$E^*$ contains a subspace isometric isomorphic to $F^*$

Question

Let $E$ a nvs and $F \neq \{0\}$ a linear subspace. By Hahn-Banach theorem we know that $F^* \neq \emptyset$ and for every $f \in F^*$ the set $E_f := \{\text{ continuous linear extensions of f } \}$ is also non-empty. So by the axiom of choice there is a function $\Phi: F^* \to E^*$ such that for every $f \in F^*$ we have $\Phi(f) \in E^*$ a continuous linear extension of $f$. My question is if it is always possible to make $\Phi$ linear? So $E^*$ always contains some subspace that is isometric isomorphic to $F^*$.

If $F$ is dense we can guarantee the unicity of the Hahn-Banach extension so the linearity it is clear (because the sum of extesions is always an extension of the sum).

Answer 1

It is not true that $E^\ast$ always contains a subspace isometrically isomorphic to $F^\ast$. See Tomasz Kania's answer: Dual of a subspace. He points out that $E = C[0, 1]$ is a Banach space with a closed subspace $F$ isometrically isomorphic to $\ell_1$ (indeed containing an isometric copy of every separable Banach space), $E^\ast$ is weakly sequentially complete and $F^\ast$ is not, yet weak sequential completeness is passed down to closed subspaces. I think the claim that $C[0, 1]^\ast$ is weakly sequentially complete is very non-obvious and is elaborated on in this other answer of Kania: https://math.stackexchange.com/a/1502668/445167 with a source in the comments.

Another possible approach is via Existence of right inverse., which states that a surjective bounded $T : X \to Y$ has a bounded right-inverse if and only if $\ker(T)$ is complemented. Perhaps this view gives a different counterexample with $T$ the restriction operation $E^\ast \to F^\ast$. Hahn-Banach shows this to be surjective.