In the snippet below, the highlighted part is the average norm, but since $1/|p_i|$ is outside the summation, it is very confusing to understand.

  • is $|p_i|$ l2-norm(as per wolfram) or l1-norm or absolute value as per wiki.

  • Should the $i$ inside summation be considered for $1/|p_i|$, which is outside the summation?

enter image description here


I agree that this notation is unclear. I would interpret it as follows:

Given that the expression is supposed to denote the average norm $|p_i|$ is likely the cardinality of the set $\{p_i\}$.

In that case the expression would just be the sum over all norms divided by the number of norms, resulting in the average norm. The authors likely use this notation because they didn't want to introduce the number $n$ of $p_{i < n}$. $|\{p_i\}|$ would be clearer, but maybe uglier to typeset.

  • $\begingroup$ Much obliged for the response, I'm confused with the usage of mod('|x|') notation, since cardinality will either take zero or a positive number. $\endgroup$ – Stephen Philip Jan 28 '20 at 4:56

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