Direct sum of normed spaces (Q1814401)

The unconditional basis constant of an unconditional basic sequence \(\{e_ k\}\) in a normed linear space is \[ K(\{e_ k\})=\sup\left\|\sum_{k=1}^ n t_ k a_ k e_ k\right\|/\left\|\sum_{k=1}^ n a_ k e_ k\right\|, \] where the supremum is taken over \((t_ 1,\dots,t_ n)\in\{-1,1\}^ n\), \((a_ 1,\dots,a_ n)\neq (0,\dots,0)\) and \(n\in\mathbb{N}\). A basic sequence is said to be \(r\)-unconditional if \(K(\{e_ k\})\leq r\). If \(\{e_ k\}\) is a 1-unconditional basic sequence in some Banach space \(X\) and \(X_ 1,X_ 2,\dots\), is a sequence of normed linear spaces then ``the direct sum of the spaces \(X_ i\) relative to the sequence \(\{e_ i\}\)'' is the set \[ \left\{x=(x_ 1,x_ 2,\dots)\in\prod_{i=1}^ \infty X_ i:\;\sum_{i=1}^ \infty \| x_ i\| e_ i\text{ is convergent in }X\right\}, \] which is a linear space, normed by \(\| x\|=\|\sum\| x_ i\| e_ i\|\). This direct sum contains natural isometric copies of each of \(X_ 1,X_ 2,\dots\), and if, for each \(i\), \(g_ i\) is an element of the copy of \(X_ i\) and \(\| g_ i\|=1\) then the sequence \(\{g_ i\}\) is equivalent to the sequence \(\{e_ i\}\). If the basic sequence \(\{e_ i\}\) is not 1-unconditional then no such construction is possible. The note establishes the following theorem: Let \(X_ 1,X_ 2,\dots,X_ n\) be infinite dimensional subspaces of a normed linear \(X\); then for any \(\varepsilon>0\) it is possible to choose elements \(x_ i\in X_ i\) with \(\| x_ i\|=1\) for \(i=1,\dots,n\) such that the unconditional basis constant of \(\{x_ i\}_ 1^ n\) is not greater than \(1+\varepsilon\).