Biggest open ball in invertible elements of a Banach algebra
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1 Biggest open ball in invertible elements of a Banach algebra D. Sukumar Geethika Indian Institute of Technology Hyderabad suku@iith.ac.in Banach Algebras and Applications Workshop-2015 Fields Institute, Toronto, Canada. August 4, 2015 Sukumar (IITH) BOBP August 4, / 11
2 Group of invertible elements in a Banach Algebra Let A be a complex unital Banach Algebra and G(A) denote the group of invertible elements of A. Theorem G(A) is an open set in A. ( ) 1 B a, a 1 G(A) Is this the biggest ball? Does there exists a non-invertible b A such that b a = 1 a 1 Sukumar (IITH) BOBP August 4, / 11
3 Biggest Open Ball Property (BOBP) Definition (An element has BOBP) An( element ) a G(A) has BOBP if the boundary of the ball 1 B a, intersects Sing(A). a 1 Definition (BOBP) A Banach algebra has BOBP if every element of G(A) has BOBP. Can we characterise all Banach Algebras which satisfy BOBP? Can we characterise all the elements of a Banach Algebra, which do not satisfy BOBP? Sukumar (IITH) BOBP August 4, / 11
4 The Banach algebra of complex numbers C ( Let z C be invertible. z 1 = 1 z. Then B 1 z, Here 0 is the singular element on the boundary. z 1 ) = B (z, z ). z
5 The Banach algebra of complex numbers C ( Let z C be invertible. z 1 = 1 z. Then B 1 z, Here 0 is the singular element on the boundary. z 1 ) = B (z, z ). 1 z 1 = z z
6 The Banach algebra of complex numbers C ( Let z C be invertible. z 1 = 1 z. Then B 1 z, Here 0 is the singular element on the boundary. z 1 ) = B (z, z ). 1 z 1 = z z 0 Sukumar (IITH) BOBP August 4, / 11
7 The uniform algebra C(X ), X compact Hausdorff Space Let f C(X ) be invertible. { } f = sup = x X f (x) inf x X { f (x) } = 1 f (x 0 ) for some x 0 X Sukumar (IITH) BOBP August 4, / 11
8 The uniform algebra C(X ), X compact Hausdorff Space Let f C(X ) be invertible. { } f = sup = x X f (x) inf x X { f (x) } = 1 f (x 0 ) for some x 0 X Consider g(x) = f (x) f (x 0 ). Then g is singular and f g = f (x 0 ) = 1 f 1 Sukumar (IITH) BOBP August 4, / 11
9 The uniform algebra C(X ), X compact Hausdorff Space Let f C(X ) be invertible. { } f = sup = x X f (x) inf x X { f (x) } = 1 f (x 0 ) for some x 0 X Consider g(x) = f (x) f (x 0 ). Then g is singular and Commutative unital C algebra f g = f (x 0 ) = 1 f 1 Since it is true for C(X ), X compact and Hausdorff, by representation theorem, any commutative unital C algebras has BOBP. For general C algebra. If x G(A) is normal, the x satisfies BOBP. Sukumar (IITH) BOBP August 4, / 11
10 Operator algebra B(H) Condition number 1 Let T B(H) be invertible such that T T 1 = 1, then T has BOBP. Sukumar (IITH) BOBP August 4, / 11
11 Operator algebra B(H) Condition number 1 Let T B(H) be invertible such that T T 1 = 1, then T has BOBP. Norm attaining T 1 Let T B(H) be invertible and T 1 is norm attaining, then T has BOBP. Sukumar (IITH) BOBP August 4, / 11
12 Operator algebra B(H) Condition number 1 Let T B(H) be invertible such that T T 1 = 1, then T has BOBP. Norm attaining T 1 Let T B(H) be invertible and T 1 is norm attaining, then T has BOBP. Example The matrix algebra M n n (C) has BOBP. Sukumar (IITH) BOBP August 4, / 11
