Background: Let $X$ be a Banach space. We know a linear map $h$ is a surjective isometry of $X$ if and only if its adjoint $h^\*$ is a surjective isometry of $X^\*$. 


In general, a linear map $g:X^\* \to X^\*$ need not have a pre-adjoint. But what if $g$ is a surjective isometry? Must there exist $f:X \to X$ such that $g=f^\*$? If we identify $X$ with its embedding in $X^{\*\*}$, this is equivalent to $g^\*(X) \subset X$. 


Sorry if this question is trivial; it seems like this should be well-known, but I haven't been able to find a reference or an easy counter-example.



Must a surjective isometry on a dual space have a pre-adjoint?
Let $X$ be the space of sequences, indexed by the nonzero integers, that tend to $0$ at $-\infty$ and to an arbitrary finite limit at $\infty$, with sup norm (a direct product of $c\_0$ and $c$). Then $X^\*$ can be identified with $\ell^1$ of $\mathbb{Z}$. If $f$ is in $\ell^1$, then the corresponding functional on $X$ sends $x$ to 
$\sum\_{n\neq0}f\_nx\_{n} + f\_0\cdot\lim\_{n\to\infty}x\_n$. The map on $\ell^1\cong X^\*$ defined by $(f\_n)\_{n\in\mathbb Z}\mapsto (f\_{-n})\_{n\in\mathbb{Z}}$ is a linear surjective isometry with no pre-adjoint. (If there were a preadjoint, it would have to send $(x\_n)\_{n\in \mathbb Z\setminus\{0\}}$ to $(x\_{-n})\_{n\in\mathbb Z\setminus\{0\}}$.)


