## Must a surjective isometry on a dual space have a pre-adjoint?

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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.

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[deleted earlier comment as it was based on a misreading] – Yemon Choi Mar 3 2011 at 22:53

## 1 Answer

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 `$(x_{-n})_{n\in\mathbb Z\setminus\{0\}}$`.)

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Much simpler than the example I was thinking of – Yemon Choi Mar 3 2011 at 21:02
Thanks, but I am curious about what your example is. – Jonas Meyer Mar 3 2011 at 21:06
Actually, I realize that I'd misread the question, so please ignore my comment! – Yemon Choi Mar 3 2011 at 22:53
Very nice example!  Thank you very much. – edward-poon Mar 11 2011 at 4:20
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