Get A First Course in Harmonic Analysis (2nd Edition) PDF

By Anton Deitmar

ISBN-10: 0387275614

ISBN-13: 9780387275611

This primer in harmonic research supplies a lean and stream-lined advent to the crucial recommendations of this pretty idea. unlike different books at the subject, a primary path in Harmonic research is solely according to the Riemann necessary and metric areas rather than the extra hard Lebesgue quintessential and summary topology. however, just about all proofs are given in complete and all primary techniques are awarded essentially. This publication introduces Fourier research, prime as much as the Poisson Summation formulation, in addition to the options utilized in harmonic research of noncommutative teams.

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THE FOURIER TRANSFORM Theta Series As an application of the Poisson summation formula we give a proof of the functional equation of the classical theta series. In Appendix A this is employed to derive the analytic continuation and the functional equation of the Riemann zeta function. Since this is a result of utmost importance to many areas of mathematics, it is included in this book. It requires, however, knowledge of complex analysis, which is why it appears only in the appendix. 1 For t > 0 let 2 e−tπk .

16 Let H be a Hilbert space and v, w ∈ H. Show that 2 2 ||v + w|| + ||v − w|| 2 2 = 2 ||v|| + 2 ||w|| . This equality is known as the parallelogram law. 17 Let H be a Hilbert space and let T : H → H be a map. , there is a map T ∗ on H such that T v, w = v, T ∗ w for all v, w ∈ H. Show that T and T ∗ are both linear. 18 Let V be a finite-dimensional Hilbert space. A linear operator A : V → V is called self-adjoint if for any two vectors v, w ∈ V we have Av, w = v, Aw . , that V has a basis consisting of eigenvectors of A.

J=1 The coefficients are unique and are given by cj = cj (h) = h, ej . The map h → (cj )j∈N gives an isometry from H to 2 (N). For h, h ∈ H we have ∞ h, h = cj (h)cj (h ), j=0 so in particular, ||h||2 = ∞ 2 j=1 |cj | . CHAPTER 2. HILBERT SPACES 34 Proof: Let h ∈ H, define cj (h) = h, ej , and for n ∈ N let sn (h) = n j=1 cj ej ∈ H. 1: 0 ≤ ||h − sn (h)||2 n = h− n cj ej , h − j=1 n = ||h||2 − cj ej j=1 |cj |2 . j=1 This implies ∞. n 2 j=1 |cj | ≤ ||h||2 for every n and therefore ∞ 2 j=1 |cj | < We therefore obtain a linear map T : H → l2 (N) mapping h to the sequence (cj (h))j .

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

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