Hello,
After reading the previous post, I still have some doubts. Let's consider everything on $R$ to avoid complications.
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Can we say that any distribution $\mu\in\mathcal{D}'(R)$ of zero order is a signed radon measure?
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Since $\mu\in\mathcal{D}'(R)$ which is non-negative on non-negative test functions $C_c^\infty(R)$ is a positive radon measure, it is natural to ask what is corresponding part for the Schwartz distribution $\mu\in\mathcal{S}'(R)$ which is non-negative on non-negative test functions $\mathcal{S}(R)$? Intuitively, it is the radon measure whose mass grows slowly at infinity. Is there a name for this measure?
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For a radon measure $\mu$ on $R$, can we apply the Lebesgue's decomposition locally with a compact set fixed (say $K=[-a,a]$)? Then $\mu_K=\mu_{ac}+\mu_{sc}+\mu_{pp}$. The absolutely continuous part $\mu_{ac}$ corresponds to an absolutely continuous function; the pure point part $\mu_{pp}$ corresponds to sum of delta functions. How about the singular continuous part $\mu_{sc}$?
EDIT: This question can also be put in the following way: Let $f$ be a singular function, what can we say $\int f\psi d x$ with $\psi\in\mathcal{D}(R)$?
Thank you for your help! 🙂
Best
Anand
Best Answer
To answer the first question: yes, at least locally. That is, given a distribution $u$ of order $0$, compactly supported for simplicity, the "order 0" condition asserts that $u$ factors through the space $C^o_c(U)$ of continuous compactly supported functions _with_the_corresponding_topology_. Then invoke the Riesz representation theorem to obtain the corresponding measure. Reduce to this case by a locally finite smooth partition of unity.
I cannot instantly answer the second question, but would try to reduce it to the compactly-supported case by a partition of unity argument.
I think the comment deals with the question about the "singular continuous" part.