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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?

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  • $\begingroup$ You might be interested in the answers to this question: mathoverflow.net/questions/4706/… $\endgroup$ Mar 10, 2010 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! $\endgroup$ Mar 13, 2010 at 14:58
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    $\begingroup$ 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, 2010 at 12:14

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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, 2010 at 8:13
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    $\begingroup$ 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, 2010 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, 2011 at 20:49
  • $\begingroup$ @deane-yang I had a quick question. In his book on distributions, Treves (1969) says that a distribution is a Radon measure if it can be shown to be continuous on $C_c^{\infty}(\Omega)$ in the topology induced by $C_c^0(\Omega)$ (Prop.21.2). He extends the continuity to $C_c^0(\Omega)$ based on the density of $C_c^{\infty}(\Omega)$ in $C_c^0(\Omega)$. However, I'm having trouble buying this, because the distribution may rely on the smoothness of a given function to 'work'. Am I missing something? $\endgroup$
    – demim00nde
    Feb 1 at 12:56
  • $\begingroup$ The key assumption is being continuous with respect to the topology of $C^0_c(\Omega$. This is a strong assumption. It is satisfied by the delta function, but not the derivative of the delta function. Indeed, the latter relies on the smoothness of the function, but that in fact is why it is NOT continuous with respect to the topology of $C^0_c(\Omega)$. In general the value of a distribution depends on the derivatives of a function up to some (finite) order. If the order is zero, then it is a Radon measure. $\endgroup$
    – Deane Yang
    Feb 1 at 19:59
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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. $\endgroup$ Mar 13, 2010 at 17:04
  • $\begingroup$ A much better explanation than mine. $\endgroup$
    – Deane Yang
    Mar 13, 2010 at 17:11
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    $\begingroup$ Sorry for perhaps sounding stupid, but: Why does the Hahn-Banach theorem not work in extending the measure here? $\endgroup$
    – Regenbogen
    Mar 14, 2010 at 0:37
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    $\begingroup$ Because it is not continuous with respect to the topology of D. $\endgroup$
    – user1688
    Mar 14, 2010 at 14:15
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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, 2011 at 9:34
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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, 2010 at 3:55
  • $\begingroup$ Are they not dual also to the subspace of continuous functions of $\mathcal D$ in its subspace topology? $\endgroup$ Mar 13, 2010 at 15:00

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