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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 $p \geq 1$

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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
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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.

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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 $x =2$.

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Sometimes $x \to 0^+$ is written $x \searrow 0$ or $x \downarrow 0$ –  GEdgar Sep 3 '11 at 14:22
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