I have seen the claim that the l0-norm ($\|X\|_0$ = support(X)) is a pseudo-norm because it does not satisfy all properties of a norm. I thought it to be triangle inequality, but am not able to show it by example. Can anyone give an example to show that the l0-norm does not satisfy the triangle inequality?


  • $\begingroup$ It does not violate triangle inequality. It's not positive homogeneous. $\endgroup$ – S.B. Jun 21 '13 at 4:03
  • $\begingroup$ This can't be the support of $X$. It must be its measure for some measure. The counting measure if $X$ is a finite-dimensional vector. The measure at use if $X$ is some measurable function. $\endgroup$ – Julien Jun 21 '13 at 15:03

is a semi-norm because it does not satisfy all properties of a norm.

This is a misuse of the term seminorm, which is defined as a norm, except that it's okay for a nonzero element to have seminorm $0$.

The cardinality of support satisfies the triangle inequality, since the support of the sum of two vectors is contained in the union of their supports.

But the property $\|\lambda x\|=|\lambda|\|x\|$ clearly fails for $l_0$: instead we have $\|\lambda x\|_0=\|x\|_0$ for all $\lambda\ne 0$.

  • $\begingroup$ Thanks, perhaps a better term is pseudo-norm? My memory was just clouded by glancing at semi-norm in some wiki articles without exactly reading through it. $\endgroup$ – Atul Deshpande Jun 21 '13 at 16:11
  • $\begingroup$ @AtulDeshpande Pseudo-norm is better, although it's also sometimes used as a synonym of "seminorm". Then there's quasinorm but it weakens the triangle inequality instead. I don't think there is a standard term for the concept obtained from norm by changing the homogeneity requirement. $\endgroup$ – ˈjuː.zɚ79365 Jun 21 '13 at 23:55

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