
Dual Banach algebras
... If E is a reflexive Banach space with the approximation property, then B(E) is Connes-amenable if and only if K (E), the algebra of compact operators, is amenable. So B(`p ) is Connes-amenable for 1 < p < ∞. ...
... If E is a reflexive Banach space with the approximation property, then B(E) is Connes-amenable if and only if K (E), the algebra of compact operators, is amenable. So B(`p ) is Connes-amenable for 1 < p < ∞. ...
Specialist Mathematics Glossary
... and a false conclusion has been reached the only thing that could be wrong is the initial assumption. Therefore the original statement is true. For example: the result 2 is irrational can be pro ...
... and a false conclusion has been reached the only thing that could be wrong is the initial assumption. Therefore the original statement is true. For example: the result 2 is irrational can be pro ...
Abstract Vector Spaces, Linear Transformations, and Their
... (4) By Theorem 1.6 span(S ∩ T ), span(S) and span(T ) are subspaces, and by Theorem 1.4 span(S) ∩ span(T ) is a subspace. Now, consider x ∈ span(S ∩ T ). There exist vectors v1 , . . . , vn ∈ S ∩ T and scalars a1 , . . . , an ∈ F such that x = a1 v1 + · · · + an vn . But since v1 , . . . , vn belong ...
... (4) By Theorem 1.6 span(S ∩ T ), span(S) and span(T ) are subspaces, and by Theorem 1.4 span(S) ∩ span(T ) is a subspace. Now, consider x ∈ span(S ∩ T ). There exist vectors v1 , . . . , vn ∈ S ∩ T and scalars a1 , . . . , an ∈ F such that x = a1 v1 + · · · + an vn . But since v1 , . . . , vn belong ...
Extension of the semidefinite characterization of sum of squares
... Both examples suggest that we should consider constraints of the form f (x) ∈ K ∀ x ∈ S, where f is a (perhaps multivariate) vector valued function, and K is some convex cone. As in the motivating one-dimensional case, this will generally be an intractable constraint, but we can try to find a tracta ...
... Both examples suggest that we should consider constraints of the form f (x) ∈ K ∀ x ∈ S, where f is a (perhaps multivariate) vector valued function, and K is some convex cone. As in the motivating one-dimensional case, this will generally be an intractable constraint, but we can try to find a tracta ...
Local convexity on smooth manifolds1,2,3 T. Rapcsák4
... extended to an open interval (t1 , t2 ), thus the Γ-convexity of the single variable C 2 ...
... extended to an open interval (t1 , t2 ), thus the Γ-convexity of the single variable C 2 ...
the angle of an operator and positive operator
... 1. Introduction. Let - 4 ^ 0 and J B ^ O be two positive bounded selfadjoint operators. The two algebraic questions which immediately arise are: (1) is A+B^O; (2) is BA j^O? The first question (and its extension to accretive operators on a Banach space) has trivially an affirmative answer; the secon ...
... 1. Introduction. Let - 4 ^ 0 and J B ^ O be two positive bounded selfadjoint operators. The two algebraic questions which immediately arise are: (1) is A+B^O; (2) is BA j^O? The first question (and its extension to accretive operators on a Banach space) has trivially an affirmative answer; the secon ...
MAT 1341E: DGD 4 1. Show that W = {f ∈ F [0,3] | 2f(0)f(3) = 0} is not
... (a) Find a basis of W . (b) Give a geometric description of W . (c) Extend your basis in (a) to a basis of R3 . Solution: (a) : Let (x, y, z) ∈ W . Then (x, y, z) = (x, y, x + y) = x(1, 0, 1) + y(0, 1, 1). Therefore, span{(1, 0, 1), (0, 1, 1)} = W . If a(1, 0, 1) + b(0, 1, 1) = 0 for some a, b ∈ R, ...
... (a) Find a basis of W . (b) Give a geometric description of W . (c) Extend your basis in (a) to a basis of R3 . Solution: (a) : Let (x, y, z) ∈ W . Then (x, y, z) = (x, y, x + y) = x(1, 0, 1) + y(0, 1, 1). Therefore, span{(1, 0, 1), (0, 1, 1)} = W . If a(1, 0, 1) + b(0, 1, 1) = 0 for some a, b ∈ R, ...
For printing
... 4.2. THE PRODUCT THEOREM. If X is a quasi-regular Baire space and Y is pseudo-complete, then X x Y is a Baire space. Proof. As in 3.3 Lemma 1, choose open subsets XP and XA of X. Then (XP x 7 ) U (XA x Γ) is a dense subspace of X x Y. Since YP xY and XAxY are open subsets of X x Y and since XP x Y i ...
... 4.2. THE PRODUCT THEOREM. If X is a quasi-regular Baire space and Y is pseudo-complete, then X x Y is a Baire space. Proof. As in 3.3 Lemma 1, choose open subsets XP and XA of X. Then (XP x 7 ) U (XA x Γ) is a dense subspace of X x Y. Since YP xY and XAxY are open subsets of X x Y and since XP x Y i ...
HW6 - Harvard Math Department
... 2. Yaglom, problem 11 on page 66. You are not required to prove the theorems. Just state them and draw accurate diagrams to illustrate them. 3. Yaglom, problem 12 on page 66. Yaglom's hint is one way to construct a proof. Another is to view this as a special case of Ceva's theorem (see problem 2). T ...
... 2. Yaglom, problem 11 on page 66. You are not required to prove the theorems. Just state them and draw accurate diagrams to illustrate them. 3. Yaglom, problem 12 on page 66. Yaglom's hint is one way to construct a proof. Another is to view this as a special case of Ceva's theorem (see problem 2). T ...
ORTHOGONAL BUNDLES OVER CURVES IN CHARACTERISTIC
... We denote by φ and ψ generators of the one-dimensional spaces Hom(F∗ OX , A) and Hom(A, B −1 ). (iii) The restriction of the quadratic form det to the subbundle F∗ OX equals the evaluation morphism F ∗ F∗ OX −→ OX . In particular the restriction of β to F∗ OX is identically zero. (iv) Let x and y be ...
... We denote by φ and ψ generators of the one-dimensional spaces Hom(F∗ OX , A) and Hom(A, B −1 ). (iii) The restriction of the quadratic form det to the subbundle F∗ OX equals the evaluation morphism F ∗ F∗ OX −→ OX . In particular the restriction of β to F∗ OX is identically zero. (iv) Let x and y be ...