By Joram Lindenstrauss

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**Extra resources for Classical Banach Spaces I: Sequence Spaces**

**Example text**

RITZ-tt zr(z)x:ll~e. P. P. Let T E L(X, Y) be compact and 1/2> e> O. The operator T*: y* -+ X* is also compact and hence there are {yt*}f=l in y** and {Xt}f=1 in X* so that lIy*lI~ 1. It follows that IIT*Y*-t~1 y:*(y*)x:ll~e, whenever IITx-l~l x:(X)Y:*II~e, whenever IIxll~ 1. This does not conclude the proof since the {Y:*}f=l are not necessarily contained in Y. We have to "push" the {yt*}f= 1 into Y. This is done by using the following lemma. 6 [90]. Let X be a Banach space, let D be a finite-dimensional subspace of X** and let £ > O.

I) The basis is boundedly complete. e. }~ 1 C X are such that lim x*(Yj) i exists for every x* E X* then there is ayE X such that x*(y)=lim X*(YI) i for every x* E X*). (iii) X does not have a subspace isomorphic to co. Proof. We assume, as we clearly may, that the unconditional constant of {x,,}:'= 1 is 1. Since Co is not w sequentially complete it is clear that (ii) => (iii). We shall prove now that (iii) => (i). Assume that the basis is not boundedly complete. xjll~ 1, for every n, but "~l a"x" does not converge.

We may clearly assume that x,#O for every i. Let {7J1}~1 be a sequence of positive scalars tending to 00 so that co L 7J,IIX,llllytll =C