Why there is no integrable complex valued function that has the following Fourier coefficients a_k = 1/k for k >= 1, and a_k = 0 otherwise?

Thanks

Alexei

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- Dec 30th 2008, 08:21 AMDifferentFourier coefficients
Why there is no integrable complex valued function that has the following Fourier coefficients a_k = 1/k for k >= 1, and a_k = 0 otherwise?

Thanks

Alexei - Dec 30th 2008, 09:01 AMConstatine11
Are you refering to complex Fourier series, or standard Fourier series (because if standard Fourier series the function is real)?

. - Dec 30th 2008, 10:00 AMDifferent
More detailed question is

Show that there is no (complex valued) Riemann integrable function F(x) with Fourier coefficient a_k = 1/k for k>=1 and a_k=0 otherwise.

This is Exercise 6 (Chapter 3) from "Fourier analysis" by Stein&Shakarchi.

...clearly, the function F(x) if exists can not be real-valued since f is real valued iff conjugate of a_k = a_{-k}, which is not true for the above a_k's. - Dec 30th 2008, 10:11 AMConstatine11
- Dec 30th 2008, 08:09 PMDifferent
- Dec 30th 2008, 08:37 PMIsomorphism
I dont know how to show "integrable" functions don't exist, but I can see one problem with these co-effs.

Observe that $\displaystyle \sum c_k$ is not finite. Of course these do not cause any problems, since Dirichlet's conditions are only sufficient conditions and not necessary.

I can see that this function $\displaystyle f(x)=\sum_{n=-\infty}^{\infty}c_n e^{\mathrm{i}nx}$ is not well defined for x = 0, because you get the harmonic series.

Does this help? - Dec 30th 2008, 11:38 PMDifferent
not really. There exist continuous functions whose Fourier series fail to converge at the point of continuity. So the fact that Fourier series diverges at some point does not solve the problem.

But I think I now how to do this problem, anyway.

Assume that an inegrable function with above coeff-s exist, then summing up coefficients gives a harmonic series which is diverging, on the other hand the same sum can be written as an $\displaystyle \sum c_k = {1\over 2\pi} \int_{0}^{2\pi} f(\theta ) \sum e^{-in\theta} d\theta $.

Sum of the geometric progression under the integral sign can be easily calculated and is an integrable function. Since $\displaystyle f(\theta) $ is integrable and the product of integrable functions is integrable we get a contradiction. - Dec 31st 2008, 12:34 AMOpalg
I'm puzzled by this question. The series $\displaystyle \sum_{k=1}^\infty \tfrac1ke^{ikx}$ converges to $\displaystyle -\log(1-e^{ix})$ everywhere in the interval [–π,π] except at x=0 (by Abel's test).

Obviously this function is not strictly speaking Riemann integrable, because it is unbounded near x=0. But it has an indefinite integral $\displaystyle (1-e^{ix})\log(1-e^{ix}) + e^{ix}$, which has a limit 1 at x=0. So it looks to me as though the improper Riemann integral of the function over the interval [–π,π] exists and is 0.

Is there something wrong with that argument? Edit: Yes there certainly is. See comments below. - Dec 31st 2008, 12:59 AMDifferent
I am not sure how you have summed up the above series, but in any case, are you sure that Fourier coeff-s of function $\displaystyle -\log(1-e^{ix})$ have the required form? i.e. a_k = 1/k for k >= 1, and a_k =0 for k<= 0 ?

- Dec 31st 2008, 09:03 AMOpalg
The series $\displaystyle \sum_{k=1}^\infty z^{k-1}$ converges to $\displaystyle \frac1{1-z}$ for |z|<1. It's legitimate to integrate term by term to get $\displaystyle \sum_{k=1}^\infty \frac{z^k}k = -\log(1-z)$ for |z|<1. By Abel's result, that equality also holds for |z|=1, except at z=1. Put $\displaystyle z=e^{ix}$ to get $\displaystyle \sum_{k=1}^\infty \frac{e^{ikx}}k = -\log(1-e^{ix})$ (except for x=0).

There's a theorem telling you that if $\displaystyle f(x) = \sum c_ke^{ikx}$ and the series $\displaystyle \sum|c_k|$ converges, then the Fourier coefficients of f are the c_k. Of course, in this example, c_k = 1/k, so the series $\displaystyle \sum|c_k|$ does not converge, and the theorem does not apply.

That was my first mistake. But the second mistake was more serious. I said that the function $\displaystyle -\log(1-e^{ix})$ had an indefinite integral. I was thinking of the fact that $\displaystyle -\log(1-z)$ has an indefinite integral with respect to z. But you don't get an indefinite integral of $\displaystyle -\log(1-e^{ix})$ with respect to x just by replacing z by $\displaystyle z=e^{ix}$.

Moral: Don't make hasty posts to MHF last thing at night. - Dec 31st 2008, 03:05 PMmr fantastic