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Another question about Fourier series convergence

  1. Sep 24, 2012 #1
    I am trying to prove a theorem related to the convergence of Fourier series. I will post my proof below, so first check it and then my question will make sense.

    Is there any flaw in my proof? Also, here I proved it for integrable functions monotonic on an interval on the left of 0. But what if the function was not monotonic on any interval around 0, and was not Lipschitz continuous either? For example, f(x) = √|x|*sin(1/x). Can I still use step functions to approximate it as in the proof for monotonic functions? I know it oscillates wildly, but by taking sufficiently small intervals, it would seem that the limit would still go to f(0-) over 2. And yet I know this is not possible since there are continuous functions whose Fourier series do not converge at some points of continuity, and my proof would lead to a contradiction if it were true for all non-monotonic functions. So is my proof flawed, and if not, what prevents it from working for functions like √|x|*sin(1/x)?
  2. jcsd
  3. Sep 24, 2012 #2
    I am trying to prove the following, under appropriate conditions for f which are mentioned:
    [tex] \lim_{n \rightarrow \infty} \frac{1}{\pi}\int_{-\pi}^{0}f(x)\frac{\sin(nx)}{x}dx = f(0-)\frac{1}{\pi}\int_{-\infty}^{0}\frac{\sin(x)}{x}dx = \frac{f(0-)}{2} [/tex]

    First I proved it for f(x) = 1:

    [tex] \lim_{n \rightarrow \infty} \frac{1}{\pi}\int_{-\pi}^{0}\frac{\sin(nx)}{x}dx = \lim_{n \rightarrow \infty} \frac{1}{\pi}\int_{-n\pi}^{0}\frac{\sin(u)}{u}du = \frac{1}{\pi}\int_{-\infty}^{0}\frac{\sin(x)}{x}dx = \frac{1}{2} [/tex]

    Then for any step function s(x) which takes on values [itex] c_1, c_2, ... ,c_n [/itex] on [itex] (0 , x_1) , (x_1 , x_2) , ... , (x_{n-1} , x_n)[/itex]. Here the only interval of interest is [itex] (0 , x_1) [/itex] since in the other intervals the limit vanishes by the Riemann-Lebesgue lemma. But on the interval of interest, s(x) is a constant multiple of 1, so the limit is the value [itex] \frac{c_1}{2} = \frac{s(0-)}{2}[/itex].

    Now consider an integrable, monotonic function on [itex] [-\pi , 0] [/itex]. By a property of integrable functions, for any [itex] \epsilon > 0 [/itex] there exist step functions such that [itex] s_1 ≤ f ≤ s_2 [/itex] and [itex] f-s_1 < \epsilon[/itex] and [itex]s_2-f < \epsilon [/itex] on [-π , 0].

    We also choose the step functions such that their constant value on the interval next to x = 0 is f(0-) (if f is monotonic around 0, we surely can do this since the maximum error will occur at the other end point, so we can choose the length of the interval so as to minimize this error).

    Then [tex] \frac{1}{\pi}\int_{-\pi}^{0}s_2(x)\frac{\sin(nx)}{x}dx - \epsilon \frac{1}{\pi}\int_{-\pi}^{0}\frac{\sin(nx)}{x}dx ≤ \frac{1}{\pi}\int_{-\pi}^{0}f(x)\frac{\sin(nx)}{x}dx ≤ \frac{1}{\pi}\int_{-\pi}^{0}s_1(x)\frac{\sin(nx)}{x}dx + \epsilon \frac{1}{\pi}\int_{-\pi}^{0}\frac{\sin(nx)}{x}dx [/tex]

    And by taking the limit as n goes to infinity,
    [tex] \frac{f(0-)}{2} - \frac{\epsilon}{2} ≤ \lim_{n \rightarrow \infty} \frac{1}{\pi}\int_{-\pi}^{0}f(x)\frac{\sin(nx)}{x}dx ≤ \frac{f(0-)}{2} + \frac{\epsilon}{2} [/tex] This inequality is true for any positive epsilon (choose the appropriate step functions) so it must be true that the limit is actually [itex] \frac{f(0-)}{2} [/itex] QED.
    Last edited: Sep 24, 2012
  4. Sep 25, 2012 #3
    Actually the last line was imprecise: what I meant is that by choosing my step functions so that epsilon is small enough, and then by choosing n large enough, I can make the integral within any given positive number error from f(0-)/2.
    Last edited: Sep 25, 2012
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