# Given Banach / normed space $X$ find $E$ such that $E^* = X$, where $^*$ denotes dual space.

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?

• $c_0$ is itself complete but not a dual of a Banach space. – Henno Brandsma Sep 21 at 18:48
• 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. – s.harp Sep 21 at 18:52
• 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. – KeeperOfSecrets Sep 21 at 21:30
• @KeeperOfSecrets So there's a term for what I'm searching for and it's predual? – Viktor Glombik Sep 21 at 21:36
• @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. – Nate Eldredge Sep 21 at 23:13