$f_n(x)= f(x+n)$ show that the limit function is uniformly continuous Let $f$ be a real-valued continuous function on $I=\{x\in \mathbb{R} | x \geq 0\}$. For a positive integer $n$ the function on $I$ is defined by
\begin{align*}f_n(x)= f(x+n)\end{align*}
Answer the following questions when the sequence of functions $\{f_n(x)\}_{n=1}^\infty$ converges uniformly on $I$

*

*The function $g$ on $I$ is defined by $g(x)=\lim _{n\xrightarrow{}{}\infty}f_n(x)$, show that $g$ is uniformly continuous on $I$

*show that $f$ is uniformly continuous on $I$
Here we need to show that $|g(x)-g(y)|<\epsilon$ whenever $|x-y|<\delta$ for all $x,y\in I$. Since $\{f_n(x)\}_{n=1}^\infty$ coverge uniformly to $g$ we have $|f_n{(x)}-g(x)|<\epsilon \implies |f(x+n)-g(x)|<\epsilon$ whenever $n>N$. How do we proceed from here.
Any hints or a solution would be appreciated.

Hints
For (1):
If $f_n\in C({\Bbb R})$, $n\geq 1$ is any uniformly convergent sequence of continuous functions, then the limit function $g$ is continuous (proof by an $\epsilon/3$-argument). This is usually part of standard undergraduate curriculum.
Look at the very definitions of $g(x)$ and $g(x+1)$ (for any fixed $x\in {\Bbb R}$) and realize that the values must be the same.
Thus, $g$ is a 1-periodic continuous function.
Conclude from this (using e.g. a compactness argument) that $g$ is uniformly continuous.
For (2): Given $\epsilon>0$ there is $N$ so that $|g(x)-f(x)|<\epsilon/3$ for $x\geq N$. Using that $g$ is uniformly continuous show there is $\delta_1$ so that $x,y\geq N, |x-y|<\delta_1 \Rightarrow |f(x)-f(y)|<\epsilon$.
Show that there is $\delta_2>0$ so that $0\leq x,y\leq N+1, |x-y|<\delta_2 \Rightarrow |f(x)-f(y)|<\epsilon$.
Finally, pick $\delta=\min\{\delta_1,\delta_2,1\}$ and conclude.
