The Wiener algebra and Wiener s lemma

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1 he Wiener algebra and Wiener s lemma Jordan Bell jordan.bell@gmail.com Department of Mathematics, University of oronto January 17, Introduction Let = R/Z. For f L 1 () we define f L 1 () = 1 For f, g L 1 (), we define (f g)(t) = 1 f g L 1 (), and satisfies Young s inequality f(t) dt. f(τ)g(t τ)dτ, t. f g L 1 () f L 1 () g L 1 (). With convolution as the operation, L 1 () is a commutative Banach algebra. For f L 1 (), we define ˆf : Z C by ˆf(k) = 1 f(t)e ikt dt, k Z. We define c 0 (Z) to be the collection of those F : Z C such that F (k) 0 as k. For f L 1 (), the Riemann-Lebesgue lemma tells us that ˆf c 0 (Z). We define l 1 (Z) to be the set of functions F : Z C such that F l1 (Z) = k Z F (k). For F, G l 1 (Z), we define (F G)(k) = j Z F (j)g(k j). 1

2 F G l 1 (Z), and satisfies Young s inequality F G l 1 (Z) F l 1 (Z) G l 1 (Z). l 1 (Z) is a commutative Banach algebra, with unity { 1 k = 0, F (k) = 0 k 0. For f L 1 () and n 0 we define S n (f) C() by S n (f)(t) = ˆf(k)e ikt, t. k n For 0 < α < 1, we define Lip α () to be the collection of those functions f : C such that For f Lip α (), we define f(t + h) f(t) sup t,h 0 h α <. f Lipα () = f C() + 2 otal variation sup t,h 0 f(t + h) f(t) h α. For f : C, we define { n } var(f) = sup f(t i ) f(t i 1 ) : n 1, 0 = t 0 < < t n =. i=1 If var(f) < then we say that f is of bounded variation, and we define BV () to be the set of functions C of bounded variation. We define f BV () = sup f(t) + var(f). t his is a norm on BV (), with which BV () is a Banach algebra. 1 heorem 1. If f BV (), then ˆf(n) var(f), n Z, n 0. n Proof. Integrating by parts, ˆf(n) = 1 f(t)e int dt = 1 hence ˆf(n) 1 n var(f). e int in df(t) = 1 e int df(t), in 1 N. L. Carothers, Real Analysis, p. 206, heorem

3 3 Absolutely convergent Fourier series Suppose that f L 1 () and that ˆf l 1 (Z). For n m, S n (f) S m (f) C() = sup ˆf(k)e ikt t m< k n m< k n ˆf(k), and because ˆf l 1 (Z) it follows that S n (f) converges to some g C(). We check that f(t) = g(t) for almost all t. We define A() to be the collection of those f C() such that ˆf l 1 (Z), and we define f A() = ˆf. l 1 (Z) A() is a commutative Banach algebra, with unity t 1, and the Fourier transform is an isomorphism of Banach algebras F : A() l 1 (Z). We call A() the Wiener algebra. he inclusion map A() C() has norm 1. heorem 2. If f : C is absolutely continuous, then ˆf(k) = o(k 1 ), k. Proof. Because f is absolutely continuous, the fundamental theorem of calculus tells us that f L 1 (). Doing integration by parts, for k Z we have F (f )(k) = 1 f (t)e ikt dt = 1 f(t)e ikt 1 f(t)( ike ikt )dt 0 = ikf (f)(k). he Riemann-Lebesgue lemma tells us that F (f )(k) = o(1), so ( ) 1 F (f)(k) = o, k. k heorem 3. If f : C is absolutely continuous and f L 2 (), then ) 1/2 f A() f L 1 () (2 + k 2 f L 2 (). k=1 Proof. First, ˆf(0) = 1 f(t)dt f L 1 (). 3

