This example uses Romande ADF for Latin letters and punctuation characters in text and math, and PX Fonts for math symbols and Greek letters. \usepackage[T1]{fontenc} \usepackage{pxfonts} \usepackage{romande} \usepackage[italic,defaultmathsizes,noasterisk]{mathastext} \renewcommand{\itshape}{\swashstyle} Typeset with mathastext 1.13 (2011/03/11).
To illustrate some Hilbert Space properties of the co-Poisson summation, we will assume K = Q. The components (aν ) of an adele a are written ap at finite places and ar at the real place. We have an embedding of the Schwartz space of test-functions on R into ∏ the Bruhat-Schwartz space on′ A which sends ψ(x) to φ(a) = p 1|ap |p ≤1 (ap ) · ψ(ar ), and we write ER (g) for the distribution on R thus obtained from E ′ (g) on A. Theorem 1. Let g be a compact Bruhat-Schwartz function on the ideles of Q. The co-Poisson summation ER′ (g) is a square-integrable function (with respect to the Lebesgue measure). The L2 (R) function ∫ ER′ (g) is equal to the constant – A× g(v)|v|–1/2 d ∗ v in a neighborhood of the origin. Proof. We may first, without changing anything to ER′ (g), replace g with its average under the action of the finite unit ideles, so that it may be assumed invariant. Any such compact invariant g is a finite linear combination of suitable multiplicative translates ∏ of functions of the type g(v) = p 1|vp |p =1 (vp ) · f (vr ) with f (t) a smooth compactly supported function on R× , so that we may assume that g has this form. We claim that: ∫ ∑ √ |φ(v)| |g(qv)| |v| d ∗ v < ∞ A×
q∈Q×
∑ Indeed q∈Q× |g(qv)| ∫ = |f (|v|)| + |f (–|v|)| is bounded above by a multiple of |v|. And A× |φ(v)||v|3/2 d ∗ v < ∞ for each Bruhat-Schwartz ∏ function on the adeles (basically, from p (1 – p–3/2 )–1 < ∞). So ∫ ∫ ∑∫ √ g(v) ∗ ′ ∗ E (g)(φ) = φ(v)g(qv) |v| d v – φ(x) dx √ dv × × A A A |v| × q∈Q ∫ ∫ ∑∫ √ g(v) ∗ ∗ ′ E (g)(φ) = φ(x) dx φ(v/q)g(v) |v| d v – √ dv × A× A |v| q∈Q× A ∏ Let us now specialize to φ(a) = p 1|ap |p ≤1 (ap ) · ψ(ar ). Each integral can be evaluated as an infinite product. The finite places contribute 0 or 1 according to whether q ∈ Q× satisfies |q|p < 1 or not. So only the inverse integers q = 1/n, n ∈ Z, contribute: ∑∫ √ dt ∫ f (t) dt ∫ ′ – ψ(x) dx ER (g)(ψ) = ψ(nt)f (t) |t| √ 2|t| R× R× |t| 2|t| R × n∈Z
We can now revert the steps, but this time on R× and we get: ∫ ∫ ∫ ∑ f (t/n) dt f (t) dt ′ ER (g)(ψ) = ψ(t) ψ(x) dx √ √ – √ 2|t| R × R× R |n| 2 |t| |t| × n∈Z √ Let us express this in terms of α(y) = (f (y) + f (–y))/2 |y|: ∫ ∫ ∞ ∫ ∑ α(y/n) α(y) ′ ER (g)(ψ) = dy – dy ψ(y) ψ(x) dx n y R 0 R n≥1 So the distribution ER′ (g) is in fact the even smooth function ∑ α(y/n) ∫ ∞ α(y) ′ ER (g)(y) = – dy n y 0 n≥1 As α(y) has compact support in R \ {0}, the summation over n ≥ 1 contains only vanishing terms for |y| small enough. So ER′ (g) is ∫ f (y) dy ∫ ∫ ∞ α(y) √ equal to the constant – 0 y dy = – R× √ 2|y| = – A× g(t)/ |t| d ∗ t |y|
in a neighborhood of 0. To prove that it is L2 , let β(y) be the smooth compactly supported function α(1/y)/2|y| of y ∈ R (β(0) = 0). Then (y , 0): ∑ 1 n ∫ ′ ER (g)(y) = β( ) – β(y) dy |y| y R n∈Z
From the usual Poisson summation formula, this is also: ∫ ∑ ∑ γ(ny) – β(y) dy = γ(ny) n∈Z
R
n,0
∫ where γ(y) = R exp(i 2πyw)β(w) dw is a Schwartz rapidly decreasing function. From this formula we deduce easily that ER′ (g)(y) is itself in the Schwartz class of rapidly decreasing functions, and in particular it is is square-integrable. □ It is useful to recapitulate some of the results arising in this proof: Theorem 2. Let g be a compact Bruhat-Schwartz function on the ideles of Q. The co-Poisson summation ER′ (g) is an even function on R in the Schwartz class of rapidly decreasing functions. It is constant,
as well as its Fourier Transform, in a neighborhood of the origin. It may be written as ∑ α(y/n) ∫ ∞ α(y) ′ ER (g)(y) = – dy n y 0 n≥1 with a function α(y) smooth with compact support away from the origin, and conversely each such formula corresponds to the coPoisson summation ER′ (g) of a compact∫ Bruhat-Schwartz function on the ideles of Q. The Fourier transform R ER′ (g)(y) exp(i2πwy) dy corresponds in the formula above to the replacement α(y) 7→ α(1/y)/|y|. Everything has been obtained previously.
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