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.

In a book from differential equations I found the following theorem, without proof and references:

Let functions $f, g: R \rightarrow R$ be continuous and $2\pi$-periodic and let $m\in N$. Assume that $$\frac{a_0}{2}+\sum_{n=1}^\infty (a_n \cos nx+b_n \sin nx),$$ $$\frac{A_0}{2}+\sum_{n=1}^\infty (A_n \cos nx+B_n \sin nx)$$ be Fourier series of $f$ and $g$ respectively, not neceserilly convergent to $f$ and $g$.

Assume that $a_0=0$ and $$(\frac{A_0}{2})^{(m)}+\sum_{n=1}^k (A_n \cos nx+B_n \sin nx)^{(m)}=\frac{a_0}{2}+\sum_{n=1}^k (a_n \cos nx+b_n \sin nx)$$ for all $x \in R$, $k\in N$. Then $g$ is of class $C^m$ and $g^{(m)}=f$.

Maybe proof of references of this theorem.

share|improve this question
This is basically a version of the Sobolev embedding theorem; I would suggest looking up that theorem in a decent functional analysis or PDE textbook. –  Paul Siegel May 9 '12 at 17:33
I think it will be simpler using complex series. Thanks to the equation, you can find a relationship between the Fourier coefficients of $f$ and $g$, which will show that $g$ is of class $C^m$. Conclude using the uniqueness of Fourier coefficients. –  Davide Giraudo May 9 '12 at 17:36
add comment

1 Answer

up vote 1 down vote accepted

We can rewrite the Fourier series as $f(x)\sim\sum_{n\in \Bbb Z}c_ne^{inx}$ and $g(x)\sim\sum_{n\in \Bbb Z}d_ne^{inx}$. We have $n^mi^md_n=c_n$ for all $n$, which proves that the Fourier coefficients of $g$ decay fast enough in order to ensure us that $g$ as a $m$-th derivative. Indeed, we have a condition on the decay the Fourier coefficients of a periodic function which can be seen by integration by parts.

Since $g^{(m)}$ and $f$ has the same Fourier coefficient and are continuous, they are equal.

share|improve this answer
add comment

Your Answer


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.