Banach Spaces by Jeffrey M. Lemm

By Jeffrey M. Lemm

Hardbound.

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Extra resources for Banach Spaces

Sample text

Given two distinct subsets A, B of T , we have t h a t i]~ - ~]i- 1. Hence the distance between any two distinct elements of the set {e A I A E (T)} is 1. If T is infinite, then this set is uncountable and gco(T) is not separable. Remark. 16 b). 3 ( 0 ) Let T be a set. If T is finite then put ~~ :: c(T)-- c0(T):: ~(T). e. 9= {A C T O T \ A is finite}, and define c ( T ) - = {x e gco(T) ] x ( ~ ) converges}, 14 1. Banach Spaces e~ := c0(T) := {x e e~(T) I lim x ( ~ ) = 0}. Then c(T) an co(T) are closed vector subspaces of e~176 and therefore Banach spaces with respect to the induced norm.

Llxll ~ p(x). e. p and II-II ~r~ equivalent. Remark. The crucial step in the above proof was applying the Bolzano- It follows that all norms on IKn are equivalent. I Weierstrass Theorem, which, like Minkowski's theorem, no longer holds when the field considered is not locally compact. For example, the "norms" ...... ~ ~+, (~,~), ~v/~ ~ + ~ , ~ ~+, (~,9), ~ ~+v~/~l are not equivalent. In fact, if ( a n ) n ~ is a sequence in ~ converging to - v / 2 then lim lan + v/-2 91 i --O, n--+O0 but lira 4 ~ n - - ~ (X) + 1 -- 45 # 0.

E) Ilxll~ ~ IlXllq _~ Ilxllv for every x e gP(T). f) If T is infinite, then the restrictions to IK (T) of the norms [1" lip, I1" IIq, [1" I1~ are pairwise non-equivalent. a), b), c), and d) ]Z T . easy to see. e) The inequality I1~11~o < II~llq is trivial. For the second inequality, we may further assume that IIXllq = 1. Then I~(t)l q _< I~(t)l" for every t C T . - 1 _< I~(t)l ~ tET whenever n 9 IN and A is a subset of T containing precisely n elements. lt~ to IK (T) are not equivalent. I The next proposition generalizes the preceding examples.