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Let $$(S,\|\cdot\|) = \{(x,y)\in \mathbb{R}^2: \|(x,y)\| =1\},$$ that is, $S$ is the collection of all norm one vectors in $\mathbb{R}^2$ with respect to the norm $\|\cdot\|.$

Question: Let $\|\cdot\|_X$ and $\|\cdot\|_Y$ be two norms on $\mathbb{R}^2$ be such that $(S,\|\cdot\|_X)$ and $(S,\|\cdot\|_Y)$ are isometric. Does there exist a bijective isometry $T:(S,\|\cdot\|_X)\to (S,\|\cdot\|_Y)$ such that $$\|Tu-\alpha Tv\|_Y \leq \|u-\alpha v\|_X$$ for all $u,v\in (S,\|\cdot\|_X)$ and all $\alpha>0?$

Note that the norms $\|\cdot\|_X$ and $\|\cdot\|_Y$ may be distinct.

I tried $\|(x,y)\|_X = |x|+|y|,$ $\|(x,y)\|_Y = \max\{|x|,|y|\}$ and $$T(x,y) = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}.$$ Note that $T$ is a rotation matrix. Clearly $T$ is a bijective isometry and satisfies the inequality.

However, I do not know whether the same holds for general $\|\cdot\|_X$ and $\|\cdot\|_Y.$

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  • $\begingroup$ Obviously in general the unit circles for the two norms are not isometric. $\endgroup$ – YCor Jan 6 at 16:23
  • $\begingroup$ @YCor I have edited my question to indicate my intention. What I want is that if two spheres are isometric, can we find a specific bijective isometry that satisfies the inequality above. $\endgroup$ – Idonknow Jan 6 at 16:42
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    $\begingroup$ I think that what you are asking is a version of the following well-known question of D. Tingley [Geom. Dedicata 22 (1987) 371–378]: For a bijective isometry $f: S_X → S_E$ between the unit spheres of two real Banach spaces $X$ and $E$, is it true that $f$ extends to a linear isometry $F: X → E$ of the corresponding spaces? There is an extensive literature on this matter, see, for example Kadets-Martin [J. Math. Anal. Appl. 396 (2012), no. 2, 441–447] $\endgroup$ – August Cleaner Jan 6 at 19:25
  • $\begingroup$ @AugustCleaner thanks for the reference. $\endgroup$ – Idonknow Jan 6 at 23:17

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