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We say a mapping $f:\mathbb R^n\to \mathbb R^n$ be 1-Lipschitz with respect to a norm $\|\cdot\|$ if $\|f(x)-f(z)\|\le\|x-z\|$ holds for all $x,z\in\mathbb R^n$. Such a mapping are sometimes called a contraction.

I want to study the existence of non-trivial distance-preserving mappings for general norm, such that the inequality becomes an equality, i.e. $\|f(x)-f(z)\|=\|x-z\|$ for all $x,z\in \mathbb R^n$.

Obviously, there are several trivial mappings that satisfy the above condition:

  • Identity mapping $f(x)=x$ and its negative counterpart $f(x)=-x$ (or generally take the negative for certain elements);
  • Addition: $f(x)=x+b$ for all $b\in\mathbb R^n$;
  • Permutation of axis, e.g. $f(x)=Ax$ where $A$ is a permutation matrix.

Besides these trivial mapping (and their composition), I wonder if there are other distance-preserving mappings given a norm $\|\cdot\|$. As an example, if $\|\cdot\|$ is the ordinary Euclidean norm, then any orthogonal transformation satisfies $\|f(x)-f(z)\|=\|x-z\|$, which includes rotation, reflection, etc. However, for other norms, I cannot easily figure out such a non-trivial mapping. So my questions are:

  • The most easiest question: for $\ell_p$ norm when $p\neq 2$, does there exist a non-trivial $f$ such that $\|f(x)-f(z)\|_p=\|x-z\|_p$? (I believe the answer is no despite the lack of a formal proof.)
  • A more general question: if the answer of the first question if no, for any norm $\|\cdot\|$ which is not $\ell_2$, does there exist a non-trivial $f$ such that $\|f(x)-f(z)\|=\|x-z\|$? (Update: as discussed in the comment, it may be hard to make this question well-defined. One can just ignore this question.)
  • A further generalization: when using different norms for the input and the output, denoted as $\|\cdot\|_I$ and $\|\cdot\|_O$ respectively, does there exist a non-trivial $f$ such that $\|f(x)-f(z)\|_O=\|x-z\|_I$? In particular, I am interested in the case when $\|\cdot\|_I$ and $\|\cdot\|_O$ are $\ell_p$ norm and $\ell_q$ norm, respectively, or at least $\|\cdot\|_I$ is the $\ell_p$ norm and $\|\cdot\|_O$ arbitrary.
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    $\begingroup$ 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. $\endgroup$
    – Narutaka OZAWA
    Commented Feb 27, 2022 at 7:07
  • $\begingroup$ 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). $\endgroup$
    – zbh2047
    Commented Feb 27, 2022 at 7:36
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    $\begingroup$ Here's the answer for the $\ell^p$-norms. You'll probably also be interested in having a look at the Ulam Mazur theorem. $\endgroup$
    – Jochen Glueck
    Commented Feb 27, 2022 at 10:23
  • $\begingroup$ @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) $\endgroup$
    – zbh2047
    Commented Feb 27, 2022 at 12:55
  • $\begingroup$ @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. $\endgroup$
    – Jochen Glueck
    Commented Feb 27, 2022 at 16:45

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