Biggest Open Ball in Invertible Elements of a Banach Algebra

Biggest open ball in invertible elements of a Banach algebra

D. Sukumar Geethika Indian Institute of Technology Hyderabad [email protected]

Banach Algebras and Applications Workshop-2015 Fields Institute, Toronto, Canada.

August 4, 2015

Sukumar (IITH) BOBP August 4, 2015 1 / 11 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, ⊆ G(A) ka−1k Is this the biggest ball? ∈ − = 1 Does there exists a non-invertible b A such that kb ak ka−1k

Sukumar (IITH) BOBP August 4, 2015 2 / 11 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 ka−1k intersects Sing(A).

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, 2015 3 / 11 | |z =

1 1 | − |z

Positive observations for BOBP The Banach algebra of complex numbers C C −1 1 , 1 , Let z ∈ be invertible. |z | = |z| . Then B z |z−1| = B (z |z|). Here0 is the singular element on the boundary.

Sukumar (IITH) BOBP August 4, 2015 4 / 11 0

z | |z =

1 1 | − |z

Sukumar (IITH) BOBP August 4, 2015 4 / 11 Positive observations for BOBP The Banach algebra of complex numbers C C −1 1 , 1 , Let z ∈ be invertible. |z | = |z| . Then B z |z−1| = B (z |z|). Here0 is the singular element on the boundary.

z | |z =

1 1 | − |z

Sukumar (IITH) BOBP August 4, 2015 4 / 11 Consider g(x) = f (x) − f (x0). Then g is singular and 1 kf − gk = |f (x )| = ∞ 0 kf −1k Commutative unital C∗ algebra 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.

Positive observations for BOBP

The uniform algebra C(X ), X compact Hausdorff Space Let f ∈ C(X ) be invertible.

1 1 1 −1 f = sup = = for some x0 ∈ X ∞ x∈X |f (x)| infx∈X {|f (x)|} |f (x0)|

Sukumar (IITH) BOBP August 4, 2015 5 / 11 Commutative unital C∗ algebra 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.

Positive observations for BOBP

The uniform algebra C(X ), X compact Hausdorff Space Let f ∈ C(X ) be invertible.

1 1 1 −1 f = sup = = for some x0 ∈ X ∞ x∈X |f (x)| infx∈X {|f (x)|} |f (x0)|

Consider g(x) = f (x) − f (x0). Then g is singular and 1 kf − gk = |f (x )| = ∞ 0 kf −1k

Sukumar (IITH) BOBP August 4, 2015 5 / 11 Positive observations for BOBP

The uniform algebra C(X ), X compact Hausdorff Space Let f ∈ C(X ) be invertible.

1 1 1 −1 f = sup = = for some x0 ∈ X ∞ x∈X |f (x)| infx∈X {|f (x)|} |f (x0)|

Consider g(x) = f (x) − f (x0). Then g is singular and 1 kf − gk = |f (x )| = ∞ 0 kf −1k Commutative unital C∗ algebra 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, 2015 5 / 11 Norm attaining T −1 Let T ∈ B(H) be invertible and T −1 is norm attaining, then T has BOBP.

Example

The matrix algebra Mn×n(C) has BOBP.

Positive observations for BOBP Operator algebra B(H)

Condition number 1 Let T ∈ B(H) be invertible such that kT kkT −1k = 1, then T has BOBP.

Sukumar (IITH) BOBP August 4, 2015 6 / 11 Example

The matrix algebra Mn×n(C) has BOBP.

Positive observations for BOBP Operator algebra B(H)

Condition number 1 Let T ∈ B(H) be invertible such that kT kkT −1k = 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, 2015 6 / 11 Positive observations for BOBP Operator algebra B(H)

Condition number 1 Let T ∈ B(H) be invertible such that kT kkT −1k = 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 Mn×n(C) has BOBP.

Sukumar (IITH) BOBP August 4, 2015 6 / 11 Proof. Let T be invertible in B(H).

Positive observations for BOBP

Theorem B(H) has BOBP.

Sukumar (IITH) BOBP August 4, 2015 7 / 11 Positive observations for BOBP

Theorem B(H) has BOBP.

Proof. Let T be invertible in B(H).

