1
$\begingroup$

I am interested in 'algebraic-density'-type properties of second adjoint operators in the algebra of bounded operator on a second dual of a Banach space. Incidentally, I have a problem with justification of the following question:

Say we have a non-quasi reflexive Banach space $V$ (that is $V^{\prime\prime}/V $ is infinite-dimensional).

Pick $y^{\prime\prime}\in V^{\prime\prime} \setminus V$. Is there a (left-) invertible operator $S\colon V\to V$ with $S^{\prime\prime} y^{\prime\prime} \in V$?

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
2
$\begingroup$

You want $S$ to be an isomorphism from $V$ onto a subspace, say $X$. (You also want $X$ to be complemented, but that is irrelevant for the answer.) This implies that $S^{''}$ is an isomorphism from $V^{''}$ onto

$X^{\perp\perp} \subset V^{''}$. So no operator like you want exists on any non reflexive Banach space.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Nitpick: you need $V''$ not $V$ in the 2nd line. $\endgroup$
    – Matthew Daws
    Commented Jan 29, 2012 at 20:34
  • $\begingroup$ And of course, for this question, this is a much (much) better answer than mine! Does the question over at math.stackexchange have a less sledge-hammer-esq counter-example? $\endgroup$
    – Matthew Daws
    Commented Jan 29, 2012 at 20:36
  • 1
    $\begingroup$ I don't think so, Matt--the SE question is much subtler. Note that every decomposable Banach space admits a decomposition of the identity like the proposer of the SE questions wants, so at the least you need a non decomposable Banach space. $\endgroup$
    – Bill Johnson
    Commented Jan 29, 2012 at 20:43
  • $\begingroup$ That last is not quite accurate--an example must have the property that it cannot be written as the direct sum of two non reflexive closed subspaces, which is formally weaker than saying that the space is indecomposable. $\endgroup$
    – Bill Johnson
    Commented Jan 29, 2012 at 23:59

You must log in to answer this question.

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