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I came across this statement in a certain lecture/paper by Witten,

"The function $\vert x \vert^{-l}$ defines a distribution (without regularization) on $\mathbb R^n$ if and only if it is locally $L^1$ iff $l."

I would be glad if someone can explain the above.

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    @YemonChoi May be this is the basic thing that you are alluding to. I am not familiar with idea of "distribution (without regularization)" and "locally $L^1$" and hence the question.2012-01-21

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Note that for $\delta>0$: $\int_{\{\delta \leq |x|\leq 1\}}|x|^{-l}dx=s_n\int_{\delta}^1r^{n-1}r^{-l}dr=s_n\int_{\delta}^1r^{n-l-1}dr,$ and it has a limit $\delta\to 0$ if and only if $n-l-1>-1$ hence $n>l$. So the function $|x|^{-l}$ is locally integrable on $\mathbb R^n$ if and only if $n>l$. If $f$ is locally integrable then it defines a distribution by $\langle T_f,\varphi\rangle=\int_{\mathbb R^n}|x|^{-l}\varphi(x)dx$, and if $f$ is not locally integrable, we don't have a distribution on $\mathbb R^n$, since $T_f$ is not well defined, for examle for a $\varphi$ which is equal to $1$ on a neighborhood of $0$.

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    I think it's indeed a convention, since it's what I red I any book.2012-01-24