I have a problem:
Let f be a continuous function on the unit circle (Γ):
Γ={eiθ:θ∈[0,2π]}
Assume that f≠0 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.
=================================================
- 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)<+∞
==========================================
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
No comments:
Post a Comment