# a complex borel measure, whose Fourier transform goes to zero

I have $\mu$ a complex borel measure on $\mathbb(R)$, whose Fourier transform goes to zero as $\xi$ goes to $\infty$. I need to prove that $|\mu|$(singleton) = 0. Should I approach $\mu$ by a sequence of absolutely continuous (to lebesgue measure) measures in the weak star topology, or is there a simpler way to do this?

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Where is $\mu$ defined? Is it on $\Bbb R^{n}$, on $\Bbb R$, or is on the $n$-torus. – Benjamin Hayes May 7 '11 at 9:32
on $\mathbb{R}$, sorry forgot to add it – jessica May 7 '11 at 9:37
The setting reminds me of Wiener's lemma. – Mark May 7 '11 at 11:51

Wiener's lemma/theorem says that

$\sum_{x \in \mathbb{R}} |\mu ( \{ x \}|^2 = \lim_{\lambda \to \infty} \frac{1}{2 \lambda} \int^{\lambda}_{- \lambda} |\hat{\mu} ( \xi ) |^2 d \xi$

for every finite Borel measure on $\mathbb{R}$ (the theorem is sometimes formulated for measures on the torus, or on $\mathbb{R}^n$ or some other locally compact abelian group. Unfortunately, there are also many unrelated theorems which go by this name).

From here you can just apply l'Hospital's theorem (or play with limits however you like) and get that $\mu$ has no atoms. The previous answers allude to this theorem. You can find it in Katznelson's "An Introduction to Harmonic Analysis", for instance.

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Nice! I vaguely remembered something like this when typing my last post, but couldn't quite remember what the exact statement was. – Benjamin Hayes May 7 '11 at 16:08

I'll try to write my suggestions assuming you are working on $\Bbb R^{n}.$

Here's what I would suggest, although this doesn't constitute a full solution. Write

$$\mu=\mu_{a}+\mu_{sc}+\mu_{d}$$

where $\mu_{a}$ is absolutely continuous with respect to Lebesgue measure, $\mu_{sc}$ is singular with respect to Lebesgue measure, but non-atomic and $\mu_{d}$ is purely atomic.

We have to argue that $\mu_{d}=0.$ By Riemann Lebesgue, we know that

$$\widehat{\mu_{a}}(\xi)\to 0$$ as $|\xi|\to \infty.$

Write $\mu_{d}=\sum_{n}c_{n}\delta_{x_{n}}$ with $x_{n}\ne x_{m}$ for $n\ne m.$

Then $$\widehat{\mu_{d}}(\xi)=\sum_{n}c_{n}e^{2\pi i \xi\cdot x_{n}}$$

and

$$\frac{1}{2^{d}T_{1}\cdots T_{d}}\int_{-T_{d}}^{T_{d}}\cdots\int_{-T_{1}}^{T_{1}}|\widehat{\mu_{d}}(\xi)|^{2}\prod_{j}d\xi_{j}=$$ $$\sum_{n}|c_{n}|^{2}+\frac{1}{2^{d}T_{1}\cdots T_{d}}\sum_{n\ne m}c_{n}\overline{c_{m}}\sin(2\pi (x_{n}-x_{m})T)$$

from here one can see that the above average tends to $\sum_{n}|c_{n}|^{2}$ as $T_{j}\to \infty$ for all $j,$ this makes it impossible for $\widehat{\mu_{d}}(\xi)\to 0$ as $|\xi|\to \infty.$

Now if one can do some analysis on $\widehat{\mu_{sc}}(\xi)$ as $|\xi|\to \infty,$ this may solve the problem.

In one dimension one can try to write $\mu_{sc}=dF,$ with $F$ a continuous function of bounded variation, it may be that some integration by parts trick works here but I'm not seeing it.

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Zygmund (Trigonometric Series, Vol. 2, Section 16.4.19) considers the Fourier-Stieltjes transform

$$\varphi (x)=\frac1{\sqrt{2\pi }}\int\limits_{-\infty }^\infty e^{-ixy}dF(y)$$ and looks for conditions on $\varphi$, under which $F$ is continuous. If I understand correctly, this is equivalent to your question. The necessary and sufficient condition is $$\int\limits_{-\omega}^\omega |\varphi (\xi )|^2d\xi =o(\omega ),\quad \text{as \omega \to \infty},$$ or, equivalently $$\int\limits_{-\omega}^\omega |\varphi (\xi )|d\xi =o(\omega ),\quad \text{as \omega \to \infty}.$$ Both conditions are different from a mere decay of the Fourier transform at infinity.

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I think I got a good one:

clealy, it's enough to prove $\mu$(singleton) = 0. Consider the function $f_{\epsilon}(x) = \frac{1}{\epsilon}$ if 0 < x < $\infty$ and 0 otherwise.

Notice that as $\epsilon$ goes to $\infty$, the limit of the fourier transform of the function above is characteristic function of the singleton 0.

writing $\mu$(a) as an integral, DCT, duality relation, and a small change of variable magically does the trick.

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