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 \(V+\mathbb Re\), choose its value at \(e\) in \[\bigcap_{a\in V}[L(a)-\left\lVert a-e\right\rVert,\ L(a)+\left\lVert a-e\right\rVert].\] The intervals intersect pairwise, since \(L(a)-L(b)\le\left\lVert a-b\right\rVert\le\left\lVert a-e\right\rVert+\left\lVert b-e\right\rVert\). 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 \(|L(a)-L(e)|\le\left\lVert a-e\right\rVert\); rescaling proves the norm bound on \(V+\mathbb Re\). 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 \[\left\lVert(a_n)\right\rVert_{\mathcal U}=\lim_{\mathcal U}\left\lVert a_n\right\rVert.\] 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 \(|\left\lVert a_n\right\rVert-1|\), whose ultralimit is zero.

Applying \(f\) termwise to unit representatives defines \(F:S_{Q_X}\to S_{Q_Y}\). The identity \[\lim_{\mathcal U}\left\lVert f(a_n)-f(b_n)\right\rVert =\lim_{\mathcal U}\left\lVert a_n-b_n\right\rVert\] proves both independence of representative and distance preservation. Applying \(f^{-1}\) termwise gives the inverse of \(F\), so it is surjective. For any fixed \(q\in[0,1]\), \[\left\lVert F([a_n])-qF([b_n])\right\rVert-\left\lVert[a_n]-q[b_n]\right\rVert =\lim_{\mathcal U}D_q(a_n,b_n)\in[-M,M].\] 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 3 to define \(p=p(v)>0\) by \(x(v)=ty+pv\in S_X\). Define successively \[ f(x(v))=tb+sw,\qquad f(z)=tb-uw\in S_Y,\qquad r=\left\lVert z-ty\right\rVert,\quad g(v)=\frac{ty-z}{r}, \tag{7}\] 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: \(v_0=(x_0-ty)/\left\lVert x_0-ty\right\rVert\) 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 \[s+u=\left\lVert x(v)-z\right\rVert\le p+r,\qquad r-u\le M.\] Since \(s=p+M\), both inequalities are equalities. Thus \[ r=u+M,\qquad \left\lVert pv+rg(v)\right\rVert=p+r. \tag{8}\] 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 \(v\), \[p(l)+r=\left\lVert x(l)-z\right\rVert =\left\lVert f(x(l))-f(z)\right\rVert\le s(l)+u,\] 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 \(\operatorname{conv}\{v_0,\ldots,v_n\}\subset E\). 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, \[1=\varphi(l)\le\left\lVert l\right\rVert\le1.\] 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 \(C=\overline{\operatorname{conv}}\{v_0,v_1,\ldots\}\). 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), \[\begin{align*} r\left\lVert g(v)-g(v')\right\rVert &\le |r-r'|+\left\lVert z-z'\right\rVert\\ &\le |r-r'|+|u-u'|+u\left\lVert w-w'\right\rVert,\\ s\left\lVert w-w'\right\rVert &\le |s-s'|+\left\lVert f(x(v))-f(x(v'))\right\rVert\\ &\le |s-s'|+|p-p'|+p\left\lVert v-v'\right\rVert. \end{align*}\] Using \(s=p+M\) and \(r=u+M\) yields \[ \left\lVert g(v)-g(v')\right\rVert \le\frac{up}{rs}\left\lVert v-v'\right\rVert +\frac{2|u-u'|}{r}+\frac{2u|p-p'|}{rs}. \tag{9}\] 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 \[\frac{up}{rs} =\left(1-\frac Mr\right)\left(1-\frac Ms\right) \le\gamma:=\left(1-\frac{M}{U}\right)^2<1.\] 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 5 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 \[\Delta_x:=D_{t-sh}(x,y)=s\bigl(\left\lVert w+hb\right\rVert-\left\lVert(p/s)v+hy\right\rVert\bigr).\] At radius \(t+uh\), the negative of the signed defect for \(z,y\) is \[\Delta_z:=-D_{t+uh}(z,y)=u\bigl(\left\lVert(r/u)v+hy\right\rVert-\left\lVert w+hb\right\rVert\bigr).\] 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 \(sr-up=Md\), \[\frac{u\Delta_x+s\Delta_z}{d}=MK.\] 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 \(\left\lVert ty+pv\right\rVert=1\) 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 \[\varphi(v)=\frac{1-t\varphi(y)}p\ge\frac\epsilon p\] because \(\left\lVert pv+ty\right\rVert=1\). If \(\theta<\beta\), take a support \(\psi\) at \(v+\beta y\). The two support inequalities give \[\theta\bigl(\psi(y)-\varphi(y)\bigr) \le\varphi(v)-\psi(v) \le\beta\bigl(\psi(y)-\varphi(y)\bigr).\] 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 7 gives \[ \frac{pu}{d\epsilon}\ge1. \tag{12}\] Let \(L\) be any support at \(v\), and write \(j=L(y)\). For a support \(\varphi\) at \(v+\theta y\), Lemma 7 gives \[ B\ge\varphi(v)=\left\lVert v+\theta y\right\rVert-\theta\varphi(y) \ge1+\theta j-\theta. \tag{13}\] In fact \(B>1+\theta j-\theta\). If equality held throughout (13), then \(\left\lVert v+\theta y\right\rVert=L(v+\theta y)\), so \(L\) would also support at \(v+\theta y\). But Lemma 7 applies to every support there and would give \(1=L(v)\le B<1\). Therefore \[1-j>\frac{1-B}{\theta}=\frac{pu}{d\epsilon}\ge1.\] 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 \[ (p',s',r',u')=(u,r,s,p),\quad (v',w',y',b')=(-w,-v,b,y). \tag{14}\] 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 \[ \left\lVert sw+\eta b\right\rVert-\left\lVert pv+\eta y\right\rVert\le M=s-p. \tag{15}\] A support at \(w\) shows \(\left\lVert sw+\eta b\right\rVert>s\). Also \(\left\lVert pv+\eta y\right\rVert<p\) 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 \(L_n(v)\le1\), \[p\le pL_n(v)+\eta_n L_n(y)\quad\Longrightarrow\quad L_n(y)\ge0.\] 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 \[pL(v)=\lim_n\bigl(\left\lVert pv+\eta_n y\right\rVert-\eta_n L_n(y)\bigr)=p.\] Thus \(L\) supports \(v\) and is nonnegative on \(y\), a contradiction. The two strict norm inequalities contradict (15). ◻

