Fourier series of functions with points of discontinuity

In summary: Not true: C[a,b] is not closed under infinite addition. All truncated Fourier series are continuous!
  • #1
jorgdv
29
0
If you have a function with countable discontinuities on an interval, I know that the Fourier series will converge to that function without those discontinuities. But how could you explain that formally? If the basis of the Fourier series span the space L^2[a,b], that would include functions with countable point-discontinuities, right?
 
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  • #2
The Fourier series for L^2 functions will converge to the function at all points of continuity and will converge to the average value at the discontinuities.
 
  • #3
mathman said:
The Fourier series for L^2 functions will converge to the function at all points of continuity and will converge to the average value at the discontinuities.

Not really. There is a continuous function whose Fourier series does not converge. What you say is only true with some additional conditions, for example a Lipschitz condition or a differentiable condition.
 
  • #4
micromass said:
Not really. There is a continuous function whose Fourier series does not converge. What you say is only true with some additional conditions, for example a Lipschitz condition or a differentiable condition.

It depends on what you mean by convergence. I was talking about convergence in the L^2 norm.
 
  • #5
mathman said:
It depends on what you mean by convergence. I was talking about convergence in the L^2 norm.

Then it's still wrong. The Fourier series of a function always converges to the function in the [itex]L^2[/itex]-norm. Doesn't matter what the discontinuities are.

http://en.wikipedia.org/wiki/Riesz–Fischer_theorem
 
  • #6
What I was thinking is that in the L2 space there is an equivalence relation such that if the Lebesgue integral of the diference is 0, then they are equivalent. However, the functions in the trigonometric basis of Fourier are contained in C[a,b], and because C[a,b] is closed under addition, the infinite linear combination with real coefficients will also be contained in C[a,b]. So the Fourier series will converge to the continuous equivalent function in the L2 space. Is that right?
 
  • #8
jorgdv said:
What I was thinking is that in the L2 space there is an equivalence relation such that if the Lebesgue integral of the diference is 0, then they are equivalent. However, the functions in the trigonometric basis of Fourier are contained in C[a,b], and because C[a,b] is closed under addition, the infinite linear combination with real coefficients will also be contained in C[a,b]. So the Fourier series will converge to the continuous equivalent function in the L2 space. Is that right?

What do you mean with "infinite linear combination" and what do you mean with "converge". The answers to your question depend on that. There are multiple ways to interpret convergence or summation of functions.
 
  • #9
With "infinite linear combination" I mean an infinite sum of elements contained in the space, in this case, scaled by real numbers (each element). And with "converge" in that context I meant pointwise.
 
  • #10
However, the functions in the trigonometric basis of Fourier are contained in C[a,b], and because C[a,b] is closed under addition, the infinite linear combination with real coefficients will also be contained in C[a,b].

Not true: C[a,b] is not closed under infinite addition. All truncated Fourier series are continuous!
 

What is a Fourier series?

A Fourier series is a way to represent a periodic function as a sum of sine and cosine functions. It is named after mathematician and physicist Joseph Fourier and has many applications in engineering, physics, and mathematics.

What is a point of discontinuity?

A point of discontinuity is a point where a function is not continuous. This means that the function has a jump or break at that point, and its value may suddenly change.

Can a function with points of discontinuity have a Fourier series?

Yes, a function with points of discontinuity can still have a Fourier series. However, the Fourier coefficients may be different at these points compared to a function that is continuous.

How are Fourier coefficients affected by points of discontinuity?

Fourier coefficients are affected by points of discontinuity because they are calculated by integrating the function over one period. If the function has a point of discontinuity, the integral will be affected and may result in different coefficients.

Are there any special techniques for finding Fourier coefficients of functions with points of discontinuity?

Yes, there are special techniques for finding Fourier coefficients of functions with points of discontinuity. These techniques involve splitting the function into pieces and calculating the coefficients separately for each piece.

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