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Is $E$ a normed banach space and is $F\subseteq E$ a closed, complemented subspace of $E$, then is $F^\bot=\{f\in E'\colon f_{|F}= 0\}$ a closed, complemented subspace of $E'$.

I do not know if I translate "complemented" right. So here is the definition:

Is $E$ a banach space, then is a closed linear subspace $F\subseteq E$ "complemented", if it exits a closed, linear subspace $G\subseteq E$ such that $E=F+G$ and $G\cap F=\{0\}$.

I appreciate any kind of help. Thanks in advance.

Edit: There was a mistake in the task, which corrected now. $E$ is a banach space.

Cornman
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  • I think you mean that $F$ has a closed complement subspace, also called a topological complement. – Gribouillis Dec 18 '17 at 13:18
  • Thank you. I edit it. – Cornman Dec 18 '17 at 13:20
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    I think you destroyed the question: it should probably be $F$ has a closed complement subspace, instead of $F$ is a closed complement subspace. Same thing for $F^\bot$. – Gribouillis Dec 18 '17 at 13:40
  • Yes, the wording of the question is very confusing. I think a functional analysis expert would know what it is supposed to be, but I can't for the life of me figure out what it means. – jdods Dec 18 '17 at 13:45
  • Excuse me for the confusion. Is the phrasing better now? What the task should say is, that $F$ has the desired property which we want to show for $F^\bot$ – Cornman Dec 18 '17 at 14:10
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    I think you mean "complemented" subspace. A subspace $F$ of a topological vector space is called complemented if it has a complement in the topological sense, i.e. there is a subspace $G$ of $E$ such that the map $(f,g) \mapsto f+g$ from $F\times G$ to $E$ is a topological isomorphism. – Daniel Fischer Dec 18 '17 at 14:19
  • Yes, that should be it! – Cornman Dec 18 '17 at 21:23
  • There was a mistake in the given task. – Cornman Dec 21 '17 at 10:10

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