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Would you Please give an example of a function which is not piecewise continuous, but has Fourier series?

It means that the coefficient in the Euler-Fourier formulas can be computed. In fact, the definite integrals exist.

6-0
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  • If you are only requiring that the coefficients make sense, then you can take any $f \in L^2$, even wildly discontinuous functions. It is when you require more convergence than just convergence in $L^2$, such as convergence almost everywhere, that you need to require regularity. – Ian Feb 03 '15 at 21:22
  • Yes, It is piecewise continuous. – 6-0 Feb 03 '15 at 21:22
  • Ian No, I just want a function as I stated in my question. – 6-0 Feb 03 '15 at 21:25
  • The characteristic function, Do the difinite integrals (Riemann) exist? I don,t think so. – 6-0 Feb 03 '15 at 21:26
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    They exist in the sense of Lebesgue integration. Coming up with a Riemann integrable example is more difficult, since a Riemann integrable function is continuous almost everywhere. (This is called Lebesgue's criterion for Riemann integrability.) You might consider a Riemann integrable function whose set of discontinuities is a Cantor set of measure zero. This just amounts to building en.wikipedia.org/wiki/Volterra's_function with the usual Cantor set replacing the "fat" (positive measure) Smith-Volterra-Cantor set. – Ian Feb 03 '15 at 21:29
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    You might also consider http://en.wikipedia.org/wiki/Thomae%27s_function , which is discontinuous at every rational number and is Riemann integrable (again by Lebesgue's criterion). – Ian Feb 03 '15 at 21:33
  • (Of course, Thomae's function is also 0 almost everywhere, so its Fourier series converges to the zero function, just like the characteristic function of the rationals does.) – Ian Feb 03 '15 at 21:38
  • Thanks, Ian, for following up here and teasing out the question that OP actually wanted answered. If OP had mentioned Riemann integration, I might have written a more elaborate answer. :) – John Hughes Feb 04 '15 at 03:15
  • Thank you John and Ian for your answers. – 6-0 Feb 04 '15 at 13:36

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Consider the characteristic function $c$ of the rationals; it's 1 at every rational, 0 at every irrational. Its FT is the constant function 0. Because $c$ is equal, almost everywhere, to the constant function $0$, they have the same FT.

John Hughes
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