2 Hölder’s Theorem and its consequences

An Archimedean ordered group is a triple $(G,\cdot , <)$ such that $(G,\cdot )$ is a group, $<$ is a linear order on $G$ , and:

1. (1) $<$ is both left and right invariant with respect to the group operation, meaning that for all $a,b,c \in G$ :Footnote ⁶

   $$\begin{align*}a < b \implies ca < cb \ \text{ and } \ ac < bc; \end{align*}$$

2. (2) whenever $1 < a < b$ , there is an $n \in \mathbb {N}$ such that $b < a^n$ .

We also say that an (un-ordered) group $(G,\cdot )$ is Archimedean if there exists a linear order $<$ on $G$ such that $(G,\cdot ,<)$ is an Archimedean ordered group.

Hölder’s Theorem is relevant for the study of continuous group actions on the reals.

Theorem 11 (Folklore; special case of Theorem 4 of [Reference Farb and Shalen2])

Suppose $I$ is an open interval of real numbers and $V$ is a subgroup of $\mathrm{Homeo}^+(I)$ that acts freely on $I$ . Then the binary relation $\prec $ on $V$ defined by

$$\begin{align*}\varphi \prec \psi \ :\iff \ \ \forall x \in I \ \varphi(x) < \psi(x) \end{align*}$$

is an Archimedean order on $V$ . Consequently, by Hölder’s Theorem, $V$ is abelian.

Since Theorem 11 is so central to the paper (in the key Lemma 13), we give a sketch of the proof. Fix any $x_0 \in I$ , and define the relation $\prec _{x_0}$ on $V$ by: $\varphi \prec _{x_0} \psi $ if $\varphi (x_0) < \psi (x_0)$ . The relation $\prec _{x_0}$ is clearly transitive and antisymmetric. It is total, because if $\varphi ,\psi \in V$ and $\varphi (x_0) = \psi (x_0)$ , then $\psi ^{-1} \circ \varphi $ fixes $x_0$ , and hence (since $V$ acts freely on $I$ ) $\psi ^{-1} \circ \varphi = \text {id}$ ; i.e., $\varphi =\psi $ . So $\prec _{x_0}$ is a linear order on $V$ . The Intermediate Value Theorem, together with the assumption that $V$ acts freely on $I$ , ensure that $\prec _{x_0}$ is the same as the relation $\prec $ defined in the statement of Theorem 11. This relation is also left and right invariant with respect to composition (on $V$ ). Finally, to see that $\prec _{x_0}$ is Archimedean, suppose $\text {id} \prec _{x_0} \varphi \prec _{x_0} \psi $ with $\varphi ,\psi \in V$ . Then $\text {id} \prec \varphi $ , and hence $\langle \varphi ^n(x_0) \ : \ n \in \mathbb {N} \rangle $ is a strictly increasing sequence. So it either converges to a point in $I$ , or to “ $+\infty $ ” (the right endpoint of $I$ ). It cannot converge to a point in $I$ , because if it did, that limit would be a fixed point of the continuous (nonidentity) function $\varphi $ , contradicting that $\varphi \in V$ and $V$ acts freely. So $\lim _{n \to \infty } \varphi ^n(x_0) = \infty $ ; hence, there is some $n$ such that $\psi (x_0) < \varphi ^n(x_0)$ . Then $\psi \prec _{x_0} \varphi ^n$ , which is equivalent to saying $\psi \prec \varphi ^n$ .

3 Proof of Theorem 5

In this section, we prove Theorem 5. If $I$ is an open interval of real numbers, the lower and upper endpoints of $I$ will be denoted $-\infty $ and $\infty $ , respectively. If $S$ is a nonempty set, $\boldsymbol {{}^I S}$ denotes the set of total functions from $I$ to $S$ , and $\boldsymbol {{}^{\underset {\smile }{I}} S}$ denotes the set of functions of the form $f: (-\infty ,t_f) \to S$ for some $t_f \in I$ . Uppercase and lowercase letters will be used for members of ${}^I S$ and ${}^{\underset {\smile }{I}} S$ , respectively, but note that if $F \in {}^I S$ and $t \in I$ , then $F|_{(-\infty ,t)}$ is a member of ${}^{\underset {\smile }{I}} S$ .

The notions of good and anonymous $S$ -predictors in this context are the obvious generalizations of the definitions given in the introduction (which were for the special case $I=\mathbb {R}$ ); i.e., a good $S$ -predictor is a function $\mathcal {P}: {}^{\underset {\smile }{I}} S \to S$ such that for all $F \in {}^I S$ ,

$$\begin{align*}\left\{ t \in I \ : \ \mathcal{P}\Big( F|_{(-\infty,t)} \Big) \ne F(t) \right\} \text{ has Lebesgue measure zero.} \end{align*}$$

If $U \subseteq \mathrm{Homeo}^+(I)$ , we say that a function $\mathcal {P}$ with domain ${}^{\underset {\smile }{I}} S$ is $U$ -anonymous if $\mathcal {P}(f) = \mathcal {P}(f \circ \varphi )$ for every $f \in {}^{\underset {\smile }{I}} S$ and every $\varphi \in U$ .

