Exercise 14, § 5
Topological Vector Spaces, IV, p. 69
Let be a Banach space satisfying the first axiom of countability. A sequence of elements of is said to be a Banach basis if the following property is satisfied : for every , there exists a unique sequence of scalars such that , where the series on the right hand side is convergent. a) Show that the family is total and free. Let be the vector subspace (closed) of generated by the for indices , and let be the projector from onto defined by . Show that the are continuous linear mappings and that . (Consider the norm on defined by ; show that is complete for this norm and deduce that it is equivalent to the given norm on (cf. I, p. 17, th. 1).) Show that for every pair of integers , the norm of the projection of onto , which is parallel to the vector subspace generated by the for indices such that , is bounded by a number independent of . b) Conversely, let be a total sequence of elements of , which is a free family and is such that the norms of the projections for are bounded by a number independent of and . Show that is a Banach basis of . (First prove that the sequence is topologically free and define the projectors ; then for all . Next observe that if is the distance of a point from , then .)[^1]
[^1]: It is clear that the existence of a Banach basis in implies that satisfies the first axiom of countability. But there are examples of Banach spaces satisfying the first axiom of countability in which there does not exist a Banach basis (P. Enflo, Acta Math., t. CXXX (1973), p. 309-317).
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