# If two spheres are isometric, does there exist a bijective isometry $T:S\to S$ with $\|Tu-\alpha Tv\|_Y \leq \|u-\alpha v\|_X$ for all $\alpha>0?$

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 $$\|\cdot\|_Y.$$

• Obviously in general the unit circles for the two norms are not isometric. – YCor Jan 6 at 16:23
• @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. – Idonknow Jan 6 at 16:42
• 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] – August Cleaner Jan 6 at 19:25
• @AugustCleaner thanks for the reference. – Idonknow Jan 6 at 23:17