Lowercase Greek letters are reserved for elements of $\mathrm{Homeo}^+(I)$ . If $\varphi \in \mathrm{Homeo}^+(I)$ and $n \in \mathbb {Z}$ , $\varphi ^n$ denotes the $n$ -fold composition of $\varphi $ if $n \ge 1$ , $\text {id}_I$ if $n=0$ , and the $|n|$ -fold composition of $\varphi ^{-1}$ if $n < 0$ .

If $F:I \to S$ and $\varphi \in \mathrm{Homeo}(I)$ , we say that $\boldsymbol {F}$ is $\boldsymbol {\varphi }$ -invariant if $F = F \circ \varphi $ , and $\boldsymbol {F}$ is $\boldsymbol {\varphi }$ -invariant before $\boldsymbol {t}$ if

$$\begin{align*}F \restriction (-\infty,t) = F \circ \varphi \restriction (-\infty,t). \end{align*}$$

We say that $\boldsymbol {F}$ is past $\boldsymbol {\varphi }$ -invariant if there is some $t \in I$ such that $F$ is $\varphi $ -invariant before $t$ .

Proof Since $\varphi $ is invertible, the relation

$$\begin{align*}x\sim z \ :\iff \ \exists n \in \mathbb{Z} \ \varphi^n(x)=z\end{align*}$$

is an equivalence relation on $I$ . Suppose $x \sim z$ ; without loss of generality, say $n \ge 0$ and $\varphi ^n(x) = z$ . Then, because $\varphi $ is order-preserving, the sequence $\Big ( \varphi ^k(x) \Big )_{k \ge 0}$ is monotonic (increasing if $x \le \varphi (x)$ , and decreasing if $x \ge \varphi (x)$ ). In particular, if both $x$ and $z$ are less than $t$ , then so is $\varphi ^k(x)$ for every integer $k$ between $0$ and $n$ . Since $F$ is $\varphi $ -invariant before $t$ , it follows that $F(x) = F(\varphi ^k(x))$ for each $k$ between $0$ and $n$ ; in particular, $F(x)=F(z)$ .

Hence, for any equivalence class $C$ , if $C \cap (-\infty ,t) \ne \emptyset $ , then the restriction of $F$ to $C \cap (-\infty ,t)$ is constant; let $s_C$ denote this constant value. Now, fix any $s^* \in S$ , and define $H:I \to S$ by

$$\begin{align*}H(y)= \begin{cases} s_{[y]}, & \text{ if } [y] \cap (-\infty,t) \ne \emptyset, \\ s^*, & \text{ otherwise,} \end{cases} \end{align*}$$

where $[y]$ denotes the equivalence class of $y$ . Then $H$ extends $F|_{(-\infty ,t)}$ , and, since $[y]=[\varphi (y)]$ for every $y \in I$ , $H$ is $\varphi $ -invariant.

The next lemma is the key use of free actions and Hölder’s Theorem.

Proof By assumption, there is a $\psi \in V$ such that $\psi \ne \text {id}$ and $F$ is $\psi $ -invariant. Now, suppose $\varphi \in V$ and $F$ is past $\varphi $ -invariant; recall this means there is some $t \in I$ such that $F$ is $\varphi $ -invariant before $t$ (i.e., if $z < t$ , then $F(\varphi (z)) = F(z)$ ). Fix any $x \in I$ ; we want to show $F(\varphi (x)) = F(x)$ . Since $\psi \ne \text {id}$ and $V$ acts freely, $\psi $ has no fixed points. Let $\prec $ be the linear order on $V$ given by Theorem 11. Then either $\psi \prec \text {id}$ or $\psi ^{-1} \prec \text {id}$ . Without loss of generality,Footnote ⁸ we can assume that $\psi \prec \text {id}$ . Then $\langle \psi ^n(x) \ : \ n \in \mathbb {N} \rangle $ is a strictly decreasing sequence. Since $\psi $ is continuous and has no fixed points, it follows that $\lim _{n \to \infty } \psi ^n(x) = -\infty $ . So, for some $n \in \mathbb {N}$ , $\psi ^n(x) < t$ . Then

$$ \begin{align*} F(x) & = F\big( \psi^n(x) \big) & \text{ (because }F \text{ is }\psi\text{-invariant)} \\& = F\big( \varphi(\psi^n(x)) \big) & \text{ (because }\psi^n(x) < t \text{ and } F \text{ is } \varphi\text{-invariant before }t)\\& = F\big( \psi^n(\varphi(x)) \big) & \text{ (because } V \text{ is abelian, by Theorem }11) \\& = F(\varphi(x)) & \text{ (because }F \text{ is }\psi\text{-invariant).} \end{align*} $$

A commutator in a group is an element of the form $aba^{-1} b^{-1}$ . If $U$ is a group, $\boldsymbol {[U,U]}$ denotes the commutator subgroup of U, which is the subgroup generated by the commutators. At one point, we will use the basic group-theoretic fact that

$$ \begin{align*} [U,U] \text{ is a normal subgroup of } U. \end{align*} $$

We now commence with the proof of Theorem 5. Fix any nonempty set $S$ for the remainder of this section. Assume that $I$ is an open interval of real numbers and $U$ is a subgroup of $\mathrm{Homeo}^+(I)$ such that:

1. (1) $[U,U]$ acts freely on $I$ , and

2. (2) each nonidentity member of $U$ has at most one fixed point.

