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Let $T : X \to Y$ be a linear isometry between normed spaces $X,Y$.

Must the dual map $T^* : Y^* \to X^*$ be an isometry?

Doug
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    Does your definition of isometry include surjectivity? – Daniel Fischer Aug 05 '13 at 21:26
  • No, it does not. – Doug Aug 05 '13 at 21:29
  • Might be helpful to see that the dual map in this case is just the restriction map to the range of $T$. Like Daniel said below, if the range of $T$ is not dense then the dual map is not injective (although surjectivity holds by Hanh-Banach). For example take $T$ to be the unilateral shift on $l^2$. – Michael Aug 06 '13 at 01:09
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    @Michael : How can we show surjectivity by Hanh-Banach ? – niki Sep 15 '14 at 13:20

1 Answers1

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Since an isometry need not be surjective, the answer is no. If the range of $T$ is not dense, then $T^\ast$ has nontrivial kernel, since

$$\operatorname{ker} T^\ast = \left(\operatorname{im} T\right)^\perp,$$

but an isometry is injective. If however the range is dense, then $T^\ast$ is an isometry.

Daniel Fischer
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  • Is this argument only valid for inner product spaces? – dafinguzman Feb 23 '14 at 06:11
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    @dafinguzman No, here $V^\perp$ denotes the annihilator of $V\subset Y$ in $Y^\ast$, $V^\perp = {\lambda\in Y^\ast : V\subset \ker\lambda}$. For Hilbert spaces, that corresponds to the orthogonal complement under the Riesz map, which probably is the reason for the use of the same notation. But as $T^\ast \colon Y^\ast\to X^\ast$ denotes the dual map, and $V^\perp$ the annihilator here, no inner product is involved. – Daniel Fischer Feb 23 '14 at 09:53