4 Next, because f is absolutely continuous, by the fundamental theorem of calculus we have f L 1 (), and for k Z, F (f )(k) = ikf (f)(k). Using the Cauchy-Schwarz inequality, and since F (f )(0) = 0, f A() = ˆf(0) + k 0 ˆf(k) = ˆf(0) + k 1 F (f )(k) k 0 f L1 () + k 2 k 0 1/2 F (f )(k) 2 k 0 1/2 = f L 1 () + (2 ) 1/2 k 2 F (f ) l 2 (Z). k=1 By Parseval s theorem we have F (f ) l2 (Z) = f L2 (), completing the proof. We now prove that if α > 1 2, then Lip α() A(), and the inclusion map is a bounded linear operator. 2 heorem 4. If α > 1 2, then Lip α() A(), and for any f Lip α () we have with f A() c α f Lipα (), c α = /2 ( 3 Proof. For f : C and h R, we define ) α α. f h (t) = f(t h), t, which satisfies, for n Z, F (f h )(n) = 1 = 1 f(t h)e int dt f(t)e in(t+h) dt = e inh F (f)(n). hus F (f h f)(n) = (e inh 1) ˆf(n), n Z. (1) Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 34, heorem 4

5 For m 0 and for n Z such that 2 m n < 2 m+1, let h m = 3 2 m. hen 3 = 2m 3 2 m nh m < 2 m m = 4π 3. If n > 0 this implies that π 3 nh m < 2 3 and so e inhm 1 = 2 sin nh m 2 sin π 2 3 = 3, and if n < 0 this implies that and so 3 < nh m 2 π 3 e inhm 1 3. his gives us ˆf(n) 2 3 ˆf(n) 2 2 m n <2 m+1 2 m n <2 m+1 e inhm 1 2 ˆf(n) 2 2 m n <2 m+1 e inhm 1 2 ˆf(n) 2. n Z Using (1) and Parseval s theorem we have e inhm 1 2 ˆf(n) 2 = F (f hm f) 2 l 2 (Z) = f h m f 2 L 2 (), and thus n Z 2 m n <2 m+1 ˆf(n) 2 f hm f 2 L 2 (). Furthermore, for g L () we have g L 2 () g L (), so 2 m n <2 m+1 ˆf(n) 2 f hm f 2 L () f 2 Lip α () h2α m ( ) 2α = 3 2 m f 2 Lip α (). 5

6 By the Cauchy-Schwarz inequality, because there are 2 m+1 nonzero terms in m+1 ˆf(n), 2 m n <2 ˆf(n) (2 m+1 ) 1/2 ˆf(n) 2 2 m n <2 m+1 2 m n <2 m+1 2 m+1 2 1/2 ( ) α 3 2 m f Lipα () ) α f. Lipα() = 2 m( 1 2 α) 2 1/2 ( 3 hen, since α > 1 2, As ˆf(n) = ˆf(0) + ˆf(n) n Z m=0 2 m n <2 m+1 ˆf(0) + 2 m( 1 α) ( ) α 2 2 1/2 f 3 Lipα() m=0 ( ) α = ˆf(0) + 2 1/2 f 3 Lipα() 2 m( 1 α) 2 m=0 ( ) α = ˆf(0) + 2 1/2 1 f 3 Lipα() α ˆf(0) f L 1 () f L () f Lip α (), we have for all f Lip α () that ˆf(n) c α f Lipα (), completing the proof. n Z We now prove that if α > 0, then BV () Lip α () A(). 3 heorem 5. If α > 0 and f BV () Lip α (), then and f A() f h f 2 L 2 () 1 h1+α f Lipα () var(f), h > 0. 3 Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 35, heorem 6

