Timeline for A property for maps between metric spaces
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 18, 2023 at 14:26 | review | Close votes | |||
| Oct 19, 2023 at 22:15 | |||||
| Oct 18, 2023 at 13:56 | answer | added | Joseph Van Name | timeline score: 3 | |
| Oct 18, 2023 at 13:37 | vote | accept | gm01 | ||
| Oct 18, 2023 at 9:24 | comment | added | Vladimir Zolotov | More generally, for $f:R^n \rightarrow R^n$ surjective. Any $3$ points sitting on the line(= not sitting on any shere) will end up mapping into the same line(= not on any sphere). Thus, $\phi(x/2) = \phi(x)/2$. | |
| Oct 18, 2023 at 8:50 | comment | added | Vladimir Zolotov | @gm01 In the case $n = 2$ and if $f$ is assumed to be onto this implies that any $3$ points sitting on the line will end up mapping into the same line. Which implies $\phi(x/2) = \phi(x)/2$ in Pietro's terms. | |
| Oct 18, 2023 at 7:28 | comment | added | Pietro Majer | Suppose $X$ is “connected by folding rules”, meaning that for any $r>0$ and $x,x’$ in $X$ there are $x=x_0,x_1,\dots, x_m=x’$ such that $d(x_i,x_{i+1})=r$ (this is the case of any normed space $X$). Then since such an $f$ maps isosceles triangles to isosceles triangles, it also verifies $d(f(x),f(y))=\phi(d(x,y))$ for some increasing function $\phi$ (and subadditive, at least for the case of $X$ a normed space). | |
| Oct 18, 2023 at 6:40 | comment | added | Pietro Majer | So in general such an $f:X\to Y$ maps any sphere $\{y\in X: d(x,y)=r\}$ of $X$ to a sphere of $Y$ centered at $f(x)$. | |
| Oct 18, 2023 at 5:00 | comment | added | gm01 | I think this argument can be generalized to $\mathbb{R}^n$. Indeed, if $f: \ \mathbb{R}^n \rightarrow \mathbb{R}^n$ has this property, then it has to preserve circumcenters of simplices. By this I mean that if $c$ is the circumencter of a (non-degenerate) simplex with vertices $v_1, ..., v_{n+1}$, then $f(c)$ is the circumcenter of the simplex with vertices $f(v_1), ..., f(v_{n+1})$. Does this imply that $f$ is affine? | |
| Oct 18, 2023 at 4:43 | comment | added | YCor | @PietroMajer no, it's much stronger. It indeed implies monotone, say, non-decreasing (up to compose with $x\mapsto -x$). Taking $t>0$, $y=x-t$ and $z=x+t$ and then exchanging $y$ and $z$ shows $f(x+t)-f(x)=f(x)-f(x-t)$. By continuity, this seems to force $f$ to be affine. I.e., in this case, we have only similarities (considering constant as $0$-similarities). | |
| Oct 17, 2023 at 19:24 | comment | added | Pietro Majer | In the case $X=Y=\mathbb R$, this property is "$f$ is monotone", isn't it? | |
| Oct 17, 2023 at 17:26 | answer | added | Iosif Pinelis | timeline score: 1 | |
| Oct 17, 2023 at 13:56 | comment | added | YCor | One can wonder which 3-ary relations $R$ arise in this way from a metric space. Obvious conditions: are $R(x,x,y)$, $R(x,y,y)$ hold for all $x,y$, transitivity of the binary relation $R(x,-,-)$ for each $x$, and $R(x,y,x)$ implies $x=y$. | |
| Oct 17, 2023 at 13:49 | comment | added | YCor | Note that this means preserving the 3-ary relation $R(x,y,z)$ defined by "$d(x,y)\le d(x,z)$". A metric space can be characterized by a countable number of binary relations (namely the relations $d(x,y)\le r$ for $r$ ranging over rationals), but not, as far as I know, a single relation. | |
| Oct 17, 2023 at 13:48 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals, added tag
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| Oct 17, 2023 at 13:35 | history | asked | gm01 | CC BY-SA 4.0 |