Uniform convergence criterion in unbounded domains

Solution 1:

It is true for any domain and it can be viewed as the definition of uniform convergence. Maybe you want to recover the $\varepsilon$-version of the definition:

Let $D$ be any domain, not necessarily compact. If $f_n$ converges to $f$ uniformly, then for any $\varepsilon>0$, there exists an $N>0$ such that $|f_n(x)-f(x)|<\varepsilon/2$ for any $x\in D$ and any $n\geq N$. By taking the supremum over all $x\in D$, you get $\|f_n-f\|_\infty\leq\varepsilon/2<\varepsilon$, which implies $\|f_n-f\|_\infty\to 0$ as $n\to \infty$.

Conversely, when you have $\|f_n-f\|_\infty\to 0$ as $n\to \infty$, it means for any $\varepsilon>0$, there exists an $N>0$ such that $\sup_{x\in D} |f_n(x)-f(x)|<\varepsilon$, in particular $|f_n(x)-f(x)|<\varepsilon$ for any $x\in D$. Then you recover the $\varepsilon$-version of defition of uniform convergence.