Sunday, 23 September 2018

functional analysis - Prove that the Fourier series of frac1f is absolutely convergent




I have a problem:




Let f be a continuous function on the unit circle (Γ):



Γ={eiθ:θ[0,2π]}



Assume that f0 on Γ, and the Fourier series of f is absolutely convergent on Γ.



Prove that the Fourier series of 1f is absolutely convergent on Γ.





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  • I've tried to use the definition of n=0an is absolutely convergent:




A real or complex series n=0an is said to converge absolutely if n=0|an|=L for some real number L.







So let the Fourier series of f be given by:
f(x)=n=aneinθ



We want to show that n=|aneinθ|f(θ)<+



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



Any help will be appreciated! Thanks!


Answer



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
http://www.ams.org/journals/proc/1975-048-01/S0002-9939-1975-0365002-8/S0002-9939-1975-0365002-8.pdf


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