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A Zeros of Fourier transform

  1. Jan 18, 2017 #1

    for a function f∈L2(ℝ), are there known necessary and sufficient conditions for its Fourier transform to be zero only on a set of Lebesgue measure zero?
  2. jcsd
  3. Jan 21, 2017 #2


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  4. Mar 1, 2017 #3

    Thanks for the reply. This is not a homework problem.

    I don't see unfortunately any clear connection between Plancherel theorem and the zeros of a Fourier transform. Maybe someone else could point out other possible directions to approach the problem?

    It would be even ok to restrict the domain of the original question to finding only sufficient conditions.
  5. Mar 1, 2017 #4


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    I originally misread the question as: "zero except on a set of Leb. measure zero" so that it would of necessity have zero total L2 norm. But I now see that isn't what you said. I still am inclined to to think about the problem in linear algebraic terms. The L2 norm is preserved under Fourier Transform... hmmm. I would have to think on this further.
  6. Mar 2, 2017 #5


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    I don't know the answer, but just playing around, I found that a simple example of a function that does have zeros in an extended region is:

    [itex]f(x) = \frac{sin(Kx)}{x}[/itex]

    The FT [itex]\tilde{f}(k)[/itex] is (if I've done it correctly) constant in the region [itex]-K < k < +K[/itex], and zero everywhere else.
  7. Mar 2, 2017 #6
    Hi stevendaryl,

    yes, I think the fact that the FT of a sinc function is rectangle function (and vice-versa) is a well-known result. Despite that, it is a useful remark. In fact, the rectangle function is one example of function whose FT has zeros on a set of Lebesgue measure zero.

    Now, one interesting thing is that the rectangle function has compact support.

    Could it be so that "compact support in time domain" => "FT with zeros on a set of Lebesgue measure zero" ?
  8. Mar 2, 2017 #7
    Yes, if an ##L^2## function is compactly supported, its Fourier transform is an entire function (i.e. analytic function on the whole complex plane), and thus it can have at most countably many isolated zeroes, accumulating to ##\infty##.

    If an ##L^2## function is supported on a half-line, then its Fourier transform (on the real line) can vanish only on a set of Lebesgue measure zero, that is a standard fact from the theory of Hardy spaces.

    But I am not aware of a necessary and sufficient condition, I think a "simple" necessary and sufficient condition is impossible.
  9. Mar 3, 2017 #8
    Very interesting and satisfactory answer!
    I didn't know about the relationship between compact support and analyticity of the FT.

    I am still wondering two things:

    1) Is there an actual difference in this case between "countably many isolated zeroes" and "zero in a set of Lebesgue measure zero"? Is it so that the former property implies the latter but not vice-versa?

    2) How does the above statement generalize to FT's of compactly supported 2-dimensional functions?
  10. Mar 8, 2017 #9
    Yes, definitely. The sets of zero measure can have quite complicated structure, they can consists of uncountably many points. For example, the classical Cantor 1/3 set has measure 0, and is uncountable.

    Some of it generalizes to any dimension ##n##. The Fourier transform of a compactly supported function in ##\mathbb R^n## is still an analytic function (you need to assume that your function is in ##L^1##, which for compactly supported functions follows from the assumption that ##f\in L^2##). The zero set of such function for ##n\ge 2## does not generally consists of countably many isolated points, but is what is called an "analytic variety". But it has measure zero, and it posesses a nice structure.

    For example, if you fix ##n-1## variables, and vary one, say ##x_k##, then the function will be zero at most at countably many values of ##x_k##: this immediately implies that the zero set has measure 0.
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