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\hypersetup{pdftitle={A positive solution to Tingley's problem},pdfauthor={OpenAI},
  pdfsubject={Sphere isometries of real Banach spaces},
  pdfkeywords={Tingley's problem, sphere isometry, Banach space}}

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\title{A positive solution to Tingley's problem}
\author{OpenAI}
\date{September 23, 2026}

\begin{document}
\maketitle

\begin{abstract}
Every surjective isometry between the unit spheres of two real Banach spaces
extends uniquely to a surjective real-linear isometry of the spaces. This gives
an affirmative answer to Tingley's problem in full generality, with no
restriction on dimension or geometric regularity. If distances between points
of different radii could fail to be preserved, we magnify that failure into a
maximal positive defect in a possibly larger pair of spaces. Extremal chords
are then aligned by a fixed-point argument for a map that strictly contracts
Kuratowski's measure of noncompactness; a support estimate and a convexity
argument finally rule the aligned configuration out.
\end{abstract}

\section{Introduction and main result}\label{sec:intro}

Let $X$ be a real normed space. Its \emph{unit sphere} is
$S_X=\{x\in X:\norm{x}=1\}$, equipped with the distance coming from the norm.
Tingley's problem \cite{Tingley1987} asks:
\begin{center}
\emph{if $f:S_X\to S_Y$ is a surjective isometry, must it extend to a
real-linear isometry of $X$ onto $Y$?}
\end{center}
A surjective sphere isometry always maps antipodal points to antipodal points,
so the only possible linear extension is the \emph{radial map}
\[
T(0)=0,\qquad T(x)=\norm{x}\,f\!\left(\frac{x}{\norm{x}}\right)\ (x\ne0).
\]
This map matches $f$ on the sphere and preserves distances between points at the
same radius. The whole difficulty is a single point: the metric of the sphere
knows nothing directly about pairs of points at \emph{different} radii, yet the
extension must preserve those distances too.

\begin{theorem}[Main theorem]\label{thm:main}
Let $X,Y$ be nonzero real Banach spaces and let $f:S_X\to S_Y$ be a surjective
isometry. Then $T$ above is a surjective real-linear isometry of $X$ onto $Y$,
and it is the unique linear extension of $f$.
\end{theorem}

The spaces are arbitrary: no separability, reflexivity, smoothness or strict
convexity is assumed, and the dimension may be infinite. Applied to complex
spaces regarded as real spaces, the conclusion is real linearity only.

\paragraph{Background.} The result belongs to the extension theory of metric
isometries. Mazur and Ulam \cite{MazurUlam1932} proved that a surjective isometry
between normed spaces is affine, and Mankiewicz \cite{Mankiewicz1972} extended
isometries of convex bodies with nonempty interior. A sphere has empty ambient
interior, so neither applies directly. Positive solutions were first obtained for
concrete classes: $C_0$-spaces \cite{Wang1994}, $\ell^p(\Gamma)$ and $L^p$ with
$p\ne2$ \cite{Ding2003,Tan2012Lp}, then, through the \emph{Mazur--Ulam property}
of Cheng and Dong \cite{ChengDong2011}, for operator algebras
\cite{FernandezPoloPeralta2018VN,Mori2018,MoriOzawa2020}. Facial methods
\cite{Tanaka2014}, low dimensions \cite{Banakh2022} and specific families
\cite{Fakhoury2025,HuangZhu2026Haagerup,HuangZhu2026Schatten} cover further cases.
The present argument is uniform: it isolates the largest possible failure to
preserve a cross-radius distance and shows that this failure cannot exist.

\paragraph{How to read the proof.} The argument has five moves.
(1) We measure the worst error in preserving a distance between two radii and
show that, if it is positive, it becomes an \emph{attained} maximum after passing
to an ultrapower (Section~\ref{sec:attainment}).
(2) From that maximum we manufacture two ``extremal chords'', one in $S_X$ and one
in $S_Y$, whose endpoints line up around common base points (Section~\ref{sec:alignment}).
(3) A return map on sphere directions, which pairs the far end of one chord with
the far end of the other, strictly shrinks Kuratowski's measure of noncompactness;
Darbo's theorem then gives a fixed direction, i.e.\ truly aligned chords.
(4) Support estimates and a convexity comparison show that aligned chords are
impossible (Section~\ref{sec:comparison}).
(5) With the defect zero, the radial map is an isometry, and the classical
Mazur--Ulam reflection trick upgrades it to a linear map (Section~\ref{sec:extension}).

