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

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 '11 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 '11 at 21:02
Thanks, but I am curious about what your example is. –  Jonas Meyer Mar 3 '11 at 21:06
Actually, I realize that I'd misread the question, so please ignore my comment! –  Yemon Choi Mar 3 '11 at 22:53
Very nice example! Thank you very much. –  user13391 Mar 11 '11 at 4:20