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 2There 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.