Define the following subsets of ${}^I S$ :

• • $\text {Inv}_{[U,U]}:=$ the set of functions in ${}^I S$ that are invariant with respect to some nonidentity member of $[U,U]$ .

• • $\text {Inv}_{U}:=$ the set of functions in ${}^I S$ that are invariant with respect to some nonidentity member of $U$ .

By the Axiom of Choice, there exists a wellordering $\triangleleft $ of ${}^I S$ such that $\text {Inv}_{[U,U]}$ functions are listed first, then functions in $\text {Inv}_U \setminus \text {Inv}_{[U,U]}$ , then the rest of ${}^I S$ .Footnote ⁹ The reason for these requirements will become apparent in the proof of Claim 14.

Recall from the discussion of anonymity in Section 1 that if $f \in {}^{\underset {\smile }{I}} S$ and $\varphi \in \mathrm{Homeo}^+(I)$ , $f \circ \varphi $ is understood to have domain $\big (-\infty , \varphi ^{-1}(t_f) \big )$ . Define $\boldsymbol { f \circ U}$ to be the set of functions of the form $f \circ \varphi $ , where $\varphi \in U$ . Let $\boldsymbol {\overline {f \circ U}}$ denote the collection of all (total) $F: I \to S$ such that $F$ extends at least one member of $f \circ U$ . Note that a given $F \in \overline {f \circ U}$ may possibly extend more than one member of $f \circ U$ ;Footnote ¹⁰ in fact, dealing with this issue is the heart of the argument. Let $F_{f \circ U}$ denote the $\triangleleft $ -least member of the set $\overline {f \circ U}$ .

Part (2) of the following claim is the key use of Lemma 13, which was a consequence of Hölder’s Theorem.

Define

$$\begin{align*}\mathcal{P}: {}^{\underset{\smile}{I}} S \to S \end{align*}$$

by

(2) $$ \begin{align} \mathcal{P}(f):= F_{f \circ U}(\underbrace{\varphi^{-1}(t_f)}_{\substack{\text{Right end-} \\ \text{point of} \\ \text{dom}\big( f \circ \varphi \big) }}), \end{align} $$

where $\varphi $ is any member of $U$ witnessing that $F_{f \circ U}$ is an element of $\overline {f \circ U}$ , i.e., where $\varphi $ is any member of $U$ such that $F_{f \circ U}$ extends $f \circ \varphi $ . The next claim says that $\mathcal {P}$ is well defined, in the sense that the expression above does not depend on the choice of the $\varphi $ .

Claim 15 The definition of $\mathcal {P}(f)$ in (2) does not depend which $\varphi \in U$ we choose to witness $F_{f \circ U} \in \overline {f \circ U}$ . That is, if $\varphi _1$ and $\varphi _2$ are both in $U$ , and $F_{f \circ U}$ extends both $f \circ \varphi _1$ and $f \circ \varphi _2$ , then

$$\begin{align*}F_{f \circ U}\big( \varphi_1^{-1}(t_f) \big) = F_{f \circ U}\big( \varphi_2^{-1}(t_f) \big). \end{align*}$$

Proof We follow the proof in [Reference Bajpai and Velleman1, Theorem 5] very closely (which was for the special case where $I=\mathbb {R}$ and $U$ was the group of increasing affine functions).

For the remainder of the proof of the claim, we abbreviate $F_{f \circ U}$ by $F$ , and omit the “ $\circ $ ” symbol in compositions. Let $x_1=\varphi ^{-1}_1(t_f)$ and $x_2=\varphi ^{-1}_2(t_f)$ ; so the domain of $f \varphi _1$ is $(-\infty ,x_1)$ , the domain of $f \varphi _2$ is $(-\infty ,x_2)$ , and $F$ extends both of them. We must show that $F(x_1)=F(x_2)$ . Clearly, this holds if $x_1=x_2$ , so suppose from now on that $x_1 \ne x_2$ . In particular,

(3) $$ \begin{align} \varphi^{-1}_1 \varphi_2 \ \text{ is a nonidentity member of } U, \end{align} $$

because it moves $x_2$ to $x_1$ . Since $F$ extends $f \varphi _1$ before $x_1$ and extends $f \varphi _2$ before $x_2$ , it follows that:Footnote ¹¹

