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I'm reading here about sequence of functions in Calculus II book,

and there's a theorem that says:

A sequence of functions $\{f_n(x)\}_0^\infty$ converges uniformly to $f(x)$ in domain $D$ $\iff$ $\lim_{n \to \infty} \sup_{x \in D} |f_n(x) - f(x)| = 0.$

I really serached a lot , in Google, Wikipedia and Youtube,

And I'm still having difficulties to understand what is sup.

I'll be glad if you can explain me. thanks in advance!

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  • $\begingroup$ I suggest you start with wiki article on supremum. $\endgroup$ Commented Jan 7, 2014 at 8:46
  • $\begingroup$ @TZakrevskiy I read, there's no reason I would lie .. $\endgroup$
    – Billie
    Commented Jan 7, 2014 at 8:51
  • $\begingroup$ Supremums are defined whenever you have a partially ordered set $A$. You can probably google it. $\endgroup$
    – Pedro
    Commented Jan 7, 2014 at 11:48

3 Answers 3

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supremum means the least upper bound. Let $S$ be a subset of $\mathbb{R}$

$$ x = \sup(S) \iff ~ x \geq y~\forall y \in S \mbox{ and } \forall \varepsilon > 0, x - \varepsilon \mbox{ is not an upper bound of } S $$

You may also define $\sup(S) = +\infty$ when $S$ is not bounded above.

The reason why we have supremum instead of simply maximum is that in some subset of $\mathbb{R}$, we do not have maximum element, let's take an open interval $(0,1)$ as an example, $\max\{(0,1)\}$ does not exist, but $\sup\{(0,1)\} = 1$.

Supremum of a nonempty subset having an upper bound always exists by the completeness property of the real numbers.

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The supremum of set $A\subseteq\mathbb{R}$ is the unique $y\in\mathbb{R}\cup\left\{ \infty\right\} $ with:

1) $a\leq y$ for each $a\in A$.

2) If $z<y$ then some $a\in A$ exists with $z<a$.

An element $y$ that suffices 1) is an upper bound of $A$. If it also suffices 2) the it is unique and is the least upper bound of $A$. Supremum and least upper bound are the same thing.

In your case $A_n=\left\{ \left|f_{n}\left(x\right)-f\left(x\right)\right|:x\in D\right\} $ and $y_n:=\sup A_n\in \mathbb R$.

On the right side of $\iff$ it is stated that $y_n$ converges to $0$.

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Your set $A_n=\{|f_n(x)-f(x)|\colon x \in D\}$ is the set of the real numbers that are the distances between the nth function and the limit function. Then the least upper bound (read the supremum) of $A_n$ is the largest distance between the nth function $f_n(x)$ and the limit function $f(x)$.

The sequence converges uniformly if the supremum of $A_n$ tends to $0$ as $x\to\infty$.

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