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Recall that a closed subspace $Y$ of a Banach space $X$ is weakly complemented if the set $$Y^{\bot}:= \{ f\in X^*| f(y) = 0 \forall y\in Y\}$$ is a complemented subspace of $ X^*$. For example, $c_0$ is a weakly complemented subspace of $l_{\infty}$.

Question: Is there a Banach space $X$ such that there is a weak${}^*$-closed subspace $Y$ which is weakly complemented but not complemented in $X$.

edited Mar 6 2012 at 23:43

Yemon Choi
8,608●1●20●45

asked Mar 6 2012 at 21:18

Denis Poulin
174●4

Bill Johnson · Accepted Answer · 2012-03-07 02:01:23Z UTC

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No. You get $Y^{**}=Y^{\perp\perp}$ complemented in $X^{**}$ and $Y$, being a dual space, is norm one complemented in $Y^{**}$.

answered Mar 7 2012 at 2:01

Bill Johnson
14.5k●1●33●60

	Dear Professor Johnson, that is a nice observation. Thank you. – Denis Poulin Mar 7 2012 at 2:28
	As an aside, this would seem to show that one can relax the hypotheses that $X$ is a dual space and $Y$ a weakly complemented, weak-star closed subspace to: $X$ is complemented in its second dual, $Y\subseteq X$ is complemented in its second dual, and $Y^{}$ is complemented in $X^{}$. (This might be of interest when $X$ is, for instance, the predual of a von Neumann algebra.) – Yemon Choi Mar 7 2012 at 2:57
	In fact, we don't even need the assumption that $X$ is complemented in its second dual. – Yemon Choi Mar 7 2012 at 2:58
	Right, Yemon: If X is complemented in any dual space, then it is complemented in its bidual. – Bill Johnson Mar 7 2012 at 21:24

weak*-closed subspaces

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