(4) $$ \begin{align} F \text{ is } \varphi^{-1}_1 \varphi_2 \text{-invariant before } x_2\text{, and } \varphi^{-1}_2 \varphi_1 \text{-invariant before } x_1. \end{align} $$

Since $F=F_{f \circ U}$ extends $f \circ \varphi _2$ , $F$ is $\varphi _1^{-1} \varphi _2$ -invariant before $\varphi _2^{-1}(t_f)=x_2$ , and both $\varphi _2$ and $\varphi _1^{-1} \varphi _2$ are in $U$ , part (1) of Claim 14 implies

(5) $$ \begin{align} F \in \text{Inv}_U. \end{align} $$

So there is some $\psi \in U$ such that $\text {id}_I \ne \psi $ and $F = F \psi $ . Without loss of generality—since both $\psi $ and $\psi ^{-1}$ are both order-preserving and $F$ is invariant with respect to both—we can assume that

(6) $$ \begin{align} \psi(x_1) \le x_1. \end{align} $$

Still following the proof of [Reference Bajpai and Velleman1, Theorem 5], we consider two cases. The assumption that nonidentity members of $U$ have at most one fixed point is used in Case 1, and the assumption that $[U,U]$ acts freely on $I$ is used in Case 2.

Case 1: $\psi $ commutes with $\varphi _2^{-1} \varphi _1$ . We first claim that $\psi (x_1) < x_1$ . If not, then by (6), $x_1$ is a fixed point of $\psi $ . Then

$$\begin{align*}\psi(\underbrace{x_2}_{\varphi_2^{-1} \varphi_1(x_1)}) = \psi \varphi_2^{-1} \varphi_1 (x_1) \overset{\text{by case}}{=} \varphi_2^{-1} \varphi_1 \underbrace{\psi(x_1)}_{x_1} = x_2, \end{align*}$$

so $x_2$ is another fixed point of $\psi $ . This contradicts that nonidentity members of $U$ have at most one fixed point.

So $\psi (x_1) < x_1$ . Then

$$ \begin{align*} F (x_1) & = F \psi (x_1) & \text{ (because }F \text{ is }\psi\text{-invariant)} \\& = F \varphi_2^{-1} \varphi_1 \psi(x_1) & \text{ (because }\psi(x_1) < x_1 \text{ and }(4)) \\ & = F \psi\varphi_2^{-1} \varphi_1(x_1) & (\psi \text{ commutes with }\varphi_2^{-1} \varphi_1 \text{ by the Case)} \\ & = F \psi(x_2) & \\ & = F(x_2) & \text{ (because }F \text{ is }\psi\text{-invariant)}. \end{align*} $$

Case 2: $\psi $ does not commute with $\varphi _2^{-1} \varphi _1$ . Let $\tau := \varphi _2^{-1} \varphi _1$ , and consider the commutator $\psi ^{-1} \tau \psi \tau ^{-1}$ . By our case,

(7) $$ \begin{align} \beta:=\psi^{-1} \tau \psi \tau^{-1} \text{ is a nonidentity member of } [U,U]. \end{align} $$

First, we show that

(8) $$ \begin{align} F \text{ is } \beta \text{-invariant before } x_2. \end{align} $$

To see this, first observe that $\tau (x_1) = x_2$ . Suppose $z < x_2$ . Then $\tau ^{-1} (z) < x_1$ , and by (6), $\psi \tau ^{-1}(z) < x_1$ , and hence (since $\tau (x_1) = x_2$ ) $\tau \psi \tau ^{-1}(z) < x_2$ . Then

$$ \begin{align*} F \beta (z) & = F \psi^{-1} \tau \psi \tau^{-1}(z) \\ & = F \tau \psi \tau^{-1}(z) & (F \text{ is } \psi\text{-invariant)} \\ & = F \underbrace{\varphi_1^{-1} \varphi_2}_{\tau^{-1}} \tau \psi \tau^{-1}(z) & \text{(by }(4)\text{ and since }\tau \psi \tau^{-1}(z) < x_2) \\ & = F \psi \tau^{-1}(z) & \\ & = F \tau^{-1}(z) & (F \text{ is }\psi\text{-invariant)} \\ & = F \varphi_1^{-1} \varphi_2 (z) & \\ & = F(z) & \text{(by }(4)\text{ and }z < x_2), \end{align*} $$

which concludes the proof of (8). By (7), (8), and part (2) of Claim 14,

(9) $$ \begin{align} F \text{ is (fully) } \beta \text{-invariant.} \end{align} $$

Since $\text {id} \ne \beta \in [U,U]$ and $[U,U]$ acts freely on $I$ , in particular, $x_2$ is not a fixed point of $\beta $ . Let $\alpha $ denote whichever one of $\beta $ , $\beta ^{-1}$ sends $x_2$ to an output strictly less than $x_2$ . Then:

