0
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ What do you exactly mean by lower bound? $\endgroup$ – Davide Giraudo Jun 5 '12 at 9:48
  • 1
    $\begingroup$ BTW welcome to Math.SE! $\endgroup$ – AD. Jun 5 '12 at 10:03
  • $\begingroup$ I mean the following: $ |\hat f(n)|\ge g(n)$ for all $n\in \mathbb{N}$, where $g\in o(n!)$ for example. $\endgroup$ – Lenava Jun 5 '12 at 11:00
  • $\begingroup$ more precisely i am concerned about the coefficients of a function $f^{-1}$, where f is a polynomial. $\endgroup$ – Lenava Jun 5 '12 at 11:24
  • 1
    $\begingroup$ 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 $n\ne 0$. $\endgroup$ – user31373 Jun 5 '12 at 15:10
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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