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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?

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  • $\begingroup$ 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. $\endgroup$
    – Ningxin
    Commented Apr 7, 2016 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
    Commented Apr 8, 2016 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
    Commented Apr 8, 2016 at 2:54
  • $\begingroup$ $\Omega$ is locally compact. $\endgroup$
    – Ningxin
    Commented Apr 8, 2016 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 $\endgroup$
    – Albert
    Commented Sep 7, 2016 at 16:15

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It's actually easier than it might seem.

Just notice that if $T$ is a positive distribution then $(T,\phi)\geq(T,\psi)$ for $\phi\geq \psi$ (since $\phi-\psi\geq 0$ implies $(T,\phi-\psi)\geq 0$). Therefore, for each $K\subset\subset D$, $$\sup_{\phi\in D_K,\,||\phi||_{C^0}\leq 1}(T,\phi)\leq(T,\eta_K) <\infty,$$ where $\eta_K\in D$ is some positive cut-off on $K$. This means that $T$ has order $0$. Thus it can be extended to $C_c(\Omega)$ and so, by Riesz's Theorem, it is a measure.

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