(10) $$ \begin{align} \alpha \in [U,U], \ \alpha(x_2) < x_2, \text{ and } F \text{ is } \alpha \text{-invariant.} \end{align} $$

Next, we claim that

(11) $$ \begin{align} F \text{ is } \tau^{-1} \alpha \tau \text{-invariant before } x_1. \end{align} $$

Suppose $z < x_1$ . Then

$$ \begin{align*} F \tau^{-1} \alpha \tau(z) & = F \varphi_1^{-1} \varphi_2 \underbrace{\alpha \overbrace{\varphi_2^{-1} \varphi_1 (z)}^{<x_2}}_{\substack{<x_2 \text{ because } \\ \alpha(x_2) \le x_2 }} & \\ & = F \alpha \varphi_2^{-1} \varphi_1(z) & \text{ by }(4) \\ & = F \varphi_2^{-1} \varphi_1(z) & (F \text{ is } \alpha\text{-invariant)} \\ & = F(z) & \text{by }(4)\text{ and }z < x_1, \end{align*} $$

which concludes the proof of (11).

Recall that commutator subgroups are always normal; hence, $\tau ^{-1} \alpha \tau $ is an element of $[U,U]$ . Then, by (11), the fact that $F$ extends $f \varphi _1$ , and since $\varphi _1 \in U$ and $\tau ^{-1} \alpha \tau \in [U,U]$ , part (2) of Claim 14 ensures that

(12) $$ \begin{align} F \text{ is (fully) } \tau^{-1} \alpha \tau \text{-invariant.} \end{align} $$

Finally,

$$ \begin{align*} F(x_1) & =F \tau^{-1} \underbrace{\alpha \overbrace{\tau (x_1)}^{x_2}}_{<x_2 \text{ by (10)}} & \text{by }(12) \\ & = F \alpha \tau (x_1)& \text{ by }(4) \\ & = F \tau(x_1) & F \text{ is }\alpha\text{-invariant} \\ & = F(x_2).\\[-32pt] & \end{align*} $$

With Claim 15 in hand, we can now finish the proof of Theorem 5. To see the $U$ -anonymity of $\mathcal {P}$ , suppose $f,g \in {}^{\underset {\smile }{I}} S$ , $\tau \in U$ , $\tau (t_g) = t_f$ , and $g = f \circ \tau $ . Then $g \in f \circ U$ , and it follows that $f \circ U = g \circ U$ and hence

(13) $$ \begin{align} F_{f \circ U} = F_{g \circ U}. \end{align} $$

Let $F$ denote this common function. Let $\varphi \in U$ witness that $F \in \overline {g \circ U}$ ; i.e., $F$ extends $g \circ \varphi = f \circ \tau \circ \varphi $ . So $F = F_{f \circ U}$ extends $f \circ (\tau \circ \varphi )$ , and since $\tau \circ \varphi \in U$ , Claim 15 ensures that $\mathcal {P}(f) = F\Big ( (\tau \circ \varphi )^{-1}(t_f) \Big )$ . Then

(14) $$ \begin{align} \mathcal{P}(g) = F \big( \varphi^{-1}(\underbrace{t_g}_{\tau^{-1}(t_f)}) \big) = F\Big( (\tau \circ \varphi)^{-1} (t_f) \Big)= \mathcal{P}(f). \end{align} $$

To see that $\mathcal {P}$ is good (this argument first appeared in Hardin–Taylor [Reference Hardin and Taylor5]), suppose $H: I \to S$ . Let

$$\begin{align*}B:= \left\{ t \in \mathbb{R} \ : \ \mathcal{P}\left( H|_{(-\infty,t)} \right) \ne H(t) \right\}. \end{align*}$$

By Fact 9 (from page 5), to see that $B$ has measure zero—in fact, is countable and nowhere dense—it suffices to show that $B$ is a wellordered subset of $\mathbb {R}$ under the usual ordering $<$ on $\mathbb {R}$ ; and for that it suffices to find an order-preserving embedding from $(B,<)$ into $\left ( {}^I S, \triangleleft \right )$ .Footnote ¹² Define

$$\begin{align*}e: (B,<) \to \left( {}^I S, \triangleleft \right), \ \ t \mapsto F_{ H|_{(-\infty,t)} \circ U}. \end{align*}$$

To see that $e$ is order-preserving, suppose $s < t$ are both in $B$ ; we will show that $e(s) \triangleleft e(t)$ by showing $e(s) \trianglelefteq e(t)$ and $e(s) \ne e(t)$ . We first prove the nonstrict inequality, i.e., that

$$\begin{align*}F_{ H|_{(-\infty,s)} \circ U} \trianglelefteq F_{ H|_{(-\infty,t)} \circ U}. \end{align*}$$

Since $F_{ H|_{(-\infty ,s)} \circ U}$ is, by definition, the $\triangleleft $ -least member of $\overline {H|_{(-\infty ,s)} \circ U} $ , it suffices to show that

(15) $$ \begin{align} F_{ H|_{(-\infty,t)} \circ U} \ \in \ \overline{H|_{(-\infty,s)} \circ U}. \end{align} $$

Now, $F_{ H|_{(-\infty ,t)} \circ U}$ , being a member of $\overline {H|_{(-\infty ,t)} \circ U}$ , extends $H|_{(-\infty ,t)} \circ \varphi $ for some $\varphi \in U$ . However, since $s < t$ (and $\varphi $ is order-preserving), it also extends $H|_{(-\infty ,s)} \circ \varphi $ . This verifies (15).

