A positive solution to Tingley’s problem

Proof. Start with the norming functional on the line through a nonzero vector, or the zero functional at zero. To extend a functional L from a subspace V to V+\R e, choose its value at e in

The intervals intersect pairwise, since L(a)-L(b)\le\norm{a-b}\le\norm{a-e}+\norm{b-e}. They have a common point: they are closed intervals, have the finite intersection property, and the interval indexed by zero is compact. The chosen value gives |L(a)-L(e)|\le\norm{a-e}; rescaling proves the norm bound on V+\R e. Finitely many such extensions suffice. Finally, a support at \alpha v+\beta w must take value one at both unit vectors, since both weights are positive.□

Proof. Continuity of the norm and boundedness of the unit ball give contact. Every point strictly between an interior point of a convex ball and a point of that ball is interior: the corresponding convex combination contains a ball of positive radius about it. This proves uniqueness. The bounds follow from the triangle inequality. If v_n\to v, every convergent subsequence of the bounded scalar sequence \rho(v_n) has limit \rho(v) by continuity of the norm and uniqueness of contact; hence the whole sequence converges.□

Proof. Choose triples approaching the absolute supremum. Pass to a subsequence of one sign and, if necessary, replace the map by its inverse. After a further subsequence the radius parameters converge to t\in[0,1]. The uniform Lipschitz bound allows us to replace them all by t, retaining convergence of the signed defects to M. By (3), M/2\le t\le1-M/2.

We describe the limit construction to justify attainment. Choose a free ultrafilter \mathcal U on \mathbb N: a maximal proper filter containing all cofinite sets. It exists by the maximal principle, since a union of a chain of such filters is again a proper filter. We use the usual set-theoretic framework with choice here. For any subset, either it or its complement belongs to \mathcal U. Every bounded real sequence has a unique limit along \mathcal U. Indeed, in a compact interval containing the sequence, the closures of its values on sets in \mathcal U have the finite intersection property; their intersection is nonempty and the ultrafilter property makes it a singleton. These limits preserve sums, inequalities, and continuous real functions.

On bounded sequences in X use the seminorm

Let Q_X be the quotient by its kernel, with the induced norm, and define Q_Y in the same way. Constant sequences embed X,Y isometrically, so these quotients are nonzero. Every unit vector in either quotient has a representative of unit vectors: normalize each nonzero term and replace zero terms by a fixed unit vector. The change has norm |\norm{a_n}-1|, whose ultralimit is zero.

Applying f termwise to unit representatives defines F:S_{Q_X}\to S_{Q_Y}. The identity

proves both independence of representative and distance preservation. Applying f^{-1} termwise gives the inverse of F, so it is surjective. For any fixed q\in[0,1],

Since the pair and q were arbitrary, the bound holds for every pair and every radius. The chosen maximizing sequences give equality at t.

Complete both quotient spaces. Their unit spheres are the metric completions of their previous spheres, by density and normalization. Extending the induced isometry and its inverse gives inverse maps by continuity and density, hence a surjective isometry of these completed spheres. Continuity preserves the defect bound and the selected equality. These spaces and this map have the asserted properties.□

Proof. Starting from a direction at ty, we map its sphere contact to S_Y and take the opposite sphere contact on the line through tb. Pull this opposite endpoint back to S_X, and reverse its direction from ty. A fixed direction for this return map will align the two chords.

For v\in S_X, use Lemma 2.2 to define p=p(v)>0 by x(v)=ty+pv\in S_X. Define successively

where w\in S_Y and u>0 is determined by radial contact in direction -w. These quantities are continuous in v, and their positive radii lie in [1-t,1+t].

Let E=\{v\in S_X:s(v)-p(v)=M\}. It is closed in X, since S_X is closed and s-p is continuous on S_X. It is nonempty: v_0=(x_0-ty)/\norm{x_0-ty} satisfies x(v_0)=x_0 by uniqueness of radial contact, and belongs to E by (4). For v\in E, distance preservation and the global defect bound give

Since s=p+M, both inequalities are equalities. Thus

In particular v and g(v) share a support. Moreover, every unit direction l sharing a support with g(v) belongs to E: for the fixed z,r,u associated with v,

so s(l)-p(l)\ge r-u=M. The reverse inequality is the global defect bound. In particular g(v)\in E.

