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So for a sequence of functions $\{ f_n\}_{n \in \mathbb{N}}$ then
$f_n$ converges pointwise to some function $f$ if $$\forall x \in I, \forall \epsilon > 0, \exists N > 0 :n\geq N \implies |f_n(x) - f(x)| < \epsilon.$$
Is an equivalent condition for this:
$f_n\to f$ pointwise on $(a,b)$ if for each fixed $x\in(a,b)$, $|f_n(x)-f(x)|\to 0$ as $n\to\infty$

and for uniform convergence,
$$\forall\epsilon > 0, \exists N>0 : n\geq N \implies |f_n(x) -f(x)| < \epsilon \quad \forall x \in I.$$ Is this equivalent to:
$f_n\to f$ uniformly on $(a,b)$ if $\sup_{a< x<b}|f_n(x)-f(x)|\to 0$ as $n\to\infty$ ?

I've searched around and every PDF file I see defines the definitions with the former for each.

Source:https://math.stackexchange.com/a/597777/424197

Natash1
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  • Look over your definitions. You have missed some parts. – md2perpe Jun 02 '17 at 11:51
  • Thanks. Just fixed it. – Natash1 Jun 02 '17 at 12:34
  • You are correct about the equivalent statements. One defines a norm $|f|u := \sup f(x)$ and then has uniform convergence if $|f_n - f|_u \to 0$. Often this norm is instead denoted $|f|\infty$ and functions that are equal on a set of measure (think: length, area, volume) zero are considered being the same function. – md2perpe Jun 02 '17 at 13:10
  • Thanks. I think I kind of get that (though, probably not fully because I haven't really learnt anything about measure theory and I've only encountered norms in the context of inner products). It seems like considering the supremum of a norm and showing that it doesn't approach 0 is much easier than finding specific values of $N$,$\epsilon$,$x$ when proving that a function does not converge uniformly? – Natash1 Jun 02 '17 at 13:13
  • Just to make sure: The second set of definitions are not iff statements right? – Natash1 Jun 03 '17 at 13:21
  • Where should "iff" be or not be, do you mean? – md2perpe Jun 03 '17 at 13:28
  • Sorry in the definition:
    $f_n\to f$ pointwise on $(a,b)$ if for each fixed $x\in(a,b)$, $|f_n(x)-f(x)|\to 0$ as $n\to\infty$
    instead of "if for each..." should it be "iff for each.."? (Most likely not since the equivalent definition isn't?)     – Natash1 Jun 03 '17 at 13:31
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    The 'if' in a definition "X is said to be A if P(X)" should be thought of as 'iff'. If P(X) then we say that X is A, and if X is A that means that P(X). A definition is not a theorem, it's an introduction of a new notation or a new term as an abbreviation for a (usually) lengthy description. – md2perpe Jun 03 '17 at 14:04
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    For uniform convergence on an interval $I$, I prefer this sort of syntax:

    $\forall\varepsilon > 0 , \exists N \in \mathbb{N} : \forall x \in I \left(n\geq N \implies \left|, f_n(x) -f(x) \right| < \varepsilon \right)$

     – M A Pelto Jun 03 '17 at 23:58
  • Thanks guys. Last question: Is this "equivalent" definition of the uniform convergence ever untrue in any setting? (Maybe in some other topological space, I haven't really learnt much on topology, so just curious) – Natash1 Jun 04 '17 at 14:47

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