Fourier Series Convergence Criterion

In summary: 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 conditionThis 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. He seems to be saying that if there are infinitely many points of discontinuity in a function, then we can fill an interval with these points so that the sum of the integrals of the function over these intervals doesn't exceed a certain value. However, this argument doesn't work if the function has infinitely many points of discontinuity.
  • #1
Chacabucogod
56
0
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 continuous 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.
 
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  • #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, don't 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 doesn't 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, that's 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, continuous function take their maximum on compact intervals and outside of $I_i$ its Zero anyways.

tho, sry i don't know how the $ ..$ environment 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 don't hesitate

best regards
 
  • #3
Chacabucogod said:
given any square integrable function f(x)
"The function [itex] f(x)[/itex] can have only a finite number of points of discontinuity.
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.
 

What is the Fourier Series Convergence Criterion?

The Fourier Series Convergence Criterion is a mathematical tool used to determine whether a Fourier series, which is a representation of a periodic function as a sum of sine and cosine terms, converges to the original function.

What is the significance of the Fourier Series Convergence Criterion?

The Fourier Series Convergence Criterion is important because it allows us to determine the accuracy of a Fourier series approximation and whether it can be used to accurately represent a function.

How is the Fourier Series Convergence Criterion calculated?

The Fourier Series Convergence Criterion is calculated using the Dirichlet Test, which evaluates the convergence of a series based on the behavior of the coefficients and the periodicity of the function.

What are the conditions for convergence in the Fourier Series Convergence Criterion?

The conditions for convergence in the Fourier Series Convergence Criterion are that the coefficients of the series must be bounded and the function must be piecewise continuous and periodic.

What happens if the Fourier Series Convergence Criterion is not satisfied?

If the Fourier Series Convergence Criterion is not satisfied, it means that the Fourier series does not accurately represent the function and cannot be used for approximation. In this case, alternative methods such as truncating the series or using a different type of series may be necessary.

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