Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I just came across with the following question.. suppose we are given a periodic function of period $2\pi$. We define $a_n$ and $b_n$ to be the Fourier coefficients of $f$. To be precise, we have $$a_n:=\frac 1\pi\int_0^{2\pi}f(x)\cos nx\mathrm dx\quad\;b_n:=\frac 1\pi\int_0^{2\pi}f(x)\sin nx\mathrm dx,$$ so that $$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left[a_n\cos(nx)+b_n\sin(nx)\right]$$ Now it is clear that if $f\in C^\infty(\mathbb R)$ then $$\lim_{n\to\infty}n^{2k}(a_n^2+b_n^2)=0,\:\forall k>0.$$ Is the converse also true?

share|improve this question
1  
Yes. High decay of Fourier coefficients means that both the Fourier series and the series obtained by differentiating term-by-term converge uniformly. Hence the Fourier series for $f$ can be differentiated term by term. This process can be repeated as many times as we wish and so $f$ is infinitely differentiable. (Please check this assertion carefully, I'm answering in a hurry and have a high probability of being wrong.) –  Giuseppe Negro Sep 4 '12 at 19:39
add comment

1 Answer 1

up vote 1 down vote accepted

Yes, the converse is true. To see that more easily, I will write the series with the exponential form, namely, $\sum_{n\in\Bbb Z}c_ne^{inx}$. The condition gives that $|c_n|^2n^{2k}\to 0$ for each $k\geq 1$. In particular, the sequence $\{|c_n|\}$ is summable, hence we can write $f(x)=\sum_{n\in\Bbb Z}c_ne^{inx}$. Now we can show by induction that $$f^{(d)}(x)=\sum_{n\in\Bbb Z}(in)^dc_ne^{inx}.$$ It's true for $d=0$, and to jump from $d$ to $d+1$, we can use Talor's formula and the fact that the series $\sum_{n\in\Bbb Z}|c_n|n^{d+1}$ is convergent.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.