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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 a non-reflexive Banach space, such as the Lebesgue space () of all bounded sequences, Riesz’s lemma does not hold for =. [ 5 ] However, every finite dimensional normed space is a reflexive Banach space, so Riesz’s lemma does holds for α = 1 {\displaystyle \alpha =1} when the normed space is finite-dimensional, as will now be shown.
A Banach space is reflexive if and only if its closed unit ball is (, ′)-compact; this is known as James' theorem. [3] If is a reflexive Banach space, then every bounded sequence in has a weakly convergent subsequence.
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 ...
The direct method may often be applied with success when the space is a subset of a separable reflexive Banach space. In this case the sequential Banach–Alaoglu theorem implies that any bounded sequence ( u n ) {\displaystyle (u_{n})} in V {\displaystyle V} has a subsequence that converges to some u 0 {\displaystyle u_{0}} in W {\displaystyle ...