Lattices and Periodic Functions

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1 Lattices and Periodic Functions Definition L.1 Let f(x) be a function on IR d. a) The vector γ IR d is said to be a period for f if f(x+γ) = f(x) for all x IR d b) Set P f = { γ IR d γ is a period for f } Ifγ,γ P f thenγ+γ P f andifγ P f then γ P f (subx = z γ intof(x+γ) = f(x)). Furthermore, the zero vector 0 IR d is always in P f. Thus P f is a (commutative) group under addition and γ 1,, γ p P f n 1 γ 1 + +n p γ p P f for all p IN and n 1,,n p Z Example L.2 a) If f(x,y) = sin ( 2πx l 1 ) cos ( 2πy l 2 ), then Pf = { (ml 1,nl 2 ) m,n Z }. b) If f(x,y) = sin ( 2πx l 1 ), then Pf = { (ml 1,y) m Z, y IR }. c) If f(x,y) = sin ( 2πx l 1 ) sinhy, then Pf = { (ml 1,0) m Z }. To exclude functions, as in Example L.2.b, that are constant in some direction, it suffices to require that 0 be an isolated point of P f. That is, to require that there be a number r > 0 such that every nonzero γ P f obeys γ r. Proposition L.3 If P is an additive subgroup of IR d and 0 is an isolated point of P, then there are d d and independent vectors γ 1,,γ d IR d such that P = { n 1 γ 1 + +n d γ d n 1,,n d Z } c Joel Feldman All rights reserved. November 21, 2009 Lattices and Periodic Functions 1

2 Proof: Claim 1. P has a shortest nonzero element. Proof of Claim 1: Define r = inf { γ } γ P, γ 0. If there were no shortest element, there would be a sequence of vectors β 1,β 2 in P with lim i β i = r and r < β i 2r for every i = 1,2,. Because the closed ball of radius 2r is compact, the sequence has a limit point and hence has a Cauchy subsequence. In particular, there are β i and β in the sequence, with β i β with β i β < r 2. But this is impossible, because β i β would be a nonzero element of P with length smaller than r. Claim 2. Let γ 1 be a shortest nonzero element of P and set P 1 = { γ P } γ γ1. Then P 1 = { nγ } 1 n Z. Proof of Claim 2: If xγ 1 P with x not an integer, then (x [x])γ 1 (where [ ] denotes integer part) is a nonzero element of P with length strictly smaller than the length of γ 1. If P = P 1, we have finished. Otherwise continue with Claim 3. Denote by IP 1 orthogonal proection in IR d onto the line { } xγ 1 x IR and by IP 1 = 1l IP 1 orthogonal proection perpendicular to the line { } xγ 1 x IR. Then P \P 1 has an element whose distance from the line { } xγ 1 x IR is a minimum, i.e. that minimizes IP 1 γ. Proof of Claim 3: Define r 1 = inf { IP 1 γ γ P \P 1 }. If there were no minimizing element, there would be a sequence of vectors β 1,β 2 in P with 2r 1 IP 1 β 1 > IP 1 β 2 > IP 1 β 3 > > r 1 Because IP 1 β i = IP 1 (β i +nγ 1 ) for all n, we may assume, without loss of generality, that IP 1 β i γ 1 for every i. Because { x IR d IP 1 x 2r 1, IP 1 x γ 1 } is compact, the sequence has a limit point and hence has a Cauchy subsequence. In particular, there are β i and β in the sequence, with β i β with β i β < r. But this is impossible, 2 because β i β would be a nonzero element of P with length smaller than r. Claim 4. Let γ 2 be an element of P \P 1 that minimizes IP 1 γ and set P 2 = P { x 1 γ 1 +x 2 γ 2 x1,x 2 IR } Then P 2 = { n 1 γ 1 +n 2 γ 2 n1,n 2 Z }. c Joel Feldman All rights reserved. November 21, 2009 Lattices and Periodic Functions 2

3 Proof of Claim 4: If x 1 γ 1 +x 2 γ 2 P withx 2 not an integer, then γ = x 1 γ 1 +(x 2 [x 2 ])γ 2 is an element of of P \ P 1 with IP 1 γ = x 2 [x 2 ] IP 1 γ 2 < IP 1 γ 2. So x 2 must be an integer. But then (x 1 γ 1 +x 2 γ 2 ) x 2 γ 2 = x 1 γ 1 P and, by Claim 2, x 1 must be an integer as well. If P = P 2, we have finished. Otherwise continue with.... To exclude functions, as in Example L.2.c, that are mixed periodic/non periodic, we shall assume that d = d. Let γ 1,, γ d IR d be d linearly independent vectors and set Γ = { n 1 γ 1 + +n d γ d n1,,n d Z } Γ is called the lattice generated by γ 1,, γ d. Problem L.1 The set of generators for a lattice are not uniquely determined. Let Γ be generated by d linearly independent vectors γ 1,, γ d IR d. Let Γ be generated by d linearly independent vectors γ 1,, γ d IRd. Prove that Γ = Γ if and only there is a d d matrix A with integer matrix elements and deta = 1 such that γ i = d A i,γ. Problem L.2 Let γ 1,, γ d IR d be d linearly independent vectors. Prove that there are two constants C and c, depending only on γ 1,, γ d such that c x x1 γ x d γ d C x for all x IR d. We ll now find a bunch of functions that are periodic with respect to Γ. Consider f(x) = e ib x. This function has period γ if and only if e ib (x+γ) = e ib x for all x IR d. This is the case if and only if e ib γ = 1 and this is the case if and only if b γ 2π Z. Definition L.4 Let Γ be a lattice in IR d. The dual lattice for Γ is Γ # = { b IR d b γ 2π Z for all γ Γ } Remark L.5 Let γ 1,, γ d IR d be linearly independent and denote by Γ the lattice that they generate. A vector b IR d is an element of Γ # if and only if b γ 2π Z for all 1 d c Joel Feldman All rights reserved. November 21, 2009 Lattices and Periodic Functions 3

