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Proof Fourier Series Converge

  1. Aug 19, 2013 #1
    Hey. I'm looking for a proof of:
    Theorem: If [itex]f \in C^1(\mathbb{T})[/itex], then the Fourier series converges to f uniformly (and hence also pointwise.)

    I have looked around for it, googled, etc, but I only found proofs which used theorem they did not prove. (Or I misunderstood what they said.)
    I'd really like to truly understand why they converge, be it uniformly or pointwise. If anyone could either link me to a proof or do it, it'd be great. Thanks.
     
  2. jcsd
  3. Aug 19, 2013 #2

    micromass

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    Check Folland's "Fourier analysis and applications", Theorem 2.5
     
  4. Aug 28, 2013 #3
    Sorry, I don't know what C^1(T) is. Are these complex functions? And what domain is T?
     
  5. Sep 8, 2013 #4
    Completely answered my question. Thanks a lot!

    C^k is the set of functions such that: There exist continuous derivatives of 0th, 1st, 2nd... and kth order.
    C^1(T) probably means that f is periodic of period 2π or something of the sorts. The number of senseful meanings is not that big.
     
  6. Sep 8, 2013 #5
    Thanks Swimmingly. Then[tex] C^1[/tex] would be real functions that have just one continuous derivative and you guess T means they are periodic. I would guess if they are not periodic one could construct examples where the Fourier series wouldn't converge at all. There is always the famous [tex]{x^2sin1/x}[/tex] which has exactly one continuous derivative at 0.
     
    Last edited: Sep 8, 2013
  7. Sep 8, 2013 #6

    Office_Shredder

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    Bold faced T usually refers to a torus - in this case I assume the one dimensional torus (which is the circle, equivalently we are discussing periodic functions on the real line)
     
  8. Sep 8, 2013 #7
    Office_Shredder- Thanks for the clarification. He most likely meant periodic functions on the real line. I personally am more than willing to consider functions on the unit circle -- quite often it helps.

    Also, for all, please substitute [tex]x^3sin1/x[/tex] for [tex]x^2sin1/x[/tex] as per my previous post. The first does have one continuous derivative at x = 0. The second has only a discontinuous derivative at x = 0.
     
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