Register to reply

Fourier series of functions with points of discontinuity

Share this thread:
jorgdv
#1
Nov17-12, 06:17 AM
P: 12
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?
Phys.Org News Partner Science news on Phys.org
An interesting glimpse into how future state-of-the-art electronics might work
Tissue regeneration using anti-inflammatory nanomolecules
C2D2 fighting corrosion
mathman
#2
Nov17-12, 03:26 PM
Sci Advisor
P: 6,059
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
#3
Nov17-12, 03:35 PM
Mentor
micromass's Avatar
P: 18,246
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
#4
Nov17-12, 05:24 PM
Sci Advisor
P: 6,059
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.
micromass
#5
Nov17-12, 05:27 PM
Mentor
micromass's Avatar
P: 18,246
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
#6
Nov18-12, 05:42 AM
P: 12
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
#7
Nov18-12, 03:33 PM
Sci Advisor
P: 6,059
http://en.wikipedia.org/wiki/Converg...Fourier_series

Above appears to be a good summary, particularly the section on pointwise convergence.
micromass
#8
Nov18-12, 04:07 PM
Mentor
micromass's Avatar
P: 18,246
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
#9
Nov18-12, 06:42 PM
P: 12
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
#10
Nov19-12, 03:11 PM
Sci Advisor
P: 6,059
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!


Register to reply

Related Discussions
Fourier series odd and even functions Calculus & Beyond Homework 4
Fourier series odd and even functions General Math 2
Discontinuity at certain points Calculus & Beyond Homework 2
Jump discontinuity with fourier series Calculus & Beyond Homework 2
Fourier series and even/odd functions Calculus & Beyond Homework 2