The concept of the dual space is well known and in some cases computably up to an isometric isomorphism, i.e. $c_0^* \cong \ell_1$ or $\ell_p \cong \ell_q$.

What if we are given a space $X$ and want to find an space $E$ such that $E^* = X$ (or $\cong$)? What constrains do we have to put on $X$ to find such a space $E$?

Could this be a way to show that given $X$, it must be complete, since the dual space of a normed space is complete?

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    $\begingroup$ $c_0$ is itself complete but not a dual of a Banach space. $\endgroup$ – Henno Brandsma Sep 21 at 18:48
  • $\begingroup$ Re your point about using existence of a predual to verify completeness: If you find Cauchy sequences inelegant, then you might define: A normed space is Banach iff the canonical injection $X\to X^{**}$ has closed image. $\endgroup$ – s.harp Sep 21 at 18:52
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    $\begingroup$ An interesting neccesery condition is that $B_X$ must have extremal points (by Krein-Milman and Banach-Alaoglu theorems). You can use this to show that $L_1$ has no predual. $\endgroup$ – KeeperOfSecrets Sep 21 at 21:30
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    $\begingroup$ @KeeperOfSecrets So there's a term for what I'm searching for and it's predual? $\endgroup$ – Viktor Glombik Sep 21 at 21:36
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    $\begingroup$ @HennoBrandsma: The Krein-Milman argument suggested by KeeperOfSecrets also works to show $C[0,1]$ is not a dual, because the only extreme points of its ball are the constants $1$ and $-1$, and the ball is certainly not the closed convex hull of those. $\endgroup$ – Nate Eldredge Sep 21 at 23:13

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