13 Theorem B(H) has BOBP. Sukumar (IITH) BOBP August 4, / 11
14 Theorem B(H) has BOBP. Proof. Let T be invertible in B(H). Sukumar (IITH) BOBP August 4, / 11
15 Theorem B(H) has BOBP. Proof. Let T be invertible in B(H). Let T = V T be the polar decomposition. As T is invertible, V is unitary and T also invertible. Sukumar (IITH) BOBP August 4, / 11
16 Theorem B(H) has BOBP. Proof. Let T be invertible in B(H). Let T = V T be the polar decomposition. As T is invertible, V is unitary and T also invertible. As T is self-adjoint { } T 1 T = 1 1 = sup = 1 λ σ( T ) λ λ 0 for some λ 0 σ( T ) Sukumar (IITH) BOBP August 4, / 11
17 Theorem B(H) has BOBP. Proof. Let T be invertible in B(H). Let T = V T be the polar decomposition. As T is invertible, V is unitary and T also invertible. As T is self-adjoint { } T 1 T = 1 1 = sup = 1 λ σ( T ) λ λ 0 for some λ 0 σ( T ) Note that T λ 0 is not invertible. Sukumar (IITH) BOBP August 4, / 11
18 Theorem B(H) has BOBP. Proof. Let T be invertible in B(H). Let T = V T be the polar decomposition. As T is invertible, V is unitary and T also invertible. As T is self-adjoint { } T 1 T = 1 1 = sup = 1 λ σ( T ) λ λ 0 for some λ 0 σ( T ) Note that T λ 0 is not invertible. Consider the operator S = V ( T λ 0 I) = V T λ 0 V = T λ 0 V. Sukumar (IITH) BOBP August 4, / 11
19 Theorem B(H) has BOBP. Proof. Let T be invertible in B(H). Let T = V T be the polar decomposition. As T is invertible, V is unitary and T also invertible. As T is self-adjoint { } T 1 T = 1 1 = sup = 1 λ σ( T ) λ λ 0 for some λ 0 σ( T ) Note that T λ 0 is not invertible. Consider the operator S = V ( T λ 0 I) = V T λ 0 V = T λ 0 V. Then S is not invertible and S T = λ 0 = 1 T 1 Sukumar (IITH) BOBP August 4, / 11
20 Negative observations for BOBP ( ) 1 B a, is not the biggest open ball around a contained in G(A) a 1 There exists a G(A) such that for every b singular a b > 1 a 1. Sukumar (IITH) BOBP August 4, / 11
21 Negative observations for BOBP Banach function algebra C 1 [0, 1] C 1 [0, 1] equipped with the norm f = f + f f C 1 [0, 1]. Sukumar (IITH) BOBP August 4, / 11
22 Negative observations for BOBP Banach function algebra C 1 [0, 1] C 1 [0, 1] equipped with the norm f = f + f f C 1 [0, 1]. Let f (x) = e x be an invertible element. Here f 1 1 = e x + 1 e x = 2 Sukumar (IITH) BOBP August 4, / 11
23 Negative observations for BOBP Banach function algebra C 1 [0, 1] C 1 [0, 1] equipped with the norm f = f + f f C 1 [0, 1]. Let f (x) = e x be an invertible element. Here f 1 1 = e x + 1 e x = 2 Let g be singular. That is g(x 0 ) = 0 for some x 0 [0, 1]. Then f g = f g + f g f (x 0 ) g(x 0 ) = e x 0 > 1 2 = 1 f 1 Sukumar (IITH) BOBP August 4, / 11
24 Negative observations for BOBP The group algebra l 1 (Z 2 ) Wiener algebra A(T) The set of all complex valued functions on [ π, π] with absolutely convergent Fourier series, that is, functions of the form f (t) = with f = n= c n <. c n e int, t [ π, π] n= Sukumar (IITH) BOBP August 4, / 11
25 Summary BOBP C(X ), X compact T 2 Commutative C algebra M n (C) B(H) Does not have BOBP C 1 [0, 1] l 1 (Z 2 ) A(T) - Wiener algebra Thank you Sukumar (IITH) BOBP August 4, / 11
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