\paragraph{Notation.} Supports are finite-dimensional throughout
(Lemma~\ref{lem:support}). We write $\co$ and $\clco$ for convex hull and closed
convex hull, and drop the norm subscript when the space is clear. Section
numbers are referenced in the text; all auxiliary results are proved, the
fixed-point inputs in Appendix~\ref{sec:appendix}.

\section{Supports and radial contact}\label{sec:supports}

A \emph{support} at a vector $a$ in a finite-dimensional normed space is a
linear functional $\varphi$ with $\norm{\varphi}\le1$ and $\varphi(a)=\norm{a}$.
The next lemma is finite-dimensional Hahn--Banach \cite{Banach1929}; we prove it
because it is used constantly.

\begin{lemma}\label{lem:support}
Supports exist and extend, with the same norm bound, to any prescribed
finite-dimensional enlargement of their domain. If unit vectors $v,w$ and numbers
$\alpha,\beta>0$ satisfy $\norm{\alpha v+\beta w}=\alpha+\beta$, then $v$ and $w$
share a support: a norm-bounded functional equal to $1$ at both.
\end{lemma}
\begin{proof}
Start with the norming functional on the line through a nonzero vector. To extend
$L$ from a subspace $V$ to $V+\R e$, pick its value at $e$ in
\[
\bigcap_{a\in V}\,[\,L(a)-\norm{a-e},\ L(a)+\norm{a-e}\,].
\]
The intervals meet pairwise because $L(a)-L(b)\le\norm{a-b}\le\norm{a-e}+\norm{b-e}$;
they have the finite intersection property, and the interval indexed by $a=0$ is
compact, so they share a point. That value gives $|L(a)-L(e)|\le\norm{a-e}$ and
hence the norm bound on $V+\R e$; finitely many extensions reach any
finite-dimensional enlargement. Finally, a support $\varphi$ at
$\alpha v+\beta w$ has $\varphi(\alpha v)\le\alpha$ and $\varphi(\beta w)\le\beta$
with total $\alpha+\beta$, forcing equality, so $\varphi(v)=\varphi(w)=1$.
\end{proof}

\begin{lemma}\label{lem:radial}
Let $\norm{c}<1$. Each ray $c+\R_+v$, $v\in S_X$, meets $S_X$ in a unique point
$c+\rho(v)v$. The radius $\rho(v)$ is continuous and satisfies
$1-\norm{c}\le\rho(v)\le1+\norm{c}$.
\end{lemma}
\begin{proof}
The norm is continuous and the unit ball is bounded, so a first contact exists.
Every point strictly between an interior point of the ball and a point of the ball
is interior (a small ball sits inside the appropriate convex combination), which
forces uniqueness. The bounds are the triangle inequality. For continuity, a
convergent subsequence of the bounded scalars $\rho(v_n)$ must converge to
$\rho(v)$ by continuity of the norm and uniqueness of contact, so the whole
sequence does.
\end{proof}

\section{Attaining the maximal defect}\label{sec:attainment}

For a surjective sphere isometry put
\begin{equation}\label{eq:defect}
D_q(x,y)=\norm{f(x)-qf(y)}-\norm{x-qy},\qquad
M=\sup_{x,y\in S_X,\ 0\le q\le1}|D_q(x,y)|.
\end{equation}
Because $f$ is bijective, the inverse map has defect
\begin{equation}\label{eq:inverse-defect}
D_q^{f^{-1}}\!(f(x),f(y))=-D_q(x,y),
\end{equation}
so the inverse has the same absolute bound. Each $D_q(\cdot,\cdot)$ is
$2$-Lipschitz in $q$ uniformly in $x,y$, and $D_0=D_1=0$. Hence
\begin{equation}\label{eq:defectbound}
|D_q(x,y)|\le2\min(q,1-q),\qquad 0\le M\le1.
\end{equation}

