The Wiener Algebra and Wiener's Lemma

The Wiener algebra and Wiener’s lemma

Jordan Bell [email protected] Department of Mathematics, University of Toronto January 17, 2015

1 Introduction

Let T = R/2πZ. For f ∈ L1(T) we define 1 Z kfk 1 = |f(t)|dt. L (T) 2π T For f, g ∈ L1(T), we define 1 Z (f ∗ g)(t) = f(τ)g(t − τ)dτ, t ∈ . 2π T T f ∗ g ∈ L1(T), and satisfies Young’s inequality

kf ∗ gk 1 ≤ kfk 1 kgk 1 . L (T) L (T) L (T)

With convolution as the operation, L1(T) is a commutative Banach algebra. For f ∈ L1(T), we deﬁne fˆ : Z → C by 1 Z fˆ(k) = f(t)e−iktdt, k ∈ . 2π Z T

We deﬁne c0(Z) to be the collection of those F : Z → C such that |F (k)| → 0 as 1 ˆ |k| → ∞. For f ∈ L (T), the Riemann-Lebesgue lemma tells us that f ∈ c0(Z). We deﬁne `1(Z) to be the set of functions F : Z → C such that X kF k 1 = |F (k)|. ` (Z) k∈Z

For F,G ∈ `1(Z), we deﬁne X (F ∗ G)(k) = F (j)G(k − j). j∈Z

1 F ∗ G ∈ `1(Z), and satisﬁes Young’s inequality

kF ∗ Gk 1 ≤ kF k 1 kGk 1 . ` (Z) ` (Z) ` (Z) `1(Z) is a commutative Banach algebra, with unity ( 1 k = 0, F (k) = 0 k 6= 0.

1 For f ∈ L (T) and n ≥ 0 we deﬁne Sn(f) ∈ C(T) by X ˆ ikt Sn(f)(t) = f(k)e , t ∈ T. |k|≤n

For 0 < α < 1, we deﬁne Lipα(T) to be the collection of those functions f : T → C such that |f(t + h) − f(t)| sup α < ∞. t∈T,h6=0 |h|

For f ∈ Lipα(T), we deﬁne |f(t + h) − f(t)| kfk = kfk + sup . Lipα(T) C(T) α t∈T,h6=0 |h| 2 Total variation

For f : T → C, we define ( n ) X var(f) = sup |f(ti) − f(ti−1)| : n ≥ 1, 0 = t0 < ··· < tn = 2π . i=1 If var(f) < ∞ then we say that f is of bounded variation, and we define BV (T) to be the set of functions T → C of bounded variation. We define kfk = sup |f(t)| + var(f). BV (T) t∈T This is a norm on BV (T), with which BV (T) is a Banach algebra.1 Theorem 1. If f ∈ BV (T), then var(f) |fˆ(n)| ≤ , n ∈ , n 6= 0. 2π|n| Z Proof. Integrating by parts, 1 Z 1 Z e−int 1 Z fˆ(n) = f(t)e−intdt = − df(t) = e−intdf(t), 2π 2π −in 2πin T T T hence 1 |fˆ(n)| ≤ var(f). 2π|n|

1N. L. Carothers, Real Analysis, p. 206, Theorem 13.4.

2 3 Absolutely convergent Fourier series

Suppose that f ∈ L1(T) and that fˆ ∈ `1(Z). For n ≥ m,

X ˆ ikt X ˆ kSn(f) − Sm(f)kC( ) = sup f(k)e ≤ |f(k)|, T t∈ T m<|k|≤n m<|k|≤n

ˆ 1 and because f ∈ ` (Z) it follows that Sn(f) converges to some g ∈ C(T). We check that f(t) = g(t) for almost all t ∈ T. We deﬁne A(T) to be the collection of those f ∈ C(T) such that fˆ ∈ `1(Z), and we deﬁne kfk = fˆ . A(T) 1 ` (Z) A(T) is a commutative Banach algebra, with unity t 7→ 1, and the Fourier transform is an isomorphism of Banach algebras F : A(T) → `1(Z). We call A(T) the Wiener algebra. The inclusion map A(T) ⊂ C(T) has norm 1.

Theorem 2. If f : T → C is absolutely continuous, then fˆ(k) = o(k−1), |k| → ∞.

Proof. Because f is absolutely continuous, the fundamental theorem of calculus tells us that f 0 ∈ L1(T). Doing integration by parts, for k ∈ Z we have 1 Z F (f 0)(k) = f 0(t)e−iktdt 2π T 1 2π 1 Z = f(t)e−ikt − f(t)(−ike−ikt)dt 2π 0 2π T = ikF (f)(k).