4 Example L.6 Let e 1,,e d be the standard basis of IR d. That is, e has all components zero, except for the th, which is one. Choosing l 1,,l d > 0 and γ = l e, Γ = { (n 1 l 1,,n d l d ) n1,,n d Z } Then (x 1,,x d ) is in Γ # if and only if Thus (x 1,,x d ) γ = l x 2π Z x 2π l Z Γ # = { ( n 1 2π l 1,,n d 2π l d ) n1,,n d Z } Example L.7 Let Γ = { n(1,0)+m(π,1) n,m Z } Then Γ # = { n(0,2π)+m(2π, 2π 2 ) } n,m Z Since [ n (1,0)+m (π,1) ] [n(0,2π)+m(2π, 2π 2 ) ] = 2π ( n m+m n ) every vector of the form n(0,2π)+m(2π, 2π 2 ) with m,n Z is indeed in Γ #. To verify that only vectors of this form are in Γ #, let z = x(0,2π)+y(2π, 2π 2 ) be any vector in IR 2. (Certainly, (0,2π) and (2π, 2π 2 ) form a basis for IR 2.) For z to be in Γ # it is necessary that z (1,0) = 2πy 2π Z z (π,1) = 2πx 2π Z which forces x,y Z. Problem L.3 Let Γ be generated by γ 1,, γ d IR d (assumed linearly independent) and let = { d t γ 0 t 1 for all 1 d } be the parallelepiped with the γ s as edges. Prove that if b Γ #, then { d d x e ib x [γ = 1, γ d ] if b = 0 0 if b 0 where [γ1, γ d ] is the volume of [γ1, γ d ]. By Problem L.1, the volume [γ1, γ d ] is independent of the choice of generators. That is, if Γ is also generated by γ 1,, γ d IRd, then [γ1, γ d ] = [γ 1, γ d ]. Consequently, it is legitimate to define = [γ1, γ d ]. Hence { 1 d d x e ib x 1 if b = 0 = 0 if b 0 c Joel Feldman All rights reserved. November 21, 2009 Lattices and Periodic Functions 4

5 Proposition L.8 If γ 1,, γ d IR d are linearly independent and Γ = { n 1 γ 1 + +n d γ d n1,,n d Z } then there exist d linearly independent vectors b 1,, b d IR d such that Γ # = { n 1 b 1 + +n d b d n1,,n d Z } Proof: For each 1 i d V i = { x 1 γ 1 + +x d γ d x1,,x d IR, x i = 0 } is a d 1 dimensional subspace of IR d. So V 1 is a one dimensional subspace of IR d. Let B i be any nonzero element of V i and define b i = 2π γ i B i B i Note that γ i B i cannot vanish because then γ i would have to be in V i, i.e. would have to be a linear combination of γ, i. Denote B = { n 1 b 1 + +n d b d n1,,n d Z } As b i γ = { 2π if i = 0 if i we have that b i Γ # and hence B Γ #. If x 1 b x d b d = 0 then (x 1 b x d b d ) γ = 2πx = 0 for every 1 d. So the b i s are linearly independent and every vector in IR d may be written in the form x 1 b 1 + +x d b d. If x 1 b 1 + +x d b d Γ #, then (x 1 b 1 + +x d b d ) γ = 2πx 2π Z so that x Z for every 1 d. Hence γ # B. From now on, we fix d linearly independent vectors γ 1,, γ d IR d, set Γ = { n 1 γ 1 + +n d γ d n1,,n d Z } The set of all C functions on IR d that are periodic with respect to Γ is denoted C ( IR d /Γ ). We have already observed that f(x) = e ib x is in C ( IR d /Γ ) if and only in b Γ #. c Joel Feldman All rights reserved. November 21, 2009 Lattices and Periodic Functions 5

