# A brief account of the historical development of by Emily Coddington

By Emily Coddington

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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.