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I have a problem:

Let $f$ be a continuous function on the unit circle $(\Gamma)$:

$$\Gamma=\{e^{i\theta}: \theta\in [0, 2 \pi]\}$$

Assume that $f \ne 0$ on $\Gamma$, and the Fourier series of $f$ is absolutely convergent on $\Gamma$.

Prove that the Fourier series of $\dfrac{1}{f}$ is absolutely convergent on $\Gamma$.


  • I've tried to use the definition of $\sum_{n=0}^\infty a_n$ is absolutely convergent:

A real or complex series $\sum_{n=0}^\infty a_n$ is said to converge absolutely if $\sum_{n=0}^\infty \left|a_n\right| = L$ for some real number $L$.

So let the Fourier series of $f$ be given by: $$f(x) = \sum_{n=-\infty}^{\infty} a_{n}e^{in \theta}$$

We want to show that $$\sum_{n=-\infty}^{\infty} |a_{n}e^{in \theta}| \to f(\theta)< +\infty$$


But I still have no solution :( . Can anyone help me!

Any help will be appreciated! Thanks!

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A pretty proof was given by Gelfand – Mlazhinka Shung Gronzalez LeWy Nov 1 '13 at 14:53
Yes, It's great link! thanks ABC. I'll write it here, Please you check my post. – kimtahe6 Nov 1 '13 at 15:18
Since the theorem $3.1$, we have $\lambda - f$ is invertible. But I think that we need to show the Theorem 1.1. (Wiener’s Theorem) (in your link)??? How can we show it? Thanks! – kimtahe6 Nov 1 '13 at 15:38
In that note, the very last paragraph is what would be the proof of Theorem 1.1. After all the work, the proof reduces to: If $f$ is not invertible there is a non-zero functional $\phi$ such that $\phi(f)=0$. All the functionals in this algebra are evaluations at certain point, so $0=\phi(f)=f(t_0)$ for some $t_0$, contradiction. – Mlazhinka Shung Gronzalez LeWy Nov 1 '13 at 15:48
Yes, it is not a simple proof (if you begin from scratch). Direct proofs, which existed before Gelfand's, are not so simple either. There are many direct proofs. Online, you can find many searching for 'Weiner's theorem'. – Mlazhinka Shung Gronzalez LeWy Nov 1 '13 at 16:20
up vote 2 down vote accepted

The standard proof ever since Gelfand uses commutative Banach agebras, but this requires a certain amount of preparations. A short proof was given by Newman in 1975. The paper is only two pages long, but still too long to summarize. Anyway, the paper is accessible freely at

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Thanks M. Mueger! It's very useful to me. – kimtahe6 Nov 2 '13 at 1:49
In Newman's proof: $$\sum_{n \ne 0}n^2|a_n|^2=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}|f'(x)|^2 \rm d x$$ I think that we use the [Parseval's theorem] We apply for $$A(x)=B(x)=f'(x)=\sum_{n=-\infty}^{+\infty}(ina_n)\cdot e^{inx}$$ then $$\sum_{n=-\infty}^{+\infty}(ina_n)\overline{(ina_n)}=\dfrac{1}{2\pi}\int_{-\pi}‌​^{\pi} f'(x) \overline{f'(x)} \rm d x$$ Whence $$\sum_{n \ne 0}n^2|a_n|^2=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}|f'(x)|^2 \rm d x$$ Do I have the correct change? [Parseval's theorem]: – kimtahe6 Nov 3 '13 at 16:25
Yes, this the way to prove it. – M. Mueger Nov 5 '13 at 10:31

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