Scottish Book Problem 155

This is one of the few problems from the Scottish book that are still open. In modern terminology, the problem is:

[Correction May 2026: Corrected sphere to closed ball]

Let $X$ and $Y$ be Banach spaces, and let $U\colon X\to Y$ be a bijective map with the following property: For every $x_0\in X$ there exists $\varepsilon>0$ such that for the closed ball $B(x_0,\varepsilon) := \{ x\in X : \|x-x_0\| \leq \varepsilon \}$ , the restriction $U|_{B(x_0,\varepsilon)}$ is isometric. Does it follow that $U$ is isometric?

It is noted in the Scottish Book that the answer is „yes“ whenever $U^{-1}$ is continuous, which is automatic if $Y$ is finite-dimensional, or if $Y$ has the property that for any two elements $y_1,y_2\in Y$ satisfying $y_2\neq 0$ and $\|y_1+y_2\|=\|y_1\|+\|y_2\|$ there exists $\lambda\geq 0$ such that $y_1=\lambda y_2$ .

[Update May 2026] In [1], Mori solved the problem under the additional assumption that X is separable, and even under the weaker assumption of surjectivity instead of bijectivity. He also explains that sphere in the Scottish book [2] means closed ball. (What one would call a sphere nowadays, namely the set $\{ x\in X : \|x-x_0\| = \varepsilon \}$ , is called the surface of a sphere in the Scottish book.)

[1] Mori. On the Scottish book problem 155 by Mazur and Sternbach. C. R. Math. Acad. Sci. Paris 362 (2024), 813–816.

[2] The Scottish Book. Mathematics from the Scottish Café with selected problems from the new Scottish Book. Second edition. Including selected papers presented at the Scottish Book Conference held at North Texas University, Denton, TX, May 1979. Edited by R. Daniel Mauldin. Birkhäuser/Springer, Cham, 2015. xvii+322 pp.

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