Simple explanation of uniform norm / sup-norm?
For finite vectors $x$, the sup norm of $x$ is the greatest element of $x$ (greatest in absolute value of course to make complex elements ordered and to not distinguish the sign). Example: $(1,-1,2i)$ has greatest element $|2i|=2$.
For infinite vectors $x$ (that is, a sequence), the sup norm is the supremum of $x$: that is the lowest upper bound of $x$. If you draw the sequence of absolute values of elements of $x$, just visualize a straight horizontal line falling down on top of these elements (staying horizontal of course and beginning far above the sequence). If it stops, it cannot have any elements above of it, but there cannot another horizontal line on top of the elements that lays lower than the first one, else the first one could fall down even more. (Note that the line does not have to touch any of the elements: an example is $x_n=1-\frac{1}{n}$. The horizontal line is the line with height $1$.)
For functions, this is the supremum of all elements of its domain evaluated in $f$ (absolute value), so it is pretty much the same as the previous one, but then possibly with uncountably many elements $f(x)$. Compare $f(x)=1-\frac{1}{x}$ with the previous example.
Another example are $f(x)=\sin(x)$, where the supremum of $|\sin(x)|$ is equal to its the maximum. Keep it mind that the sequences and functions must be bounded in order to use the sup norm.
In the case of a finite-dimensional space $\ \mathbb R^n\ $ the sup-norm is simply the max-norm. The balls of such a space with the max-norm are cubes which have edges parallel to coordinate axes.
In applications, we often make a bunch $\ A\ $ of experiments, each has a numeric value, and an associated error. The said bunch and errors result in the respective vectors $\ x\ e\in\mathbb R^A.\ $ We often need to decide, based on the individual errors, what is the overall error for the whole given bunch. Depending on the nature of applications, errors are defined differently. One of the common ways is the uniform error, it is the max-norm $\ ||e||_\infty.$
Let $\ X\ $ be a non-empty set $\ (X\ne\emptyset).\ $ Given a function $\ f:X\rightarrow\mathbb R,\ $ we say that real numbers $\ B\,\ C\in\mathbb R\ $ are a lower and an upper bound of $\ f \,\ \Leftarrow:\Rightarrow$
$$ \forall_{x\in X}\quad B \le f(x) $$ and $$ \forall_{x\in X}\quad f(x) \le C $$
One or both bounds may be not well-defined (infinite).
Next, we define function $\ |f|:X\rightarrow\mathbb R\ $ as follows:
$$ \forall_{x\in X}\quad |f|(x)\ :=\ |f(x)| $$
A function $\ f:X\rightarrow\mathbb R\ $ is bounded $\ \Leftarrow:\Rightarrow\ f\,\ $ satisfies any of the following four equivalent conditions:
- $\ f\ $ admits both an upper and a lower bound;
- $\ |f|\ $ admits an upper bound;
- $\ \exists_{C\in\mathbb R}\quad |f|\le C $
- $\ \exists_{C\in\mathbb R}\forall_{x\in X}\quad |f|(x)\le C\ $
Thus $\ f\ $ is bounded $\ \Leftrightarrow\ f\ $ admits both a lower and an upper bound. Finally, the final elementary definition here: given $\ f:X\rightarrow\mathbb R,\ $ we define $\ \sup (f)\ $ as the least upper bound of $\ f\ $ (when it exists), i.e. as a real number $\ s\in\mathbb R\ $ such that:
$$ \forall_{x\in X}\quad f(x)\le s $$ and $$ \forall_{t<s}\exists_{x\in X}\quad f(x)>t $$
(In the case of a finite $\ X,\ $ we have $\ \sup(f) = \max(f)).$
Let $\ \mathcal B(X)\ $ be the set of all bounded functions $\ f:X\rightarrow\mathbb R.\ $ This set deserves to be called space since it has many nice properties, in particular, it is an algebra over $\mathbb R,\ $ and it is a superspace for several more famous spaces. We define a norm $\ ||f||_\infty\ $ called the sup-norm or the uniform norm or $L^\infty$-norm of $\ f\in\mathcal B(X),\ $ as follows
$$ ||f||_\infty\ :=\ \sup(|f|) $$