Fourier series of functions with points of discontinuity


by jorgdv
Tags: convergence, discontinuity, fourier, series
jorgdv
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#1
Nov17-12, 06:17 AM
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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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mathman
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#2
Nov17-12, 03:26 PM
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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.
micromass
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#3
Nov17-12, 03:35 PM
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Quote Quote by mathman View Post
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.

mathman
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#4
Nov17-12, 05:24 PM
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Fourier series of functions with points of discontinuity


Quote Quote by micromass View Post
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.
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#5
Nov17-12, 05:27 PM
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Quote Quote by mathman View Post
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%E...ischer_theorem
jorgdv
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#6
Nov18-12, 05:42 AM
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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?
mathman
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#7
Nov18-12, 03:33 PM
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http://en.wikipedia.org/wiki/Converg...Fourier_series

Above appears to be a good summary, particularly the section on pointwise convergence.
micromass
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#8
Nov18-12, 04:07 PM
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Quote Quote by jorgdv View Post
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.
jorgdv
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#9
Nov18-12, 06:42 PM
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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.
mathman
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#10
Nov19-12, 03:11 PM
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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!


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