The Poisson Summation Formula and an Application to Number Theory Jason Payne Math 248- Introduction Harmonic Analysis, February 18, 2010

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1 The Poisso Summatio Formula ad a Applicatio to Number Theory Jaso Paye Math 48- Itroductio Harmoic Aalysis, February 8, This talk will closely follow []; however some material has bee adapted to a slightly more geeral case usig (without metio theorems foud i, for example, []. The applicatio will closely follow [3]. Essetially the idea behid the mai topic of this talk stems from the questio of how oe ca take a fuctio f L (R, ad maufacture from it a periodic fuctio. There are two approaches- the first is to restrict ˆf to the lattice Z ad create the Fourier series Sf(x = ˆf(κe κ Z πiκ x ; the secod is to average, i some sese, the fuctio over all periods i Z, i.e. to geerate the series P f(x = k Z τ kf(x. The Poisso Summatio Formula shows that both methods ot oly work, but they give the same result. Before gettig to this theorem, we will require the followig lemma ad the subsequet techical theorem. Lemma. If f, g L (R, the ( ˆfĝˇ= f g. Proof. By Hölder s iequality ad the Placherel theorem, we have that ˆfĝ L ˆf L ĝ L < ; hece ˆfĝ L, so ( ˆfĝˇ is well-defied. Now, for each x R, we defie h x (y = g(x y. Observe that ĥ(ξ = e πiξ y g(x ydy = e πiξ x R e πiξ (x y g(x ydy R = e πiξ x R e πiξ z g(zdz = e πiξ x E ξ, g = e πiξ x g, E ξ Thus, sice F is uitary o L, = e πiξ x R g(ze πiξ z dz = e πiξ x ĝ(ξ. (f g(x = f(yg(x ydy = f(yh(ydy = f, h = ˆf, ĥ R R = ˆf(ξ ĥ(ξdξ = ˆf(ξĝ(ξe πiξ x dξ = ( ˆfĝˇ(x, R R Theorem. Suppose that Φ L ( C ( (where is either R or T, Φ( =, ad φ = ˇΦ L (. Give f L ( + L (, for t > set f t (x = ˆf(ξΦ(tξe πiξ x dξ. (a If f L p (, for p <, the f t L p ( ad f t f L p as t. (b If f is bouded ad uiformly cotiuous, the so is f t, ad f t f uiformly as t. (c Suppose also that φ(x C(+ x ɛ for some C, ɛ >. The f t (x f(x for every x i the Lebesgue set of f.

2 Proof. Write f = f + f where f L ad f L. The, from the Hausdorff- Youg Iequality, we have that ˆf L, ˆf L. Moreover, sice Φ L C, we have that Φ L = Φ = Φ + Φ, E E c where E = {x Φ(x ɛ} is compact sice Φ C. Therefore Φ L m(e max x E { Φ } + ɛ Φ L <, where the last iequality comes from the followig facts: m(e < sice E is a compact subset of R, ad is therefore bouded; max x E { Φ } < sice Φ is cotiuous ad E is compact; Φ L < sice Φ L. Hece Φ L L, ad so f t coverges absolutely for every x. Moreover, we have that φ t (ξ = t φ(t ξe πix ξ dξ = t t φ(ze πitx z dz = ˇΦ(tx = Φ(tx, where the last equality is justified sice Φ L, so we ca apply the Fourier Iversio Theorem. I particular, φ t (ξdξ = φ t ( = Φ( =. From here, sice φ, f L, Youg s Iequality implies that f φ L, ad sice ˆf L ad Φ L, Hölder s Iequality implies that f Φ L. Therefore, applyig the Fourier Iversio Theorem, ˆf (ξφ(tξe πiξ x dξ = ˆf (ξ φ t (ξe πiξ x dξ = (f φ t ˆ(ξe πiξ x dξ = (f φ t (x. Additioally, we kow that ˇφ = Φ L L, so by the Placherel theorem ad the iversio theorem ˆˇφ = φ L ; hece, by Lemma, we have that f (ξφ(tξe πiξ x dξ = ˆf (ξ φ t (ξe πiξ x dξ = (f φ t ˆ(ξe πiξ x dξ = ( f φ t ˇ(x = (f φ t (x. Fially, by the liearity of F ad, we have that f t = f φ t, ad the assertios follow from the approximate idetity theorem. Theorem (The Poisso Summatio Formula. Suppose f C(R satisfies f(x C( + x ɛ ad ˆf(ξ C( + ξ ɛ for some C, ɛ >. The f(x + k = ˆf(κe πiκ x, k Z κ Z where both series coverge absolutely ad uiformly o T. x =, f(k = ˆf(κ. k Z κ Z I particular, takig

