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It is well known from the rudiments of Fourier Analysis, signals (whether time or space dependent) cannot display details that are smaller than the shortest period of their Fourier components. Can someone provide a mathematical proof?

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    $\begingroup$ This would be better suited to math.SE. As a first step though I suggest you take the derivative of the Fourier transform... $\endgroup$ – lemon Aug 17 '17 at 8:48
  • $\begingroup$ No, you can't provide a mathematical proof, because what you claim is false. Sampled data can only represent a signal with a finite bandwidth (that is intuitively obvious, because in any time interval you only have a finite number of data points to represent the signal) but there is no requirement that the bandwidth must include 0Hz. If you know the bandwidth for some other reason, the same sample rate can represent frequencies between 1,000,000Hz and 1,000,100Hz just as precisely as frequencies between 0Hz and 100Hz. $\endgroup$ – alephzero Aug 17 '17 at 21:51
  • $\begingroup$ @lemon Fourier transforms and Fourier series are two different things. The OP seems to be asking about Fourier series where at least this is a common misunderstanding. The claim made in the question is certainly not true for Fourier transforms of functions. $\endgroup$ – alephzero Aug 17 '17 at 21:54
  • $\begingroup$ Look at: en.wikipedia.org/wiki/Superoscillation $\endgroup$ – David Aug 18 '17 at 7:08
  • $\begingroup$ @David, you are referring to the correct wiki page. The claim made in my question is, in fact, from Yakir Aharonov's work entitled as The Mathematics of Superoscillations given here arxiv.org/abs/1511.01938. Unfortunately, I do not understand the mathematics of this claim. $\endgroup$ – Seeker Aug 18 '17 at 18:18

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