\begin{lemma}[Attainment]\label{lem:attainment}
If some surjective sphere isometry has $M>0$, there are nonzero real Banach spaces,
a surjective sphere isometry (still called $f:S_X\to S_Y$), and $x_0,y\in S_X$,
$0<t<1$, with
\begin{equation}\label{eq:maximal}
|D_q(a,c)|\le M\quad(a,c\in S_X,\ 0\le q\le1),
\qquad
\norm{f(x_0)-tf(y)}-\norm{x_0-ty}=M>0.
\end{equation}
\end{lemma}
\begin{proof}
Take triples approaching the supremum, pass to a subsequence of one sign and, if
needed, replace $f$ by $f^{-1}$. A further subsequence has radius parameters
converging to $t\in[0,1]$; the $2$-Lipschitz bound lets us replace them all by
$t$ while the signed defects still tend to $M$. By \eqref{eq:defectbound},
$M/2\le t\le1-M/2$, so $0<t<1$.

We build the limit as a Banach ultrapower \cite{DacunhaCastelle1972}. Fix a free
ultrafilter $\mathcal U$ on $\mathbb N$; every bounded real sequence has a unique
limit along $\mathcal U$, and these limits respect sums, inequalities and
continuous functions. On bounded sequences in $X$ set
$\norm{(a_n)}_{\mathcal U}=\lim_{\mathcal U}\norm{a_n}$, let $Q_X$ be the quotient
by the kernel, and define $Q_Y$ likewise; constant sequences embed $X,Y$
isometrically, so the quotients are nonzero. Every unit vector of $Q_X$ or $Q_Y$
has a representative made of unit vectors (normalize each nonzero term, replace
zeros by a fixed unit vector).

Apply $f$ termwise to unit representatives to get $F:S_{Q_X}\to S_{Q_Y}$. The
identity $\lim_{\mathcal U}\norm{f(a_n)-f(b_n)}=\lim_{\mathcal U}\norm{a_n-b_n}$
shows $F$ is well defined and preserves distances, and applying $f^{-1}$ termwise
gives its inverse, so $F$ is a surjective sphere isometry. For fixed
$q\in[0,1]$,
\[
\norm{F([a_n])-qF([b_n])}-\norm{[a_n]-q[b_n]}=\lim_{\mathcal U}D_q(a_n,b_n),
\]
which lies in $[-M,M]$ since each term does; the chosen maximizing sequences give
equality at $t$. Completing both quotients extends $F$ and $F^{-1}$ by continuity
and density to a surjective sphere isometry of the completions, and continuity
keeps the bound and the equality.
\end{proof}

It remains only to rule out the attained configuration \eqref{eq:maximal}, retained
through Sections~\ref{sec:alignment} and~\ref{sec:comparison}.

\section{Aligning extremal chords}\label{sec:alignment}

For a bounded set $H$ in a normed space, the Kuratowski measure of noncompactness
\cite{Kuratowski1930} is
\[
\chi(H)=\inf\{\delta>0:\ H\ \text{is covered by finitely many sets of diameter }\le\delta\}.
\]
We use the strict-contraction fixed-point theorem of Darbo \cite{Darbo1955}, proved
in Appendix~\ref{sec:appendix}.

\begin{theorem}\label{thm:condensing}
Let $C$ be a nonempty closed bounded convex subset of a Banach space and
$g:C\to C$ continuous with $\chi(g(H))\le\gamma\chi(H)$ for all $H\subset C$ and
some $0\le\gamma<1$. Then $g$ has a fixed point.
\end{theorem}

\begin{proposition}[Aligned chords]\label{prop:alignment}
Under \eqref{eq:maximal}, set $b=f(y)$ and fix $y,t$. There are endpoints
$x,z\in S_X$, directions $v\in S_X$, $w\in S_Y$, and numbers $p,r,s,u>0$ with
\begin{equation}\label{eq:chords}
x=ty+pv,\quad z=ty-rv,\quad f(x)=tb+sw,\quad f(z)=tb-uw,
\end{equation}
and
\begin{equation}\label{eq:radii}
s-p=r-u=M,\qquad d:=p+r=s+u=\norm{x-z}\le2.
\end{equation}
All four radii lie in $[1-t,1+t]$.
\end{proposition}
\begin{proof}
Fix a direction at $ty$, map its sphere contact to $S_Y$, take the opposite sphere
contact on the line through $tb$, pull it back to $S_X$, and reverse its direction
from $ty$. A fixed direction of this return map aligns the two chords.

