composition of two uniformly continuous functions. 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?


$f(g(x))$ is uniformly continuous.
$f(g(x))$ is continuous but not uniformly continuous.
$f(g(x))$ 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.
Option $(1):$ True:

Choose $\epsilon&gt;0.$ 

Then $\exists~\delta_1&gt;0$ such that $|x_1-x_2|&lt;\delta_1$$\implies|f(x_1)-f(x_2)|&lt;\epsilon.$ 

For above $\delta_1&gt;0~\exists~\delta&gt;0$ such that $|x_1-x_2|&lt;\delta$$\implies|g(x_1)-g(x_2)|&lt;\delta_1.$ 

Consequently, $|x_1-x_2|&lt;\delta$$\implies|(f\circ g)(x_1)-(f\circ g)(x_2)|&lt;\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 $\mathbb R.$
I don't think that composition of two uniform continuous function is uniform continuous......
my argument is.. let F be a function from $(0,1)$ to $(0,1)$ and $F(x)=x$
now let G be another function from $[0,1]$ to $[0,1]$ and $G(x)=x^2$ then G compose F is defined as from $(0,1)$ to $[0,1]$ and G compose f is x^2 which is not uniform in $(0,1)$. but F and G are uniform.