MathOverflow is a question and answer site for professional mathematicians. 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

On the one hand, Wikipedia suggests that every distribution defines a Radon measure:

On the other hand, Terry Tao and LK suggest not:

Can someone please clarify this for me?

share|cite|improve this question
You might be interested in the answers to this question:… – Tom Leinster Mar 10 '10 at 19:00
Wikipedia does not suggest that every distribution defines a Radon measure, it says that every distribution which is non-negative on non-negative functions is positive Radon measure, and this is a rather different statement! – Mariano Suárez-Alvarez Mar 13 '10 at 14:58
Yes, see my comment on Deane Huang's post below. My error was assuming that every distribution is the difference of two positive distributions. This holds (as far as I remember) for signed measures. – Tom Ellis Mar 14 '10 at 12:14

Do you mean this sentence:

Conversely, essentially by the Riesz representation theorem, every distribution which is non-negative on non-negative functions is of this form for some (positive) Radon measure.

The condition that the distribution be non-negative for non-negative functions is non-trivial. Not every distribution satisfies this, so not every distribution is a Radon measure.

The fundamental examples are the delta function at a point (which is a measure) and its derivatives (which are not measures).

share|cite|improve this answer
Perhaps I should have been clearer: does every distribution correspond to a signed measure? – Tom Ellis Mar 12 '10 at 8:13 says "For an example of a distribution .. which does not arise from a measure, consider the derivative of the Dirac distribution. (As a functional, it maps a test function f to −f′(0).)" -- so my understanding of signed measures is not deep enough: there's something magic about measures that makes every signed measure the difference of two measures. The equivalent result is clearly not true for distributions! – Tom Ellis Mar 12 '10 at 8:49
I love the sentence “The condition that the distribution be non-negative for non-negative functions is non-trivial.” It seems that it should be simplifiable by some sort of elimination of double negations, but any such ‘simplification’ radically changes its meaning. – L Spice Jun 21 '11 at 20:49

I think the decisive point is continuity with respect to different topologies. Let $C$ be the space of continuous functions of compact support and $D$ the space of smooth functions of compact support. The inclusion $D\hookrightarrow C$ is a continuous map when you give both spaces the corresponding inductive limit topology. That means, that every continuous linear functional of $C$, i.e., each Radon-measure, defines a continuous linear functional on $D$, i.e., a distribution. But not every distribution extends to a continuous linear map on $C$. Examples are the derivatives of the Dirac distribution. The line in Wikipedia relates to an important property of linear functionals on $C$: if such a functional is positive, i.e., if it maps functions $f\ge 0$ to numbers $\ge 0$, then it is AUTOMATICALLY CONTINUOUS. This is a very important fact, though it is not hard to prove.

share|cite|improve this answer
It is worth noting that the first part of this answer generalizes, in the sense that $D$ is dense in just about any function space you can think of, with continuous inclusion. E.g., $L^p$, Sobolev spaces, and so forth. And hence the duals of such function spaces can be considered to consist of distributions. – Harald Hanche-Olsen Mar 13 '10 at 17:04
A much better explanation than mine. – Deane Yang Mar 13 '10 at 17:11
Sorry for perhaps sounding stupid, but: Why does the Hahn-Banach theorem not work in extending the measure here? – Regenbogen Mar 14 '10 at 0:37
Because it is not continuous with respect to the topology of D. – Anton Mar 14 '10 at 14:15
up vote 2 down vote accepted

This is a summary of what I've learned about this question based on the answers of the other commenters.

[*] Any positive distribution defines a positive Radon measure.

I had naively assumed a result for distributions like The Hahn Decomposition Theorem[1] for measures, i.e. I assumed that a distribution could be expressed as the difference of two positive distributions. If it could be, then applying Theorem [*] would yield the result that any distribution is a signed measure.

However, this is not the case. The derivative of the delta function, i.e. δ', satisfies δ'(f) = -f'(0). This is not a measure. I can't find any way of proving it's not the difference of two positive distributions, other than by contradiction using the above result.


share|cite|improve this answer
δ' is not continuous on the space of continuous functions. Would this show that δ' is not a signed measure? – timur Jul 15 '11 at 9:34

Measures are dual to continuous functions, whereas distributions are derivatives of them.

share|cite|improve this answer
well, at least in original definition, distributions are dual to smooth functions... – Yemon Choi Mar 11 '10 at 3:55
Are they not dual also to the subspace of continuous functions of $\mathcal D$ in its subspace topology? – Mariano Suárez-Alvarez Mar 13 '10 at 15:00

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.