I've read that Kahane and Salem show that if $\mu$ is any measure supported on the ternary Cantor set, then $\hat{\mu}(\xi) \not\to 0$ as $|\xi| \to \infty$, however I have been unable to find a reference for this result. Does anyone know where I can find this result? Thanks in advance.
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$\begingroup$ They have 5 joint papers together and a book, Kahane, Jean-Pierre; Salem, Raphaël, Ensembles parfaits et séries trigonométriques. (French), Actualités Sci. Indust., No. 1301 Hermann, Paris 1963 192 pp. Second edition. With notes by Kahane, Thomas W. Körner, Russell Lyons and Stephen William Drury. Hermann, Paris, 1994. 245 pp. ISBN: 2-7056-6193-X. This book (w/o page number) seems to be the typical reference. The updated Zygmund (Trigonometric Series) cites it, as does Geometry of Sets and Measures in Euclidean Spaces (Mattila). $\endgroup$– ThisNameForSaleCommented Dec 4, 2013 at 0:52
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$\begingroup$ Typically you see Kahane and Salem mentioned together for the opposite question: when does a set support a measure that decays as fast as possible? These are called Salem sets and Kahane constructed examples using Brownian motion. The question you're asking sounds like it has to do with lacunary Fourier series, so maybe try their book (in French) referenced here: en.wikipedia.org/wiki/Set_of_uniqueness. Mockenhaupt has a paper (Salem sets and Restriction …, GAFA, 2000) that mentions that the ternary Cantor set has Fourier dimension 0 (and mentions Kahane and Salem) but without reference $\endgroup$– Brendan MurphyCommented Dec 4, 2013 at 0:55
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$\begingroup$ Also this article by Kahane mentions more about sets of uniqueness and Cantor sets. He doesn't prove the fact you want, but he cites the book mentioned by ThisNameForSale for more info: ams.org/journals/bull/1964-70-02/S0002-9904-1964-11080-6/… $\endgroup$– Brendan MurphyCommented Dec 4, 2013 at 1:05
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$\begingroup$ Thank you for the comments. Is there an english version of "ensembles parfaits et séries trigonométriques"? I tried "perfect sets and trigonometric series", to no avail. $\endgroup$– user43622Commented Dec 4, 2013 at 1:54
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