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A Banach space is super-reflexive if all Banach spaces finitely representable in are reflexive, or, in other words, if no non-reflexive space is finitely representable in . The notion of ultraproduct of a family of Banach spaces [ 14 ] allows for a concise definition: the Banach space X {\displaystyle X} is super-reflexive when its ultrapowers ...
In mathematics, more specifically in functional analysis, a Banach space (/ ˈ b ɑː. n ʌ x /, Polish pronunciation:) is a complete normed vector space.Thus, a Banach space is a vector space with a metric that allows the computation of vector length and distance between vectors and is complete in the sense that a Cauchy sequence of vectors always converges to a well-defined limit that is ...
Tsirelson space, a reflexive Banach space in which neither nor can be embedded. W.T. Gowers construction of a space X {\displaystyle X} that is isomorphic to X ⊕ X ⊕ X {\displaystyle X\oplus X\oplus X} but not X ⊕ X {\displaystyle X\oplus X} serves as a counterexample for weakening the premises of the Schroeder–Bernstein theorem [ 1 ]
In mathematics, the Milman–Pettis theorem states that every uniformly convex Banach space is reflexive.. The theorem was proved independently by D. Milman (1938) and B. J. Pettis (1939).
The topological dual of -Banach space deduced from by any restriction scalar will be denoted ′. (It is of interest only if is a complex space because if is a -space then ′ = ′. James compactness criterion — Let X {\displaystyle X} be a Banach space and A {\displaystyle A} a weakly closed nonempty subset of X . {\displaystyle X.}
If X is a reflexive Banach space, then every completely continuous operator T : X → Y is compact. Somewhat confusingly, compact operators are sometimes referred to as "completely continuous" in older literature, even though they are not necessarily completely continuous by the definition of that phrase in modern terminology.
The unit sphere can be replaced with the closed unit ball in the definition. Namely, a normed vector space is uniformly convex if and only if for every < there is some > so that, for any two vectors and in the closed unit ball (i.e. ‖ ‖ and ‖ ‖) with ‖ ‖, one has ‖ + ‖ (note that, given , the corresponding value of could be smaller than the one provided by the original weaker ...
Indeed, the elements of define a pointwise bounded family of continuous linear forms on the Banach space := ′, which is the continuous dual space of . By the uniform boundedness principle, the norms of elements of S , {\displaystyle S,} as functionals on X , {\displaystyle X,} that is, norms in the second dual Y ″ , {\displaystyle Y'',} are ...