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For a $C^n[-\pi,\pi]$ function $f$ we have that $|\hat{f}(k)|\in O(1/k^n)$. This implies that if $f$ is $C^\infty[-\pi,\pi]$ then its $k-th$ Fourier coefficient decays faster than any $1/k^n$, $n\geq0.$ My questions:

$\bullet$ There is a name for this order of convergence?(something like "Schwarz order" if there exists such a denomination)

$\bullet$ Can't we say more than this? Y mean, there are a stronger result that states that these coefficients decay at an exponential rate (for example)?

David C. Ullrich
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  • There is a bit more that can be said if $f(t)$ is analytic in a complex neighborhood of the real line, e.g. $f(t)=\sqrt{1- m\sin(t)^2}$ with $|m|<1$. In that case, one indeed has exponentially-fast decay of the Fourier components in the desired sense. For discussion/proof of this result, see Trethefen and Weideman's 2014 SIAM excellent review article "The Exponentially Convergent Trapezoidal Rule" especially eq. (3-11) and Theorem 3.2. – Semiclassical May 26 '16 at 02:51

1 Answers1

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Well first, what you say about $C^\infty([-\pi,\pi])$ is not true. For example, $f\in C^\infty([-\pi,\pi])$ if $f(t)=t$, but the coefficients of that function are not $O(1/k^2)$.

Let's talk about $2\pi$-periodic functions - that's what you meant, or what you should have meant. I'd just say the coefficients were rapidly decreasing, and specify what that meant if it mattered.

No, there's no more that can be said. It's easy to see that if $\hat f(k)=O(1/k^{n+2})$ then $f\in C^n$ (somewhat stronger results in this direction exist that are not quite as easy). Hence if $\hat f(k)=O(1/k^n)$ for all $n$ then $f\in C^\infty$.

David C. Ullrich
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  • I have a question that may be unrelated. Do the standard decay results hold (for ex the ones in Grafakos's classical Fourier analysis, chapter 3, section 3.3 third edition, for smooth, and BV functions) when dealing with generalized Fourier series, i.e., Fourier series with general orthonormal basis, with the basis having nice decay properties? – Siddhartha Apr 04 '23 at 17:25