3 Proof. We begi by otig that the reciprocal polyomial growth coditio o f ad ˆf esure the absolute ad uiform covergece of the series ivolved. Ideed, f(x + k f(x+k C (+ x+k ɛ (+ x+k ɛ <, k Z k Z k Z R ad similarly for the series ivolvig ˆf. The uiform covergece of P f = τ k f(x = f(x + k k Z k Z implies that P f C(T. But T is compact ad has fiite measure, so M = max x T P f exists ad is fiite; hece T P f M m(t <, so P f L (T. Now cosider Φ. We have that Φ( =, T Φ = m(t <, ad {x T Φ(x ɛ} = { if ɛ > T < ɛ, which are both compact, so Φ L (T C (T. Thus, Φ satisfies the coditios of Theorem, so we have from parts (b ad (a that (P f t = ˆf(κEκ P f κ Z uiformly ad i L (T. Fially, ote that the uiform covergece esures that we have poitwise covergece as well, i.e. f(x + k = ˆf(κe πiκ x, k Z κ Z Now that we have prove the cetral object of study of this talk, we will move o to a applicatio i the field of aalytic umber theory ad the study of the Riema zeta fuctio. Defiitio. We defie the Jacobi theta fuctio as ψ(x = e πx. = We will apply the Poisso Summatio Formula to the Gauss kerel, but first we eed to verify that it satisfies the hypotheses of this theorem. Lemma. Let Φ(x = e π x The, for each t >, Φ t satisfies the hypotheses of Theorem. Proof. C =, ɛ = /. 3

4 Theorem 3. Let ψ(x be as above. The ψ(x satisfies the fuctioal equatio ψ(x + = ( (ψ +. x x Proof. We begi by applyig Poussi s summatio formula to the fuctio Φ t to obtai t = k Z k Z e kπ/t = k Z Φ t (k = κ Z Φ( tκ = k Z Φ t (κ Φ( tκ = k Z e κ πt. From here, simply ote that the terms i both summatios are symmetric with respect to the idex, so we have ( (ψ + = ψ(x +, x x To see how this fuctioal equatio is useful i studyig the Riema zeta fuctio, we will first require the followig two classical results. Lemma 3 (Euler s Reflectio Formula. Proof. We begi by recallig that Γ(sΓ( s = π si(πs. thus Γ(s = seγs = (( + s e s/ ; Γ(sΓ( s = s e γs e γs (( + s e s/ (( s e s/ = s ( s = = However, from the fuctioal equatio for the Gamma fuctio we have Γ(sΓ( s = s = ( s. Fially, we make use of the product formula for si(s/s: so that si(πs = πs ( s = Γ(sΓ( s = π si(πs., Lemma 4 (The Duplicatio Formula. ( Γ(sΓ s + = s πγ(s. 4

5 Proof. For this proof we will first itroduce the beta fuctio B(s, ω defied below B(s, ω = Γ(s + ω Γ(sΓ(ω. It is well kow that this has the followig itegral represetatio From this we have B(s, ω = λ s ( λ ω dλ. B(s, s = λ s ( λ s dλ = / λ s ( λ s dλ. The by makig the chage of variable λ = / / µ we have B(s, s = ( 4 s ( 4 µ dµ / µ = s ( µ s µ / dµ = s B s,. Usig the origial defiitio of the beta fuctio o both sides we have Γ(s (Γ(s = Γ ( s + s Γ(s π, from which the desired result follows immediately. Now equipped with these results, we ca prove oe of the key theorems i the study of the Riema zeta fuctio. Theorem 4 (Fuctioal Equatio for the Riema Zeta Fuctio. ( π ζ(s = s π s si s Γ( sζ( s. Proof. Suppose that σ >, the Γ ( s s π = s/ I particular, for σ > we have that Γ ( s ζ(s π s/ = = x s/ e πx dx = For simplicity of otatio we will write x s/ e πx dx. x s/ = e πx dx. Thus, we have that ψ(x = e πx. = ζ(s = πs/ Γ ( s Next, usig Lemma, we obtai = Γ ( s ζ(s π s/ = x s/ ( x ψ ( x x s/ ψ(xdx + x s/ ψ(xdx. + x dx + 5 x s/ ψ(xdx x s/ ψ(xdx

6 = s s + = s(s + x s/ 3/ ψ ( dx + x x s/ ψ(xdx (x s/ / + x s/ ψ(xdx Fially, we ote first that the remaiig itegral is coverget for all values of s, so by the Priciple of Aalytic Cotiuatio, we have that the above formula holds for all s. Next ote that the right had side of the formula is uchaged by the chage of variable s = s, ad hece we have or equivaletly, Γ ( s ζ(s = Γ π s/ Now, applyig Lemma, we have ( s ζ( s, π ( s/ ζ(s = π s / Γ ( s Γ ( s ζ( s. ( π ( ζ(s = π s 3/ si s Γ s Γ ( s Fially, by applyig Lemma 3, we have ( π ζ(s = s π s si s Γ( sζ( s, ζ( s. Refereces [] G.B. Follad. Real Aalysis: Moder Techiques ad Their Applicatios. Wiley, 999. [] W. Rudi. Fourier Aalysis o Groups. Joh Wiley & Sos, 96. [3] E.C. Titchmarsh ad D.R. Heath-Brow. The Theory of the Riema Zeta Fuctio. Oxford Uiversity Press, 7. 6