For $v\in S_X$ define $p=p(v)>0$ by $x(v)=ty+pv\in S_X$ (Lemma~\ref{lem:radial}).
Then set successively
\begin{equation}\label{eq:return}
f(x(v))=tb+sw,\qquad f(z)=tb-uw\in S_Y,\qquad
r=\norm{z-ty},\qquad g(v)=\frac{ty-z}{r},
\end{equation}
where $w\in S_Y$ and $u>0$ is the radial contact in direction $-w$. All these
quantities are continuous in $v$, and each radius lies in $[1-t,1+t]$.

Let $E=\{v\in S_X:\ s(v)-p(v)=M\}$, closed in $X$. It is nonempty: with
$v_0=(x_0-ty)/\norm{x_0-ty}$, uniqueness of radial contact gives $x(v_0)=x_0$, and
\eqref{eq:maximal} puts $v_0\in E$. For $v\in E$, distance preservation and the
bound \eqref{eq:maximal} give
\[
s+u=\norm{x(v)-z}\le p+r,\qquad r-u\le M.
\]
Since $s=p+M$, both are equalities:
\begin{equation}\label{eq:equality}
r=u+M,\qquad \norm{pv+rg(v)}=p+r.
\end{equation}
Thus $v$ and $g(v)$ share a support.

\emph{Propagation.} If a unit $l$ shares a support with $g(v)$, then $l\in E$.
Indeed, for the fixed $z,r,u$ attached to $v$, the shared support gives
$\norm{x(l)-z}\ge p(l)+r$ (both summands are supported to the value $1$ along
$l$ and $g(v)$), while distance preservation gives
\[
\norm{x(l)-z}=\norm{f(x(l))-f(z)}\le s(l)+u.
\]
Hence $s(l)-p(l)\ge r-u=M$, and the bound \eqref{eq:maximal} makes it equal, so
$l\in E$. In particular $g(v)\in E$.

\emph{An invariant convex set.} Start with $v_0\in E$ and suppose
$\co\{v_0,\dots,v_n\}\subset E$. Take a strictly positive convex combination $a_n$
of $v_0,\dots,v_n$ and set $v_{n+1}=g(a_n)$. A common support $\varphi$ of
$a_n,v_{n+1}$, extended to the finite span with the $v_i$, equals $1$ at each $v_i$
by strict positivity, so for every $l$ in the enlarged hull,
$1=\varphi(l)\le\norm{l}\le1$. Thus $l$ is a unit vector sharing $\varphi$ with
$g(a_n)$ and lies in $E$ by propagation.

Choose the coefficients so that the $a_n$ are dense in $C=\clco\{v_0,v_1,\dots\}$:
enumerate finite rational probability vectors, mix each (zero-padded) with the
uniform vector on its list using weights $\lambda_n\to0$, and note the mixture
moves the combination by at most $2\lambda_n$ since all $v_i$ are unit. Then $C$
is nonempty closed bounded convex with $C\subset E\subset S_X$, and
$g(a_n)=v_{n+1}\in C$ together with continuity gives $g(C)\subset C$. Only this
constructed set need be separable; the ambient spaces are unrestricted.

