with Cauchy Kernel on L 2 Takanori Yamamoto Hokkai-Gakuen University ICM Satellite Conference on Operator Algebras and Applications Cheongpung, KOREA (AUG 8-12, 2014)
1. Introduction and Problem T = {z : z = 1} : the unit circle in the complex plane C. For 0 < p, L p = L p (T, dx/2π) : the usual Lebesgue space, H p = H p (T, dx/2π) : the usual Hardy space. If p = 2, then f, g = 1 π ( 1 π 1/2 f (e ix )ḡ(e ix )dx, and f = f (e ix ) dx) 2. 2π π 2π π For f L 1 the Cauchy s principal value integral (Sf )(z) := 1 f (τ) πi τ z dτ exists for almost all z in T. T
Let P = (I + S)/2, Q = (I S)/2, where I denotes the identity operator. Let H 2 = L 2 H 2. Then P : L 2 H 2 is an orthogonal projection. Q : L 2 H 2 is an orthogonal projection. Let α L. T α : H 2 H 2, T α f = Pαf (f H 2 ) T α : H 2 H 2, T α f = Qαf (f H 2 ) H α : H 2 H 2, H α f = Qαf (f H 2 ) H α : H 2 H 2, H α f = Pαf (f H 2 ) M α : L 2 L 2, M α f = αf (f L 2 ) Let α, β L. S α,β : L 2 L 2, S α,β f = αpf + βqf (f L 2 ) Then M α = S α,α.
Problem When S α,β is self-adjoint? When S α,β is normal? When S α,β is hyponormal?
Since Hᾱf, g = f, H α g, (f H 2, g H 2 ), H α = H ᾱ. Lemma 2.1. Let α and β be in L. Then the operators S α,β and Sα,β have the matrix forms ( ) ( ) T S α,β = α Hβ, Sα,β H α Tβ = Tᾱ Hᾱ on L 2 = H β T β on L2 H 2 H 2.
2. Self-adjoint operator S α,β Theorem 2.1. Let α and β be functions in L. TFAE 1 S α,β is self-adjoint. 2 α and β are real-valued functions and α β is a real constant. 3 α = f + f + p and β = f + f + q for some f zh 2 and some p, q R. Theorem 2.2. Let α and β be functions in L. TFAE 1 S α,β is positive. 2 α and β are non-negative functions and α β is a real constant.
3. Normal operator S α,β We need some lemmas to prove the main theorems. Lemma 3.1. Let α, β L and φ = α β. Then 1 Sα,β S α,β S α,β Sα,β ( H α Hᾱ H β H β = T φ H β H α T φ 2 S α,β is normal if and only if H β T φ T φ Hᾱ H β H β H α Hᾱ ). H α Hᾱ = H β H β, H β H β = H α Hᾱ, H α T φ = T φ H β.
Proof of Lemma 3.1. (1): (1,1) element is TᾱT α + HᾱH α T α Tᾱ H β H β = T α 2 T α Tᾱ H β H β = H α Hᾱ H β H β. (2,2) element is H β H β + T β T β H α Hᾱ T β T β = T β 2 H α Hᾱ T β T β = H β H β H α Hᾱ. (2,1) element is H βt α + T βh α H α Tᾱ T β H β = H α β H α Tᾱ T β H β = T φ H β H α T φ. (1,2) element is Tᾱ H β + Hᾱ T β T α Hᾱ H β T β = Hᾱβ T α Hᾱ H β T β = H β T φ T φ Hᾱ. (2): Since (1,2) element is the adjoint of the (2,1) element, T φ H β = H α T φ if and only if H β T φ = T φ Hᾱ.
By the proof of the Cowen theorem [Cow], the first author and Takahashi [NakTak] proved the equivalence of (1) and (3) in the following lemma by the Sarason theorem ([Sar] and p.230 of [Nik]). We have had an another proof which does not use the Sarason theorem. Lemma 3.2. Let α and β be L functions. TFAE. 1 H αh α = H β H β (That is HᾱH α = H βh β ). 2 H α H α = H β H β (That is H α Hᾱ = H β H β). 3 α cβ H for some c T.
Lemma 3.3. Let α be functions in L. Suppose F, f zh 2 satisfy α ˆα(0) = F + f. Then F, f 1<p< L p. Lemma 3.4. If g L 2, then Qḡ = zp( zg) = Pg ĝ(0).
