By Zhao W., Liu P.

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**Example text**

2B R The Fourier inversion formula for f holds ( f is integrable) and it holds everywhere ( f is continuous). In particular, the nth Fourier coefficient of is equal to f − The formal Fourier series of n . 2B is therefore 1 2B f n∈Z n n e−2iπ 2B ν . e. 9), that is is almost everywhere equal to its Fourier series. This proves (†). 9), a normally convergent series of bounded and continuous functions). 1 Fourier Theory in L 1 1 f˜(t) = 2B 43 f n∈Z = R n∈Z n 2B n R h(ν) e2iπ ν(t− 2B ) dν n n 1 e−2iπ ν 2B f 2B 2B h(ν) e2iπ νt dν.

13 Let f : R → C be an integrable function such that for some α > 1, f (t) = O 1 1 + |t|α (|t| → ∞), and f (ν) = O 1 1 + |ν|α (|ν| → ∞), then the strong Poisson summation formula holds for all T ∈ R+ \{0}. 21. ) with rapid convergence, or to obtain some remarkable formulas. Here is a typical example. Let a > 0. The ft of the function f (t) = e−2πa|t| is f (ν) = π(a 2a+ν 2 ) . 1 Fourier Theory in L 1 41 is a continuous function with bounded variation, we have by the strong Poisson summation formula e−2πa|n| = n∈Z The left-hand side is equal to as 1 πa +2 a n≥1 π(a 2 +n 2 ) .

1 Fourier Theory in L 1 39 Proof The ft g of the (integrable) function g is g(ν) e−2iπ νt dt . g(t) = R By the inversion formula, the last integral is equal to g(−t). Since g is assumed continuous, this equality holds everywhere, and in particular for t = nT . By the weak version of the Poisson sum formula, T g(nT ) = T g(− nT ) is the nth Fourier coefficient of n∈Z g(ν + n/T ). Therefore if (b) is true, then (a) is necessarily true. Conversely, if (a) is true, then the sequence {T g(− nT )}n∈Z is the sequence of Fourier coefficients of two functions, the constant function equal to T g(0), and n∈Z g(ν + n/T ), and therefore the two functions must be equal almost everywhere.