I was thinking the following problem while reading some functional analysis notes.
Is it possible to characterize the Hahn-Banach extensions (meaning, extensions with the same norm) of a functional in a Banach space $E$ to the double continuous dual $E^{**}$?
My intuition tells me that there aren't many. This is due to two facts:
Theorem 1 (Goldstine) The unit ball $B_E$ of $E$ is $w^*$-dense in the unit ball of $E^{**}$, $B_{E^{**}}$
Theorem 2 There is a canonical (H-B) extension of a functional $\varphi \in E^*$ given by the natural map $E^* \to E^{***}$. Even more so, this map splits.
Are there examples of spaces and functionals that admit many H-B extensions to its double dual?
Just for reference, the first example one would think is $c_0$ which has $c_0^{**}=\ell^\infty$. But in this case, it is well known that the H-B extension is always unique.
Edit: I'll extend the results I know a little in order to attain a possible answer.
Theorem 3 (Phelps) Given a closed subspace $Y$ of a Banach space $X$, every functional on $Y$ has a unique norm-preserving extension if and only if the distance from a functional $f \in X^*$ to $Y^\perp$ is attained uniquely, in the sense that there exists a unique $g \in Y^\perp$ such that $$ d(f,Y^\perp) = \|f-g\|$$
Form this is an easy exercise that the Hahn-Banach extension is always unique if and only if the dual space is strictly convex.
Using theorem 3 and theorem 2, it is also possible to observe the following:
The triple dual, $E^{***}$ splits $$ E^{***} = E^* \oplus E^\perp$$ Then every functional on $E$ admits unique H-B extension to $E^{**}$ if and only if the norm on $E^{***}$ satisfies $$\|f\|_{E^{***}} = \|f|_E\|_{E^*} + \| f - f_E \|_{E^\perp} $$
Is this always the case? Thanks in advance.
It took me some time to come back to this question but I've found an answer that somewhat satisfies me.
An immediate corollary is