Lemma 3.5. Let α and β be functions in L and let φ = α β. Suppose F, f, G, g zh 2 satisfy α ˆα(0) = F + f and β ˆβ(0) = G + ḡ. If H α T φ = T φ H β, then Gg = Ff. Lemma 3.6. Let α and β be functions in L and let φ = α β. Suppose F, f, G, g zh 2 satisfy α ˆα(0) = F + f and β ˆβ(0) = G + ḡ. If S α,β is normal, then there exists c, d T such that g = cf, G = df, (1 cd)ff = 0 and either α cβ C or α dβ C holds.
Theorem 3.1. Let α, β L, and φ = α β. Suppose F, f, G, g zh 2 satisfy α ˆα(0) = F + f and β ˆβ(0) = G + ḡ. TFAE. 1 S α,β is normal. 2 There exist constants c, d such that c = 1, α = cβ + d and T φ H β H α T φ = 0, where φ = α β. 3 There exist constants c, d such that c = 1, α = cβ + d and (c 1) β 2 + d β c dβ H. 4 There exist constants c, d such that c = 1, α = cβ + d and (c 1) α 2 + dᾱ c dα H.
Theorem 3.2. Let α and β be functions in L. Suppose α β is a non-constant function. Then the following are equivalent. 1 S α,β is normal. 2 There exist a non-constant unimodular function ψ in L and constants a, b, c such that a 0, c = 1, c 1, α = caψ + b and β = aψ + b.
Brown-Halmos theorem T α is normal. α = au + b for some real valued function u L and a, b C. For a T the subset (BH)(a) is defined by (BH)(a) = {af + f + b : f zh 2, b C}. By Lemma 3.1, (BH)(a) (BH). If a, b T and a b, then (BH)(a) (BH)(b) = C.
Theorem 3.3. Let α and β be non-constant functions in L. Suppose α β is a non-zero constant. Then the following are equivalent. 1 S α,β is normal. 2 α (BH)((α β)/(ᾱ β)). 3 There exist f zh 2 and p, q C with p q such that α = p q p q f + f + p, β = p q p q f + f + q.
4. Hyponormal Operator S α,β If S α,β is hyponormal, then S α,β (u + v) 2 Sα,β (u + v) 2, for all u H 2, v H 2. Hence, (T α + H α )u + ( H β + T β )v 2 (Tᾱ + H β)u + ( Hᾱ + T β)v 2. Let φ = α β. By Lemma 3.1, if S α,β is hyponormal, then ( Hα [Sα,β, Hᾱ S H ) β H β H β T α,β ] = φ T φ Hᾱ 0. T φ H β H α T φ H β H β H α Hᾱ
[S α,β, S α,β] is diagonal if and only if T φ H β H α T φ = 0. By Lemma 3.5, if [S α,β, S α,β] is diagonal, then (Qα)(Qᾱ) = (Qβ)(Q β). S α,β is hyponormal if and only if ( H α Hᾱ H β H β)u, u + (H β H β H α Hᾱ)v, v for all u H 2, v H 2. +2Re ( T φ H β H α T φ)u, v 0,
By the proof of the Cowen theorem ([Cow]), Lemma 4.1. Let α and β be functions in L. TFAE. 1 H αh α H β H β (That is HᾱH α H βh β ). 2 H α H α H β H β (That is H α Hᾱ H β H β). 3 There exists a k H with k 1 such that kα β belongs to H. The proof of Lemma 3.2 does not use Sarason lifting theorem. The proof of Lemma 4.1 uses Sarason lifting theorem. Lemma 3.2 follows from Lemma 4.1.
Lemma 4.2. Let α and β be functions in L. TFAE. 1 S α,β is hyponormal and [S α,β, S α,β] is diagonal. 2 There exist functions k and h in H with k 1 and h 1 such that both kᾱ β and hβ α belong to H, and T φ H β H α T φ = 0.
