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Everybody knows that the Fourier coefficients of an $L^1$ function converge to zero. For an $L^2$ function we can say much more: $(\hat f(k))_{k\in\mathbb Z}\in\ell^2$. Therefore, it is reasonable to conjecture that we can say even more for an $L^\infty$ function. My first guess was that $\hat f(k) = O(1/k)$, but now I doubt that. Does anbody know more about this and/or an example of $f\in L^\infty$ such that $\hat f(k)\neq O(1/k)$?

Friedrich Philipp
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  • $\displaystyle f(x) = \sum_{n=1}^\infty \frac{e^{i n^{1/2}}}{n^a} e^{ i n x}$ with $a \in (0,1)$ converges since it is very similar to $\displaystyle g(x) =\int_1^\infty e^{i \tau^{1/2}} \tau^{-a} e^{i x \tau} d\tau = \int_1^\infty e^{i t} t^{-2a} e^{i x t^2} 2t dt = - \int_1^\infty \frac{d}{dt}(e^{i t} t^{-2a}) \frac{e^{i x t^2}}{ix} dt$ and I bet it is bounded when $a > 1/2$ – reuns Jun 07 '16 at 06:15
  • @user1952009 Thanks, but how do you justify your "guess/bet"? – Friedrich Philipp Jun 07 '16 at 06:18
  • the last integral converges absolutely, the only one problem being the $1/x$, but at $x=0$ : $\int_1^\infty e^{it} t^{1-2a} dt$ is known to converge when $a > 1/2$. so it leaves us with the problem of proving $g(x)$ is continuous at $0$, for proving it is bounded – reuns Jun 07 '16 at 06:27
  • $L^{\infty}[0,2\pi]\subset L^2[0,2\pi]\cap L^1[0,2\pi]$. So everything you know about the two spaces on the right applies to $L^{\infty}$. In that sense it buys you more. – Disintegrating By Parts Jun 07 '16 at 15:16
  • @TrialAndError What do you mean by "it buys you more"? We have $L^\infty\subset L^2\subset L^1$. This was in particular a motivation for my question. – Friedrich Philipp Jun 07 '16 at 23:36

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