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Elon Lindenstrauss explains in his talk at the MSRI in Fall 2008 (the relevant comment is at minute 41 of the video) that the set of large Fourier coefficients of a probability measure $\mu$ on the torus ${\mathbb T}^n$ respects the additive structure. More precisely, he defines

$$A_{\delta} := \lbrace b \in {\mathbb Z}^n \mid |\hat \mu(b)| \geq \delta \rbrace$$

and says that it is "morally" true that $A_{\delta} - A_{\delta} \subset A_{\delta^2}$. (Here, the difference of two subsets is defined to be the set of all possible differences of elements in the respective subsets.) The precise statement (according to Lindenstrauss) is a consequence of the Balog-Szemeredi-Gowers Lemma.

Can someone provide the precise statement or give some hint how the lemma can be used to obtain bounds on Fourier coefficients?

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This set is often called the $delta$-spectrum of A, and is discussed at some length in section 4.6 of Tao and Vu, Additive Combinatorics. I can't recall anything related to your question there, however, and would be very interested myself to find out more. –  Thomas Bloom Aug 25 '10 at 11:33
    
Chapter 2 of these notes dpmms.cam.ac.uk/~bjg23/papers/icmsnotes.pdf, and in particular Chang's theorem, should answer your question. –  Kevin O'Bryant Aug 25 '10 at 13:50
    
@Kevin: Maybe the results in Chapter 2 are related, but for the moment I do not see how they help. First of all, there is no set $A$ in my question; secondly, I am talking about Fourier coefficients of a measure on ${\mathbb T}^n$, not ${\mathbb Z}/n{\mathbb Z}$. –  Andreas Thom Aug 25 '10 at 16:33
    
The link is currently broken since the MSRI is revamping its video section. Temporary link. –  Greg Graviton Nov 9 '10 at 16:53
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I think I figured it out myself. What was meant is that for every finite subset $S$ of $A_{\delta}$ one has $$| \lbrace (n,m) \in S \times S \mid n-m \in A_{\delta^2/2} \rbrace | \geq \frac{\delta^2 |S|^2}{2}.$$ This follows from the proof of the second part of Lemma 4.37 in Tao and Vu, Additive Combinatorics. (There, the inequality is stated incorrectly as $\leq$.)

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Absolutely right. I believe you can use this iteratively to prove Chang's inequality, mentioned by Kevin O'Bryant above. There is a paper of Bourgain (Lambda(p) subsets of squares, perhaps) where, instead of recalling an appropriate reference to this theory, he merely says "by linearisation". –  Ben Green Aug 29 '10 at 16:13
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