Suppose $X$ is a nonreflexive Banach space, $Z$ a closed subspace of $X^*$, and $f$ a bounded functional on $Z$ with the property that there exists nonzero $x\in X$ such that $f(z^*)=z^*(x)$ for all $z^*\in Z$. Can $f$ be always extended to a functional on $X^*$ having the same property? If the answer is negative in general as I suspect, is there any characterization of the subspaces $Z$ (or/and functionals $f$) for which this would be true?
1 Answer
I do not understand your question: it seems that $x$ considered as a functional on $X^*$ will be the desired extension.
EDIT (in connection with the comment of Markus below): If you look for a normpreserving extension, then the answer is different. For all nonquasireflexive Banach spaces $X$ there exist total nonnorming subspaces $Z$ in $X^*$. Total means that they are weak$^*$ dense in $X^*$. Nonnorming means that the function $\sup\{z^*(x):z^*\in B(Z)\}$ takes arbitrarily small positive values of the unit sphere of $X$. This implies that there exist vectors in $X$ whose norms, as functionals on $Z$, can be very small, but the unique weak$^*$ continuous (by weak$^*$ density) extensions to $X^*$ have norm $1$.
See my survey for suitable references. I think that checking this literature one can answer majority of questions in this direction, at least in the separable case. For example, one can show that there is a normpreserving extension for each $x$ if and only if $Z$ is $1$norming.

$\begingroup$ I had a HahnBanach extension in mind. Is it clear that $x$ and $f$ must have the same norm? $\endgroup$– MarkusMar 29, 2017 at 5:36