An invariant convex set. Start with v_0\in E. Suppose \co\{v_0,\ldots,v_n\}\subset E. Choose a strictly positive convex combination a_n of these vectors, and set v_{n+1}=g(a_n). A common support \varphi for a_n,v_{n+1}, extended to their finite span with the v_i, takes value one at each v_i by strict positivity. For every l in the enlarged convex hull,

Thus l is a unit vector sharing \varphi with v_{n+1}=g(a_n), and belongs to E by the preceding propagation rule.

Choose the coefficients so that the a_n are dense in C=\clco\{v_0,v_1,\ldots\}. Explicitly, enumerate all finite rational probability vectors on initial lists of indices, repeat each arbitrarily late, and at stage n use one whose list is contained in \{0,\ldots,n\}. Mix its zero-padded coefficients with the uniform positive vector on that list of n+1 indices, using a weight 0<\lambda_n<1 tending to zero. This changes the prescribed combination by at most 2\lambda_n, because every v_i is unit. Each prescribed rational combination is therefore a limit of a_n’s. Such combinations are dense in C, proving the assertion. The set C is nonempty, closed, bounded, convex, and contained in E\subset S_X. Since g(a_n)=v_{n+1}\in C, continuity gives g(C)\subset C. Only this constructed set is separable; the ambient spaces are unrestricted.

A strict decrease of noncompactness. Compare v,v'\in E, marking their associated variables by primes. From (7),

Using s=p+M and r=u+M yields

Set a=1-t>0 and U=1+t. All radii lie in [a,U], and r=u+M, s=p+M give 0<M<r,s. Hence

Fix a finite cover of H\subset C by sets of diameter at most \delta. For \eta>0, intersect its parts with H and subdivide them according to finite partitions of the ranges of p and u into intervals of length at most \eta. Since r,s\ge a and u/r<1, the last two terms in (9) sum to at most 4\eta/a on each resulting part. The images of these parts cover g(H) with diameters at most \gamma\delta+4\eta/a. Letting \eta\downarrow0 for this fixed cover and then taking the infimum over diameter covers gives \chi(g(H))\le\gamma\chi(H).

Theorem 4.1 now gives v\in C with g(v)=v. Set x=x(v); this endpoint may differ from x_0. Substitution in (7) gives (5). Equation (8) and distance preservation give (6).□

Proof. At radius t-sh, the signed defect for x,y is

At radius t+uh, the negative of the signed defect for z,y is

Both radii lie in [0,1], so both defects are at most M. The choice of h makes at least one radius an endpoint, where its defect is zero. The weighted sum u\Delta_x+s\Delta_z cancels the common target norm; since sr-up=Md,

The inequality t/s\le\epsilon/u is equivalent to B\le\epsilon. In that case \Delta_x=0 and K\le s/d=1-B; otherwise \Delta_z=0 and K\le u/d=B. At B=\epsilon both defects vanish, so K=0 and the stated upper bound remains valid.

The supported vector is nonzero: if v+\theta y=0, then \theta=1 and v=-y, whereas \norm{ty+pv}=1 forces p=1+t and hence \theta\le t/p<1.

Since r/u>p/s>0, a support at v+\theta y, hence also at (p/s)v+hy, bounds the secant from below by \varphi(v). It remains to prove \varphi(v)\ge\epsilon/p. Write \beta=t/p, so 0<\theta\le\beta. If \theta=\beta, then

because \norm{pv+ty}=1. If \theta<\beta, take a support \psi at v+\beta y. The two support inequalities give

Thus \psi(y)\ge\varphi(y) and \varphi(v)\ge\psi(v)\ge\epsilon/p.□

Proof. First suppose B>\epsilon. Then h=\epsilon/u and Lemma 5.1 gives

Let L be any support at v, and write j=L(y). For a support \varphi at v+\theta y, Lemma 5.1 gives

In fact B>1+\theta j-\theta. If equality held throughout (13), then \norm{v+\theta y}=L(v+\theta y), so L would also support at v+\theta y. But Lemma 5.1 applies to every support there and would give 1=L(v)\le B<1. Therefore

Every support at v is strictly negative on y. Applying one to rv-ty=-z gives r-tj\le1, so u<r<1. Equation (12), namely Au\ge\epsilon, then gives A>\epsilon.

