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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).

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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: math.stackexchange.com/questions/282271/… –  Jonas Meyer Oct 18 '13 at 6:07

1 Answer 1

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 $p\searrow 0$.

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