Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I would like to know the interpretation of limit expression which calculates the number of non-zero entries in vector $x$ in the expression below: $$\lVert x\rVert_0 = \#\{i \mid x[i] \neq 0\} = \lim_{p\to 0^+} \left(\sum_{i=1}^{m} |x[i]|^p\right),$$ where $x[i]$ is the $i$th coordinate of $x$ in $\mathbb{R}^m$ and $p \geq 1$

share|cite|improve this question
I think that the restriction $p \geq 1$ can't be right. You need $p$ to get close to zero from the right. – Ben Blum-Smith Aug 30 '11 at 15:52
up vote 14 down vote accepted

$\lim_{p \to 0^{+}}$ indicates that the limit is meant to be taken only from the positive direction; it's a one-sided limit.

share|cite|improve this answer

As Qiaochu Says $\lim_{x \to 0+}$ means that $x$ approaches to $0$ from the Positive side. It shall be clear once you see an example.

Example so that you can understand better. Consider $f: \mathbb{R} \to \mathbb{R}$ Let $f(x) = \lfloor{x\rfloor}$. We prove that the point at which $f$ is not continuous at $2$. Let us consider the right hand limit $\lim_{h \to 0+} f(2)$ and the left hand limit $\lim_{h \to 0-} f(2)$.

  • Right hand Limit: $\displaystyle\lim_{h \to 0+} f(2) = \displaystyle\lim_{h \to 0} f(2+h) = \lim_{h \to 0}\lfloor{2+h\rfloor} =2$.

  • Left Hand Limit: $\displaystyle\lim_{h \to 0-} f(2)= \displaystyle\lim_{h \to 0} f(2-h) =\lim_{h \to 0} \lfloor{2-h\rfloor} = 1$.

Hence $f$ is not continuous at $x =2$.

share|cite|improve this answer
Sometimes $x \to 0^+$ is written $x \searrow 0$ or $x \downarrow 0$ – GEdgar Sep 3 '11 at 14:22
@GEdgar I thought $x \downarrow 0$ means that $x$ is approaching $0$ from above in a (strictly) decreasing manner, whereas $x \rightarrow 0^+$ does not necessarily imply that the values of $x$ are strictly decreasing as $x$ approaches $0$; depending on the function, it could be that $x$ is generally nonincreasing. I guess what I'm asking is the notation $\downarrow$ restricted to strictly decreasing values of $x$? – Benedict Voltaire Sep 15 '15 at 19:59

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.