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 $l<n$."

I would be glad if someone can explain the above.

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Did you mean to write $|x|^{-l}$? –  Nate Eldredge Jan 14 '12 at 21:30
On what space is this function defined? ${\mathbb R}^n$? In that case, isn't this just the fact that when $l$ is small relative to $n$, then that function is integrable? (Switch to polar coordinates) –  user16299 Jan 14 '12 at 21:36
@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. –  Anirbit Jan 21 '12 at 21:41

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 $0$.
Thanks for the reply. What is the $s_n$ in your first line? Can you kindly explain as to exactly what is the definition of "local integrability" and why is that necessary for something to be a distribution? I am not familiar with this theory and hence my question. –  Anirbit Jan 21 '12 at 21:44
$s_n$ is the area of the surface of the unit ball in $\mathbb R^n$. A function $f$ is locally integrable if $f\mathbf 1_K$ is integrable for all compact $K$. If $f$ is non-negative and not locally integrable you will have a problem, since you can take a compact on which $f$ is not integrable and a test function which is equal to $1$ on this compact. –  Davide Giraudo Jan 21 '12 at 21:51
But on the RHS of $<T_f,\phi>$ you seem to be integrating $\vert x\vert ^{-l}\phi(x)$ on the whole of $\mathbb{R}^n$, doesn't that somehow seem stronger than just "local integrability"? And where does being $L^1$ enter the discussion as was said in the initial comment? –  Anirbit Jan 22 '12 at 21:51
In fact when we integrate $|x|^{-l}\phi(x)$ on $\mathbb R^n$, we take the integral over a compact since $\phi$ has a compact support. If $f$ is locally integrable, then for a test function $\phi$ the expression $\int_{\mathbb R^n}f(x)\phi(x)dx$ makes sense, since we integrate over a compact and $\phi$ is bounded on this compact. –  Davide Giraudo Jan 23 '12 at 12:52
Okay. So for the definition of distributions you are restricting to those $\phi$ that have compact support? Is that the convention about defining distributions? –  Anirbit Jan 24 '12 at 20:53