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!
On
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 $0$.
On
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 $x\to\infty$.
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.