6 Remark L.9 Here is the story (at least in short form) behind the notation C ( IR d /Γ ). We have already observed that IR d is a group (under addition) and that Γ is a subgroup of of IR d. As IR d is abelian, all subgroups are normal and the set of equivalence classes under the equivalence relation x y x y Γ is itself a group, denoted, as usual IR d /Γ. Precisely, the equivalence class of x IR d is [x] = { y IR d } x y IR d and IR d /Γ = { [x] x IR d }. The group operation in IR d /Γ is [x]+[y] = [x+y] As well as being a group, IR d /Γ can also be turned into a smooth manifold, called a d dimensional torus. If O is any open subset of IR d with the property that no two points of O are equivalent under, then the map ξ O : O IR d /Γ x [x] is one to one. Its inverse is a coordinate map for IR d /Γ. If Γ is generated by γ 1,,γ d and X is any point in IR d, { X +t 1 γ 1 + +t d γ d 0 < t < 1 for all 1 d } is one possible choice of O. The notation C ( IR d /Γ ) designates the set of smooth (that is, C ) functions on the manifold IR d /Γ. Theorem L.10 (Fourier Series) A function f : IR d C is in C ( IR d /Γ ) if and only if Furthermore, in this case, f(x) = 1 ˆfb e ib x with b 2n ˆf b = ˆfb < d d x e ib x f(x) for all n IN Proof of if : Suppose that we are given ˆf b, b Γ # obeying b Γ b 2n ˆfb < # for all n IN. In particular b Γ ˆfb < so the series 1 ˆf # b e ib x converges absolutely and uniformly to some continuous function that is periodic with respect to Γ. Call the function f(x). Furthermore for any i 1, i d IN ( d x e ib x = b e ib x b Σ ˆfb c Joel Feldman All rights reserved. November 21, 2009 Lattices and Periodic Functions 6

7 so the series 1 x that f(x) is C. Furthermore d d x e ib x f(x) = by Problem L.3. e ib x also converges absolutely and uniformly. This implies = 1 = 1 = ˆf b d d x e ib x [ c Γ # 1 ] ˆfc e ic x c Γ # d d x e i(c b) x ˆfc d d x ˆf b + 1 c Γ # c b d d x e i(c b) x ˆfc Proof of only if : Now suppose that we are given f C ( IR d /Γ ). Define Then for any i 1, i d IN ( d so that, by Problem L.2, ( d ˆf b = b = b = = = d d x e ib x f(x) d d x d d x sup x b )e ib x f(x) e )f(x) ib x x d d x e ib x ( 1+ b d+1 d 1+ b d+1 x b [ sup (1+ b d+1 ( ) d [ sup (1+ b d+1 ( ) d x f(x)) < f(x)) b ] b ] n Z d 1 1+ b d (c n ) d+1 < c Joel Feldman All rights reserved. November 21, 2009 Lattices and Periodic Functions 7

8 Hence, by the only if part of this Theorem, that we have already proven, g(x) = 1 ˆfb e ib x is a C function and ˆf b = d d x e ib x g(x) We ust have to show that g(x) = f(x). Here is one proof that g(x) = f(x). By (L.1) d d x e ib x [g(x) f(x)] = 0 (L.1) for all b Γ #. Consequently, d d x ϕ(x)[g(x) f(x)] = 0 for any function ϕ P(Γ # ) where P(Γ # ) is the set of all functions that are finite linear combinations of the e ib x s with b Γ #. Consequently, d d x ϕ(x)[g(x) f(x)] = 0 for any function ϕ P(Γ # ) where P(Γ # ) is the set of all functions that are uniform limits of sequences of functions in P(Γ # ). But by the Stone Weierstrass Theorem [Walter Rudin, Principles of Mathematical Analysis, Theorem 7.33], P(Γ # ) is the set of all continuous functions that are periodic with respect to Γ. In particular, the complex conugate of g(x) f(x) is in P(Γ # ). Hence d d x g(x) f(x) 2 = 0 so that g(x) = f(x) for all x. One may also build Problem L.5, below, into a second proof that g(x) = f(x). Just make a change of variables so that Γ is replaced by 2π Z d and apply Problem L.5.b, once in each dimension. Problem L.4 Let Γ be generated by γ 1,, γ d IR d (assumed linearly independent) and also by γ 1,, γ d IRd (also assumed linearly independent). Recall that = { d t γ 0 t 1 for all 1 d } c Joel Feldman All rights reserved. November 21, 2009 Lattices and Periodic Functions 8

9 is the parallelepiped with the γ s as edges. Let, for y IR d, y+ = { y+ d t γ 0 t 1 for all 1 d } be the translate of by y. Let f(x) be periodic with respect to Γ. Prove that d d x f(x) = y+ d d x f(x) = d d x f(x) [γ 1, γ d ] We denote d d x f(x) = IR d /Γ d d x f(x) Problem L.5 Let f C 1 (IR) be periodic of period 2π. Set c n = 2π 0 e inx f(x) dx and (S M f)(θ) = 1 2π M n= M c n e inθ a) Prove that S M f(θ) = 1 2π 2π f(θ+x) sin(m+1/2)x 0 sin(x/2) dx. b) Prove that S M f(θ) converges to f(θ) as M. c Joel Feldman All rights reserved. November 21, 2009 Lattices and Periodic Functions 9

The Contraction Mapping Theorem and the Implicit and Inverse Function Theorems

The Contraction Mapping Theorem and the Implicit and Inverse Function Theorems The Contraction Mapping Theorem Theorem The Contraction Mapping Theorem) Let B a = { x IR d x < a } denote the open ball of