We record the inverse configuration explicitly. For f^{-1} use f(z) as the first endpoint, b as the base direction, and f(x) as the opposite endpoint. The new parameters are

By (2), the inverse retains the bound in (4), and its defect at (f(z),b,t) is s'-p'=r-u=M. The new configuration therefore has the same hypotheses, with (A',B')=(B,A). The preceding argument thus says that A>\epsilon implies B>\epsilon and that every support at -w is strictly negative on b. Equivalently, every support at w is strictly positive on b.

Consequently, if either strict inequality holds, both support-sign conclusions hold. For 0<\eta<t, the defect bound at radius t-\eta gives

A support at w shows \norm{sw+\eta b}>s. Also \norm{pv+\eta y}<p for all sufficiently small positive \eta. Otherwise take \eta_n\downarrow0 with these norms at least p, and supports L_n at pv+\eta_n y on \operatorname{span}\{v,y\}. Since L_n(v)\le1,

A convergent subsequence in the compact dual unit ball of this finite-dimensional span has a limit L with L(y)\ge0. The support identities give

Thus L supports v and is nonnegative on y, a contradiction. The two strict norm inequalities contradict (15).□

Proof. By Lemma 5.2, A,B\le\epsilon. Apply Lemma 5.1 both to the original configuration and to (14). This gives

The smaller of the two products is strictly below 1/4: if A\le B, then A<1/2 and A(1-B)\le A(1-A)<1/4; interchange the letters otherwise. As d\le2, (16) implies \epsilon<1/2, so A\le\epsilon<t and 1-B\ge t. Consequently,

Consider the convex functions

The unit lengths of x,f(x) give

By (17) and monotonicity of secant slopes,

On the other hand, u/\epsilon\in[0,s/t], so convexity gives G(u/\epsilon)\le1/t. But sphere distance preservation gives

a contradiction.□

Proof of Theorem 1.1. If the defect of the original sphere map were positive, Lemma 3.1 and Proposition 5.3 would contradict one another. Thus M=0. For a=\alpha x and c=\beta y, where x,y are unit and \alpha\ge\beta>0, put q=\beta/\alpha. The radial map satisfies

Exchanging the points handles the other ordering; distances involving zero follow from norm preservation. For b\in Y\setminus\{0\}, choose y\in S_X with f(y)=b/\norm{b}. Then T(\norm{b}y)=b. Hence T is a surjective metric isometry fixing zero.

For completeness we prove the affine step of Mazur–Ulam [13] in the short reflection presentation of Nica [16], who explicitly credits Väisälä and Vogt [22, 23]. If F:E\to F_0 is an onto isometry of real normed spaces and a,c\in E, let

The triangle inequality gives \delta\le\norm{a-c}/2. Reflection R(z)=2m'-z is an onto isometry of F_0. Its conjugate H=F^{-1}RF interchanges a,c and has midpoint defect

Apply the same construction repeatedly to the resulting onto self-isometries of E. Their midpoint defects double but obey the same bound \norm{a-c}/2, so \delta=0. Thus every onto isometry preserves midpoints.

