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 $f$?
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2$\begingroup$ en.wikipedia.org/wiki/Radon–Nikodym_theorem $\endgroup$– Ryan BudneyCommented Feb 8, 2010 at 2:35
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1$\begingroup$ 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.) $\endgroup$– Yemon ChoiCommented Feb 8, 2010 at 2:43
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3 Answers
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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.
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$\begingroup$ Do you have any references where I could learn the details of this? Thank you very much! $\endgroup$ Commented Feb 8, 2010 at 2:55
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1$\begingroup$ 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. $\endgroup$ Commented Feb 8, 2010 at 3:14
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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.
answered Feb 8, 2010 at 3:25
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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").