Proof. By Lemma 8, \(A,B\le\epsilon\). Apply Lemma 7 both to the original configuration and to (14). This gives \[ \epsilon\le dA(1-B),\qquad \epsilon\le dB(1-A). \tag{16}\] 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, \[ \frac p t<\frac r\epsilon,\qquad \frac s t\ge\frac u\epsilon. \tag{17}\]

Consider the convex functions \[F(\lambda)=\left\lVert y+\lambda v\right\rVert,\qquad G(\lambda)=\left\lVert b+\lambda w\right\rVert\qquad(\lambda\ge0).\] The unit lengths of \(x,f(x)\) give \[F(0)=G(0)=1,\qquad F(p/t)=G(s/t)=1/t>1.\] By (17) and monotonicity of secant slopes, \[F(r/\epsilon)\ge1+\frac{r/\epsilon}{p/t}(1/t-1)>1/t.\] On the other hand, \(u/\epsilon\in[0,s/t]\), so convexity gives \(G(u/\epsilon)\le1/t\). But sphere distance preservation gives \[\epsilon F(r/\epsilon)=\left\lVert y-z\right\rVert =\left\lVert b-f(z)\right\rVert=\epsilon G(u/\epsilon),\] a contradiction. ◻

Proof of Theorem 1. If the defect of the original sphere map were positive, Lemma 4 and Proposition 9 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 \[\left\lVert T(a)-T(c)\right\rVert =\alpha\left\lVert f(x)-qf(y)\right\rVert =\alpha\left\lVert x-qy\right\rVert=\left\lVert a-c\right\rVert.\] 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 \(f(y)=b/\left\lVert b\right\rVert\). Then \(T(\left\lVert b\right\rVert y)=b\). 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 \[m=(a+c)/2,\qquad m'=(F(a)+F(c))/2, \qquad \delta=\left\lVert F(m)-m'\right\rVert.\] The triangle inequality gives \(\delta\le\left\lVert a-c\right\rVert/2\). 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 \[\left\lVert H(m)-m\right\rVert=\left\lVert RF(m)-F(m)\right\rVert=2\delta.\] Apply the same construction repeatedly to the resulting onto self-isometries of \(E\). Their midpoint defects double but obey the same bound \(\left\lVert a-c\right\rVert/2\), 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 \(\operatorname{conv}H_i\) 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 \(\operatorname{conv}H\) has the form \[\sum_{i=1}^N\lambda_i z_i,\qquad z_i\in\operatorname{conv}H_i, \quad\lambda_i\ge0,\quad\sum_i\lambda_i=1,\] 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 \(\operatorname{conv}H\). 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 \(\chi(\operatorname{conv}H)\le\delta+R\eta\). 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 \(\Delta_m=\{a\in\mathbb R^{m+1}:a_i\ge0,\ \sum_i a_i=1\}\). 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 \[\lambda_i(x)= \frac{\max\{\delta-\left\lVert P_i-g(x)\right\rVert,0\}} {\sum_j\max\{\delta-\left\lVert P_j-g(x)\right\rVert,0\}}.\] 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 11 there is a fixed coefficient vector. Its convex combination \(x\in K\) satisfies \[\left\lVert x-g(x)\right\rVert \le\sum_i\lambda_i(x)\left\lVert P_i-g(x)\right\rVert<\delta,\] 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 5. Set \(C_0=C\) and \(C_{n+1}=\overline{\operatorname{conv}}(g(C_n))\). These are nonempty closed bounded convex subsets of \(C\) satisfying \(C_{n+1}\subset C_n\). By Lemma 10, \[\chi(C_{n+1})=\chi(g(C_n))\le\gamma\chi(C_n),\] 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 12. ◻