\emph{Strict contraction.} For $v,v'\in E$, with primes marking associated
quantities, \eqref{eq:return} gives
\[
r\norm{g(v)-g(v')}\le|r-r'|+\norm{z-z'}
\le|r-r'|+|u-u'|+u\norm{w-w'},
\]
\[
s\norm{w-w'}\le|s-s'|+\norm{f(x(v))-f(x(v'))}
\le|s-s'|+|p-p'|+p\norm{v-v'}.
\]
Using $s=p+M$, $r=u+M$ yields
\begin{equation}\label{eq:contraction}
\norm{g(v)-g(v')}
\le\frac{up}{rs}\norm{v-v'}+\frac{2|u-u'|}{r}+\frac{2u\,|p-p'|}{rs}.
\end{equation}
With $a=1-t>0$, $U=1+t$, all radii lie in $[a,U]$ and $M<r,s$, so
\[
\frac{up}{rs}=\Big(1-\frac Mr\Big)\Big(1-\frac Ms\Big)\le\gamma:=\Big(1-\frac MU\Big)^2<1.
\]
Cover $H\subset C$ by finitely many sets of diameter $\le\delta$; intersect with
$H$ and subdivide according to finite partitions of the ranges of $p$ and $u$ into
intervals of length $\le\eta$. On each piece the last two terms of
\eqref{eq:contraction} sum to at most $4\eta/a$, so $g(H)$ is covered by pieces of
diameter $\le\gamma\delta+4\eta/a$. Letting $\eta\downarrow0$ and taking the
infimum gives $\chi(g(H))\le\gamma\chi(H)$.

Theorem~\ref{thm:condensing} now yields $v\in C$ with $g(v)=v$. Setting
$x=x(v)$ and substituting in \eqref{eq:return} gives \eqref{eq:chords}; the
equalities \eqref{eq:equality} and distance preservation give \eqref{eq:radii}.
\end{proof}

\section{The chord contradiction}\label{sec:comparison}

Keep the configuration of Proposition~\ref{prop:alignment} and set
\begin{equation}\label{eq:AB}
A=\frac pd,\qquad B=\frac ud,\qquad \epsilon=1-t.
\end{equation}
Then $A,B>0$ and, since $r=u+M$, $s=p+M$, one checks $r=d(1-A)$, $s=d(1-B)$,
hence $A+B=1-M/d<1$. Every radius lies in $[1-t,1+t]$.

\begin{lemma}[Secant estimate]\label{lem:secant}
Put $h=\min\{t/s,\epsilon/u\}$, $\theta=hs/p$, and
\[
K=\frac{\norm{(r/u)v+hy}-\norm{(p/s)v+hy}}{r/u-p/s}.
\]
For every support $\varphi$ at $v+\theta y$,
\begin{equation}\label{eq:K}
\frac{\epsilon}{p}\le\varphi(v)\le K\le
\begin{cases}1-B,& B\le\epsilon,\\ B,& B>\epsilon.\end{cases}
\end{equation}
\end{lemma}
\begin{proof}
At radius $t-sh$ the defect of the pair $x,y$ is
$\Delta_x=s\big(\norm{w+hb}-\norm{(p/s)v+hy}\big)$; at radius $t+uh$ the negative
defect of $z,y$ is $\Delta_z=u\big(\norm{(r/u)v+hy}-\norm{w+hb}\big)$. Both radii
lie in $[0,1]$, so $|\Delta_x|,|\Delta_z|\le M$. The choice of $h$ makes one
radius an endpoint, so one of $\Delta_x,\Delta_z$ is $0$. Adding,
$u\Delta_x+s\Delta_z=su\big(\norm{(r/u)v+hy}-\norm{(p/s)v+hy}\big)=K(sr-up)$,
and $sr-up=Md$. Hence $(u\Delta_x+s\Delta_z)/d=MK$.

Since $t/s\le\epsilon/u\iff B\le\epsilon$: in that case $\Delta_x=0$ and
$\Delta_z\le M$ give $K\le s/d=1-B$; otherwise $\Delta_z=0$ and
$\Delta_x\le M$ give $K\le u/d=B$. (At $B=\epsilon$ both are zero.)

The supported vector $v+\theta y\ne0$: $v+\theta y=0$ would force $\theta=1$ and
$v=-y$, so $p=1+t$ and $\theta\le t/p<1$, a contradiction. Because
$r/u>p/s>0$, a support at $v+\theta y$ is also a support at
$(p/s)v+hy$, so it bounds the secant from below by $\varphi(v)$, giving
$\varphi(v)\le K$. For the lower bound write $\beta=t/p$, so $0<\theta\le\beta$.
If $\theta=\beta$ then $v+\theta y=x/p$ has norm $1/p$, so
$\varphi(v)=(1-t\varphi(y))/p\ge\epsilon/p$. If $\theta<\beta$, take a support
$\psi$ at $v+\beta y=x/p$; the support inequalities give
\[
\theta(\psi(y)-\varphi(y))\le\varphi(v)-\psi(v)\le\beta(\psi(y)-\varphi(y)),
\]
forcing $\psi(y)\ge\varphi(y)$ and hence
$\varphi(v)\ge\psi(v)=(1-t\psi(y))/p\ge\epsilon/p$.
\end{proof}

