Aligned chords

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This lesson turns an attained positive defect into a rigid configuration: a line through the point $ty$ inside the unit ball of $X$ whose chord is mapped by $f$ to a chord of the unit sphere of $Y$ through $t f(y)$, with the lengths of the four half-chords related by the defect (Proposition 5.1). The configuration is a fixed point of a return map on directions [OpenAI-T, Section 4]. Starting from a direction $v$ at $ty$, go to the sphere, apply $f$, take the opposite end of the chord through $tf(y)$ in $Y$, pull it back by $f^{-1}$, and read off the direction from that point back to $ty$. Extremality of the defect propagates along common supports and produces a closed convex set of directions that the return map preserves. On this set the return map contracts the measure of noncompactness, and Darbo's theorem gives a fixed direction.

We use: from Supports, radial contact and the Mazur–Ulam theorem, Lemma 1.1 (common supports), Lemma 2.1 (radial contact) and the defect of Section 4; from Ultrapowers and Darbo's fixed-point theorem, Proposition 3.1 (attainment) and Theorem 6.1 (Darbo).

Standing assumptions. $X,Y$ are real Banach spaces, $f:S_X\to S_Y$ is a surjective isometry, $M>0$, and there are $x_0,y\in S_X$ and $0<t<1$ with the conditions (0.1): $|D_q(a,c)|\le M$ for $a,c\in S_X$ and $0\le q\le1$, and $D_t(x_0,y)=M$, as Proposition 3.1 of the previous lesson provides. Put $b=f(y)\in S_Y$ and $\epsilon=1-t$.

1. The return map

For $v\in S_X$ define, in this order:

$p=p(v)>0$ with $x(v)=ty+pv\in S_X$, by Lemma 2.1 of the first lesson with $c=ty$;
$s=s(v)=\|f(x(v))-tb\|$ and $w=w(v)=(f(x(v))-tb)/s\in S_Y$, so that $f(x(v))=tb+sw$;
$u=u(v)>0$ with $tb-uw\in S_Y$, by Lemma 2.1 of the first lesson in $Y$ with $c=tb$ and the direction $-w$;
$z=z(v)=f^{-1}(tb-uw)\in S_X$, $r=r(v)=\|z-ty\|$, and $g(v)=(ty-z)/r\in S_X$, so that $z=ty-r\,g(v)$.

Lemma 1.1. The maps $p,s,w,u,z,r,g$ are well defined and continuous on $S_X$, and $p,s,u,r\in[\epsilon,1+t]$. For every $v\in S_X$, $\begin{aligned} &s-p=D_t(x(v),y)\\ &\quad \le M,\\ &r-u=-D_t(z(v),y)\\ &\quad \le M. \end{aligned}\tag{1.1}$

Proof. For unit vectors $e$ and $\|c\|=t$, $\|e-c\|\in[1-t,1+t]$; this bounds $s$ and $r$, and Lemma 2.1 of the first lesson bounds $p$ and $u$. In particular $s,r>0$, so $w$ and $g$ are defined. Continuity follows from the continuity of $p$ and of the radial contact in $Y$ (Lemma 2.1 of the first lesson), of $f$ and $f^{-1}$, and of the norm. Finally $\begin{aligned} &D_t(x(v),y)=\|tb+sw-tb\|\\ &\qquad-\|ty+pv-ty\|=s-p \end{aligned}$ and $D_t(z(v),y)=u-r$, and (0.1) bounds both. $\square$

2. The extremal set and propagation

Let $E=\{v\in S_X:s(v)-p(v)=M\}$.

Lemma 2.1. $E$ is closed, and $v_0=(x_0-ty)/\|x_0-ty\|$ lies in $E$.

Proof. $E$ is closed by continuity of $s-p$ on the closed set $S_X$. The point $x_0=ty+\|x_0-ty\|v_0$ lies on $S_X$, so $x(v_0)=x_0$ by uniqueness of radial contact, and $s(v_0)-p(v_0)=D_t(x_0,y)=M$. $\square$

Lemma 2.2. Let $v\in E$. Then $r-u=M$, $\|x(v)-z(v)\|=p+r=s+u$, and $v$ and $g(v)$ share a support.

Lemma 2.3 (propagation). Let $v\in E$, and let $l\in S_X$ share a support with $g(v)$. Then $l\in E$. In particular $g(E)\subseteq E$.

Proof. Keep $z=z(v)$, $r=r(v)$, $u=u(v)$, $w=w(v)$, and let $\varphi$ be a common support of $l$ and $g(v)$. Since $x(l)-z=p(l)l+r\,g(v)$, $\begin{aligned} &p(l)+r=\varphi\bigl(x(l)-z\bigr)\\ &\quad \le\|x(l)-z\|\le p(l)+r, \end{aligned}$ so $\|x(l)-z\|=p(l)+r$. On the other side, $\|x(l)-z\|=\|f(x(l))-f(z)\|$, which is $\|s(l)w(l)+uw\|\le s(l)+u$. Hence $s(l)-p(l)\ge r-u=M$, and (1.1) gives equality. By Lemma 2.2, $g(v)$ shares a support with itself, so $g(v)\in E$. $\square$

3. An invariant convex set

Lemma 3.1. There is a nonempty closed convex set $C\subseteq E$ with $g(C)\subseteq C$.

