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A closed subspace $E$ of a Banach space $X$ is said to complemented if there exists a closed subspace $F$ of $X$ such that $E+F=X$ and $E\cap F=\{0\}$. I was thinking whether $X$ is complemented in $X^{**}$?

Eric Wofsey
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NewB
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1 Answers1

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No. For example $c_0$ is not complemented in its double dual $l^\infty$ (with detailed reasons here). But it is true that $X^*$ is always complemented in $X^{***}$. (and by this $c_0$ cannot be isomorphic to a dual space)

Nick
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    Can you remark how one can see $X^$ being complemented in $X^{}$? Is this an iff statement? (ie does there exist a Banach space without a pre-dual but that is complemented in $X^{}$?) – s.harp Aug 24 '19 at 19:38
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    The transposed of the inclusion into the second dual is a projection. $L^1 [0,1]$ is not a dual space but complemented in its bidual. – Jochen Aug 26 '19 at 18:12
  • @s.harp: On the one hand, you have the natural embedding $i : X^* \hookrightarrow X^{*}$. On the other hand, the natural embedding $j : X \hookrightarrow X^{}$ induces the natural projection ("pull-back") $p : X^{*} \twoheadrightarrow X^$. The fact that $X^$ is both a subspace and a quotient space (isomorphic to $X^{} / \ker p$) of $X^{**}$ shows that it is a direct summand of it. – Alex M. Dec 24 '24 at 19:18