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Let $M$ be a closed subspace of a Banach space $X$. Then we can identify $(X/M)^*$ with $M^\perp$ and $M^*$ with $X^*/M^\perp$.

Indeed, if $Q^*:X\to X/M$ is the quotient map, then $Q^*:M^*\to X^*$ is a linear isometry with range $M^\perp$. Moreover, if $J:M\to X$ is the embedding, then $J^*:X^*\to M^*$ has kernel $M^\perp$ and induces a surjective linear isometry $\widehat{J^*}:X^*/M^\perp\to M^*$.

Let $M$ and $N$ be closed subspaces of a Banach space $X$ with $N\subset M$. Then we can identify $(M/N)^*$ and $N^\perp/M^\perp$, I think.

Which operator induces the identification a similar way as in the above cases $N=\{0\}$ and $M=X$?

Added June 17: Is there a reference for this identification?

cross-posted to m.se

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  • $\begingroup$ I know little functional analysis, but how are you defining $M^\perp$ for a Banach space? Don't you need a scalar product for that to make sense? $\endgroup$ Commented Jun 16, 2021 at 8:46
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    $\begingroup$ @DenisNardin I'd guess $M^\perp$ is the annihilator of $M$ in this case. $\endgroup$ Commented Jun 16, 2021 at 9:17
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    $\begingroup$ This does not belong to MO. $\endgroup$
    – user160032
    Commented Jun 16, 2021 at 12:38
  • $\begingroup$ No answers in Mathematics. $\endgroup$ Commented Jun 16, 2021 at 13:38
  • $\begingroup$ This question has only been on MSE for 4 days... give people some time. I suggest that if no one answers in a week or two, then re-post here. $\endgroup$
    – Amir Sagiv
    Commented Jun 20, 2021 at 1:53

1 Answer 1

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As is so often the case, the isomorphism is the only reasonable map you can write down in the general case:

$N^\perp / M^\perp \to (M/N)^\ast, f+M^\perp \mapsto (m+N \mapsto f(m))$

All that's left to prove is that it actually works ;-)

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