The Riemann-Lebesgue lemma tells us that F (f 0)(k) = o(1), so

1 F (f)(k) = o , |k| → ∞. k

Theorem 3. If f : T → C is absolutely continuous and f 0 ∈ L2(T), then

∞ !1/2 X −2 0 kfk ≤ kfk 1 + 2 k kf k 2 . A(T) L (T) L (T) k=1 Proof. First, Z ˆ 1 |f(0)| = f(t)dt ≤ kfk 1 . 2π L (T) T

3 Next, because f is absolutely continuous, by the fundamental theorem of calculus we have f 0 ∈ L1(T), and for k ∈ Z, F (f 0)(k) = ikF (f)(k).

Using the Cauchy-Schwarz inequality, and since F (f 0)(0) = 0, X kfk = |fˆ(0)| + |fˆ(k)| A(T) k6=0 X = |fˆ(0)| + |k|−1|F (f 0)(k)| k6=0  1/2  1/2 X −2 X 0 2 ≤ kfk 1 + |k| |F (f )(k)| L (T)     k6=0 k6=0

∞ !1/2 X −2 0 = kfk 1 + 2 k kF (f )k 2 . L (T) ` (Z) k=1

0 0 By Parseval’s theorem we have kF (f )k 2 = kf k 2 , completing the proof. ` (Z) L (T)

1 We now prove that if α > 2 , then Lipα(T) ⊂ A(T), and the inclusion map is a bounded linear operator.2

1 Theorem 4. If α > 2 , then Lipα(T) ⊂ A(T), and for any f ∈ Lipα(T) we have

kfk ≤ cα kfk , A(T) Lipα(T) with 2π α 1 c = 1 + 21/2 . α 1 −α 3 1 − 2 2 Proof. For f : T → C and h ∈ R, we deﬁne

fh(t) = f(t − h), t ∈ T, which satisﬁes, for n ∈ Z, Z 1 −int F (fh)(n) = f(t − h)e dt 2π T 1 Z = f(t)e−in(t+h)dt 2π T = e−inhF (f)(n). Thus −inh ˆ F (fh − f)(n) = (e − 1)f(n), n ∈ Z. (1) 2Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 34, Theorem 6.3.

4 For m ≥ 0 and for n ∈ Z such that 2m ≤ |n| < 2m+1, let 2π h = · 2−m. m 3 Then 2π 2π 2π 4π = 2m · · 2−m ≤ |nh | < 2m+1 · · 2−m = . 3 3 m 3 3 If n > 0 this implies that π nh 2π ≤ m < 3 2 3 and so nhm π √ |e−inhm − 1| = 2 sin ≥ 2 sin = 3, 2 3 and if n < 0 this implies that 2π nh π − < m ≤ − 3 2 3 and so √ |e−inhm − 1| ≥ 3. This gives us X X |fˆ(n)|2 ≤ 3|fˆ(n)|2 2m≤|n|<2m+1 2m≤|n|<2m+1 X ≤ |e−inhm − 1|2|fˆ(n)|2 2m≤|n|<2m+1 X ≤ |e−inhm − 1|2|fˆ(n)|2. n∈Z Using (1) and Parseval’s theorem we have

X −inhm 2 ˆ 2 2 2 |e − 1| |f(n)| = kF (fh − f)k 2 = kfh − fk 2 , m ` (Z) m L (T) n∈Z and thus X ˆ 2 2 |f(n)| ≤ kfh − fk 2 . m L (T) 2m≤|n|<2m+1 ∞ Furthermore, for g ∈ L ( ) we have kgk 2 ≤ kgk ∞ , so T L (T) L (T)

X ˆ 2 2 |f(n)| ≤ kfh − fk ∞ m L (T) 2m≤|n|<2m+1 ≤ kfk2 · h2α Lipα(T) m 2π 2α = kfk2 . 3 · 2m Lipα(T)

5 By the Cauchy-Schwarz inequality, because there are ≤ 2m+1 nonzero terms in P ˆ 2m≤|n|<2m+1 |f(n)|,

 1/2 X m+1 1/2 X 2 |fˆ(n)| ≤ (2 )  |fˆ(n)|  2m≤|n|<2m+1 2m≤|n|<2m+1 α m+1 2π ≤ 2 2 kfk 3 · 2m Lipα(T) α m 1 −α 1/2 2π = 2 ( 2 ) · 2 · kfk . 3 Lipα(T)

1 Then, since α > 2 ,

∞ X X X |fˆ(n)| = |fˆ(0)| + |fˆ(n)| n∈Z m=0 2m≤|n|<2m+1 ∞ α X m 1 −α 1/2 2π ≤ |fˆ(0)| + 2 ( 2 ) · 2 · kfk 3 Lipα(T) m=0 α ∞ 1/2 2π X m 1 −α = |fˆ(0)| + 2 kfk 2 ( 2 ) 3 Lipα(T) m=0 α ˆ 1/2 2π 1 = |f(0)| + 2 kfk 1 Lipα(T) −α 3 1 − 2 2 As ˆ |f(0)| ≤ kfk 1 ≤ kfk ∞ ≤ kfk , L (T) L (T) Lipα(T) we have for all f ∈ Lipα(T) that X ˆ |f(n)| ≤ cα kfk , Lipα(T) n∈Z completing the proof.

