Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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

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 $|\widehat{f}(n)|\geqslant d_n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.