What is the l1-normalization of some data? From this page and in this paper (first paragraph of chapter 2.1) there is the term of "$l_1$-normalization" or absolute normalization of a vector (i.e. some data). The scope is to turn the data into a distribution (PMF) that sums up to 1. The definition on the page is:

It may be defined as the normalization technique that modifies the
dataset values in a way that in each row the sum of the absolute
values will always be up to 1. It is also called Least Absolute
Deviations.

For example $v=[1,2,3]^T$. Does the $l_1$-normalization simply mean:
$$
l_1(v_i)=\frac{|v_i|}{\sum_{j=1}^n |v_j|} \Leftrightarrow l_1(v)=[\frac{1}{6},\frac{2}{6},\frac{3}{6}]^T
$$
? Also, is this "better" in terms of keeping the original proportions than the min-max or softmax normalization?
 A: In general, the normalization of a vector $v$ with respect to a norm $\| \cdot \|$ is given by $y = \frac{v}{\| v \|}$. This new vector $y$ has the properties that (1) it has norm one, meaning that $\| y \| = 1$, and (2) it has the same direction as the original vector $v$, meaning that $v$ is proportionate to $y$.
The term $\ell_1$ normalization just means that the norm being used is the $\ell_1$ norm $\| v \|_1 = \sum_{i = 1}^n |v_i|$. This means that your formula is somewhat mistaken, as you shouldn't be taking the absolute values of the $v_i$'s in the numerator. And note that in general, $\ell_1$ normalization does not make a vector into a pmf because the normalized vector can have negative entries.
Vector normalization always preserves the original proportions, regardless of the norm used, because
$$
\frac{v_i}{v_j} = \frac{v_i / \| v \|}{v_j / \| v \|},
$$ but min-max and softmax normalizations do not. Try it with the vector $v = (1, 2, 3)$:
$$\text{min-max}(v) = \left( \frac{1 - 1}{3 - 1}, \frac{2 - 1}{3 - 1}, \frac{3 - 1}{3 - 1}\right) = \left(0, \frac{1}{2}, 1\right) \\
\text{softmax}(v) = (0.09, 0.24, 0.67)
$$
which clearly don't have the same proportions as $(1, 2, 3)$.
