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Fourier series of functions with points of discontinuity |
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| Nov17-12, 06:17 AM | #1 |
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Fourier series of functions with points of discontinuity
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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| Nov17-12, 03:26 PM | #2 |
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Recognitions:
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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.
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| Nov17-12, 03:35 PM | #3 |
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| Nov17-12, 05:24 PM | #4 |
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Recognitions:
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Fourier series of functions with points of discontinuity |
| Nov17-12, 05:27 PM | #5 |
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http://en.wikipedia.org/wiki/Riesz%E...ischer_theorem |
| Nov18-12, 05:42 AM | #6 |
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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?
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| Nov18-12, 03:33 PM | #7 |
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Recognitions:
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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. |
| Nov18-12, 04:07 PM | #8 |
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| Nov18-12, 06:42 PM | #9 |
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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.
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| Nov19-12, 03:11 PM | #10 |
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Recognitions:
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| convergence, discontinuity, fourier, series |
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