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Absolute norm

Let $X$ and $Y$ be Banach spaces. Let $Z=X\times Y$ a norm $\|\cdot\|_N$ on $Z$ is called absolute if there is a function $N\colon R^2\rightarrow R$ such that $$ \|(x,y)\|_N=N((\|x\|, \|y\|)) \qquad \text{ for all } z=(x,y)\in Z. $$

For example, the $\ell_p$-norms are absolute norms.

1-unconditional sum

Let $E$ be a Banach space with a 1-unconditional normalized Schauder basis. We can think of the elements of $E$ as sequences with the property that $$ \|(a_1,a_2,\dots)\|_E=\|(|a_1|,|a_2|,...\|_E \qquad \text{ for all } (a_j)\in E. $$ Note that $E$ is naturally endowed with the structure of a Banach lattice with respect to the pointwise operations.

Suppose that $X_1, X_2,\dots$ are Banach spaces. Their $E$-sum $X=(X_1, X_2, \dots)_E$ consists of all sequences $(x_j)$ with $x_j\in X_j$ and $(\|{x_j}\|)\in E$ with the norm $\|(x_j)\|=\|(\|x_j\|)\|_E$.


Let $Z=X_1\times X_2\times...$. Can I equip $Z$ with an absolute norm? If so is this norm equivalent to equipping $Z$ with an 1-unconditional norm?

Thanks in advance!

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If you mean the unreatricted direct product, then I think the answer is no: just take each X_i to be one-dimensional. –  Yemon Choi Jan 6 '13 at 18:43
Also on stackexchange: math.stackexchange.com/q/271686 –  Martin Jan 6 '13 at 22:01
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1 Answer

up vote 1 down vote accepted

What Yemon says is correct. The "right" space to use is the space $Y$ of all elements in $Z$ such that only finitely many terms are non zero--you can always complete at the end.

Unconditional sums can be much more complicated than absolute sums. In an absolute sum, if you have linear operators $T_n$ on $X_n$ s.t. $\sup_n \|T_n\| < \infty$ and define $T$ on $Y$ by $Tx = (T_n x(n))$ for $x=(x(n))$ in $Y$, then $T$ is a bounded linear operator on $Y$. This is not true for unconditional sums. In fact, the Kalton-Peck space is (the completion of) an unconditional sum of a sequence of 2-dimensional spaces (which can all be taken to be two dimensional Hilbert spaces) and yet does not have an unconditional basis!

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