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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 $X$.

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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 $Y^{**}$.

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  • $\begingroup$ Dear Professor Johnson, that is a nice observation. Thank you. $\endgroup$
    – Denis Poulin
    Commented Mar 7, 2012 at 2:28
  • $\begingroup$ 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.) $\endgroup$
    – Yemon Choi
    Commented Mar 7, 2012 at 2:57
  • $\begingroup$ In fact, we don't even need the assumption that $X$ is complemented in its second dual. $\endgroup$
    – Yemon Choi
    Commented Mar 7, 2012 at 2:58
  • $\begingroup$ Right, Yemon: If X is complemented in any dual space, then it is complemented in its bidual. $\endgroup$
    – Bill Johnson
    Commented Mar 7, 2012 at 21:24

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