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Tsirel son, It is impossible to imbed of into an arbitrary Banach space , Funkcional. MR [3] Florent Baudier, Metrical characterization of super-reflexivity and linear type of Banach spaces , Arch. MR [4] Florent P. Baudier, Quantitative nonlinear embeddings into Lebesgue sequence spaces , J. MR [5] F.

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Lancien, Tight embeddability of proper and stable metric spaces , Anal. Spaces 3 , MR [6] F. Baudier, N.

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  5. Kalton, and G. Lancien, A new metric invariant for Banach spaces , Studia Math. MR [8] B. Ferenczi, A uniformly convex hereditarily indecomposable Banach space , Israel J. MR [13] Steven C. MR [14] T. Johnson, A uniformly convex Banach space which contains no , Compositio Math. Godefroy, G.

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    Lancien, and V. MR [16] Robert C.

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    7. James, Bases and reflexivity of Banach spaces , Ann. MR [17] Robert C. Lovasoa Randrianarivony, does not coarsely embed into a Hilbert space , Proc. MR [19] N. Kalton, Coarse and uniform embeddings into reflexive spaces , Quart. Oxford 58 , no. MR [20] N. Kalton, Lipschitz and uniform embeddings into , Fund.

      MR [21] Nigel J. Lovasoa Randrianarivony, The coarse Lipschitz geometry of , Math. MR [22] Nigel J. MR [25] A. MR [29] Piotr W. When the scalars are real, this map is an isometric isomorphism. The extreme points of P K are the Dirac measures on K. In the commutative Banach algebra C K , the maximal ideals are precisely kernels of Dirac mesures on K ,.

      For every normed space X , there is a natural map,. It is indeed isometric, but not onto. If F X is surjective , then the normed space X is called reflexive see below. Using the isometric embedding F X , it is customary to consider a normed space X as a subset of its bidual.

      Here are the main general results about Banach spaces that go back to the time of Banach's book Banach and are related to the Baire category theorem. Therefore, a Banach space cannot be the union of countably many closed subspaces, unless it is already equal to one of them; a Banach space with a countable Hamel basis is finite-dimensional.

      The Banach—Steinhaus theorem is not limited to Banach spaces. This result is a direct consequence of the preceding Banach isomorphism theorem and of the canonical factorization of bounded linear maps. The normed space X is called reflexive when the natural map. This is a consequence of the Hahn—Banach theorem. Further, by the open mapping theorem, if there is a bounded linear operator from the Banach space X onto the Banach space Y , then Y is reflexive.

      Hilbert spaces are reflexive. More generally, uniformly convex spaces are reflexive, by the Milman—Pettis theorem. However, Robert C. Furthermore, this space J is isometrically isomorphic to its bidual. When X is reflexive, it follows that all closed and bounded convex subsets of X are weakly compact. In a Hilbert space H , the weak compactness of the unit ball is very often used in the following way: every bounded sequence in H has weakly convergent subsequences.

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      Weak compactness of the unit ball provides a tool for finding solutions in reflexive spaces to certain optimization problems. For example, every convex continuous function on the unit ball B of a reflexive space attains its minimum at some point in B. The following theorem of Robert C. James provides a converse statement. On every non-reflexive Banach space X , there exist continuous linear functionals that are not norm-attaining. The Banach space X is weakly sequentially complete if every weakly Cauchy sequence is weakly convergent in X. It follows from the preceding discussion that reflexive spaces are weakly sequentially complete.

      An orthonormal sequence in a Hilbert space is a simple example of a weakly convergent sequence, with limit equal to the 0 vector.

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      For every weakly null sequence in a Banach space, there exists a sequence of convex combinations of vectors from the given sequence that is norm-converging to 0. The unit ball of the bidual is a pointwise compact subset of the first Baire class on K. This applies to separable reflexive spaces, but more is true in this case, as stated below.

      The weak topology of a Banach space X is metrizable if and only if X is finite-dimensional. This applies in particular to separable reflexive Banach spaces. Although the weak topology of the unit ball is not metrizable in general, one can characterize weak compactness using sequences. A Banach space X is reflexive if and only if each bounded sequence in X has a weakly convergent subsequence.

      Banach spaces with a Schauder basis are necessarily separable , because the countable set of finite linear combinations with rational coefficients say is dense. They are called biorthogonal functionals.

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      Most classical separable spaces have explicit bases. The Schauder system is a basis in the space C [0, 1]. Since every vector x in a Banach space X with a basis is the limit of P n x , with P n of finite rank and uniformly bounded, the space X satisfies the bounded approximation property. The first example by Enflo of a space failing the approximation property was at the same time the first example of a separable Banach space without a Schauder basis. Robert C. James characterized reflexivity in Banach spaces with a basis: the space X with a Schauder basis is reflexive if and only if the basis is both shrinking and boundedly complete.

      Let X and Y be two K -vector spaces. There are various norms that can be placed on the tensor product of the underlying vector spaces, amongst others the projective cross norm and injective cross norm introduced by A. Grothendieck in In general, the tensor product of complete spaces is not complete again. Grothendieck proved in particular that [47]. Let X be a Banach space.

      When X has the approximation property , this closure coincides with the space of compact operators on X. For every Banach space Y , there is a natural norm 1 linear map. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Prove that every Hamel basis of X is uncountable. Ask Question. Asked 6 years, 11 months ago. Active 2 months ago.

      Viewed 13k times. Can anyone help how can I solve the above problem? Martin Sleziak An algebraic basis?

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      It's important at this point. Every finite-dimensional normed space is complete, see PlanetMath. A complete subspace of a normed space is closed.