Exercise 4 of Chapter 6 in Rudin's Functional Analysis states that every "positive" distribution $\Lambda\in D^{'}(\Omega)$, i.e, $\Lambda\psi\geq 0$ whenever $\psi\in D(\Omega)$, is a positive measure in $\Omega$, where $\Omega\subseteq\mathbb{R}^d$ open and $ D^{'}(\Omega)$ is the space of test functions on it.

My question is that does problem directly follow from the Riesz Representation theorem that says every positive linear functional on the space of compactly supported smooth functions on a locally compact Hausdorff space is a positve Radon measure?

  • $\begingroup$ The Riesz Representation theorem is for positive linear functionals on $C_c(\Omega)$. Hence you need to either extend $\Lambda$ to $C_c(\Omega)$ (maybe $\Lambda f = \sup{\Lambda \psi: \psi \leq f}$ is an option), or prove Urysohn's lemma and partition of unity with smooth functions in place of continuous ones. $\endgroup$ – Qiyu Wen Apr 7 '16 at 22:42
  • $\begingroup$ @QiyuWen That's what I thought at first. But in Rudin it explicitly said that "$\psi\in D(\Omega)$ if and only if $\psi\in C^{\infty}(\Omega)$ and the support of $\psi$" is a compact subset of $\Omega. $ $\endgroup$ – dezdichado Apr 8 '16 at 2:46
  • $\begingroup$ @QiyuWen So I am wondering that if the implication is not trivial, then it has to do something with $\Omega$ not being locally compact? $\endgroup$ – dezdichado Apr 8 '16 at 2:54
  • $\begingroup$ $\Omega$ is locally compact. $\endgroup$ – Qiyu Wen Apr 8 '16 at 3:44
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    $\begingroup$ @dezdichado the implication is not trivial because the topology on smooth functions here is not the compact-open topology but the topology of test functions for distribution (which is horribly complicated, but you might think of it as induced by semi norms of sup on compacts of all derivatives). so think of it as proving that a distribution of order more than one can never be positive $\endgroup$ – Glougloubarbaki Sep 7 '16 at 16:15

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