TITLE: Must a surjective isometry on a dual space have a pre-adjoint?
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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 $(x_{-n})_{n\in\mathbb Z\setminus\{0\}}$.)