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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

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  • $\begingroup$ Look over your definitions. You have missed some parts. $\endgroup$ – md2perpe Jun 2 '17 at 11:51
  • $\begingroup$ Thanks. Just fixed it. $\endgroup$ – Natash1 Jun 2 '17 at 12:34
  • $\begingroup$ 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. $\endgroup$ – md2perpe Jun 2 '17 at 13:10
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    $\begingroup$ 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. $\endgroup$ – md2perpe Jun 3 '17 at 14:04
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    $\begingroup$ For uniform convergence on an interval $I$, I prefer this sort of syntax: $\forall\varepsilon > 0 \, \exists N \in \mathbb{N} : \forall x \in I \left(n\geq N \implies \left|\, f_n(x) -f(x) \right| < \varepsilon \right)$ $\endgroup$ – Matt A Pelto Jun 3 '17 at 23:58

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