Let $f : \mathbb{R} \rightarrow \mathbb{R}$ and $g : \mathbb{R} \rightarrow \mathbb{R}$ are two uniform continuous functions. Which of the following options are correct and why?
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$f(g(x))$ is uniformly continuous.
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$f(g(x))$ is continuous but not uniformly continuous.
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$f(g(x))$ is continuous and bounded.
My attempt:
Every uniformly continuous function maps a Cauchy sequence to a Cauchy sequence. (Here I have a doubt, as the converse may not be true). So if $\{x_n\}$ be a Cauchy sequence, $\{f(x_n)\}$ and $\{g(f(x_n))\}$ both will be Cauchy sequence. So $g(f(x))$ will be uniformly continuous, i.e. 1 is true.
Composite function of two continuous functions will be continuous. As 1 is true, 2 is false.
$f(x) = x $ is uniformly continuous. $g(x) = \log(x)$ is uniformly continuous in $[1,\infty)$. So $g(f(x)) = \log(x)$ is uniformly continuous in $[1, \infty)$, but not in $\mathbb{R}$.
Thank you for your help.
Best Answer
Option $(1):$ True:
Choose $\epsilon>0.$
Then $\exists~\delta_1>0$ such that $|x_1-x_2|<\delta_1$$\implies|f(x_1)-f(x_2)|<\epsilon.$
For above $\delta_1>0~\exists~\delta>0$ such that $|x_1-x_2|<\delta$$\implies|g(x_1)-g(x_2)|<\delta_1.$
Consequently, $|x_1-x_2|<\delta$$\implies|(f\circ g)(x_1)-(f\circ g)(x_2)|<\epsilon.$
Options $(2),~(3):$ Not necessarily true:
Take $f(x)=g(x)=x$ on $\mathbb R.$
Then $f,g$ are uniformly continuous on $\mathbb R.$
Note $(f\circ g)(x)=x$ is unbounded and uniformly continuous on $\mathbb R.$