Let $\mu$ be a complex Borel measure on $\mathbb R$ such that $\int_a ^{b} e^{itx} d\mu (x) \to 0$ as $|t| \to \infty$ whenever $a<b$. Can we conclude that $\mu$ is absolutely continuous with respect to Lebesgue measure?
Asked
Active
Viewed 131 times
2
-
My guess is that $\mu$ need not be absolutely continuous but I am unable to construct a counter-example. – Kavi Rama Murthy Jan 14 '18 at 11:55
-
I agree the answer is surely not, altough I don't have a counterexample. Note the hypothesis does imply that $\mu$ is continuous, which is to say that the measure of any singleton is $0$; see for example WIener's Lemma – David C. Ullrich Jan 14 '18 at 16:48
2 Answers
2
No. The counterexamples are the Rajchman measures. I don't know a simple construction, but the Wikipedia article has some references.
If you know a construction of a Rajchman measure on the unit circle, you can transfer it to $\mathbb R$ by extending periodically then multiplying by an integrable function with compact Fourier transform.
We then need that if $\mu$ is Rajchman then so is $1_{[a,b]}\mu.$ In general, given a Rajchman measure $\mu,$ any $\nu\ll\mu$ is also Rajchman, because we can take a sequence $f_n$ with compactly supported Fourier transforms such that $f_n\to d\mu/d\nu$ in $L^1(\mu).$
Dap
- 25,286
-
Thank you Dap! It will take a while for me to study your answer but I am approving your answer and bounty worth 50 reputation. – Kavi Rama Murthy Jan 17 '18 at 09:15
-
Cool. The fact that if $\hat\mu\to0$ and $\nu<<\mu$ then $\hat\nu\to0$ is something I should have known but didn't. On the circle we just approximate $d\nu/d\mu$ by trigonometric polynomials (which explains to me roughly why the approximation you suggest is possible). – David C. Ullrich Jan 17 '18 at 22:06
1
Partial answer: Let $\mu$ be the Cantor measure; that is, the "uniform" measure on the Cantor set. It would be enough to show that $\hat\mu(t)\to0$ at infinity.
Why? Fix $a,b$ and let $d\nu=\chi_{(a,b)}d\mu$. Since $(a,b)$ is the union of the triadic intervals contained in $(a,b)$ it follows that $||\nu-\nu_n||\to0$, where each $\nu_n$ is a finite linear combination of translates of $\mu$. Hence $\hat\nu_n(t)\to0$, and $\hat\nu_n\to\hat\nu$ uniformly, so $\hat\nu\to0$.
Edit: Alas $\hat\mu(t)\not\to0$ at infinity, because $\hat\mu(t)=\prod_k\cos(2\pi t/3^k)$, which makes it clear that $\hat\mu(3^n)\not\to0$. I'm going to leave this here anyway, because one can imagine using the same argument with a different "fractal measure" in place of $\mu$.
David C. Ullrich
- 89,985