Lemma 4.3. Let α and β be functions in L. If S α,β is hyponormal, then there exist functions k and h in H with k 1 and h 1 such that both kᾱ β and hβ α belong to H. Cowen theorem([cow], [NakTak]) Let E(α) = {k H : k 1 and kᾱ α H }. Then T α is hyponormal. E(α) is not empty. By Lemma 4.3, if S α,ᾱ is hyponormal, then there exists a function k in H with k 1 such that kᾱ α H. By the Cowen Theorem ([Cow]), this implies that T α is hyponormal.
Theorem 4.1. Let α C and β L. TFAE. 1 S α,β is hyponormal. 2 S α,β is hyponormal and [Sα,β, S α,β] is diagonal. 3 β H Theorem 4.2. Let α L and β C. TFAE. 1 S α,β is hyponormal. 2 S α,β is hyponormal and [Sα,β, S α,β] is diagonal. 3 α H.
Theorem 4.3. Main Theorem Let α and β be non-constant functions in L such that α β is a non-zero constant. Then the following are equivalent. 1 S α,β is hyponormal. 2 S α,β is normal. 3 [S α,β, S α,β] is diagonal. 4 There exist f zh 2 and p, q C with p q such that α = p q p q f + f + p, β = p q p q f + f + q.
Lemma 4.4. Let α H and β L. TFAE. 1 [S α,β, S α,β] is diagonal. 2 Tφ H β H α T φ = 0. 3 Either β H or β, α β β 2 H. Proof. By Lemma 3.5, if T φ H β H α T φ = 0, then (Qα)(Qᾱ) = (Qβ)(Q β). If β H, then β, φ H and 0 = ( T φ H β H α T φ)p = T φ H βp = QφQ βp = QφQ βp + QφP βp = Qφ βp = Q(α β β 2 )P.
If α H and β H, then S α,β is hyponormal and [S α,β, S α,β] is diagonal. Theorem 4.4. Main Theorem Let α H, β L and β H. TFAE. 1 S α,β is hyponormal and [S α,β, S α,β] is diagonal. 2 Both β and α β β 2 are in H, and there exists a k such that kᾱ β H and k 1. If α, β H, then Nakazi([Nak]) proved that S α,β is hyponormal is also equivalent. Example 1 α = z 2, β = z. 2 α = n j=1 a j z j, a n 0, β = a n z. [Nak]
Lemma 4.5. Let α, β L and β H. TFAE. 1 [S α,β, S α,β] is diagonal. 2 Tφ H β H α T φ = 0. 3 Either α H or ᾱ, α β α 2 H. Proof. By Lemma 3.5, if T φ H β H α T φ = 0, then (Qα)(Qᾱ) = (Qβ)(Q β). If α H, then ᾱ, φ H and 0 = ( T φ H β H α T φ)p = H α T φp = QαP φp = Qα φp = Q(α β α 2 )P.
Theorem 4.5. Main Theorem Let α, β L, β H and α H. TFAE. 1 S α,β is hyponormal and [S α,β, S α,β] is diagonal. 2 Both ᾱ and α β α 2 are in H, and there exists an h such that hβ α H and h 1. If ᾱ, β H, then Nakazi([Nak]) proved that S α,β is hyponormal is also equivalent. Example 1 α = z, β = z 2. 2 α = b n z, b n 0, and β = n j=1 b j z j. [Nak]
Norm inequality related to Theorem 4.3. For a function r satisfying 0 r 1 and T rdm > 0, we define the Helson-Szegő type set (HS)(r): (HS)(r) = {γe u+ṽ ; γ is a positive constant, u, v are real functions, u L 1, v L, v π/2, r 2 e u + e u 2 cos v} By the Cotlar-Sadosky lifting theorem, Theorem 4.7. Let a, b, c, d L. TFAE. (1) For all f L 2, S a,b f S c,d f. (2) a c, b d, and either (a) or (b) holds. (a) ad bc = 0. (b) log a b c d L 1, and there is a V in (HS)(r) such that (a b c d)/v is in H 1/2, where r = ad bc / a b c d.
Corollary 4.1. Let α, β L. TFAE. (1) For all f L 2, S α,β f Sᾱ, βf. (2) For all f L 2, S α,β f = Sᾱ, βf. (3) α β ᾱβ is a constant. Corollary 4.2. Let α, β L, and α β = c for some non-zero constant c. TFAE. (1) For all f L 2, S α,β f S α,β f. (2) For all f L 2, S α,β f = S α,β f. (3) For all f L 2, S α,β f Sᾱ, βf. (4) For all f L 2, S α,β f = Sᾱ, βf. (5) α c ᾱc is a constant.