7 Proof. For N 1 and h = N, f h f 2 L 2 () = 1 = 1 = 1 0 N f h (t) f(t) 2 dt jh j=1 (j 1)h h N j=1 = 1 h 0 0 f h (t) f(t) 2 dt f jh (t) f (j 1)h (t) 2 dt N f jh (t) f (j 1)h (t) 2 dt j=1 1 f h f L () 1 f h f L () h 0 j=1 h 0 N f jh (t) f (j 1)h (t) dt var(f)dt. As f Lip α (), f h f L () hα f Lipα (), hence f h f 2 L 2 () 1 h1+α f Lipα () var(f). 4 Wiener s lemma For k 1, using the product rule (fg) = f g + fg we check that C k () is a Banach algebra with the norm f Ck () = k f (j) C(). If f C k () and f(t) 0 for all t, then the quotient rule tells us that ( f 1 ) (t) = f (t) f(t) 2, using which we get 1 f Ck (). hat is, if f C k () does not vanish then f 1 = 1 f Ck (). If B is a commutative unital Banach algebra, a multiplicative linear functional on B is a nonzero algebra homomorphism B C, and the collection B of multiplicative linear functionals on B is called the maximal ideal space of B. he Gelfand transform of f B is Γ(f) : B C defined by Γ(f)(h) = h(f), h B. 7

8 It is a fact that f B is invertible if and only if h(f) 0 for all h B, i.e., f B is invertible if and only if Γ(f) does not vanish. We now prove that if f A() and does not vanish, then f is invertible in A(). We call this statement Wiener s lemma. 4 heorem 6 (Wiener s lemma). If f A() and f(t) 0 for all t, then 1/f A(). Proof. Let w : A() C be a multiplicative linear functional. he fact that w is a multiplicative linear functional implies that w = 1. Define u(t) = e it, t, for which u A() = 1. We define λ = w(u), which satisfies λ w u A() = 1 and because u 1 A() = 1 we have λ 1 = w(u 1 ) and λ 1 w u 1 A() = 1, hence λ = 1. hen there is some t w such that λ = e itw. For n Z, w(u n ) = λ n = e intw. If P (t) = n N a ne int is a trigonometric polynomial, then w(p ) = w a n u n = a n w(u) n = a n e intw = P (t w ). (2) n N n N n N For g A(), if ɛ > 0, then there is some N such that g S N (g) A() < ɛ. Using (2) and the fact that g C() g A(), w(g) g(t w ) w(g) w(s N (g)) + w(s N (g)) S N (g)(t w ) + S N (g)(t w ) g(t w ) = w(g S N (g)) + S N (g)(t w ) f(t w ) w g S N (g) A() + S N (g) g C() w g S N (g) A() + g S N (g) A() < 2ɛ. Because this is true for all ɛ > 0, it follows that w(g) = g(t w ). Let be the maximal ideal space of A(). hen for w there is some t w such that w(f) = f(t w ), hence, because f(t) 0 for all t, Γ(f)(w) = w(f) = f(t w ) 0. hat is, Γ(f) does not vanish, and therefore f is invertible in A(). It is then immediate that f 1 (t) = 1 f(t) for all t, completing the proof Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 239, heorem 8

9 he above proof of Wiener s lemma uses the theory of the commutative Banach algebras. he following is a proof of the theorem that does not use the Gelfand transform. 5 Proof. Because f A(), f defined by f (t) = f(t), t, belongs to A(). Let g = f 2 f 2 = ff C() f 2 A(), C() which satisfies 0 < g(t) 1 for all t. As 1 f = f f = 2 1/f A() it suffices to show that 1 g A(). Because g is continuous and g(t) 0 for all t, δ = inf g(t) > 0; t f f 2 C() g, to show that if δ = 1 then g = 1, and indeed 1 g A(). Otherwise, g 1 C() = 1 δ < 1. his implies that g is invertible in the Banach algebra C() and that g 1 = (1 g)j in C(). Let h = 1 g A(). For ɛ > 0, there is some N such that h S N (h) A() < ɛ. Now, if P is a trigonometric polynomial of degree M then using the Cauchy-Schwarz inequality and Parseval s theorem, P A() = ˆP l 1 (Z) (2M + 1) 1/2 ˆP l2 (Z) = (2M + 1) 1/2 P L 2 () (2M + 1) 1/2 P L (). Furthermore, for j 1, P j is a trigonometric polynomial of degree jm. he binomial theorem tells us, with P = S N (h) and r = h P, h k = (P + r) k = k ( ) k P j r k j, j 5 Karlheinz Gröchenig, Wiener s Lemma: heme and Variations. An Introduction to Spectral Invariance and Its Applications, p. 180, 5.2.4, in Brigitte Forster and Peter Massopust, eds., Four Short Courses on Harmonic Analysis, pp