\begin{lemma}\label{lem:large}
Neither $A>\epsilon$ nor $B>\epsilon$ holds.
\end{lemma}
\begin{proof}
Assume $B>\epsilon$. Then $h=\epsilon/u$ and \eqref{eq:K} (with $K\le B$) gives
\begin{equation}\label{eq:product}
\frac{pu}{d\epsilon}\ge1.
\end{equation}
Let $L$ be any support at $v$ and write $j=L(y)$. For a support $\varphi$ at
$v+\theta y$, \eqref{eq:K} and the definition of a support give
$B\ge\varphi(v)=\norm{v+\theta y}-\theta\varphi(y)\ge1+\theta j-\theta$. The
inequality is strict: equality would make $L$ a support at $v+\theta y$, and
\eqref{eq:K} applied to $L$ would give $1=L(v)\le B<1$. Hence
\[
1-j>\frac{1-B}{\theta}=\frac{pu}{d\epsilon}\ge1,
\]
so $j<0$: every support at $v$ is strictly negative on $y$. Applying one to
$rv-ty=-z$ gives $r-tj\le1$, hence $u<r<1$, and \eqref{eq:product} then gives
$A>\epsilon$.

The inverse map fits the same hypotheses. For $f^{-1}$ use first endpoint $f(z)$,
base direction $b$, opposite endpoint $f(x)$; the parameters become
\begin{equation}\label{eq:inverse}
(p',s',r',u')=(u,r,s,p),\qquad (v',w',y',b')=(-w,-v,b,y),
\end{equation}
and by \eqref{eq:inverse-defect} the bound \eqref{eq:maximal} is retained, with
defect $s'-p'=r-u=M$, so $(A',B')=(B,A)$. Thus $A>\epsilon$ implies
$B>\epsilon$ and that every support at $-w$ is negative on $b$, i.e.\ every
support at $w$ is positive on $b$. So if either $A$ or $B$ exceeds $\epsilon$,
both sign conclusions hold.

Under these sign conclusions, for $0<\eta<t$ the bound at radius $t-\eta$ reads
\begin{equation}\label{eq:perturb}
\norm{sw+\eta b}-\norm{pv+\eta y}\le M=s-p.
\end{equation}
A support at $w$ gives $\norm{sw+\eta b}>s$. Also $\norm{pv+\eta y}<p$ for all
small $\eta>0$: otherwise $\eta_n\downarrow0$ with $\norm{pv+\eta_n y}\ge p$ and
supports $L_n$ on $\operatorname{span}\{v,y\}$ would satisfy
$L_n(v)\le1$ and $\eta_n L_n(y)\ge0$, so $L_n(y)\ge0$; a subsequential limit $L$
in the compact dual ball has $L(y)\ge0$ and $pL(v)=p$, i.e.\ $L$ supports $v$ while
$L(y)\ge0$, contradicting $j<0$. The two strict inequalities contradict
\eqref{eq:perturb}.
\end{proof}

\begin{proposition}\label{prop:no-defect}
An attained positive maximum as in \eqref{eq:maximal} is impossible.
\end{proposition}
\begin{proof}
By Lemma~\ref{lem:large}, $A,B\le\epsilon$. Lemma~\ref{lem:secant} applied to the
configuration and to \eqref{eq:inverse} gives
\begin{equation}\label{eq:small}
\epsilon\le dA(1-B),\qquad \epsilon\le dB(1-A).
\end{equation}
Since $A+B<1$, the smaller product satisfies $\min\{A(1-B),B(1-A)\}<1/4$; as
$d\le2$, \eqref{eq:small} gives $\epsilon<1/2$, so $\epsilon<t$, and
$A\le\epsilon<t$, $1-B\ge t$. Consequently
\begin{equation}\label{eq:ratios}
\frac pt<\frac r\epsilon,\qquad \frac st\ge\frac u\epsilon.
\end{equation}