Proof of Lemma 3.2 without any lifting theorem Lemma 3.2. Let α and β be L functions. TFAE. 1 H αh α = H β H β (That is HᾱH α = H βh β ). 2 H α H α = H β H β (That is H α Hᾱ = H β H β). 3 α cβ H for some c T. Proof of (2) (3): Suppose α ˆα(0) = F + f and β ˆβ(0) = G + ḡ for some F, f, G, g H 2 0, and let h = f + ḡ. Then h L 2, so that T h is a densely defined operator on H 2. Let Q 1 denote the rank one orthogonal projection of L 2 onto C z such that (Q 1 f )(z) = ˆf ( 1) z (f L 2 ).
By the calculation, for all trigonometric polynomial k, three equalities: T h Qk = T h Qk, T z T h T h T z Qk T h T h Qk = (zq 1 T h T z Qk)(Q h), T z T h T h T z Qk T h T h Qk = (zq 1 T h T z Qk)(Qh) hold, it follows that and T h T h = T h T h, (zq 1 T h T z Qk)(Q h) = (zq 1 T h T z Qk)(Qh). Let a = zq 1 T h T z Qk and let b = zq 1 T h T z Qk. Then both a and b are constants such that aq h = bqh. Hence Q(a h bh) = 0. This implies that a h bh H.
Since h = Qα + Qβ, this implies that Thus Therefore a(qα + Qβ) b(qα + Qβ) H. Q(aα bβ) = aqα bqβ H 2 H 2 = {0}. aα bβ H, for some constants a, b C. This implies that ah α = bh β. By (2), this implies that a 2 H α H α = b 2 H β H β = b 2 H α H α, so that ( a 2 b 2 )H α H α = 0. By (2), if α H, then β H, so that (3) holds with any constant c. Thus we can assume that α H. Then H α 0, so that H α H α 0. Since ( a 2 b 2 )H α H α = 0, this implies that a = b.
By (2), if a = b = 0, then Q 1 T h T z Qk = Q 1 T h T z Qk = 0, so that for all n 1, h, z n = h, z n = 0. This implies that h C. Since h = f + ḡ, this implies that h = 0, so that f = ḡ H 2 H 2 = {0}. Since f = g = 0, it follows that α, β H, so that (3) holds with any constant c. Thus we can assume that both a and b are non-zero constants. Let c = b/a. Then This implies (3). α cβ = 1 a (aα bβ) H, c = 1.
References [BroHal] A. Brown and P.R. Halmos, Algebraic properties of Toeplitz operators, J. Reine Angew. Math. 213(1963/1964), 89 102. [Cow] C.C. Cowen, Hyponormality of Toeplitz operators, Proc. Amer. Math. Soc. 103(1988), 809 812. [Dou] R.G. Douglas, Banach Algebra Techniques in Operator Theory, 2nd. ed. Springer-Verlag, 1998. [GohKru] I. Gohberg and N. Krupnik, One-Dimensional Linear Singular Integral Equations, vols. I, II, Birkhäuser, Basel, 1992. [MarRos] R.A. Martínez-Avendaño and P. Rosenthal, An introduction to operators on the Hardy-Hilbert space, Springer, 2007. [Nak] T. Nakazi, Hyponormal singular integral operators near to normal ones, preprint.
References [NakTak] T. Nakazi and K. Takahashi, Hyponormal Toeplitz operators and extremal problems of Hardy spaces, Trans. Amer. Math. Soc. 338 (1993), 753 767. [NakYam] T. Nakazi and T. Yamamoto, Normal singular integral operators with Cauchy kernel on L 2, Integr. Equ. Oper. Theory 78 (2014), 233-248. [Nik] N.K. Nikolski, Operators, Functions, and Systems: An Easy Reading. Vol. 1, Amer. Math. Soc., Providence, 2002. [Sar] D. Sarason, Generalized interpolation in H, Trans. Amer. Math. Soc. 127 (1967), 179 203. [Yam] T. Yamamoto, Majorization of singular integral operators with Cauchy kernel on L 2, Ann. Funct. Anal. 5 (2014), 101 108.
THANK YOU