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?

  • $\begingroup$ You might be interested in the answers to this question: mathoverflow.net/questions/4706/… $\endgroup$ – Tom Leinster Mar 10 '10 at 19:00
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    $\begingroup$ 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! $\endgroup$ – Mariano Suárez-Álvarez Mar 13 '10 at 14:58
  • $\begingroup$ 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. $\endgroup$ – 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).

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  • $\begingroup$ Perhaps I should have been clearer: does every distribution correspond to a signed measure? $\endgroup$ – Tom Ellis Mar 12 '10 at 8:13
  • $\begingroup$ ncatlab.org/nlab/show/distribution 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! $\endgroup$ – Tom Ellis Mar 12 '10 at 8:49
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    $\begingroup$ 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. $\endgroup$ – LSpice 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.

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    $\begingroup$ 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. $\endgroup$ – Harald Hanche-Olsen Mar 13 '10 at 17:04
  • $\begingroup$ A much better explanation than mine. $\endgroup$ – Deane Yang Mar 13 '10 at 17:11
  • $\begingroup$ Sorry for perhaps sounding stupid, but: Why does the Hahn-Banach theorem not work in extending the measure here? $\endgroup$ – Regenbogen Mar 14 '10 at 0:37
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    $\begingroup$ Because it is not continuous with respect to the topology of D. $\endgroup$ – user1688 Mar 14 '10 at 14:15

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.

[1] http://en.wikipedia.org/wiki/Hahn_decomposition_theorem

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  • $\begingroup$ δ' is not continuous on the space of continuous functions. Would this show that δ' is not a signed measure? $\endgroup$ – timur Jul 15 '11 at 9:34

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

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

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