Sphere covered by finite-dimensional faces, normed spaces: every nonexpansive bijection onto the ball is linear

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Let $X$ and $Y$ be real normed spaces. Neither space is assumed complete or finite-dimensional, and they may live in different universes. Write $B_X=\{x\in X:\|x\|\le 1\}$ and $S_X=\{x\in X:\|x\|=1\}$, and define $B_Y$ and $S_Y$ in the same way. Let $F\colon X\to Y$ be an arbitrary function defined on all of $X$. Assume the following four hypotheses.

1. (Injective on the ball.) For all $x,x'\in B_X$, $F(x)=F(x')$ implies $x=x'$.
2. (Onto the ball.) Every $y\in B_Y$ equals $F(x)$ for some $x\in B_X$, that is, $B_Y\subseteq F(B_X)$; the inclusion $F(B_X)\subseteq B_Y$ is not assumed.
3. (Nonexpansive on the ball.) $\|F(x)-F(x')\|\le\|x-x'\|$ for all $x,x'\in B_X$.
4. (Sphere covered by finite-dimensional sphere faces.) For every $y\in S_Y$ there is a set $C\subseteq Y$ with $y\in C$ such that: (a) $C$ is nonempty; (b) $C$ is convex; (c) $C\subseteq S_Y$; (d) $C$ is an extreme subset of $B_Y$, meaning $C\subseteq B_Y$ and, whenever $y_1,y_2\in B_Y$, $c\in C$ and $c=\lambda y_1+(1-\lambda)y_2$ for some $0<\lambda<1$, both $y_1$ and $y_2$ belong to $C$; (e) the linear subspace $\operatorname{span}\{c-c' : c,c'\in C\}$ (the direction space of the affine hull of $C$) is finite-dimensional.

Conclusion: there exists a map $T\colon X\to Y$ that is linear, bijective and norm-preserving, $\|T(x)\|=\|x\|$ for all $x\in X$, such that

Nothing is assumed or asserted about the values of $F$ outside $B_X$. No hypothesis says that $F$ maps $S_X$ into $S_Y$, and no covering hypothesis is made on $S_X$.

Flags

Lean conventions. None changes the meaning. Set.SurjOn F B_X B_Y means $B_Y\subseteq F(B_X)$, not a map into $B_Y$. LipschitzOnWith 1 is phrased with extended distances but is the ordinary inequality of hypothesis 3. vectorSpan ℝ C is the direction space of the affine hull, so (e) says the affine hull of $C$ is finite-dimensional, which is equivalent to the linear span of $C$ being finite-dimensional. X ≃ₗᵢ[ℝ] Y is a surjective linear isometry, so the conclusion includes that $X$ and $Y$ are isometrically isomorphic.

Splitting the bijection (purpose note). The choice is correct. Hypotheses 2 and 3 alone give $F(B_X)\subseteq B_Y$: if $z\in B_X$ had $\|F(z)\|>1$, the point $u=-F(z)/\|F(z)\|\in S_Y$ would have a preimage $p\in B_X$, and $\|F(z)-F(p)\|=\|F(z)\|+1>2\ge\|z-p\|$ would contradict hypothesis 3. Hence hypotheses 1 to 3 together say exactly that the restriction of $F$ to $B_X$ is a nonexpansive bijection of $B_X$ onto $B_Y$, with no loss against a bijection formulation. The same two hypotheses also give $F(0)=0$: if $F(0)\ne 0$, a preimage $p\in B_X$ of $-F(0)/\|F(0)\|$ gives $\|F(0)-F(p)\|=1+\|F(0)\|>1\ge\|p\|$.

Degenerate cases. If $X=Y=\{0\}$ the statement holds trivially. If $X=\{0\}\ne Y$, hypothesis 2 fails, and if $Y=\{0\}\ne X$, hypothesis 1 fails, so both cases are vacuous. Hypothesis 4 holds automatically when $Y$ is finite-dimensional (take $C$ to be the minimal face of $B_Y$ containing $y$) and when $Y$ is strictly convex (take $C=\{y\}$). The hypotheses are satisfiable, for example by $F=\mathrm{id}$ on $\mathbb R$, so the statement is not vacuous.

Hypotheses, one at a time.

• Hypothesis 1 is needed: $X=(\mathbb R^2,\|\cdot\|_\infty)$, $Y=\mathbb R$, $F(z)=z_1$ satisfies 2 to 4 (faces $\{\pm1\}$), but no linear bijection $\mathbb R^2\to\mathbb R$ exists.
• Hypothesis 2 is needed: $X=Y=\mathbb R$, $F(t)=t/2$ satisfies 1, 3 and 4 but agrees with no isometry on $[-1,1]$.
• Hypothesis 3 is needed: $X=Y=\mathbb R$, $F(t)=t^3$ satisfies 1, 2 and 4.
• Hypothesis 4 is not known to be removable: without it the statement asserts that every nonexpansive bijection between the unit balls of two real normed spaces is the restriction of a linear isometry, which contains the open plasticity conjecture.
• Inside hypothesis 4, conjuncts (a), (b) and (c) are redundant: any extreme subset of $B_Y$ containing $y\in S_Y$ contains the minimal face of $y$, which is nonempty, convex and contained in $S_Y$, so hypothesis 4 is equivalent to "the minimal face of every $y\in S_Y$ has finite-dimensional affine hull". This redundancy is harmless.

Name. The name describes the statement only in part. exists_linearIsometryEquiv alone suggests merely that $X$ and $Y$ are isometrically isomorphic, while the statement says that $F$ itself coincides on $B_X$ with such an isomorphism. sphere_covered does not say which sphere (it is $S_Y$) or by what (sphere faces with finite-dimensional affine hull). A name such as exists_linearIsometryEquiv_eqOn_closedBall_of_sphere_covered_by_finiteDimensional_faces would be more exact; this is a minor point.

Truth. I believe the statement is true as written. The route I see: card 2 applies to every $x\in S_X$ once $F(S_X)\subseteq S_Y$ is known, homogeneity on diameters through sphere points then gives homogeneity on every diameter $[-x,x]$ with $x\in B_X$, card 3 makes $F$ an isometry of $B_X$ onto $B_Y$ fixing $0$, and Mankiewicz's extension theorem (valid in real normed spaces) turns it into a linear isometry $T$ with $T(B_X)=B_Y$, hence surjective. The steps I did not reconstruct independently are the derivation of $F(S_X)\subseteq S_Y$ from hypothesis 4 and card 2 for faces of positive dimension, both of which appear to need a dimension argument on the finite-dimensional faces.