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Given that a Fourier Transform of a sequence has norm 1 for (almost) all of the unit circle, prove that the sequence has only one non-zero value. (The Fourier transform of a sequence $f(n)$ is $\widehat{f}(\xi)=\sum_{n\in\mathbb{Z}}f(n)e^{-2\pi i \xi n}$.) I have tried extracting individual coefficients or using Parseval's Identity but I am very very limited in my knowledge of the Fourier Transform. I assume the proof of this fact is very simple but I am clueless. After Spencer's hint I have that $\widehat{f}(\xi)$ multiplied by its conjugate is $1$ everywhere but I don't know how this finishes the problem.

user290819
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  • Considering you have been on this site before, you should know that just asking a question without showing any working is frowned upon. What have you tried? – Matthew Cassell Feb 20 '16 at 02:31
  • I have tried extracting individual coefficients or using Parseval's Identity but I am very very limited in my knowledge of the Fourier Transform. I assume the proof of this fact is very simple but I am clueless. – user290819 Feb 20 '16 at 02:33
  • Edit your original post to include your working. – Matthew Cassell Feb 20 '16 at 02:33
  • Mattos do you have a solution? – user290819 Feb 20 '16 at 02:37
  • I'm going to provide a small hint. Try some example sequences and compute the norm of the resulting Fourier transforms. What difference do you notice when you have 1 nonzero coefficient versus 2 or 3 nonzero coefficients? If you don't understand what I mean by this, then that is part of the problem. When you've tried these out update the question with your relevant work/ideas. – Spencer Feb 20 '16 at 03:24
  • Spencer I don't exactly understand your hint. If it amounts to the fact that the norm now has some cross terms and that these can't sum to zero (the average of such terms is trivially zero), I understand this outline but I don't understand how to show this rigorously. – user290819 Feb 20 '16 at 03:30
  • @user290819, the system doesn't ping me with you're reponse unless you initiate the message with "@Spencer". The exception to this rule is that the OP of a question gets pinged whenever a comment is made on the question. – Spencer Feb 20 '16 at 03:46
  • To answer your question the fact you observed can be shown formally for any finite sequence. Just use the distributive property. – Spencer Feb 20 '16 at 03:47
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    In general the statement is not true. $e^{i 3.1\xi}$ has a norm of $1$ but when expanded in an infinite Fourier series it has more than two nonzero coefficients. Are there any special conditions on the sequence $f(n)$? – Spencer Feb 20 '16 at 03:48
  • @Spencer The question stems from the following question on AoPS. http://artofproblemsolving.com/community/c7h1197793_finding_all_functions_from_z_to_r_with_some_property My question is simply how does one reach the final conclusion in post #4. I believe that is equivalent to my question but due to my naïveté I might be missing something. – user290819 Feb 20 '16 at 04:02
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    @user290819 : Kent on AoPS made a mistake, as Spencer proved. for what he said to be true, you need that $f(n)$ has a finite support. tell him that with $f(n) = \int_0^1 e^{i \sin(2 \pi x)} e^{2 i \pi n x} dx$ he'll even get a sequence $f(n)$ which is exponentially decreasing when $n\to \pm \infty$ and whose autocorrelation is perfectly defined and is $= \delta(n)$ – reuns Feb 20 '16 at 05:08

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