Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f\in L^{1}(\mathbb T)$ and define the Fourier coefficient of $f$ : $\hat{f}(n)=\frac{1}{2\pi} \int _{-\pi}^{\pi} f(t) e^{-int} dt; (n\in \mathbb Z)$ and we put, $$A(\mathbb T):= \{f\in L^{1}(\mathbb T): \hat{f}\in \ell^{1}(\mathbb Z), \ \text {that is,} \ \sum_{n\in \mathbb Z} |\hat{f}(n)| < \infty \}.$$

Result of Katznelson: If $F$ is defined on on $[-1, 1]$ and composition of $F$ and $f$, $F(f) \in A(\mathbb T)$ whenever $f\in A(\mathbb T)$ and $f(\mathbb T)\subset [-1,1]$, then $F$ must be analytic on $[-1,1 ].$

As a corollary to this result, there exist $f\in A(\mathbb T)$ such that $|f|$ does not belong to $A(\mathbb T)$.

On the other hand, Beurling has shown the the following:

Result of Beurling: If $f\in A(\mathbb T)$ such that $|\hat{f}(\pm n)| \leq c_{n}, \ (n=0,1,2,...), $ where $c_{n}\downarrow 0$ and $\sum_{n=0}^{\infty}c_{n} < \infty,$ then $|f|\in A(\mathbb T).$

I read the above result in the book (by R. E. Edwards, Fourier series, A Modern Introduction, Volume-1; p.178); in which he state this result without proof, for further reading. I am unable to find the proper reference for the same, in web search.

My Request: If possible, please, can you give me a proper reference book for the result of Beurling or in which paper of the Beurling this result has been appear ? (I guess this must be appear between the years 1955 and 1965)

Thanks a lot;

share|improve this question

1 Answer 1

up vote 1 down vote accepted

This is Theorem V (page 16) from:

A. Beurling, On the spectral synthesis of bounded functions. Acta Math. 81 (1948).

In fact, Beurling proves the stronger statement:

Theorem Let $f(x) = \sum_{n=-\infty}^{\infty} a_n e(nx)$ (with $a_0=0$) have an absolutely convergent Fourier series such that $|a_{\pm n}| \leq a_n^{*}$ where $a_n^{*}$ is a non-increasing sequence in $\ell^{1}$. Moreover, assume that $g(x) = \sum_{n=-\infty}^{\infty} b_n e(nx)$ (with $b_0=0$) is a contraction of $f$. Then the Fourier series of $g(x)$ converges absolutely and $\sum_{n=-\infty}^{\infty}|b_n| \ll \sum_{n=-\infty}^{\infty} a_n^{*}$.

share|improve this answer
@MK; Thanks a lot, I got it; (The result you have pointed out to me , ah!, I find that result extremely beautiful, thanks again) –  Inquisitive Feb 14 at 16:03

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.