To prove that

$$\begin{align*}F_{ H|_{(-\infty,s)} \circ U} \ne F_{ H|_{(-\infty,t)} \circ U}, \end{align*}$$

suppose toward a contradiction that they are equal; let $F$ denote the common function. Let $\varphi \in U$ witness that $F \in \overline {H|_{(-\infty ,t)} \circ U}$ ; so $F$ extends $H|_{(-\infty ,t)} \circ \varphi $ . Since $s < t$ and $\varphi $ is order-preserving, $F$ also extends $H|_{(-\infty ,s)} \circ \varphi $ . Then, by Claim 15 and the fact that $F = F_{H|_{(-\infty ,s)} \circ U}$ ,

(16) $$ \begin{align} \mathcal{P}\Big(H|_{(-\infty,s)} \Big) = F\Big( \varphi^{-1}(s) \Big). \end{align} $$

Now, the left side of (16) is not equal to $H(s)$ , because $s \in B$ . So,

(17) $$ \begin{align} F\Big( \varphi^{-1}(s) \Big) \ne H(s). \end{align} $$

On the other hand, $F$ extends $H|_{(-\infty ,t)} \circ \varphi $ , whose domain is $(-\infty , \varphi ^{-1}(t))$ ; and since $s < t$ and $\varphi $ (and hence $\varphi ^{-1}$ ) is order-preserving, $\varphi ^{-1}(s) < \varphi ^{-1}(t)$ . So $F$ and $H|_{(-\infty ,t)} \circ \varphi $ agree on the input $\varphi ^{-1}(s)$ . Then $F\big (\varphi ^{-1}(s)\big ) = H|_{(-\infty ,t)} \circ \varphi \big ( \varphi ^{-1}(s) \big ) = H\big ( \varphi (\varphi ^{-1}(s)) \big ) = H(s)$ , contradicting (17).

This concludes the proof of Theorem 5.

4 Proof of Corollary 6

In this section, we prove parts (C) and (E) of Corollary 6. The remaining parts should be clear. The following lemma proves part (C).

Proof Suppose toward a contradiction that there is some nonidentity $\varphi \in I$ and some $x_1 < x_2$ in $I$ such that both $x_1$ and $x_2$ are fixed points of $\varphi $ . Since $U$ acts transitively on $I$ , there is some $\tau \in U$ such that $\tau (x_1) = x_2$ . Then $x_2$ is a fixed point of $\tau \varphi \tau ^{-1} \varphi ^{-1}$ . We claim that $\tau \varphi \tau ^{-1} \varphi ^{-1} \ne \text {id}_I$ , which will contradict the assumption that $[U,U]$ acts freely on $I$ .

Suppose toward a contradiction that $\tau \varphi \tau ^{-1} \varphi ^{-1} = \text {id}_I$ ; i.e., that

(18) $$ \begin{align} \tau \varphi \tau^{-1}=\varphi. \end{align} $$

For $n \ge 3$ , set $x_n:= \tau (x_{n-1})$ . Since $\tau $ is increasing and $\tau (x_1) = x_2$ , the sequence $(x_n)_n$ is a strictly increasing sequence in $I$ . We prove by induction that each $x_n$ is a fixed point of $\varphi $ , which will contradict that $\varphi $ has only finitely many fixed points. That $x_1$ and $x_2$ are fixed points is by assumption. Now, if $x_n$ is a fixed point of $\varphi $ , then

$$ \begin{align*} \varphi(x_{n+1}) & = \tau \varphi \tau^{-1} (x_{n+1}) & \text{by }(18) \\ &= \tau \varphi \tau^{-1} \big( \tau(x_n) \big) & \text{definition of }x_{n+1} \\ & = \tau \varphi(x_n) & \\ & = \tau(x_n) & \text{Induction Hypothesis} \\ & = x_{n+1}.\\[-32pt] \end{align*} $$

The “at most one” in the conclusion of Lemma 16 is best possible, since when $I=\mathbb {R}$ and $U$ is the group of increasing affine functions, the commutator subgroup $[U,U]$ is exactly the group of shift functions (i.e., affine functions of slope 1). And in that case, $U$ acts transitively on $\mathbb {R}$ , $[U,U]$ acts freely on $\mathbb {R}$ , and members of $U \setminus [U,U]$ have exactly one fixed point.

