A nonexpansive ball bijection that keeps the sphere is homogeneous at every preimage of a point of a finite-dimensional face

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Let $X$ and $Y$ be real normed spaces, neither assumed complete or finite-dimensional. 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:

1. (Injective on the ball.) For all $x,x'\in B_X$, $F(x)=F(x')$ implies $x=x'$.
2. (Onto the ball.) $B_Y\subseteq F(B_X)$; 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 to sphere.) $\|F(x)\|=1$ whenever $\|x\|=1$, that is, $F(S_X)\subseteq S_Y$.
5. $C\subseteq Y$ is a set that is nonempty, convex, contained in $S_Y$, and an extreme subset of $B_Y$: $C\subseteq B_Y$ 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 $C$.
6. The linear subspace $\operatorname{span}\{c-c':c,c'\in C\}$ is finite-dimensional.
7. $x\in X$ satisfies $\|x\|\le 1$.
8. $F(x)\in C$.

Conclusion: for every real $t$ with $-1\le t\le 1$,

The statement holds for every $F$, $C$ and $x$ meeting these hypotheses; nothing is said about $F$ outside $B_X$ (all points $tx$ in the conclusion lie in $B_X$).

Flags

Lean conventions. None changes the meaning; the remarks on Set.SurjOn, LipschitzOnWith and vectorSpan made for the previous card apply here too. The conclusion is a universally quantified equation over $t\in[-1,1]$, not a hypothesis.

Splitting the bijection (purpose note). The choice is correct, by the argument given for the previous card: hypotheses 2 and 3 give $F(B_X)\subseteq B_Y$ and $F(0)=0$, so hypotheses 1 to 3 say exactly that $F$ restricted to $B_X$ is a nonexpansive bijection onto $B_Y$. Further consequences used below: every point of $B_X$ that $F$ sends into $S_Y$ lies in $S_X$ (if $F(p)=u\in S_Y$ and $F(q)=-u$ with $q\in B_X$, then $2\le\|p-q\|\le\|p\|+\|q\|$), so hypotheses 7 and 8 force $\|x\|=1$. With hypothesis 4 one also gets $\|F(z)\|=\|z\|$ for all $z\in B_X$ (the bound $\le$ from $F(0)=0$, the bound $\ge$ from $\|F(z/\|z\|)\|=1$ and $\|F(z/\|z\|)-F(z)\|\le 1-\|z\|$).

Degenerate cases. If $Y=\{0\}$ then $S_Y=\emptyset$, so no nonempty $C\subseteq S_Y$ exists and the statement is vacuous. Faces of dimension $0$ are the singletons $\{e\}$ with $e$ an extreme point of $B_Y$; then hypothesis 6 holds trivially. The hypotheses are satisfiable, for example by $F=\mathrm{id}$ on $\mathbb R$, $C=\{1\}$, $x=1$. Hypothesis 6 restricts only the one face $C$, not the rest of $S_Y$.

Hypotheses, one at a time.

• Hypothesis 1: for singleton faces the statement stays true without it, provided hypothesis 4 is kept (argument below); for faces of positive dimension I cannot say.
• Hypothesis 2 is needed: $X=\mathbb R$, $Y=(\mathbb R^2,\|\cdot\|_\infty)$, $F(t)=(t,|t|)$, $C=\{(1,1)\}$, $x=1$ satisfies all other hypotheses, but $F(-1)=(-1,1)\ne-F(1)$.
• Hypothesis 3 is needed: $X=Y=\mathbb R$, $F(t)=t^3$, $C=\{1\}$, $x=1$.
• Hypothesis 4: for singleton faces the statement stays true without it, provided hypothesis 1 is kept (argument below); for faces of positive dimension I cannot say, and I know no argument deriving hypothesis 4 from hypotheses 1 to 3 in infinite dimensions. Since the previous card does not assume hypothesis 4, it may be removable here as well if the face argument does not use it.
• Hypotheses 5 and 8 are not known to be removable. Dropping 5 allows $C=\{F(x)\}$ for every $x\in B_X$, and dropping 8 leaves $C$ unrelated to $x$; either way the statement would assert homogeneity on every diameter of $B_X$, which together with card 3 makes every such $F$ an isometry, an open statement in general.
• Hypothesis 6 likewise: every point of $S_Y$ lies in a sphere face (its minimal face), so without 6 the statement together with card 3 would give the plasticity conclusion for every $F$ satisfying 1 to 4, which is open.
• Hypothesis 7 is needed: $X=Y=\mathbb R$, $F(t)=t$ for $|t|\le 1$ and $F(t)=1$ otherwise, $C=\{1\}$, $x=2$; then $F(1)=1\ne\tfrac12 F(2)$.
• Within hypothesis 5, nonemptiness follows from 8, the inclusion $C\subseteq B_Y$ from $C\subseteq S_Y$, and convexity can be dropped by applying the convex case to the minimal face of $F(x)$, which lies in $C$; these redundancies are harmless.

The singleton case. Let $C=\{e\}$, so $F(x)=e$. For $0<a<1$, $\|F(ax)\|\le a$ and $\|e-F(ax)\|\le 1-a$, so $e=a\cdot\frac{F(ax)}{a}+(1-a)\cdot\frac{e-F(ax)}{1-a}$ is a convex combination of points of $B_Y$, and extremality gives $F(ax)=ae$. Choose $v\in B_X$ with $F(v)=-e$; then $\|x-v\|=2$ and likewise $F(av)=-ae$ for $a\in[0,1]$. For $m=\tfrac12(x+v)$, $\|F(m)-e\|\le 1$ and $\|F(m)+e\|\le 1$, so $e=\tfrac12(e+F(m))+\tfrac12(e-F(m))$ forces $F(m)=0$. Then $m=0$ follows either from injectivity and $F(0)=0$ or from norm preservation (which uses hypothesis 4), so $v=-x$ and $F(-ax)=-aF(x)$.

Name. map_smul_of_map_mem_face describes the statement accurately; it does not mention that the face is finite-dimensional or that $t$ ranges over $[-1,1]$, which is acceptable.

Truth. I believe the statement is true as written. I verified it fully for faces of dimension $0$, and for $X=Y$ finite-dimensional it follows from the classical fact that a nonexpansive bijection of a compact metric space onto itself is an isometry, combined with $F(0)=0$ and Mankiewicz's theorem. For faces of positive dimension in infinite-dimensional spaces, the elementary estimates only give $F(tx)\in tC$ for $t\in[0,1]$, and pinning $F(tx)=tF(x)$ needs a further argument (presumably invariance of domain on the finite-dimensional face) that I did not reconstruct; I found no counterexample.