# Uniform convergence criterion in unbounded domains [closed]

We see that for a sequence of functions $$\langle f_n \rangle$$ be defined on a compact set $$D \subset \mathbb R$$, $$f_n \rightrightarrows f ~~~\text{if and only if}~~~||f_n-f||_\infty \rightarrow 0$$ where $$'\rightrightarrows'$$ indicates the uniform convergence and $$\displaystyle{||f_n-f||_\infty=\sup_{x}\{|f_n(x)-f(x)|~:~x \in D\}}$$ ,

Can we extend the this necessary and sufficient condition for uniform convergence in to a semi infinite domain or even into $$\mathbb R$$? Any difficulties with the extension?

• This holds for any domain (whether it is a subset of reals or not) almost by definition. Jan 4, 2022 at 16:47
• That is the definition of uniform convergence. Jan 4, 2022 at 16:48

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.
• @peek-a-boo Yeah, you are right. I should make it $\varepsilon/2$ to avoid the nuance. Jan 4, 2022 at 17:11