Next, we prove part (E) of Corollary 6. Suppose $\varphi \in \mathrm{Homeo}^+(I)$ , and that the set $C$ of fixed points of $\varphi $ has Lebesgue measure zero. Since $\varphi $ is continuous, $C$ is closed. For $t \in I \setminus C$ , let $a(t)$ denote the largest member of $C \cap (-\infty ,t)$ if that intersection is nonempty; otherwise, set $a(t)=-\infty $ . Closure of $C$ , together with the assumption that $t \notin C$ , ensures $a(t)$ is well defined. Similarly, let $b(t)$ be the smallest element of $C \cap (t,\infty )$ if that intersection is nonempty, and $b(t)=+\infty $ otherwise. Finally, if $t \in I \setminus C$ , set $J(t):= \big ( a(t), b(t) \big )$ .

Let

$$\begin{align*}\mathcal{J}:= \big\{ J(t) \ : \ t \in I \setminus C \big\}, \end{align*}$$

and notice $\mathcal {J}$ is a pairwise disjoint collection of open intervals; hence,

(19) $$ \begin{align} \mathcal{J} \text{ is countable.} \end{align} $$

Furthermore, for each $J \in \mathcal {J}$ , $\varphi |_{J}$ is an element of $\mathrm{Homeo}^+(J)$ , and moreover is fixed-point-free, since $J=(a(t),b(t))$ for some $t \in I \setminus C$ , and $C \cap (a(t),b(t))$ is empty. The following claim is a basic application of the Intermediate Value Theorem.

Suppose $J \in \mathcal {J}$ . Since $\varphi |_J$ is fixed-point-free, Claim 17 ensures that the cyclic subgroup $\langle \varphi |_J \rangle $ generated by $\varphi |_J$ in $\mathrm{Homeo}^+(J)$ acts freely on $J$ . So, by Theorem 5, there exists a good, $\langle \varphi |_J \rangle $ -anonymous predictor $\mathcal {P}_J$ for functions in ${}^J S$ . Define

$$\begin{align*}\mathcal{Q}: {}^{\underset{\smile}{I}} S \to S \end{align*}$$

as follows: fix any $s_0 \in S$ . Given $f: (-\infty , t_f) \to S$ , consider cases:

• • If $t_f \in C$ , let $\mathcal {Q}(f):= s_0$ .

• • If $t_f \notin C$ , define $\mathcal {Q}(f):= \mathcal {P}_{J(t_f)} \Big ( f|_{J(t_f)} \Big )$ ; note that the domain of $f|_{J(t_f)}$ is $J(t_f) \cap (-\infty , t_f) = \big ( a(t_f), t_f \big )$ .

To see that $\mathcal {Q}$ is $\langle \varphi \rangle $ -anonymous, consider any $f:(-\infty ,t_f) \to S$ and any nonidentity $\tau \in \langle \varphi \rangle $ . We need to show that $\mathcal {Q}(f) = \mathcal {Q}\big ( f \tau \big )$ . If $t_f \in C$ , then by the claim, $t_f$ is a fixed point of both $\varphi $ and $\tau $ , and hence $t_f = \tau ^{-1}(t_f)$ is the right endpoint of the domain of $f \tau $ . So, by definition of $\mathcal {Q}$ , $\mathcal {Q}(f) = s_0 = \mathcal {Q}(f \tau )$ . If $t_f \notin C$ , then since $a(t_f)$ and $b(t_f)$ are fixed points of $\tau $ , $\tau ^{-1}(t_f)$ is an element of $ J(t_f)$ . In particular,

(20) $$ \begin{align} \tau^{-1}(t_f) \notin C, \ J\left(\tau^{-1}(t_f)\right) = J(t_f) =:J, \text{ and } (f \tau)|_{J} = f|_{J} \ \tau|_{J}. \end{align} $$

Then

$$ \begin{align*} \mathcal{Q} (f) & = \mathcal{P}_{J}\Big( f|_{J} \Big) & \text{(definition of }\mathcal{Q})\\ & = \mathcal{P}_{J}\Big( f|_{J} \ \tau|_{J} \Big) & \text{(by }\langle \varphi|_{J} \rangle\text{-anonymity of }\mathcal{P}_{J}) \\ & = \mathcal{P}_J \Big( (f\tau)_J \Big) & \text{(by }(20)) \\ & = \mathcal{Q}\big( f \tau \big) & \text{(by definition of }\mathcal{Q}, \text{ since }\tau^{-1}(t_f) \notin C). \end{align*} $$

To see that $\mathcal {Q}$ is good, fix any $H: I \to S$ , and let

$$\begin{align*}B:= \Big\{ t \in I \ : \ \mathcal{Q} \left( H|_{(\infty,t)} \right) \ne H(t) \Big\}. \end{align*}$$

Now, $B = (B \cap C) \cup (B \setminus C)$ , so it suffices to show that each set in the union is null. Since $C$ is null by assumption, $B \cap C$ is null; so it remains to show that $B \setminus C$ is also null. Now,

(21) $$ \begin{align} B \setminus C \subseteq \bigcup_{J \in \mathcal{J}} (B \cap J). \end{align} $$

Moreover, by the way we defined $\mathcal {Q}$ , $B \cap J$ is exactly the set of $t \in J$ where $\mathcal {P}_J\big ( H|_{J \cap (-\infty , t)} \big ) \ne H(t)$ , which has measure zero because $\mathcal {P}_J$ is good. Together with (19), this implies that the right side of (21) has measure zero.

