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Fourier Series Convergence Criterion

  1. Jul 24, 2014 #1
    I'm currently reading Tolstov's "Fourier Series" and in page 58 he talks about a criterion for the convergence of a Fourier series. Tolstov States:

    " If for every continous function F(x) on [a,b] and any number ε>0 there exists a linear combination
    [itex]σ_n(x)=γ_0ψ_0+γ_1ψ_1+....+γ_nψ_n[/itex] for which

    [itex] \int_a^b [F(x)-σ_n(x)]^2\,dx<ε [/itex]​

    then the system is complete.

    We begin by noting that given any square integrable function f(x), there exists a function F(x) for which

    [itex] \int_a^b [f(x)-F(x)]^2\,dx<ε [/itex]​


    This fact is quite clear geometrically, but for the reader who does not regard it as completely obvious, we give the following proof:" (Here is where I don't understand what he is doing)

    "The function [itex] f(x)[/itex] can have only a finite number of points of discontinuity. In particular, [itex] f(x)[/itex] can have only a finite number of points at which it becomes unbounded. Every such point can be included in an interval of such small length that the sum of the integrals of the function [itex] f^2(x)[/itex] over these intervals does not exceed ε/4. Define the auxiliary Θ(x) as being equal to [itex] f(x)[/itex] outside these intervals and equal zero inside the. Θ(x) is bounded and can have only a finite number of discontinuities, and obviously

    [itex] \int_a^b [f(x)-Θ(x)]^2\,dx<ε/4 [/itex]​

    Next, we include each point of discontinuity of Θ(x) in an interval of such small length that the total length l of all these new intervals satisfies the condition

    [itex] 4M^2l≤ε/4 [/itex]​

    where M is any number such that |Θ(x)|<M for a≤x≤b.

    Finally consider the continuos function F(x) which equals Θ(x) outside the intervals just described and which is "linear" inside of them. Obviously we have

    [itex] \int_a^b [f(x)-F(x)]^2\,dx≤4M^2l≤ε/4 [/itex]​
    ."

    So. Is the very first F(x) equal to the second F(X)?
    Why does he use the number ε/4? why not ε/7 or ε/122?
    If Θ(x)=f(x) when bounded, isn't the function Θ(x) not supposed to have discontinuities?
    Why is Θ(x) less than M?

    Thank you for taking the time to read this post and to answer my questions.
     
  2. jcsd
  3. Jul 25, 2014 #2
    hey

    i would say:
    the 2nd $F(x)$ is just a way you can define those $F(x)$, so
    $F(x) = \Theta(x)$, if $\Theta$ is continues, say at the Intervals $I_i$, $i\in\mathbb R$
    on the intervals $I_i$ you connect the ends on the lefthandside and on the righthandside so that is linear, so for instance if $I_i= [a,b]$ (sry bad interval names, dont interchange with the names of the theorem), $$ f(x) =\frac{ \Theta(a)-\Theta(b)}{a-b} x+\Theta(a)-a\frac{\Theta(a)-\theta(b)}{a-b}, \qquad \text{on the interval } I_i$$
    I just solve $$f(a) = d*a+c \stackrel{!}{=} \Theta(a)\quad\text{and}\quad f(b) = d*b+c \stackrel{!}{=} \Theta(b)\quad \text{for}\quad f(x)= d x+c$$

    to your 2nd question, well cause $\epsilon$ is arbitrary it doesnt matter if $< \epsilon/4$ or $\epslon / 7$ or $<10 \epsilon$, u can choose $0<\epsilon$ small enough anyway.

    3rd question, it is discontinous on the interval $[a,b]$ but not on the intervals $I_i$.

    4th question, thats not about $M$, just said that there is a positive constant $M$ that bounds $Theta$, its continuous on the intervals $I_i$ cause $I_i$ is compact, continous function take their maximum on compact intervals and outside of $I_i$ its Zero anyways.

    tho, sry i dont know how the $ ..$ enviroment from LaTeX is here in the forum, do you know who do I convert it?
    I hope its correct what i wrote and helpful, if not or if there are any furhter question pls dont hesitate

    best regards
     
  4. Jul 25, 2014 #3

    jbunniii

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    Is there an unstated hypothesis? This isn't true in general for an integrable (or square integrable) function. Even the Riemann integral (which I assume is in use here) can handle a function which has infinitely many points of discontinuity, as long as the set of such points has measure zero.

    The rest of the author's argument doesn't work if ##f## has infinitely many points of discontinuity.
     
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