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On $\mathbb{R}^n$ and $p\ge 1$ the $p$-norm is defined as $$\|x\|_p=\left ( \sum _{j=1} ^n |x_j| ^p \right ) ^{1/p}$$ and there is the $\infty$-norm which is $\|x\|_\infty=\max _j |x_j|$. It's called the $\infty$ norm because it is the limit of $\|\cdot\|_p$ for $p\to \infty$.

Now we can use the definition above for $p<1$ as well and define a $p$-"norm" for these $p$. The triangle inequality is not satisfied, but I will use the term "norm" nonetheless. For $p\to 0$ the limit of $\|x\|_p$ is obviously $\infty$ if there are at least two nonzero entries in $x$, but if we use the following modified definition $$\|x\|_p=\left ( \frac{1}{n} \sum _{j=1} ^n |x_j| ^p \right ) ^{1/p}$$ then this should have a limit for $p\to 0$, which should be called 0-norm. What is this limit?

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It gives the geometric mean of absolute values of the components. – Raskolnikov Dec 5 '10 at 18:43
In some fields, the $0$-norm is defined to be the number of nonzero entries in $x$: – Rahul Dec 5 '10 at 19:20
up vote 18 down vote accepted

When $p$ is small, $$x^p = \exp(p \log x) \approx 1 + p\log x.$$ Therefore $$\frac{1}{n} \sum_{j=1}^n x_j^p \approx 1 + p\frac{1}{n} \sum_{j=1}^n \log x_j = 1 + p \log \sqrt[n]{\prod_{i=1}^n x_j}.$$ On the other hand, we have $$(1+py)^{1/p} \longrightarrow \exp(y),$$ and so we easily get that the norm approaches the geometric mean, as Raskolnikov commented.

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