# 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?

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 $$(1, 2, 3)$$.