3 We now prove that if α > 0, then BV (T) ∩ Lipα(T) ⊂ A(T).

Theorem 5. If α > 0 and f ∈ BV (T) ∩ Lipα(T), then

2 1 1+α kfh − fk 2 ≤ h kfk var(f), h > 0. L (T) 2π Lipα(T) and f ∈ A(T). 3Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 35, Theorem 6.4.

2 1 1+α kfh − fk 2 ≤ h kfk var(f). L (T) 2π Lipα(T)

4 Wiener’s lemma

For k ≥ 1, using the product rule (fg)0 = f 0g + fg0 we check that Ck(T) is a Banach algebra with the norm

k X (j) kfkCk( ) = f . T C( ) j=0 T

If f ∈ Ck(T) and f(t) 6= 0 for all t ∈ T, then the quotient rule tells us that 0 0 f (t) f −1 (t) = − , f(t)2

1 k k using which we get f ∈ C (T). That is, if f ∈ C (T) does not vanish then −1 1 k f = f ∈ C (T). 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. The Gelfand transform of f ∈ B is Γ(f) : ∆B → C deﬁned by

Γ(f)(h) = h(f), h ∈ ∆B.

7 It is a fact that f ∈ B is invertible if and only if h(f) 6= 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(T) and does not vanish, then f is invertible in A(T). We call this statement Wiener’s lemma.4 Theorem 6 (Wiener’s lemma). If f ∈ A(T) and f(t) 6= 0 for all t ∈ T, then 1/f ∈ A(T). Proof. Let w : A(T) → C be a multiplicative linear functional. The fact that w is a multiplicative linear functional implies that kwk = 1. Define u(t) = eit, t ∈ , for which kuk = 1. We define λ = w(u), which satisfies T A(T) |λ| ≤ kwk kuk = 1 A(T) and because u−1 = 1 we have λ−1 = w(u−1) and A(T) −1 −1 |λ | ≤ kwk u = 1, A(T)

itw hence |λ| = 1. Then there is some tw ∈ T such that λ = e . For n ∈ Z, w(un) = λn = eintw .

P int If P (t) = |n|≤N ane is a trigonometric polynomial, then  

X n X n X intw w(P ) = w  anu  = anw(u) = ane = P (tw). (2) |n|≤N |n|≤N |n|≤N

For g ∈ A( ), if > 0, then there is some N such that kg − SN (g)k < . T A(T) Using (2) and the fact that kgk ≤ kgk , C(T) A(T)

+ |SN (g)(tw) − g(tw)|

= |w(g − SN (g))| + |SN (g)(tw) − f(tw)|

≤ kwk kg − SN (g)k + kSN (g) − gk A(T) C(T) ≤ kwk kg − SN (g)k + kg − SN (g)k A(T) A(T) < 2.

Because this is true for all > 0, it follows that w(g) = g(tw). Let ∆ be the maximal ideal space of A(T). Then for w ∈ ∆ there is some tw ∈ T such that w(f) = f(tw), hence, because f(t) 6= 0 for all t ∈ T,

Γ(f)(w) = w(f) = f(tw) 6= 0.

That is, Γ(f) does not vanish, and therefore f is invertible in A(T). It is then −1 1 immediate that f (t) = f(t) for all t ∈ T, completing the proof.

4Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 239, Theorem 2.9.

8 The above proof of Wiener’s lemma uses the theory of the commutative Banach algebras. The following is a proof of the theorem that does not use the Gelfand transform.5

Proof. Because f ∈ A(T), f ∗ defined by f ∗(t) = f(t), t ∈ T, belongs to A(T). Let |f|2 ff ∗ g = = ∈ A(T), kfk2 kfk2 C(T) C(T) ∗ ∗ which satisfies 0 < g(t) ≤ 1 for all t ∈ . As 1 = f = f , to show that T f |f|2 kfk2 g C(T) 1 1/f ∈ A(T) it suffices to show that g ∈ A(T). Because g is continuous and g(t) 6= 0 for all t ∈ T, δ = inf g(t) > 0; t∈T if δ = 1 then g = 1, and indeed 1 ∈ A( ). Otherwise, kg − 1k = 1 − δ < 1. g T C(T) This implies that g is invertible in the Banach algebra C(T) and that g−1 = P∞ j j=0(1 − g) in C(T). Let h = 1 − g ∈ A(T). For > 0, there is some N such that kh − SN (h)k < . Now, if P is a A(T) trigonometric polynomial of degree M then using the Cauchy-Schwarz inequality and Parseval’s theorem,

kP k = Pˆ A(T) 1 ` (Z)

