Pome

Theorem 2.5.66A928E70A605

Theorem 2.5.6.6A928E70A605 Let $T$ be a positive distribution on $\Omega $. Then there is a unique Radon measure $\mu $ on $\Omega $ such that $\langle T,\chi \rangle =\int \chi \, \mathrm d\mu $ for every test function $\chi $. Conversely, every Radon measure defines a positive distribution in this way.

Proof. Let $K\subset \Omega $ be compact and let $\eta $ be a smooth cutoff function for $K$ in $\Omega $ (lemma 2.5.1(4)). For a real test function $\chi $ with $\operatorname{Supp}\chi \subset K$, the test functions $\| \chi \| _\infty \eta \pm \chi $ are nonnegative, so positivity gives, as in (2.3.1),

\[ |\langle T,\chi \rangle |\leq \langle T,\eta \rangle \| \chi \| _\infty . \]

We use this bound to extend $T$ to $C_ c(\Omega ,\mathbb R)$. If $f\in C_ c(\Omega ,\mathbb R)$, the real test functions $f*\rho _\varepsilon $ are supported in a fixed compact set for small $\varepsilon $ and converge uniformly to $f$ (lemma 2.5.1(5)). By the displayed estimate, $\langle T,f*\rho _\varepsilon \rangle $ converges as $\varepsilon \to 0$, and its limit $\Lambda (f)$ is also the limit of $\langle T,\chi _ k\rangle $ for any real test functions $\chi _ k$ supported in a common compact set and converging uniformly to $f$. Approximating two functions of $C_ c(\Omega ,\mathbb R)$ by test functions with a common compact support, we see that $\Lambda $ is linear. If $f\geq 0$, then $f*\rho _\varepsilon \geq 0$, so $\Lambda (f)\geq 0$.

By theorem 2.3.3, there is a Radon measure $\mu $ with $\Lambda (f)=\int f\, \mathrm d\mu $. For a real test function $\chi $, the constant sequence $\chi _ k=\chi $ shows that $\Lambda (\chi )=\langle T,\chi \rangle $; applying this to $\operatorname{Re}\chi $ and $\operatorname{Im}\chi $ gives $\langle T,\chi \rangle =\int \chi \, \mathrm d\mu $ for all test functions. The measure $\mu $ is unique by proposition 2.5.3(4). Conversely, if $\mu $ is a Radon measure, then $|\int \chi \, \mathrm d\mu |\leq (1+\mu (K))p_{K,0}(\chi )$ for $\chi \in \mathcal D_ K(\Omega )$, so integration against $\mu $ is a distribution, and it is positive. $\square$

Cite this page

Comments

Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.