A First Course in Harmonic Analysis (2nd Edition) by Anton Deitmar

By Anton Deitmar

This primer in harmonic research offers a lean and stream-lined advent to the imperative techniques of this gorgeous conception. unlike different books at the subject, a primary path in Harmonic research is completely according to the Riemann imperative and metric areas rather than the extra tough Lebesgue crucial and summary topology. however, just about all proofs are given in complete and all critical recommendations are offered essentially. This publication introduces Fourier research, major as much as the Poisson Summation formulation, in addition to the options utilized in harmonic research of noncommutative teams.

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2. We have to show that T (V ) is dense in H = 2 (N). Let f ∈ 2 (N), and for n ∈ N let fn ∈ 2 (N) be given by fn (j) = f (j) if j ≤ n, 0 if j > n. Then ||f − fn ||2 = |f (j)|2 , j>n which tends to zero as n tends to infinity. So the sequence (fn ) converges to f in 2 (N). For j = 1, 2, . . , n let λj = f (j). Then fn = T (λ1 e1 + · · · + λn en ), so fn lies in the image of T , which therefore is dense in H. This concludes the existence part of the proof. For the uniqueness condition assume that there is a second isometry T : V → H onto a dense subspace.

Assume that its Fourier series converges pointwise to the function g; then ck (g)e2πikx , g(x) = k∈Z so that for x = 0 we get f (l) = g(0) = l∈Z ck (g) k∈Z 1 = g(y)e−2πiky dy k∈Z 0 1 f (y + l)e−2πiky dy. 6. THE POISSON SUMMATION FORMULA 55 Assuming that we may interchange summation and integration, this equals l+1 f (y)e−2πiky dy = k∈Z −∞ l k∈Z l∈Z ∞ f (y)e−2πiky dy = fˆ(k). k∈Z This is a formal computation, valid only under certain assumptions. We will now turn it into a theorem by giving a set of conditions that ensures the validity of those assumptions.

Proof: Since S is mapped to itself, the corollary follows from the inversion theorem. 6 Let f (x) = e−πx . Then f ∈ S and fˆ = f. 3 the function f is, up to scalar multiples, the unique solution of the differential equation f (x) = −2πxf (x). By induction one deduces that for every natural number n there is 2 2 a polynomial pn (x) such that f (n) (x) = pn (x)e−πx . Since e−πx decreases faster than any power of x as |x| → ∞, it follows that f lies in S. Then fˆ also lies in S, and we compute (fˆ) (y) = ∞ 2 (−2πix)e−πx e−2πixy dx −∞ ∞ = i 2 (e−πx ) e−2πixy dx −∞ = −2πy fˆ(y), 52 CHAPTER 3.

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A First Course in Harmonic Analysis (2nd Edition) by Anton Deitmar
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