# weak*-closed subspaces

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 $$X$$.

No. You get $Y^{**}=Y^{\perp\perp}$ complemented in $X^{**}$ and $Y$, being a dual space, is norm one complemented in $Y^{**}$.
• 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.) Mar 7, 2012 at 2:57
• In fact, we don't even need the assumption that $X$ is complemented in its second dual. Mar 7, 2012 at 2:58