Proof. Fix an enumeration $(\pi_n)_{n\ge0}$ of the pairs $(m,\beta)$, $m\ge0$, $\beta$ a probability vector with rational entries indexed by $\{0,\dots,m\}$, in which every pair occurs infinitely often; the set of pairs is countable. Put $\lambda_n=1/(n+2)$. Let $v_0$ be as in Lemma 2.1, and suppose $v_0,\dots,v_n$ have been chosen with $\operatorname{conv}\{v_0,\dots,v_n\}\subseteq E$, which holds for $n=0$. If $\pi_n=(m,\beta)$ with $m\le n$, let $\alpha^{(n)}$ be $(1-\lambda_n)\tilde\beta+\lambda_n\upsilon_n$, where $\tilde\beta$ is $\beta$ extended by zeros to $\{0,\dots,n\}$ and $\upsilon_n$ is the uniform probability vector on $\{0,\dots,n\}$; otherwise let $\alpha^{(n)}=\upsilon_n$. Put $\begin{aligned} &a_n=\sum_{i=0}^n\alpha^{(n)}_iv_i\in\operatorname{conv}\{v_0,\dots,v_n\}\\ &\quad \subseteq E,\\ &v_{n+1}=g(a_n). \end{aligned}$ All coefficients $\alpha^{(n)}_i$ are positive. By Lemma 2.2, $a_n$ and $v_{n+1}$ share a support $\varphi$. By Lemma 1.1 of the first lesson, $\varphi(v_i)=1$ for $i\le n$, hence $\varphi=1$ on $\operatorname{conv}\{v_0,\dots,v_{n+1}\}$, and every $l$ in this hull is a unit vector sharing the support $\varphi$ with $g(a_n)$. By Lemma 2.3, $l\in E$. This continues the induction.

Let $C$ be the closed convex hull of $\{v_0,v_1,\dots\}$. It is nonempty, closed, bounded and convex, and $C\subseteq E$ because $E$ is closed and contains every $\operatorname{conv}\{v_0,\dots,v_n\}$. The points $a_n$ are dense in $C$: the rational convex combinations $\sum_{i\le m}\beta_iv_i$ are dense in $C$, and for each such $(m,\beta)$ the infinitely many $n\ge m$ with $\pi_n=(m,\beta)$ give $\|a_n-\sum_i\beta_iv_i\|\le\lambda_n\|\upsilon_n-\tilde\beta\|_1$, which is at most $2\lambda_n\to0$. Since $g(a_n)=v_{n+1}\in C$ and $g$ is continuous, $g(C)\subseteq C$. $\square$

4. The return map contracts noncompactness

Lemma 4.1. For $v,v'\in E$, with the quantities of $v'$ marked by primes, $\begin{aligned} &\|g(v)-g(v')\|\le\gamma\|v-v'\|\\ &\quad +\frac2\epsilon\bigl(|u-u'|+|p-p'|\bigr),\\ &\gamma=\Bigl(1-\frac M{1+t}\Bigr)^2<1. \end{aligned}$

Proof. From $g(v)=(ty-z)/r$ and $\|ty-z'\|=r'$, $\begin{aligned} &r\|g(v)-g(v')\|\\ &\quad =\Bigl\|(ty-z)-\tfrac r{r'}(ty-z')\Bigr\|\\ &\quad \le\|z-z'\|+|r-r'|. \end{aligned}$ Since $f$ is an isometry, $\|z-z'\|=\|uw-u'w'\|$, which is at most $|u-u'|+u\|w-w'\|$. In the same way, from $w=(f(x(v))-tb)/s$, $\begin{aligned} &s\|w-w'\|\le\|f(x(v))-f(x(v'))\|\\ &\quad +|s-s'|=\|pv-p'v'\|+|s-s'|\\ &\quad \le p\|v-v'\|+|p-p'|+|s-s'|. \end{aligned}$ On $E$ we have $s=p+M$ and $r=u+M$ (Lemma 2.2), so $|s-s'|=|p-p'|$ and $|r-r'|=|u-u'|$. Combining, $\begin{aligned} &\|g(v)-g(v')\|\le\frac{up}{rs}\|v-v'\|\\ &\quad +\frac{2|u-u'|}r+\frac{2u|p-p'|}{rs}. \end{aligned}$ Here $\frac{up}{rs}=(1-\frac Mr)(1-\frac Ms)\le\gamma$ because $r,s\le1+t$, and $\frac ur<1$, $r,s\ge\epsilon$. $\square$

Lemma 4.2. For every nonempty $H\subseteq C$, $\chi(g(H))\le\gamma\chi(H)$.

