A positive solution to Tingley’s problem

1. Pass to a sphere isometry attaining the supremum

Since $D_0=D_1=0$ and the defects are uniformly $2$-Lipschitz in $q$, one has $|D_q|\le2\min(q,1-q)$. Choose a sequence approaching $M>0$, replacing $f$ by its inverse if necessary to make the defects positive. A subsequence of radius ratios converges to $t\in[M/2,1-M/2]$; the uniform estimate allows all ratios to be replaced by $t$.

Applying $f$ termwise to unit-vector representatives induces an onto sphere isometry on the ultrapower quotients. Independence of representatives follows from distance preservation, and surjectivity from applying $f^{-1}$ termwise. After completion, the bound $|D_q|\le M$ still holds for every radius ratio and every pair of unit vectors, while $D_t(x_0,y)=M$ for a selected pair. This global bound is essential for the later radial perturbations. The argument does not classify the new sphere map: it proves that such an attained configuration cannot exist.

Lemma 3.1; (1)–(4) · pp. 3–4

2. Exclude the aligned chords and recover all distances

Proposition 4.2 aligns chords through $ty$ and $tf(y)$, with corresponding endpoints and radial differences equal to $M$. Proposition 5.3 excludes this configuration using supports and convexity of the norm. Hence the original map also has $M=0$.

For $a=\alpha x$, $c=\beta y$, with $x,y\in S_X$ and $\alpha\ge\beta>0$, put $q=\beta/\alpha$. Then $$ \|T(a)-T(c)\|=\alpha\|f(x)-qf(y)\| =\alpha\|x-qy\|=\|a-c\|. $$ Reversing the radius ordering and including zero cover the remaining cases. Surjectivity of $f$ gives, for each $b\ne0$, a unit vector $y$ with $f(y)=b/\|b\|$, hence $T(\|b\|y)=b$.

Propositions 4.2, 5.3; §6 · pp. 4, 8–9

3. Double the midpoint defect to obtain real linearity and uniqueness

For an onto isometry $F$, set $m=(a+c)/2$, $m'=(F(a)+F(c))/2$, and $\delta=\|F(m)-m'\|$. Conjugating reflection $R$ about $m'$ gives $H=F^{-1}RF$, which exchanges $a,c$ and has midpoint defect $2\delta$. Every onto isometry, however, has midpoint defect bounded by $\|a-c\|/2$. Iterated doubling forces $\delta=0$.

Midpoint preservation and $T(0)=0$ yield additivity, rational homogeneity, and then real homogeneity by continuity. Any linear extension is determined by $x=\|x\|(x/\|x\|)$, proving uniqueness. Section 6 reproduces this Mazur–Ulam step using Nica’s reflection presentation.

§6, proof of Theorem 1.1 · p. 9

[Nic, Theorem (Mazur–Ulam 1932); reflection proof, p. 1]

1. Return through the opposite endpoint

For $v\in S_X$, take $x(v)=ty+p(v)v\in S_X$ and write $f(x(v))=tb+sw$. The sphere contact from $tb$ in direction $-w$ is $tb-uw=f(z)$. Define $$ r=\|z-ty\|,\qquad g(v)=\frac{ty-z}{r}. $$ Lemma 2.2 gives uniqueness and continuity of contacts, with all four radii in $[1-t,1+t]$. The extremal set $E=\{v\in S_X:s(v)-p(v)=M\}$ is nonempty and closed.

For $v\in E$, sphere distance preservation and the opposite signed defect bound give $$ s+u=\|x(v)-z\|\le p+r,\qquad r-u\le M=s-p. $$ Both inequalities must be equalities. Thus $r=u+M$ and $\|pv+rg(v)\|=p+r$, so Lemma 2.1 supplies a common support for $v,g(v)$. Every unit direction $\ell$ sharing a support with $g(v)$ satisfies $p(\ell)+r\le s(\ell)+u$ and therefore belongs to $E$. This propagation permits the next convex-set construction.

Lemmas 2.1–2.2; (7)–(8) · pp. 2–3, 5

2. Construct an invariant closed convex set inside the sphere

Starting with $v_0\in E$, choose a convex combination $a_n$ assigning strictly positive coefficients to all $v_0,\ldots,v_n$, and set $v_{n+1}=g(a_n)$. A common support of $a_n,v_{n+1}$, extended to the finite-dimensional span, equals one on each $v_i$ by positivity of the coefficients. The enlarged convex hull remains on the sphere and belongs to $E$ by propagation.

Choose $a_n$ to approximate every finite rational convex combination repeatedly. Then the $a_n$ are dense in $C=\overline{\operatorname{conv}}\{v_0,v_1,\ldots\}$. Closedness of $E$ and continuity of $g$ show that $C\subset E$ is nonempty, closed, bounded, and convex, and that $g(C)\subset C$. Only the constructed set is separable; the ambient spaces remain unrestricted.

