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Suppose $X$ is a Banach space with dual space $X^*$. If $Y$ is a closed subspace of $X$, then $Y^\perp=\{x^*\in X^*: x^*(y)=0 \text{ for all } y\in Y\}$ is a closed subspace in $X^*$. I am wondering whether this correspondence is reversible. That is, if $Y^*$ is closed in $X^*$, is it true that $Y^*\simeq (X/M)^*$ for some closed subspace $M$ in $X$?

Thank you.

user122916
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    I may be missing something here, but, for example, $c_0$ is a closed subspace of $\ell_\infty$ that is not a dual space (that is, $c_0$ has no predual). – David Mitra Apr 21 '15 at 14:30

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Yes, it is. Let $M=\{x\in X: y^*(x)=0 \text{ for all }y^*\in Y^*\}$. This is sometimes denoted ${}^\perp Y^*$ and called the pre-annihilator of $Y^*$.

Every $y^*\in Y^*$ induces a linear functional on $X/M$ in a natural way: $y^*(x+M)=y^*(x)$ is well-defined.

Conversely, if $\phi$ is a linear functional on $X/M$, then its composition with the projection $\pi:X\to X/M$ defines $\phi\circ \pi$, an element of ${}^\perp Y^*$.

It is easy to see that two maps described in the two previous paragraphs are inverses of each other.

  • I see that $\phi \circ \pi$ is an element of $X^$, but I am wondering how we can say it is actually a member of $Y^$? – user122916 Apr 21 '15 at 20:58