Proof. Let $H$ be covered by finitely many sets $H_1,\dots,H_N\subseteq H$ of diameter at most $\delta$, and let $\eta>0$. Divide $[\epsilon,1+t]$ into finitely many intervals of length at most $\eta$, and split each $H_j$ according to the intervals containing $p(v)$ and $u(v)$. On each of the finitely many resulting pieces, Lemma 4.1 gives $\|g(v)-g(v')\|\le\gamma\delta+4\eta/\epsilon$, since $C\subseteq E$. So $g(H)$ is covered by finitely many sets of diameter at most $\gamma\delta+4\eta/\epsilon$. Letting $\eta\to0$ and then $\delta\downarrow\chi(H)$ proves the claim. $\square$

5. Aligned chords

Proposition 5.1. Under the standing assumptions there are $v\in S_X$, $w\in S_Y$ and numbers $p,r,s,u\in[\epsilon,1+t]$ such that the points $\begin{gathered} x=ty+pv,\\ z=ty-rv\qquad \text{lie in }S_X,\\ f(x)=tb+sw,\\ f(z)=tb-uw, \end{gathered}\tag{5.1}$ and $\begin{aligned} &s-p=r-u=M,\\ &d:=p+r=s+u\\ &\quad =\|x-z\|\le2. \end{aligned}\tag{5.2}$

Proof. By Lemmas 3.1 and 4.2, Darbo's theorem (Theorem 6.1 of the previous lesson) applies to $g$ on $C$, in the Banach space $X$, and gives $v\in C$ with $g(v)=v$. Take $x=x(v)$, $z=z(v)=ty-r\,g(v)=ty-rv$, and $p,s,u,r,w$ as in Section 1. Since $v\in E$, Lemma 2.2 gives $s-p=r-u=M$ and $\|x-z\|=p+r=s+u$, and $\|x-z\|\le\|x\|+\|z\|=2$. $\square$

Both chords lie on lines: the chord $[z,x]$ of the unit ball of $X$ through $ty$ has direction $v$, and its image chord $[f(z),f(x)]$ through $tb$ has direction $w$. The segment from $ty$ to $x$ has length $p$, and its image segment from $tb$ to $f(x)$ is longer by exactly $M$; on the other side the segment from $ty$ to $z$ is longer than the image segment by exactly $M$. The point $x$ need not equal $x_0$.

6. Exercises

Exercise 6.1 (easy). Show that if $f$ is the restriction of a linear isometry $T$, then $g(v)=v$ for every $v\in S_X$, with $w=Tv$, $s=p$ and $u=r$.

Exercise 6.2 (medium). Show that the constant $\gamma$ of Lemma 4.1 can be replaced by $\sup_{v\in E}\frac{u(v)p(v)}{r(v)s(v)}$, and explain why the argument needs $M>0$: what happens to the estimate when $M=0$?

Exercise 6.3 (easy). Let $\varphi_n$ be the common support of $a_n$ and $v_{n+1}$ in the proof of Lemma 3.1. Show that $\varphi_n=1$ on $\overline{\operatorname{conv}}\{v_0,\dots,v_{n+1}\}$, and that this set consists of unit vectors.

Exercise 6.4 (medium). Show that a single $\varphi\in X^*$ with $\|\varphi\|\le1$ equals $1$ on all of $C$. (Use the Banach–Alaoglu theorem.)

7. Solutions

6.1. Then $f(x(v))=T(ty+pv)=tb+pTv$, so $s=p$ and $w=Tv$; the opposite contact in $Y$ is $tb-rTv=T(ty-rv)$, where $ty-rv\in S_X$ is the opposite contact of the line in $X$, so $u=r$, $z=ty-rv$, and $g(v)=v$.

6.2. The first term of the last display in the proof of Lemma 4.1 is $\frac{up}{rs}\|v-v'\|$, so the supremum over $E$ can replace $\gamma$. When $M=0$, $\frac{up}{rs}=1$ and no contraction results; positivity of $M$ makes the factor $(1-\frac Mr)(1-\frac Ms)$ strictly less than $1$, uniformly because $r,s\le1+t$.

6.3. $\varphi_n$ has norm at most $1$ and equals $1$ on $\operatorname{conv}\{v_0,\dots,v_{n+1}\}$, as shown in the proof; by continuity it equals $1$ on the closure, and $1=\varphi_n(l)\le\|l\|\le1$ there.

6.4. For each $i$, $\varphi_n(v_i)=1$ for all $n\ge i$, by Exercise 6.3. By the Banach–Alaoglu theorem the sequence $(\varphi_n)$ has a weak* limit point $\varphi$ in the unit ball of $X^*$. For $N\ge i$, the weak* closed set $\{\psi:\psi(v_i)=1\}$ contains every $\varphi_n$ with $n\ge N$, hence $\varphi$; so $\varphi(v_i)=1$ for all $i$, and $\varphi=1$ on $C$ by linearity and continuity. The argument of the lesson does not need this.

References

[OpenAI-T] OpenAI, A positive solution to Tingley's problem, OpenAI Math Release preprint, 23 September 2026, Section 4. https://github.com/openai/math/blob/main/preprints/A-positive-solution-to-Tingleys-problem-September-23-2026