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Consider a Function $f\in L^2(\mathbb{T})$. Is there any lower bound for the decay of the Fourier coefficients

$$\hat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(t) e^{-int} dt$$ known?

There are a lot of upper bounds known but i cant find anything about a lower bound.

I would appreciate if you can help me!

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What do you exactly mean by lower bound? –  Davide Giraudo Jun 5 '12 at 9:48
BTW welcome to Math.SE! –  AD. Jun 5 '12 at 10:03
I mean the following: $ |\hat f(n)|\ge g(n)$ for all $n\in \mathbb{N}$, where $g\in o(n!)$ for example. –  Lenava Jun 5 '12 at 11:00
more precisely i am concerned about the coefficients of a function $f^{-1}$, where f is a polynomial. –  Lenava Jun 5 '12 at 11:24
So, $f$ is the reciprocal of a (trigonometric or algebraic?) polynomial. This information certainly belongs in the post, because the question is trivial ("$0$ is the best lower bound you can have) without such information about $f$. As it stands, we still don't know enough to give any nontrivial bound: if $f=[1+\text{(some tiny polynomial terms)}]^{-1}$, then $\hat f(n)$ is tiny for $n\ne 0$. –  user31373 Jun 5 '12 at 15:10

1 Answer 1

Theorem 3.2.2. in Grafakos's book Classical Fourier analysis (page 176) states that given a sequence $(d_n,n\geqslant 0)$ which converges to $0$, we can find an integrable function $f$ such that $|\widehat{f}(n)|\geqslant d_n$.

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