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Hermitian Operators and Isometries on Banach Algebras of Continuous Maps with Values in Unital C∗-Algebras

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Hermitian Operators and Isometries on Banach Algebras of Continuous Maps with Values in Unital C∗-Algebras Hermitian operators and isometries on Banach algebras of continuous maps with values in unital C∗-algebras . Osamu Hatori∗ (Niigata University) ∗ Supported by JSPS KAKENHI Grant Numbers JP16K05172, JP15K04921 «Joint work with Shiho Oi» RTOTA 3 May, 2018 University of Memphis 1/ 25 General Problem . Problem . Ej : Banach space, C(Xj ; Ej ) ⊃ Aj with a Banach space norm When is it true that every surjective isometrya U : A1 ! A2 will be of the form (canonical form) [ ] (Uf )(x) = Vx (f ('(x)) ; x 2 X2 for continuous (with respect to SOT) map x 7! Vx from X2 to the space of surjective isometries on E1 to E2, and a homeomorphism '? aIn this talk an isometry on a complex Banach space always means a .complex-linear isometry. 2/ 25 The origin of the problem . The Banach-Stone theorem (1932, 1937 and later) . Suppose that U : C(X1; C) ! C(X2; C) is a surjective isometry. () U(1) is a continous function with the unit modulus 9' : X2 ! X1 : homeomorphism s.t. U(f )(x) = U(1)(x)f ('(x)); x 2 X2 2 C .for every f C(X1; ). This canonical form is also called a weighted composition operator. 3/ 25 After the Banach-Stone theorem: . Several surjective isometries are of weighted composition operators Lp spaces (Isometries between Lp[0; 1] have been already studied by Banach in his book, Lamperti (1958) and several variations) Kadison’s theorem for a unital C∗-algebra 1951,1952 (later non-unital case by Paterson and Sinclair 1972) Nagasawa’s theorem for a uniform algebra (simple counterexample for non-unital case) Spaces of analytic functions (de Leeuw-Rudin-Wermer for H1 and for a uniform algebra (1960), Forelli for Hp (1964), and ··· ) Algebras of Lipschitz functions (de Leeuw(1961) and ········· ) ··· 4/ 25 Not always of a canonical form. An example1 of Lip(fa; b; cg) with k · k = maxfk · k1; L(·)g The Wiener algebra P 1 ^ W (T) = ff 2 C(T): kf kW = −∞ jf (n)j < 1g, and it’s analytic part ^ WA(T) = ff 2 W (T): f (n) = 0; 8n < 0g. are isometric by X X X U( ^f (n)einθ)) = ^f (`(n))einθ; ^f (n)einθ 2 W (T) n2Z n2Z n2Z for any bijection ` : Z ! Z+. · · ·· and many 1N.Weaver, Isometries of noncompact Lipschitz spaces, CMB (1995) 5/ 25 Not always of a canonical form. An example1 of Lip(fa; b; cg) with k · k = maxfk · k1; L(·)g The Wiener algebra P 1 ^ W (T) = ff 2 C(T): kf kW = −∞ jf (n)j < 1g, and it’s analytic part ^ WA(T) = ff 2 W (T): f (n) = 0; 8n < 0g. are isometric by X X X U( ^f (n)einθ)) = ^f (`(n))einθ; ^f (n)einθ 2 W (T) n2Z n2Z n2Z for any bijection ` : Z ! Z+. · · ·· and many 1N.Weaver, Isometries of noncompact Lipschitz spaces, CMB (1995) 5/ 25 Not always of a canonical form. An example1 of Lip(fa; b; cg) with k · k = maxfk · k1; L(·)g The Wiener algebra P 1 ^ W (T) = ff 2 C(T): kf kW = −∞ jf (n)j < 1g, and it’s analytic part ^ WA(T) = ff 2 W (T): f (n) = 0; 8n < 0g. are isometric by X X X U( ^f (n)einθ)) = ^f (`(n))einθ; ^f (n)einθ 2 W (T) n2Z n2Z n2Z for any bijection ` : Z ! Z+. · · ·· and many 1N.Weaver, Isometries of noncompact Lipschitz spaces, CMB (1995) 5/ 25 Not always of a canonical form. An example1 of Lip(fa; b; cg) with k · k = maxfk · k1; L(·)g The Wiener algebra P 1 ^ W (T) = ff 2 C(T): kf kW = −∞ jf (n)j < 1g, and it’s analytic part ^ WA(T) = ff 2 W (T): f (n) = 0; 8n < 0g. are isometric by X X X U( ^f (n)einθ)) = ^f (`(n))einθ; ^f (n)einθ 2 W (T) n2Z n2Z n2Z for any bijection ` : Z ! Z+. · · ·· and many 1N.Weaver, Isometries of noncompact Lipschitz spaces, CMB (1995) 5/ 25 A problem on isometries between spaces of vector-valued maps e C(Xj ; Ej ) ⊃ Bj with a Banach space norm The problem is : f f If U : B1 ! B2 is a surjective isometry, then E1 ∼ E2? No! Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me. 6/ 25 A problem on isometries between spaces of vector-valued maps e C(Xj ; Ej ) ⊃ Bj with a Banach space norm The problem is : f f If U : B1 ! B2 is a surjective isometry, then E1 ∼ E2? No! Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me. 6/ 25 A problem on isometries between spaces of vector-valued maps e C(Xj ; Ej ) ⊃ Bj with a Banach space norm The problem is : f f If U : B1 ! B2 is a surjective isometry, then E1 ∼ E2? No! Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me. 6/ 25 A problem on isometries between spaces of vector-valued maps e C(Xj ; Ej ) ⊃ Bj with a Banach space norm The problem is : f f If U : B1 ! B2 is a surjective isometry, then E1 ∼ E2? No! Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me. 6/ 25 A problem on isometries between spaces of vector-valued maps e C(Xj ; Ej ) ⊃ Bj with a Banach space norm The problem is : f f If U : B1 ! B2 is a surjective isometry, then E1 ∼ E2? No! Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me. 6/ 25 A problem on isometries between spaces of vector-valued maps e C(Xj ; Ej ) ⊃ Bj with a Banach space norm The problem is : f f If U : B1 ! B2 is a surjective isometry, then E1 ∼ E2? No! Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me. 6/ 25 In this talk : E = a unital C∗-algebra, C(X; C), the algebra of marices k − k f 2 F(x) F(y) E 1g Lip(K ; E) = F C(K ; E): L(F) = supx=6 y d(x;y) < with kFk1 + L(F) 0 C1(K; E) for K = [0; 1]; T with kFk1 + kF k1 A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm Compair : k · k1 + L(·) with maxfk · k1; L(·)g for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space? 7/ 25 In this talk : E = a unital C∗-algebra, C(X; C), the algebra of marices k − k f 2 F(x) F(y) E 1g Lip(K ; E) = F C(K ; E): L(F) = supx=6 y d(x;y) < with kFk1 + L(F) 0 C1(K; E) for K = [0; 1]; T with kFk1 + kF k1 A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm Compair : k · k1 + L(·) with maxfk · k1; L(·)g for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space? 7/ 25 In this talk : E = a unital C∗-algebra, C(X; C), the algebra of marices k − k f 2 F(x) F(y) E 1g Lip(K ; E) = F C(K ; E): L(F) = supx=6 y d(x;y) < with kFk1 + L(F) 0 C1(K; E) for K = [0; 1]; T with kFk1 + kF k1 A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm Compair : k · k1 + L(·) with maxfk · k1; L(·)g for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space? 7/ 25 In this talk : E = a unital C∗-algebra, C(X; C), the algebra of marices k − k f 2 F(x) F(y) E 1g Lip(K ; E) = F C(K ; E): L(F) = supx=6 y d(x;y) < with kFk1 + L(F) 0 C1(K; E) for K = [0; 1]; T with kFk1 + kF k1 A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm Compair : k · k1 + L(·) with maxfk · k1; L(·)g for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space? 7/ 25 In this talk : E = a unital C∗-algebra, C(X; C), the algebra of marices k − k f 2 F(x) F(y) E 1g Lip(K ; E) = F C(K ; E): L(F) = supx=6 y d(x;y) < with kFk1 + L(F) 0 C1(K; E) for K = [0; 1]; T with kFk1 + kF k1 A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm Compair : k · k1 + L(·) with maxfk · k1; L(·)g for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space? 7/ 25 In this talk : E = a unital C∗-algebra, C(X; C), the algebra of marices k − k f 2 F(x) F(y) E 1g Lip(K ; E) = F C(K ; E): L(F) = supx=6 y d(x;y) < with kFk1 + L(F) 0 C1(K; E) for K = [0; 1]; T with kFk1 + kF k1 A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm Compair : k · k1 + L(·) with maxfk · k1; L(·)g for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space? 7/ 25 When E = C(Y ) By identifying C(K ; C(Y )) with C(K × Y ) by the correspondence F 2 C(K ; C(Y )) −! F(x) 2 C(Y ); 8x 2 K [ ] −! (x; y) 7! (F(x))(y) 2 C(K × Y ); we may assume that Lip(X; C(Y )) ⊂ C(X × Y ); C1(K; C(Y )) ⊂ C(K × Y ); 8/ 25 A surjective isometry on Lip(K ; C(Y )) is a weighted composition operator .
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