**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 $\|(x\_j)\|=\|(\|x\_j\|)\|\_E$. 


**Question**


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!



Absolute norms and 1-unconditional sums
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! 


