Suppose X be a linear space. My question is: How can we say that X is not reflexive in any norm. Reflexive means X is linearly isometric to its second dual X''
Reflexive means more than linearly isometric to its second dual: reflexive means that the function J:X -> X'' given by J(x)(y) = y(x) is a bijective isomorphism. (There is also the James space where there is a linear isometry X -> X'', but J is not it; this space is *not* reflexive. We could ask the question: Is X reflexive in any equivalent norm to the one given for X? (Equivalent norms give the same topology, while inequivalent ones do not.) Rephrasing the question in terms of topologies then avoids the need to refer to equivalent norms: Are the weak and weak* topologies for X' the same? If they are different, then no matter what equivalent norm we choose for X (which give equivalent norms & the same topologies for each of X' and X''), this does not change the weak and weak* topologies of X'. So, for example, if X = L^1(0,1), X' = L^\infty(0,1), and the weak and weak* topologies of X' are *not* the same, so X is *not* reflexive in any equivalent topology.
@ David, Ok. you are referring to equivalent norms. I agreed. But actually my question was for some norm., ( that norm need not be equivalent to the given norm). In short we have to see whether X is not reflexive in any norm.
If X is a Banach space with respect to one of the norms, it is no longer a Banach space if you use an inequivalent norm (either that or you have to throw some elements out of X). For any normed vector space X, X' is a Banach space, and so is X''. So for X to be reflexive, X = X'' must be a Banach space, which in turn implies that we have to stay with equivalent norms.
In fact, there is a criterion of reflexivity: that a bounded and closed set is weakly compact, almost as in finite dimension. The criterion is independent on a norm or metric.
The following geometric characterization for the reflexivity of a Banach space X holds: X is reflexive if and only if every closed convex subset of X is proximinal (see "Geometric functional analysis and its applications" by Richard B. Holmes)..
Here is an answer to a related problem. Namely, i will try to show that if a linear space X has an algebraic countable Hamel basis, then there exists a locally convex topology on X which makes X reflexive. Take as the desired topology the finest locally convex topology on X (in this topology every linear functional is continuous, so that the topological linear dual X ' equals the algebraic dual X*). This locally convex space is a direct sum of a countable collection of "lines", so it is the strict inductive limit of a sequence of finite dimensional spaces. All these spaces are reflexive, so that their inductive limit X is reflexive too (see "Topological Vector Spaces" by H. H. Schaefer, third edition, Section 5.8, page 146). It is known that this locally convex topology cannot be generated by a norm. Moreover, this topology is not metrizable, but it is complete. The conclusion is valid for any strict inductive limit of a sequence of (complete) reflexive locally convex spaces. Thus we avoid the assumption on countable algebraic dimension on X.
If one speaks about not "too big" spaces, i.e. about those spaces X whose Hamel basis cardinality does not exceed continuum, then the complete answer to your question consists of two parts.
1. If the cardinality of Hamel basis of X is exactly continuum, one can use the fact that the Hamel basis of l2 is also of continuum cardinality and define a linear bijection T: X \to l2. Then in the norm ||x|| := ||Tx|| the space X will be isomorphic to l2 and hence reflexive.
2. If the cardinality of Hamel basis of X is infinite but smaller than continuum, then X is not isomorphic to any infinite-dimensional Banach space, so in particular it is not isomorphic to a reflexive normed space. Consequently in this case X is not reflexive in any norm.
I also have to comment about David E. Stewart words "If X is a Banach space with respect to one of the norms, it is no longer a Banach space if you use an inequivalent norm". This is a typical mistake which many people erroneously keep in their memory after a Functional Analysis course. I even have seen once a Functional Analysis course in which this statement is formulated as an exercise with a (wrong) sketch of the solution. The correct statement (which is a consequence of Banach's inverse operator theorem) contains more conditions. Namely the correct statement is:
"Let X be a Banach space with respect to the original norm and let p be another norm, in which (X, p) is also complete. Then, if there is a constant C > 0 such that ||x|| is less or equal to Cp(x) for all x in X, then p is equivalent to the original norm". The condition ||x|| \le Cp(x) may be substituted by ||x|| \ge Cp(x), but it cannot be dropped.
The reason why an additional condition is unavoidable here, is that two non-isomorphic Banach spaces X, Y may have the same Hamel basis cardinality, and consequently may possess a (discontinuous) linear bijection. Say, on l1 there is a norm (non-comparable with the original one) in which the space is isomorphic to l2.
That is why this effect is not related to the question asked by Prasanth G. Narasimha-Shenoi
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