Sukumar (IITH) BOBP August 4, 2015 7 / 11 Positive observations for BOBP

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, 2015 7 / 11 Positive observations for BOBP

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

1 1 −1 −1 T = |T | = sup = for some λ0 ∈ σ(|T |) λ∈σ(|T |) |λ| |λ0|

Sukumar (IITH) BOBP August 4, 2015 7 / 11 Positive observations for BOBP

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

1 1 −1 −1 T = |T | = sup = for some λ0 ∈ σ(|T |) λ∈σ(|T |) |λ| |λ0|

Note that |T | − λ0 is not invertible.

Sukumar (IITH) BOBP August 4, 2015 7 / 11 Positive observations for BOBP 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

1 1 −1 −1 T = |T | = sup = for some λ0 ∈ σ(|T |) λ∈σ(|T |) |λ| |λ0|

Note that |T | − λ0 is not invertible. Consider the operator

S = V (|T | − λ0I) = V |T | − λ0V = T − λ0V .

Sukumar (IITH) BOBP August 4, 2015 7 / 11 Positive observations for BOBP 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

1 1 −1 −1 T = |T | = sup = for some λ0 ∈ σ(|T |) λ∈σ(|T |) |λ| |λ0|

Note that |T | − λ0 is not invertible. Consider the operator

S = V (|T | − λ0I) = V |T | − λ0V = T − λ0V .

− = | | = 1 Then S is not invertible and kS T k λ0 kT −1k

Sukumar (IITH) BOBP August 4, 2015 7 / 11 Negative observations for BOBP

, 1 ( ) B a ka−1k is not the biggest open ball around a contained in G A There exists a ∈ G(A) such that for every b singular

1 ka − bk > . ka−1k

Sukumar (IITH) BOBP August 4, 2015 8 / 11 Let f (x) = ex be an invertible element. Here

1 −1 f −1 = + = 2 x x e ∞ e ∞

Let g be singular. That is g(x0) = 0 for some x0 ∈ [0, 1]. Then 1 1 0 0 x0 kf − gk = kf − gk + f − g ≥ |f (x0) − g(x0)| = e > = ∞ ∞ 2 kf −1k

Negative observations for BOBP

Banach function algebra C1[0, 1] C1[0, 1] equipped with the norm

0 1 kf k = kf k∞ + kf k∞ ∀ f ∈ C [0, 1].

Sukumar (IITH) BOBP August 4, 2015 9 / 11 Let g be singular. That is g(x0) = 0 for some x0 ∈ [0, 1]. Then 1 1 0 0 x0 kf − gk = kf − gk + f − g ≥ |f (x0) − g(x0)| = e > = ∞ ∞ 2 kf −1k

Negative observations for BOBP

Banach function algebra C1[0, 1] C1[0, 1] equipped with the norm

0 1 kf k = kf k∞ + kf k∞ ∀ f ∈ C [0, 1].

Let f (x) = ex be an invertible element. Here

1 −1 f −1 = + = 2 x x e ∞ e ∞

Sukumar (IITH) BOBP August 4, 2015 9 / 11 Negative observations for BOBP

Banach function algebra C1[0, 1] C1[0, 1] equipped with the norm

0 1 kf k = kf k∞ + kf k∞ ∀ f ∈ C [0, 1].

Let f (x) = ex be an invertible element. Here

1 −1 f −1 = + = 2 x x e ∞ e ∞

Let g be singular. That is g(x0) = 0 for some x0 ∈ [0, 1]. Then 1 1 0 0 x0 kf − gk = kf − gk + f − g ≥ |f (x0) − g(x0)| = e > = ∞ ∞ 2 kf −1k

Sukumar (IITH) BOBP August 4, 2015 9 / 11 Negative observations for BOBP

1 The group algebra ` (Z2)

Wiener algebra A(T) The set of all complex valued functions on [−π, π] with absolutely con- vergent Fourier series, that is, functions of the form

∞ int f (t) = ∑ cne , t ∈ [−π, π] n=−∞

∞ with kf k = ∑n=−∞ |cn| < ∞.

Sukumar (IITH) BOBP August 4, 2015 10 / 11 Summary

BOBP Does not have BOBP C(X ), X compact T2 C1[0, 1] Commutative C∗ algebra 1 ` (Z2) Mn(C) A(T) - Wiener algebra B(H)

Thank you

Sukumar (IITH) BOBP August 4, 2015 11 / 11