Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

What properties does a distribution (in the generalized function sense) has to have in order to be a function. That is, when is $T(\varphi) = \int f \varphi$ for some $f$?

share|improve this question
en.wikipedia.org/wiki/Radon–Nikodym_theorem –  Ryan Budney Feb 8 '10 at 2:35
Ryan: I think a bit more detail might be needed to show how one leverages the RN-theorem for measures to get an analogous result for distributions. (Also, link error.) –  Yemon Choi Feb 8 '10 at 2:43

3 Answers 3

up vote 7 down vote accepted

First of all, $T$ must have order zero, i.e., $|T(\varphi)|\le C(K)\sup|\varphi|$ for any test function $\varphi$ supported on a compact set $K$. By Riesz representation theorem, $T$ is a measure. To be a locally integrable function, it must be absolutely continuous with respect to the Lebesgue measure. One way to express this condition: $C(K)\to 0$ as the Lebesgue measure of $K$ tends to zero, which $K$ staying within a fixed compact set.

share|improve this answer
Do you have any references where I could learn the details of this? Thank you very much! –  commonname Feb 8 '10 at 2:55
Almost any introduction to distribution theory will contain the required ingredients for this argument. Personally, I learned it first from Rudin's functional analysis book. –  Harald Hanche-Olsen Feb 8 '10 at 3:14

I haven't thought about this carefully enough, but it seems that there is some ambiguity in your question about what the integral $\int f\varphi$ is supposed to mean. As Ryan and Leonid have said: if you want the representing function $f$ to be locally integrable then the Radon-Nikodym theorem is what you need.

On the other hand, if you allow principal-value integrals (which is probably not what you want, I'm guessing, but I wasn't sure from your question) then I think

$$ \varphi \mapsto \int_{\rm p.v.} \frac{\varphi(t)}{t}\ dt $$

would be a tempered distribution that is in some sense `represented by a function', even though the function is not everywhere locally integrable.

share|improve this answer

Assuming that the question is to be understood in the sense of when a distribution is represented by a locally integrable function, here is a characterisation which is perhaps more applicable than the solution already given: for each compact $K$ and each sequence $(\phi_n)$ of test functions with support in $K$ which are uniformly bounded and converge in the $L^1$-norm to zero, $T(\phi_n) \to 0$. This is because there is a nice, complete topology on $L^\infty(K)$ for which the test functions are dense, the dual is $L^1$ and the convergence is as above. There are several explicit descriptions of this topology---as a strict topoogy, as a mixed topology or as the Mackey topology for the duality $(L^\infty,L^1)$ (see the book "Saks Spaces and Applications to Functional Analysis").

share|improve this answer

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.