≤ (2M + 1)1/2 Pˆ 2 ` (Z) 1/2 = (2M + 1) kP k 2 L (T) 1/2 ≤ (2M + 1) kP k ∞ . L (T) Furthermore, for j ≥ 1, P j is a trigonometric polynomial of degree jM. The binomial theorem tells us, with P = SN (h) and r = h − P ,

k X k hk = (P + r)k = P jrk−j, j j=0

5Karlheinz Gr¨ochenig, Wiener’s Lemma: Theme and Variations. An Introduction to Spec- tral Invariance and Its Applications, p. 180, §5.2.4, in Brigitte Forster and Peter Massopust, eds., Four Short Courses on Harmonic Analysis, pp. 175–234.

9 j 1/2 j and using this and P ≤ (2jN + 1) P ∞ , A(T) L (T) k k X k j k−j h ≤ P r A(T) j A(T) A(T) j=0 k X k j k−j ≤ P kh − SN (h)k j A(T) A(T) j=0 k X k 1/2 j k−j ≤ (2jN + 1) P ∞ j L (T) j=0 k 1/2 X k j k−j ≤ (2kN + 1) kP k ∞ j L (T) j=0 1/2 k = (2kN + 1) (kP k ∞ + ) . L (T) Because

kP k ∞ ≤ kh − SN (h)k ∞ + khk ∞ L (T) L (T) L (T)

≤ kh − SN (h)k + khk ∞ A(T) L (T)

< + khk ∞ , L (T) we have

k 1/2 k 1/2 k h ≤ (2kN + 1) (khk ∞ + 2) = (2kN + 1) (1 − δ + 2) . A(T) L (T)

δ Take some < 2 , so that 1 − δ + 2 < 1. Then with N = N(), ∞ ∞ √ X k X 1/2 k 1 1 h ≤ (2kN+1) (1−δ+2) = 2NΦ 1 − δ + 2, − , < ∞, A(T) 2 2N k=0 k=0

P∞ k where Φ is the Lerch transcendent. This implies that the the series k=0 h P∞ k converges in A(T). We check that k=0 h is the inverse of 1 − h, namely, g = 1 − h is invertible in A(T), 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

σA(f) = {λ ∈ C : f − λ 6∈ U(A)}, called the spectrum of f. It is a fact that σA(f) is a nonempty compact subset of C.

10 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. The following are equivalent:

1. A is inverse-closed in B.

2. σA(f) = σB(f) for all f ∈ A.

Proof. Assume that A is inverse-closed in B and let f ∈ A. If λ 6∈ σA(f) then f − λ ∈ U(A) ⊂ U(B), hence λ 6∈ σB(f). Therefore σB(f) ⊂ σA(f). If −1 λ 6∈ σB(f) then f − λ ∈ U(B). That is, (f − λ) ∈ B. Because A is inverse- −1 closed in B and f −λ ∈ A, we get (f −λ) ∈ A. Thus λ 6∈ σA(f), and therefore σA(f) ⊂ σB(f). We thus have obtained σA(f) = σB(f). Assume that for all f ∈ A, σA(f) = σB(f). Suppose that f ∈ A and −1 f ∈ B. That is, f ∈ U(B), so 0 6∈ σB(f). Then 0 6∈ σA(f), meaning that f ∈ U(A).

A(T) ⊂ C(T) are Banach algebras with unity 1. Wiener’s lemma states that A(T) is inverse-closed in C(T). It is apparent that for f ∈ C(T), σC(T)(f) = f(T) ⊂ C. Therefore, Lemma 7 tells us for f ∈ A(T) that σA(T)(f) = f(T). The Wiener-L´evytheorem states that if f ∈ A(T), Ω ⊂ C is an open set containing f(T), and F :Ω → C is holomorphic, then F ◦ f ∈ A(T).7 In particular, if f ∈ A(T) does not vanish, then Ω = C \{0} is an open set 1 containing f(T) and F (z) = z is a holomorphic function on Ω, and hence 1 F ◦ f(t) = f(t) belongs to A(T), which is the statement of Wiener’s lemma.

6Karlheinz Gr¨ochenig, Wiener’s Lemma: Theme and Variations. An Introduction to Spec- tral Invariance and Its Applications, p. 183, §5.2.5, in Brigitte Forster and Peter Massopust, eds., Four Short Courses on Harmonic Analysis, pp. 175–234. 7Karlheinz Gr¨ochenig, Wiener’s Lemma: Theme and Variations. An Introduction to Spec- tral Invariance and Its Applications, p. 187, Theorem 5.16, in Brigitte Forster and Peter Mas- sopust, eds., Four Short Courses on Harmonic Analysis, pp. 175–234; 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, §7.7.