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Pointwise Convergence of Fourier Series for a continuous function

  1. Sep 23, 2012 #1
    Where is the fallacy in this "proof" that the Fourier series of f(x) converges to f(x) if f is continuous at x and has period 2π? (I read in Wikipedia that a counterexample had been provided).

    Start with the Dirichlet integral for the N-th partial sum of the (trigonometric) Fourier series. After some trig manipulations, u-substitutions and applications of the Riemann-Lebesgue lemma on intervals where the integrand is bounded, it is possible to write:

    [tex] \lim_{N \rightarrow ∞}[s_N(x)-f(x)] = \lim_{N \rightarrow ∞} \frac{1}{\pi}\int_{-\delta}^{\delta}[f(x+\alpha)-f(x)]\frac{\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}}\sin (N\alpha) d\alpha. [/tex]

    for any δ > 0.

    (this part is proven in standard textbooks and the fallacy most likely isn't here, and I am convinced that all steps are true. For a reference, I read it in N. Piskounov's "Calcul Differentiel et Integral", Tome II, Ch. XVII, #9. in french).

    Now whereas the book assumes more restrictive conditions on f(x) to prove convergence of the series, I was wondering whether mere continuity was sufficient. Here's my attempt at a proof, which is supposedly false:

    Given any ε > 0, choose N = 1/ε. Then choose a δ > 0 such that |f(x+α)-f(x)| < 1/N^2 = ε^2 for all |α| < δ (we are allowed to do so if f is continuous at x, since N is a fixed number now). Use this δ in the integral formula derived above. Moreover, [itex] 0 < \frac{\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}}\sin (N\alpha) ≤ N[/itex] on some interval around 0 (assume δ is included in it).

    Then [itex] |s_N(x) - f(x)| ≤ \frac{1}{\pi}\int_{-\delta}^{\delta}|f(x+\alpha)-f(x)|\frac{\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}}\sin (N\alpha) d\alpha ≤ \frac{1}{\pi}\int_{-\delta}^{\delta}\frac{1}{N^2}N d\alpha = \frac{2\delta}{\pi N} < ε [/itex] if δ is also smaller than 1/2.

    We showed that the absolute value of the difference can be made as small as desired by requiring N to be large enough. This proves that the limit as N goes to infinity exists and that s_N(x) converges to f(x) for all x at which f is continuous.
     
  2. jcsd
  3. Sep 23, 2012 #2

    jbunniii

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    Later in your proof, you appear to assume that the equality of these two limits implies the equality of the expressions to which the limits are attached. In other words, you removed the [itex]\lim_{N \rightarrow \infty}[/itex] from both sides. But is it true that the two expressions are equal, or only their limits?
     
  4. Sep 23, 2012 #3
    Oh I am sorry.
    Actually the two expressions inside the limits are shown to be equal, and then they took the limits as N goes to infinity (which is valid since two equal things tend to the same value).

    I skipped over this whole part because it would be long to write here, and as I said, it can be found in textbooks on the subject.
     
  5. Sep 23, 2012 #4

    jbunniii

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    So you are claiming that
    [tex]
    s_N(x)-f(x) = \frac{1}{\pi}\int_{-\delta}^{\delta}[f(x+\alpha)-f(x)]\frac{\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}} \sin (N\alpha) d\alpha[/tex]
    is actually true, for every [itex]N[/itex] and for every [itex]\delta > 0[/itex]? I don't buy it.

    I don't have an explicit counterexample, but suppose I define new continuous function [itex]g[/itex] such that [itex]g(x) = 0[/itex], and such that the first [itex]N[/itex] coefficients of the Fourier series for [itex]g[/itex] are zero. If I then apply the equation above to [itex]f + g[/itex] instead of [itex]f[/itex], then the left hand side remains the same, but the right hand side could change.
     
  6. Sep 23, 2012 #5
    I don't see why the right side would change. You have basically replaced f(x) by f(x) + g(x) but g(x) = 0 so in fact you did not change anything in either side.

    Maybe I didn't understand you correctly?
    Otherwise, so you believe that the mistake is actually in this expression, and not in the following argument using delta and epsilon? Maybe you're right, and then I must have missed something while reading the proof, but I believe that I did read it correctly...

    Ok I found an online version of the text, so here's a link to the part about this integral expression for the partial sum: http://www.scribd.com/doc/26412116/...Integral-Tome2-N-Piskounov-Mir#outer_page_194

    It's $8 and $9.
    (sorry, it is in french, but I don't think the language is really important here).
     
  7. Sep 23, 2012 #6
    :surprised:

    I think I understand my mistake. You're right that only the limits are equal, since when they split the integral into three regions, the integrals go to 0 only after the limit is taken and not before, so in my proof I had to include these integrals too.

    :blushing:
     
  8. Sep 23, 2012 #7

    jbunniii

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    I meant, choose [itex]g[/itex] with [itex]g(x) = 0[/itex] (i.e. at the single point [itex]x[/itex]), but [itex]g(x+a) \neq 0[/itex] for [itex]a \neq 0[/itex]. Something like [itex]g(x) = \cos(kx)[/itex] for an appropriate [itex]k[/itex] so that:

    * [itex]g[/itex] has the same period as [itex]f[/itex], so their Fourier series use the same basis functions
    * the first N coefficients of the Fourier series for [itex]g[/itex] are nonzero, so [itex]g+f[/itex] will yield the same left-hand side as [itex]f[/itex]
    * [itex]g[/itex] is an even function, so when multiplied by the kernel in your integral, it will still be an even function, so the integral of the portion contributed by [itex]g[/itex] will be nonzero.

    I know I'm being a bit vague because I'm too lazy to come up with a concrete example that works, but I think if you flesh out the above sketch, you would have a counterexample.
    Yes, I do think so. I read through your delta-epsilon argument, and it looked OK assuming the disputed equality was correct.
     
  9. Sep 23, 2012 #8

    jbunniii

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    Yes, that makes sense.
     
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