After reading the previous post, I still have some doubts. Let's consider everything on $R$ to avoid complications.

  1. Can we say that any distribution $\mu\in\mathcal{D}'(R)$ of zero order is a signed radon measure?

  2. 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?

  3. 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

  • 2
    $\begingroup$ The singular part corresponds to a singular continuous function(!). That is: a continuous function such that for any eps>0, there exists a disjoint collection of intervals whose combined length is 2a-eps on which the variation is less than epsilon. $\endgroup$ – Anthony Quas Jul 12 '11 at 17:00
  • $\begingroup$ Thanks Prof. Anthony Quas. Then can we view the singular continuous function as a distribution? $\endgroup$ – Anand Jul 12 '11 at 18:04

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.

  • $\begingroup$ Thanks Prof. Paul Garrett. I am clear now for the first one. $\endgroup$ – Anand Jul 14 '11 at 19:40

About the second question:

The Schwartz space $\mathcal{S}(R)$ is a Frechet space, i.e. it's topology is given by a countable sequence of seminorms. Any Schwartz distribution (continuous linear functional on $\mathcal{S}(R)$) has to bounded by one of these seminorms. If the Schwartz distribution is of "order zero" then the only relevant seminorms are $\|f\|_n = \sup (1+|x|)^n |f(x)|$. If $T\in \mathcal{S}'(R)$ is bounded by $\|\ldots\|_n$, then this means $T$ is of the form $\langle T, f\rangle = \int (1+|x|)^{n} f(x) \mu(dx)$ for some finite measure on $R$.

So I would think measures which are Schwartz distributions are measures "of polynomial growth."

  • $\begingroup$ Thanks Sigurd Angenent. I totally agree with you. It is also my guess. This statement can be made further for zero order distributions in $\mathcal{O}_c'$ (distributions with fast decrease, I am using Schwartz notations). Then the measure should have fast decrease (faster than any polynomials). Thanks a lot! :-) $\endgroup$ – Anand Jul 15 '11 at 9:29

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