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In this question the term rapidly decaying sequence is used. What is the definition of a rapidly decaying sequence.(in cases of terms being purely real or complex.) How to prove that the sequence of Fourier series coefficients of a compactly supported, real valued, smooth function is rapidly decaying ? Please suggest a reference if the proof is lengthy. Also for the converse statement.

Rajesh D
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1 Answers1

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A sequence of complex numbers $\{a_n\}$ is said to be rapidly decaying if it converges to $0$ at a faster rate than any negative power of $n$: $$ \sup_n|a_n|n^k<\infty\qquad\forall k\in\mathbb{N}. $$ Let $\mathbb{T}$ denote the unit circle. If $f\in C^\infty(\mathbb{T})$, that is, $f$ is a periodic function of period $2\pi$ that admits derivatives of all orders, then the (doubly infinte) sequence of its Fourier coefficients is rapidly decaying. Indeed, integration by parts shows that $|\hat f_n|\le n^{-1}\|f'\|_\infty$; iterating the argument one gets that for all $k\in\mathbb{N}$, $|\hat f_n|\le n^{-k}\|f^{(k)}\|_\infty$.

Conversely, if the sequence of Fourier coefficients of a function $f\in L^1(\mathbb{T})$ is rapidly decaying, then the corresponding Fourier series and all the series derived from it by term by term differentiation are uniformly convergent. It follows that $f$ is equal almost everywhwre to a $C^\infty$ function.

Faster decay gives more smoothnes of $f$. See for instance Section 2 of chapter V in AN INTRODUCTION TO HARMONIC ANALYSIS by Yitzhak Katznelson.