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 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.

share|cite|improve this question
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
up vote 2 down vote accepted

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 $0$.

share|cite|improve this answer
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

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.