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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;

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up vote 2 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 $\sum_{n=-\infty}^{\infty}|b_n| \ll \sum_{n=-\infty}^{\infty} a_n^{*}$.

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@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 '14 at 16:03

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