# Existence of p=-infinity norm [closed]

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.

• It's not a norm, though. Oct 25, 2018 at 7:53
• 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 Oct 25, 2018 at 13:35
• I think this question is well suited for math.SE ... Oct 25, 2018 at 14:40

## 2 Answers

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.