Given a vector $\mathbf {x} =(x_{1},\ldots ,x_{n})$ the p-norm is defined as $$ \left\|\mathbf {x} \right\|_{p}:={\bigg (}\sum _{i=1}^{n}\left|x_{i}\right|^{p}{\bigg )}^{1/p} $$ for $p \geq 1$.

For $p=\infty$ one obtains the maximum norm $$\left\|\mathbf {x} \right\|_{\infty }:=\max \left(\left|x_{1}\right|,\ldots ,\left|x_{n}\right|\right). $$

I was wondering what happens if we consider the case $p=-\infty$?

From some numerical experiments it seems like $$ \lim_{p\rightarrow -\infty} \bigg (\sum _{i=1}^{n}\left|x_{i}\right|^{p}{\bigg )}^{1/p}= \min \left(\left|x_{1}\right|,\ldots ,\left|x_{n}\right|\right). $$ Does this result hold in general? I could not find any reference for this problem.

  • 5
    $\begingroup$ It's not a norm, though. $\endgroup$
    – Yemon Choi
    Oct 25, 2018 at 7:53
  • $\begingroup$ The other interesting limit case is $p=0$. That is, $$\lim_{p\to 0}\left(\frac{|x_1|^p+\dots+|x_n|^p}{n}\right)^{1/p}$$The geometric mean $\endgroup$ Oct 25, 2018 at 13:35
  • 2
    $\begingroup$ I think this question is well suited for math.SE ... $\endgroup$
    – Suvrit
    Oct 25, 2018 at 14:40

2 Answers 2


Yes this limit is correct (subject to Yemon Choi's comment). See for example the section on means, in the book Analytic Inequalities by Mitrinovic.


A simple relation between $\|\cdot\|_p$ and $\|\cdot\|_{-p}$ is $$ \displaystyle\left\|\frac{1}{\mathbf x}\right\|_{-p} = \frac{1}{\|\mathbf x\|_{p}} $$ where $\frac{1}{\mathbf x}$ is defined with components $\frac{1}{x_k}$. You then easily deduce your min formula from the max formula you already stated. Of course, all $x_n$ are nonzero, so that you can do negative powers.


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