# If $f$ and $g$ are uniformly continuous on $\Bbb{R}$ then $f\circ g$ is uniformly continuous on $\Bbb{R}$

Prove or disprove: If $f$ and $g$ are uniformly continuous on $\Bbb{R}$ then $f\circ g$ is uniformly continuous on $\Bbb{R}$. I think there's something crooked in my attempt. I would like to know what it is and would appreciate your replies.

$Attempt:$ Let $\epsilon>0$. $f$ is uniformly continuous and therefore $\exists \delta>0$ such that $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta$, $|f(x)-f(y)|<\epsilon$. Since $g$ is uniformly continuous, for this $\delta$ there exists $\delta _1$ such that $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta_1$, $|g(x)-g(y)|<\delta$. Taking $\delta_{f\circ g}=\min\{\delta, \delta_1\}$, we get that $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta_{f\circ g}, |f(x)-f(y)|<\epsilon$ and $|g(x)-g(y)|<\delta$. Therefore $|f(g(x))-f(g(y))|<\epsilon$ $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta_{f\circ g}$, proving $f\circ g$ is uniformly continuous on $\Bbb{R}$.

• The only crooked thing I see is that you misspelt "attempt". – David Mitra Jan 28 '15 at 23:32
• Oh, Too tired(and not native lol). Thank you for your evaluation. – Donna Jan 28 '15 at 23:33

Taking $\delta_{f\circ g}=\min\{\delta, \delta_1\}$, we get that $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta_{f\circ g}, |f(x)-f(y)|<\epsilon$ and $|g(x)-g(y)|<\delta$.
Also, you do not need the $\min$. The following works:
We get $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta_1$ that $|g(x)-g(y)|<\delta$, and thus $|f(g(x))-f(g(y))|<\epsilon$.
• Glad it was helpful. I also had to look twice to note that the $\min$ is not necessary. – quid Jan 28 '15 at 23:49