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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?

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    $\begingroup$ This holds for any domain (whether it is a subset of reals or not) almost by definition. $\endgroup$
    – peek-a-boo
    Jan 4, 2022 at 16:47
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    $\begingroup$ That is the definition of uniform convergence. $\endgroup$
    – copper.hat
    Jan 4, 2022 at 16:48

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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.

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    $\begingroup$ There is only one subtlety in the equivalence, but you haven't explained or even mentioned it. In the first paragraph, taking the supremum doesn't necessarily retain a strict inequality. It should be a weak inequality. $\endgroup$
    – peek-a-boo
    Jan 4, 2022 at 17:09
  • $\begingroup$ @peek-a-boo Yeah, you are right. I should make it $\varepsilon/2$ to avoid the nuance. $\endgroup$
    – Muduri
    Jan 4, 2022 at 17:11

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