Difference between measures and distributions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:36:54Z http://mathoverflow.net/feeds/question/17732 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17732/difference-between-measures-and-distributions Difference between measures and distributions Tom Ellis 2010-03-10T15:31:38Z 2010-03-14T13:53:10Z <p>On the one hand, Wikipedia suggests that every distribution defines a Radon measure:</p> <ul> <li><a href="http://en.wikipedia.org/wiki/Distribution_(mathematics)#Functions_as_distributions" rel="nofollow">http://en.wikipedia.org/wiki/Distribution_(mathematics)#Functions_as_distributions</a></li> </ul> <p>On the other hand, Terry Tao and LK suggest not:</p> <ul> <li><a href="http://www.math.ucla.edu/~tao/preprints/distribution.pdf" rel="nofollow">http://www.math.ucla.edu/~tao/preprints/distribution.pdf</a></li> <li><a href="http://mathoverflow.net/questions/14586/when-can-a-function-be-recovered-from-a-distribution" rel="nofollow">http://mathoverflow.net/questions/14586/when-can-a-function-be-recovered-from-a-distribution</a></li> </ul> <p>Can someone please clarify this for me?</p> http://mathoverflow.net/questions/17732/difference-between-measures-and-distributions/17733#17733 Answer by Deane Yang for Difference between measures and distributions Deane Yang 2010-03-10T15:41:54Z 2010-03-10T15:41:54Z <p>Do you mean this sentence:</p> <blockquote> <p>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.</p> </blockquote> <p>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.</p> <p>The fundamental examples are the delta function at a point (which is a measure) and its derivatives (which are not measures).</p> http://mathoverflow.net/questions/17732/difference-between-measures-and-distributions/17748#17748 Answer by Orbicular for Difference between measures and distributions Orbicular 2010-03-10T18:30:59Z 2010-03-10T18:30:59Z <p>Measures are dual to continuous functions, whereas distributions are derivatives of them.</p> http://mathoverflow.net/questions/17732/difference-between-measures-and-distributions/18062#18062 Answer by anton for Difference between measures and distributions anton 2010-03-13T14:50:33Z 2010-03-13T16:04:09Z <p>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.</p> http://mathoverflow.net/questions/17732/difference-between-measures-and-distributions/18166#18166 Answer by Tom Ellis for Difference between measures and distributions Tom Ellis 2010-03-14T13:53:10Z 2010-03-14T13:53:10Z <p>This is a summary of what I've learned about this question based on the answers of the other commenters.</p> <p>[*] Any positive distribution defines a positive Radon measure.</p> <p>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 <em>could</em> be, then applying Theorem [*] would yield the result that any distribution is a signed measure.</p> <p>However, this is not the case. The derivative of the delta function, i.e. &delta;', satisfies &delta;'(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.</p> <p>[1] <a href="http://en.wikipedia.org/wiki/Hahn_decomposition_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Hahn_decomposition_theorem</a></p>