Showing a sequence of functions converges uniformly on any bounded interval 
Question: Let $\{f_n\}$ be a sequence of continuous functions on $\mathbb{R}$.  Let $f_n \to f$ uniformly on $\mathbb{R}$.  Let $g_n(x):=f_n(x+\frac{1}{n})$ for $n=1,2,3,....$  Then $g_n \to f$ uniformly on any bounded interval $[a,b]$.  

I think its apparent that the pointwise limit of the function $g_n$ is $f$.
I'm stuck trying to show this convergence is uniform on any bounded interval.
 A: Hint 1: Decompose the problem into two easier questions:
$$
g_n(x)-f(x)=f_n\left(x+\frac{1}{n}\right)-f\left(x+\frac{1}{n}\right)+f\left(x+\frac{1}{n}\right)-f(x).
$$
Hint 2: Continuous implies uniformly continuous on compact intervals (Heine-Cantor).
Details: For all $x\in\mathbb{R}$
$$
\lvert f_n\left(x+\frac{1}{n}\right)-f\left(x+\frac{1}{n}\right)\rvert\leq \|f_n-f\|_\infty.
$$
So the left hand side term of the sum above converges uniformly to $0$ on $\mathbb{R}$, a fortiori on $[a,b]$.
Now $f$ is continuous on $[a,b+1]$, so it is uniformly continuous on this compact interval. This means that for every $\epsilon>0$, there exists $\delta>0$ such that such that $|f(y)-f(x)|\leq \epsilon$ for all $x,y$ in $[a,b+1]$ such that $|x-y|\leq\delta$. Now take $N$ large enough to have $\frac{1}{N}\leq \delta$. Then for all $n\geq N$ and all $x\in [a,b]$, we have $|x+\frac{1}{n}-x|=\frac{1}{n}\leq\frac{1}{N}\leq\delta$ and $a\leq x<x+\frac{1}{n}\leq b+1$. So 
$$
\lvert f\left(x+\frac{1}{n}\right)-f(x) \rvert\leq \epsilon.
$$
This proves that the right hand side of the sum above converges uniformly to $0$ on $[a,b]$.
So $g_n-f$, which is the sum of these two terms, converges uniformly to $0$ on $[a,b]$.
