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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 $S^{\prime\prime} y^{\prime\prime} \in V$?

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This is very similar to a math.stackexchange question:… My answer their is easily adapted to show that the answer to this question is "no". – Matthew Daws Jan 29 '12 at 18:05

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.

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Nitpick: you need $V''$ not $V$ in the 2nd line. – Matthew Daws Jan 29 '12 at 20:34
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? – Matthew Daws Jan 29 '12 at 20:36
Thanks for catching the typo, Matt. I corrected it. – Bill Johnson Jan 29 '12 at 20:36
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. – Bill Johnson Jan 29 '12 at 20:43
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. – Bill Johnson Jan 29 '12 at 23:59

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