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Feb 28, 2022 at 1:23 comment added zbh2047 Thank you! I get it.
Feb 27, 2022 at 16:45 comment added Jochen Glueck @zbh2047: For the case of complex scalars, just consider the one-dimensional space $\mathbb{C}$ itself: complex conjugation is obviously a mapping which is isometric (i.e. distance-preserving) but not linear over $\mathbb{C}$. However, every normed space over $\mathbb{C}$ can, of course, also be considered as a normed space over $\mathbb{R}$, and on this space you can apply Ulam Mazur.
Feb 27, 2022 at 12:55 comment added zbh2047 @JochenGlueck Thanks! It seems that your answer along with Ulam Mazur theorem proves both questions 1 and 3. Do I understand correctly? Also, I notice that if $\mathbb R^n$ is replaced by $\mathbb C^n$, the mapping will not be restricted to be linear. Does it mean in this case there might exist other non-trivial distance-preserving mappings for general norm? (I guess problem may be hard in this setting)
Feb 27, 2022 at 10:23 comment added Jochen Glueck Here's the answer for the $\ell^p$-norms. You'll probably also be interested in having a look at the Ulam Mazur theorem.
Feb 27, 2022 at 7:50 history edited zbh2047 CC BY-SA 4.0
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Feb 27, 2022 at 7:43 history edited zbh2047 CC BY-SA 4.0
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Feb 27, 2022 at 7:36 comment added zbh2047 Thanks for pointing in out. I am aware that it may be hard to exclude such a case. Nevertheless, I am most interested in $L_p$ norm (and also the last question).
Feb 27, 2022 at 7:07 comment added Narutaka OZAWA The case of $L^p$ is already dealt with in Banach's monograph. I'm not sure what you mean by "non-trivial". A (signed) permutation isometry is somewhat non-trivial to me, because in general it's not obvious which one is isometric w.r.t. a given norm.
Feb 27, 2022 at 6:07 history asked zbh2047 CC BY-SA 4.0