Consider the convex functions $F(\lambda)=\norm{y+\lambda v}$ and
$G(\lambda)=\norm{b+\lambda w}$ on $\lambda\ge0$. From $x=ty+pv$ and
$f(x)=tb+sw$,
\[
F(0)=G(0)=1,\qquad F(p/t)=G(s/t)=1/t>1.
\]
Convexity below a chord and \eqref{eq:ratios} give
\[
F(r/\epsilon)\ge1+\frac{r/\epsilon}{p/t}\Big(\frac1t-1\Big)>\frac1t,
\]
because $r/\epsilon>p/t$; and, as $u/\epsilon\le s/t$,
$G(u/\epsilon)\le1/t$. But $z=ty-rv$ and $f(z)=tb-uw$ give
\[
\epsilon F(r/\epsilon)=\norm{y-z}=\norm{b-f(z)}=\epsilon G(u/\epsilon),
\]
a contradiction.
\end{proof}

\section{The linear extension}\label{sec:extension}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
If the sphere isometry had positive defect, Lemma~\ref{lem:attainment} and
Proposition~\ref{prop:no-defect} would clash; hence $M=0$. For $a=\alpha x$ and
$c=\beta y$ with $x,y$ unit and $\alpha\ge\beta>0$, put $q=\beta/\alpha$. Then
\[
\norm{T(a)-T(c)}=\alpha\norm{f(x)-qf(y)}=\alpha\norm{x-qy}=\norm{a-c},
\]
and symmetry handles $\beta\ge\alpha$; distances to $0$ follow from norm
preservation. For $c\in Y\setminus\{0\}$ pick $y\in S_X$ with
$f(y)=c/\norm{c}$; then $T(\norm{c}\,y)=c$, so $T$ is a surjective isometry onto
$Y$ fixing $0$.

It remains to upgrade a surjective isometry to a linear map; this is the affine
step of Mazur--Ulam \cite{MazurUlam1932} in the reflection form of Nica
\cite{Nica2012}, who credits V\"ais\"al\"a and Vogt \cite{Vaisala2003,Vogt1973}.
Let $F:E\to E'$ be a surjective isometry of real normed spaces and $a,c\in E$. With
$m=(a+c)/2$, $m'=(F(a)+F(c))/2$ and $\delta=\norm{F(m)-m'}$, the triangle
inequality gives $\delta\le\norm{a-c}/2$. The reflection
$R(z)=2m'-z$ is a surjective isometry of $E'$, so $H=F^{-1}RF$ is a surjective
self-isometry of $E$ swapping $a,c$ with midpoint defect
$\norm{H(m)-m}=2\delta$. Iterating produces surjective self-isometries of $E$
whose midpoint defects double while staying $\le\norm{a-c}/2$; thus $\delta=0$,
and every surjective isometry preserves midpoints.

So $T((a+c)/2)=(T(a)+T(c))/2$. With $T(0)=0$ this gives first $T(2a)=2T(a)$ and
then $T(a+c)=T(a)+T(c)$; additivity plus continuity gives real homogeneity.
Finally, a linear extension is fixed on each nonzero vector by its norm and unit
direction, so it is unique.
\end{proof}

\appendix

\section{The fixed-point ingredients}\label{sec:appendix}

We prove Theorem~\ref{thm:condensing}. The measure-of-noncompactness mechanism is
Darbo's \cite{Darbo1955}; the compact case is Schauder's \cite{Schauder1930},
obtained from the simplex argument via Sperner's lemma \cite{Sperner1928}.

\begin{lemma}\label{lem:chi}
For every bounded nonempty $H$ in a normed space,
$\chi(\overline H)=\chi(H)$ and $\chi(\clco H)=\chi(H)$. If $C_1\supset C_2\supset\cdots$
are nonempty closed bounded subsets of a Banach space with $\chi(C_n)\to0$, their
intersection is nonempty and compact.
\end{lemma}
\begin{proof}
Closing a finite diameter cover preserves diameters, giving the first identity.
For the second, cover $H$ by finitely many $H_1,\dots,H_N$ of diameter $\le\delta$,
with $R$ a norm bound. Each $\co H_i$ has diameter $\le\delta$. Write a point of
$\co H$ as $\sum_i\lambda_i z_i$, $z_i\in\co H_i$, $\sum_i\lambda_i=1$, and
partition the coefficient simplex into finitely many pieces of $\ell_1$-diameter
$\le\eta$. Two sums in one piece differ by at most $\delta+R\eta$. Hence
$\chi(\co H)\le\delta+R\eta$; let $\eta\downarrow0$, minimize over $\delta$, and use
monotonicity plus the closure identity for the reverse inequality.

