16
$\begingroup$

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?

$\endgroup$
20
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.