In particular T((a+c)/2)=(T(a)+T(c))/2. Since T(0)=0, first T(2a)=2T(a) and then T(a+c)=T(a)+T(c). Rational homogeneity follows from additivity, and real homogeneity from continuity. Finally, a linear extension is determined on every nonzero vector by its norm and its unit direction, proving uniqueness.□

Proof. Closing each member of a finite diameter cover preserves its diameter and covers the closure, proving the first identity. To prove the second, cover H by finitely many nonempty subsets H_1,\ldots,H_N of diameter at most \delta, and let R bound the norms of their points. Each \co H_i has diameter at most \delta: for two convex combinations, expand their difference using products of their coefficients and apply the triangle inequality. Every point of \co H has the form

choosing arbitrary z_i when \lambda_i=0. Partition the coefficient simplex into finitely many sets of \ell_1-diameter at most \eta. For each part, take all sums admitting a representation with coefficients in that part. These sets cover \co H. Two sums in one such set differ in norm by at most \delta+R\eta, by adding and subtracting the combination with one coefficient vector and the other set of z_i’s. Thus \chi(\co H)\le\delta+R\eta. Let \eta\downarrow0, take the infimum in \delta, and use the closure identity and monotonicity for the reverse inequality.

For the intersection assertion, choose x_n\in C_n. For each positive integer k, a sufficiently late C_n has a finite cover of diameter less than 2^{-k}. Successive infinite subsequences, followed by a diagonal choice, produce a Cauchy subsequence of the x_n with indices tending to infinity. Its limit belongs to every C_n, by completeness, nesting, and closedness. Their intersection is closed and has measure of noncompactness zero. Finite covers of arbitrarily small diameter give total boundedness, so the intersection is compact.□

Proof. The zero-dimensional case is immediate. Let P be a continuous self-map of the simplex \Delta_m=\{a\in\R^{m+1}:a_i\ge0,\ \sum_i a_i=1\}. Subdivide it into simplices, and label each subdivision vertex a by an index i such that a_i>0 and a_i\ge P(a)_i. Such an index exists: otherwise summing over the positive coordinates of a would give 1<\sum_i P(a)_i=1. The label of a vertex lies among the vertices of every original face containing it.

Here is the required combinatorial fact, including its parity proof. In a finite face-compatible triangulation of an m-simplex with labels 0,\ldots,m satisfying that boundary condition, the number of small simplices carrying all labels is odd. Count incidences of small m-simplices with facets labeled 0,\ldots,m-1, modulo two. Internal facets count twice. Such a boundary facet can lie only in the original face missing label m; by induction its number is odd, starting with the single vertex in dimension zero. A small m-simplex contributes one incidence if it carries all labels, two if its vertex labels include 0,\ldots,m-1 with one of these repeated, and none otherwise. This proves the claim.

There are finite compatible subdivisions of arbitrarily small mesh: subdivide faces first and cone their subdivisions to the barycenter. The vertices of each resulting simplex are barycenters of a nested chain of faces. If two nested faces have k and l vertices, k\le l, their barycenters differ by at most (1-k/l) times the original diameter. Thus a subdivision in dimension m\ge1 reduces each diameter by a factor at most m/(m+1). Repeat.

In each subdivision choose a fully labeled small simplex. Compactness and the shrinking mesh give a subsequence whose vertices converge to one point a. For every coordinate i, the vertex with label i satisfies the corresponding inequality; continuity yields a_i\ge P(a)_i. Summing gives equality in every coordinate, so a=P(a).□

Proof. For \delta>0, choose a finite strict \delta-net P_1,\ldots,P_N\in K. For x\in K set

The denominator is positive, and these are continuous probability weights. Map the probability simplex to itself by sending coefficients (a_i) to (\lambda_i(\sum_j a_jP_j)). By Lemma A.2 there is a fixed coefficient vector. Its convex combination x\in K satisfies

since a positive weight implies the corresponding distance is less than \delta. Letting \delta\downarrow0 and taking a convergent subsequence in K gives a fixed point.□

Proof of Theorem 4.1. Set C_0=C and C_{n+1}=\clco(g(C_n)). These are nonempty closed bounded convex subsets of C satisfying C_{n+1}\subset C_n. By Lemma A.1,

so their intersection K is nonempty, compact, and convex. If x\in K, then g(x)\in g(C_n)\subset C_{n+1} for every n, so g(K)\subset K. Apply Lemma A.3.□