Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The p-norm is defined as:

$$ \ \|x\|_p=\left(|x_1|^p+|x_2|^p+\dotsb+|x_n|^p\right)^{\frac{1}{p}} $$

When $p<1$, this is no longer a "norm" because it violates the triangle inequality (- it is super additive and not subadditive). However, it is still valid to ask, what is its limit when p goes to 0?

My guess is that:

  • If all coordinates are 0, then $l_p=0$, and it remains like this when p=0.
  • If exactly one coordinate, say $x_i$, is non-zero, then $l_p=x_i$, and it remains like this when p=0.
  • If more than one coordinate (say, $x_i$ and $x_j$) are non-zero, then $l_p>x_i$, and because the exponent goes to $\infty$, the value of $l_p$ also goes to $\infty$.

Is this correct?

(Note that this is not equal to the $l_0$ "norm" = the number of nonzero elements. This is also not equal to the scaled norm, in which there is an additional $1/n$ factor).

share|cite|improve this question
If $n\gt1$, then $$ \left(|x_1|^p+|x_2|^p+\dots+|x_n|^p\right)^{1/p}\to n^{1/p}\to\infty $$ so yes, you are correct. – robjohn Oct 18 '13 at 6:01
Related:… – Jonas Meyer Oct 18 '13 at 6:07
up vote 3 down vote accepted

Your guesses are all correct. In particular, the limit is $+\infty$ if there are two or more nonzero components.

Suppose WLOG that $x_1\neq 0$ and $x_2\neq 0$, and that $|x_1|\leq |x_2|$. Then $\|x\|_p\geq 2^{1/p}|x_1|\to +\infty$ as $p\searrow 0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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