0
$\begingroup$

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!

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

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.