0
$\begingroup$

A subspace $Y$ of $X$ is said to be semi M-ideal if $\exists$ a projection $P$ (not necessarily linear) from $X^*$ to $Y^\perp$ such that $\|x^*\|=\|Px^*\|+\|x^*-Px^*\|$. And also, $P(\lambda x^*+Py^*)=\lambda Px^*+Py^*$ , $\forall x^*, y^*\in X^*$.

It is known that $ker(\mathbb{1})$ as a subspace of $\ell_1$ is a semi M-ideal, where $\mathbb{1}\in \ell_\infty$. My question is what is the corresponding projection $P:\ell_\infty\to sp\{\mathbb{1}\}$ which satisfies above norm decomposition?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

For $a=(a_n) \in \ell_\infty$ let $m(a) = (\sup a_n + \inf a_n)/2$. Then $a\mapsto P(a)=m(a) 1$ is the projection you're looking for.

Dirk

Improve this answer
$\endgroup$
3
  • $\begingroup$ Welcome to MathOverflow Dirk! $\endgroup$
    – shane.orourke
    Commented Aug 18, 2018 at 20:57
  • $\begingroup$ That's for real valued $\ell_\infty$. Will you replace it with the circumecenter of the sets of values in complex case? I haven't check... $\endgroup$
    – Uri Bader
    Commented Aug 19, 2018 at 6:20
  • 1
    $\begingroup$ @Uri The complex case is much harder, see E. Behrends's paper in Math. Ann. 290, No. 3, 463-471 (1991). $\endgroup$
    – Dirk Werner
    Commented Aug 19, 2018 at 18:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .