongers What is the correct negation? – Jake Oct 28 '18 at 22:27
• 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 $\ge \epsilon$. – user296602 Oct 28 '18 at 22:30
• @T.Bongers I made these edits. How do they look? I am also working on a direct proof at the moment... – Jake Oct 28 '18 at 22:43
• 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 $f(x), f(y) \in \mathbb{R}$? – user296602 Oct 28 '18 at 22:44

## 1 Answer

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 $$\mathbb{R}.$$