For the intersection, pick $x_n\in C_n$. For each $k$ a late $C_n$ has a finite
cover of diameter $<2^{-k}$; successive subsequences and a diagonal choice give a
Cauchy subsequence of the $x_n$ with indices going to infinity. Its limit lies in
every $C_n$, so the intersection is nonempty; it is closed with
$\chi=0$, hence totally bounded, hence compact.
\end{proof}

\begin{lemma}\label{lem:simplex}
Every continuous self-map of a finite-dimensional probability simplex has a fixed
point.
\end{lemma}
\begin{proof}
Let $P$ be a continuous self-map of
$\Delta_m=\{a\in\R^{m+1}:a_i\ge0,\ \sum_i a_i=1\}$ ($m=0$ trivial). Subdivide into
simplices and label each vertex $a$ by an index $i$ with $a_i>0$ and
$a_i\ge P(a)_i$ (it exists, else summing over positive coordinates gives
$1<\sum_i P(a)_i=1$); the label of a vertex lies among the vertices of every
original face containing it.

Sperner's parity fact: in a finite face-compatible triangulation of an $m$-simplex
with labels $0,\dots,m$ obeying that boundary condition, the number of small
$m$-simplices carrying all labels is odd. Count, mod $2$, incidences of small
$m$-simplices with facets labeled $0,\dots,m-1$: internal facets count twice; a
boundary facet lies only in the original face missing label $m$, whose count is
odd by induction (starting from the single vertex in dimension $0$). A small
$m$-simplex contributes one if fully labeled, two if it carries $0,\dots,m-1$ with
a repeat, and none otherwise.

Subdivisions of arbitrarily small mesh exist: subdivide faces, cone to the
barycenter, and note that barycenters of nested faces with $k\le l$ vertices differ
by at most $(1-k/l)$ times the diameter, so dimension $m\ge1$ shrinks diameters by
at most $m/(m+1)$ per round. Pick a fully labeled small simplex in each round; by
compactness and shrinking mesh a subsequence of vertices converges to one point
$a$. The label-$i$ vertices satisfy $a_i\ge P(a)_i$ in the limit for every $i$,
and summing forces equality, so $a=P(a)$.
\end{proof}

\begin{lemma}\label{lem:compactfixed}
A continuous self-map $g$ of a nonempty compact convex set $K$ in a normed space
has a fixed point.
\end{lemma}
\begin{proof}
For $\delta>0$ choose a finite strict $\delta$-net $P_1,\dots,P_N\in K$ and set
$\lambda_i(x)=\max\{\delta-\norm{P_i-g(x)},0\}/\sum_j\max\{\delta-\norm{P_j-g(x)},0\}$,
continuous probability weights. The simplex self-map
$(a_i)\mapsto(\lambda_i(\sum_j a_jP_j))$ has, by Lemma~\ref{lem:simplex}, a fixed
coefficient vector; its convex combination $x\in K$ satisfies
$\norm{x-g(x)}\le\sum_i\lambda_i(x)\norm{P_i-g(x)}<\delta$. Let $\delta\downarrow0$
and pass to a convergent subsequence in $K$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:condensing}]
Let $C_0=C$ and $C_{n+1}=\clco(g(C_n))$; these are nonempty closed bounded convex
sets with $C_{n+1}\subset C_n$. By Lemma~\ref{lem:chi},
$\chi(C_{n+1})=\chi(g(C_n))\le\gamma\chi(C_n)$, so $K=\bigcap_n C_n$ is nonempty,
compact and convex. If $x\in K$ then $g(x)\in g(C_n)\subset C_{n+1}$ for all $n$,
so $g(K)\subset K$; Lemma~\ref{lem:compactfixed} applies.
\end{proof}

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