10 and using this and P j A() (2jN + 1) 1/2 P j L (), Because h k A() k ( k P j) j A() r k j A() k k ( ) k P j j h S A() N (h) k j A() ( ) k (2jN + 1) 1/2 P j j L () ɛk j (2kN + 1) 1/2 k ( ) k P j L j () ɛk j = (2kN + 1) 1/2 ( P L () + ɛ)k. P L () h S N(h) L () + h L () h S N (h) A() + h L () < ɛ + h L (), we have h k A() (2kN + 1) 1/2 ( h L () + 2ɛ)k = (2kN + 1) 1/2 (1 δ + 2ɛ) k. ake some ɛ < δ 2, so that 1 δ + 2ɛ < 1. hen with N = N(ɛ), h k A() (2kN+1) 1/2 (1 δ+2ɛ) k = ( 2NΦ 1 δ + 2ɛ, 1 ) 2, 1 <, 2N k=0 k=0 where Φ is the Lerch transcendent. his implies that the the series k=0 hk converges in A(). We check that k=0 hk is the inverse of 1 h, namely, g = 1 h is invertible in A(), proving the claim. 5 Spectral theory Suppose that A is a commutative Banach algebra with unity 1. We define U(A) to be the collection of those f A such that f is invertible in A. It is a fact that U(A) is an open subset of A. We define σ A (f) = {λ C : f λ U(A)}, called the spectrum of f. It is a fact that σ A (f) is a nonempty compact subset of C. 10

11 If A B are Banach algebras with unity 1, we say that A is inverse-closed in B if f A and f 1 B together imply that f 1 A. 6 Lemma 7. Suppose that A B are Banach algebras with unity 1. he following are equivalent: 1. A is inverse-closed in B. 2. σ A (f) = σ B (f) for all f A. Proof. Assume that A is inverse-closed in B and let f A. If λ σ A (f) then f λ U(A) U(B), hence λ σ B (f). herefore σ B (f) σ A (f). If λ σ B (f) then f λ U(B). hat is, (f λ) 1 B. Because A is inverseclosed in B and f λ A, we get (f λ) 1 A. hus λ σ A (f), and therefore σ A (f) σ B (f). We thus have obtained σ A (f) = σ B (f). Assume that for all f A, σ A (f) = σ B (f). Suppose that f A and f 1 B. hat is, f U(B), so 0 σ B (f). hen 0 σ A (f), meaning that f U(A). A() C() are Banach algebras with unity 1. Wiener s lemma states that A() is inverse-closed in C(). It is apparent that for f C(), σ C() (f) = f() C. herefore, Lemma 7 tells us for f A() that σ A() (f) = f(). he Wiener-Lévy theorem states that if f A(), Ω C is an open set containing f(), and F : Ω C is holomorphic, then F f A(). 7 In particular, if f A() does not vanish, then Ω = C \ {0} is an open set containing f() and F (z) = 1 z is a holomorphic function on Ω, and hence F f(t) = 1 f(t) belongs to A(), which is the statement of Wiener s lemma. 6 Karlheinz Gröchenig, Wiener s Lemma: heme and Variations. An Introduction to Spectral Invariance and Its Applications, p. 183, 5.2.5, in Brigitte Forster and Peter Massopust, eds., Four Short Courses on Harmonic Analysis, pp Karlheinz Gröchenig, Wiener s Lemma: heme and Variations. An Introduction to Spectral Invariance and Its Applications, p. 187, heorem 5.16, in Brigitte Forster and Peter Massopust, eds., Four Short Courses on Harmonic Analysis, pp ; Walter Rudin, Fourier Analysis on Groups, Chapter 6; N. K. Nikolski (ed.), Functional Analysis I, p. 235; V. P. Havin and N. K. Nikolski (eds.), Commutative Harmonic Analysis II, p. 240,