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? 2 − = 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 2 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 j jz = 1 1 j − jz 0 Positive observations for BOBP The Banach algebra of complex numbers C C −1 1 , 1 , Let z 2 be invertible. jz j = jzj . Then B z jz−1j = B (z jzj). Here0 is the singular element on the boundary. z Sukumar (IITH) BOBP August 4, 2015 4 / 11 0 Positive observations for BOBP The Banach algebra of complex numbers C C −1 1 , 1 , Let z 2 be invertible. jz j = jzj . Then B z jz−1j = B (z jzj). Here0 is the singular element on the boundary. z j jz = 1 1 j − jz 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 2 be invertible. jz j = jzj . Then B z jz−1j = B (z jzj). Here0 is the singular element on the boundary. z j jz = 1 1 j − jz 0 Sukumar (IITH) BOBP August 4, 2015 4 / 11 Consider g(x) = f (x) − f (x0). Then g is singular and 1 kf − gk = jf (x )j = ¥ 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 2 G(A) is normal, the x satisfies BOBP. Positive observations for BOBP The uniform algebra C(X ), X compact Hausdorff Space Let f 2 C(X ) be invertible. 1 1 1 −1 f = sup = = for some x0 2 X ¥ x2X jf (x)j infx2X fjf (x)jg jf (x0)j 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 2 G(A) is normal, the x satisfies BOBP. Positive observations for BOBP The uniform algebra C(X ), X compact Hausdorff Space Let f 2 C(X ) be invertible. 1 1 1 −1 f = sup = = for some x0 2 X ¥ x2X jf (x)j infx2X fjf (x)jg jf (x0)j Consider g(x) = f (x) − f (x0). Then g is singular and 1 kf − gk = jf (x )j = ¥ 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 2 C(X ) be invertible. 1 1 1 −1 f = sup = = for some x0 2 X ¥ x2X jf (x)j infx2X fjf (x)jg jf (x0)j Consider g(x) = f (x) − f (x0). Then g is singular and 1 kf − gk = jf (x )j = ¥ 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 2 G(A) is normal, the x satisfies BOBP. Sukumar (IITH) BOBP August 4, 2015 5 / 11 Norm attaining T −1 Let T 2 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 2 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 2 B(H) be invertible such that kT kkT −1k = 1, then T has BOBP. Norm attaining T −1 Let T 2 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 2 B(H) be invertible such that kT kkT −1k = 1, then T has BOBP. Norm attaining T −1 Let T 2 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 jT j be the polar decomposition. As T is invertible, V is unitary and jT j 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 jT j be the polar decomposition. As T is invertible, V is unitary and jT j also invertible. As |T| is self-adjoint 1 1 −1 −1 T = jT j = sup = for some l0 2 s(jT j) l2s(jT j) jlj jl0j 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 jT j be the polar decomposition. As T is invertible, V is unitary and jT j also invertible. As |T| is self-adjoint 1 1 −1 −1 T = jT j = sup = for some l0 2 s(jT j) l2s(jT j) jlj jl0j Note that jT j − l0 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 jT j be the polar decomposition. As T is invertible, V is unitary and jT j also invertible. As |T| is self-adjoint 1 1 −1 −1 T = jT j = sup = for some l0 2 s(jT j) l2s(jT j) jlj jl0j Note that jT j − l0 is not invertible. Consider the operator S = V (jT j − l0I) = V jT j − l0V = T − l0V . 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 jT j be the polar decomposition. As T is invertible, V is unitary and jT j also invertible. As |T| is self-adjoint 1 1 −1 −1 T = jT j = sup = for some l0 2 s(jT j) l2s(jT j) jlj jl0j Note that jT j − l0 is not invertible. Consider the operator S = V (jT j − l0I) = V jT j − l0V = T − l0V . − = j j = 1 Then S is not invertible and kS T k l0 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 2 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 2 [0, 1]. Then 1 1 0 0 x0 kf − gk = kf − gk + f − g ≥ jf (x0) − g(x0)j = 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¥ 8 f 2 C [0, 1]. Sukumar (IITH) BOBP August 4, 2015 9 / 11 Let g be singular. That is g(x0) = 0 for some x0 2 [0, 1]. Then 1 1 0 0 x0 kf − gk = kf − gk + f − g ≥ jf (x0) − g(x0)j = 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¥ 8 f 2 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¥ 8 f 2 C [0, 1].