Proposition 4.2, invariant convex set · p. 5

3. Refine covers by radii to meet Darbo’s hypothesis

For $v,v'\in E$, the manuscript obtains $$ \|g(v)-g(v')\|\le\frac{up}{rs}\|v-v'\| +\frac{2|u-u'|}{r}+\frac{2u|p-p'|}{rs}. $$ The identities $r=u+M$, $s=p+M$ give $up/(rs)\le\gamma=(1-M/(1+t))^2<1$. The additional radial differences prevent this from directly asserting metric contraction.

Refine a finite diameter cover of $H\subset C$ according to intervals of length $\eta$ for the values of $p,u$. Since $r,s\ge1-t$, each image has diameter at most $\gamma\delta+4\eta/(1-t)$. Sending $\eta\downarrow0$ yields $\chi(g(H))\le\gamma\chi(H)$ for Kuratowski’s measure of noncompactness. The Darbo form in Theorem 4.1 now gives $g(v)=v$, so $z=ty-rv$ and the endpoint correspondences hold simultaneously. This is the alignment needed for the main theorem.

Theorem 4.1; (9), Proposition 4.2 · pp. 4–6

[Dar55, §4, Teorema (unnumbered), print pp. 90–91 / PDF pp. 8–9]

1. Move to an endpoint radius to bound a secant

Put $h=\min(t/s,\varepsilon/u)$, $\theta=hs/p$, and $$ K=\frac{\|(r/u)v+hy\|-\|(p/s)v+hy\|}{r/u-p/s}. $$ Both $t-sh,t+uh$ lie in $[0,1]$, and one is an endpoint. In the weighted sum of the corresponding defects $\Delta_x=D_{t-sh}(x,y)$ and $\Delta_z=-D_{t+uh}(z,y)$, the target norm cancels, giving $(u\Delta_x+s\Delta_z)/d=MK$. One defect vanishes and the other is at most $M$, yielding $K\le1-B$ if $B\le\varepsilon$, and $K\le B$ if $B>\varepsilon$.

Every support $\varphi$ at $v+\theta y$ bounds this secant below. Comparing it with a support at $v+(t/p)y$ gives the other bound in Lemma 5.1: $$ \frac{\varepsilon}{p}\le\varphi(v)\le K. $$ The perturbation step requires this estimate for every support, not merely a chosen one.

Lemma 5.1, (11) · p. 7

2. Exclude large ratios using support signs

If $B>\varepsilon$, Lemma 5.1 gives $pu/(d\varepsilon)\ge1$. For any support $L$ at $v$, let $j=L(y)$. Excluding equality in the support estimate gives $$ 1-j>\frac{1-B}{\theta}=\frac{pu}{d\varepsilon}\ge1, $$ so every support at $v$ is negative on $y$. Since $\|rv-ty\|=1$, one obtains $r<1$ and hence $u<1$; then $Au\ge\varepsilon$ implies $A>\varepsilon$.

For $f^{-1}$, the aligned configuration has $(p',s',r',u')=(u,r,s,p)$ and $(v',w',y',b')=(-w,-v,b,y)$, so $(A',B')=(B,A)$. Consequently, if either ratio exceeds $\varepsilon$, both do, and every support at $w$ is positive on $b$. For sufficiently small $\eta>0$, $$ \|sw+\eta b\|>s,\qquad \|pv+\eta y\|<p. $$ The defect at radius $t-\eta$ then exceeds $M=s-p$. The strict decrease in the second inequality follows from compactness of the dual unit ball of $\operatorname{span}\{v,y\}$ and the sign of every support.

Lemma 5.2, (12)–(15) · pp. 7–8

3. In the remaining regime, bounds on a preserved distance conflict

For $A,B\le\varepsilon$, apply Lemma 5.1 to the original and inverse configurations: $$ \varepsilon\le dA(1-B),\qquad\varepsilon\le dB(1-A). $$ Since $A+B<1$, the smaller product on the right is strictly below $d/4$. As $d\le2$, this gives $\varepsilon<1/2$. Thus $A\le\varepsilon<t$ and $1-B\ge t$, which imply $$ \frac pt<\frac r\varepsilon,\qquad\frac st\ge\frac u\varepsilon. $$ The convex functions $F(\lambda)=\|y+\lambda v\|$ and $G(\lambda)=\|b+\lambda w\|$ satisfy $F(0)=G(0)=1$ and $F(p/t)=G(s/t)=1/t>1$. Monotonicity of secant slopes yields $F(r/\varepsilon)>1/t$, whereas convexity gives $G(u/\varepsilon)\le1/t$. But sphere distance preservation requires $$ \varepsilon F(r/\varepsilon)=\|y-z\| =\|b-f(z)\|=\varepsilon G(u/\varepsilon), $$ a contradiction. Attainment of a positive maximal defect is excluded, completing the return to Theorem 1.1.

Proposition 5.3, (16)–(17) · pp. 8–9