5 Proof of Theorem 8

In this section, we prove Theorem 8, which strengthens Bajpai–Velleman’s Theorem 4. We remark that their proof actually showed something a bit stronger than is stated in Theorem 4 (or in their paper). Namely, their set $S=\mathbb {R}/\sim $ had the property that there is no good $S$ -predictor that is anonymous with respect to the class

(22) $$ \begin{align} C^\infty_{\text{Lipschitz}} \cap \mathrm{Homeo}^+(\mathbb{R}), \end{align} $$

where $C^\infty _{\text {Lipschitz}}$ is the set of $f \in C^\infty $ such that for all $k \in \mathbb {N}$ , $f^{(k)}$ is bounded.

We strengthen their result in two ways:

1. (1) We show that their result still holds when the class (22) is replaced by the smaller class $D^\infty _{\text {Lipschitz}}$ of infinitely Lipschitz diffeomorphisms; i.e., the set of invertible $f: \mathbb {R} \to \mathbb {R}$ such that both $f$ and $f^{-1}$ are $C^\infty $ functions, and for every $k \in \mathbb {N}$ , $f^{(k)}$ and $(f^{-1})^{(k)}$ are bounded.

2. (2) We show that even a weaker kind of prediction fails.

We first describe the weaker kind of prediction. Given a set $S$ , let $\boldsymbol {{}^{\mathbb {R}^{\circ }} S}$ denote the set of $S$ -valued functions $f$ such that $\text {dom}(f) = \mathbb {R} \setminus \{ h_f \}$ for some $h_f \in \mathbb {R}$ (so the domain of $f$ is the reals with a hole at $h_f$ ). If $F$ is a total function on $\mathbb {R}$ and $x \in \mathbb {R}$ , $F \setminus \{ x \}$ will denote the restriction of $F$ to the domain $\mathbb {R} \setminus \{x \}$ . A weak S-predictor will refer to any function

$$\begin{align*}\mathcal{P}: {}^{\mathbb{R}^{\circ}} S \to S, \end{align*}$$

and it will be called a good weak $S$ -predictor if for every total $F: \mathbb {R} \to S$ , the set

$$\begin{align*}\Big\{ x \in \mathbb{R} \ : \ \mathcal{P}\left( F \setminus \{ x \} \right) \ne F(x) \Big\} \end{align*}$$

has Lebesgue measure zero. Roughly, $\mathcal {P}$ almost always “predicts” $F(x)$ based on the past and future behavior of $F$ .

Every good $S$ -predictor yields a good weak $S$ -predictor, because if $\mathcal {P}$ is a good $S$ -predictor, define $\mathcal {Q}$ on ${}^{\mathbb {R}^{\circ }} S$ as follows: given any $f \in {}^{\mathbb {R}^{\circ }} S$ whose domain has a hole at $h_f$ , define

$$\begin{align*}\mathcal{Q}(f):= \mathcal{P}\Big( f \restriction (-\infty, h_f) \Big). \end{align*}$$

Then, for any total $F: \mathbb {R} \to S$ , we have $\mathcal {Q}\Big ( F \setminus \{ x \} \Big ) = \mathcal {P} \Big ( F \restriction (-\infty ,x) \Big )$ , which can only fail to equal $F(x)$ on a measure zero set.

If $U \subseteq \mathrm{Homeo}^+(\mathbb {R})$ , let us say that a weak $S$ -predictor $\mathcal {P}$ is $\boldsymbol {U}$ -anonymous if whenever $f,g \in {}^{\mathbb {R}^{\circ }} S$ (with holes $h_f$ , $h_g$ in their respective domains), $\varphi \in U$ , $\varphi (h_f) = h_g$ , and $f = g \circ \varphi \restriction \left ( \mathbb {R} \setminus \{ h_f \} \right )$ , then $\mathcal {P}(f) = \mathcal {P}(g)$ . Similarly, as above, any good $U$ -anonymous $S$ -predictor yields a good $U$ -anonymous weak $S$ -predictor.

Our goal is to prove the following theorem.

Our proof of Theorem 19 relies heavily on the proof of [Reference Bajpai and Velleman1, Theorem 8]. Much of the following lemma was implicit in their proof.

Proof Let $S:=\mathbb {R}/\sim $ , and suppose toward a contradiction that $\mathcal {P}:{}^{\mathbb {R}^{\circ }} S \to S$ is a $U$ -anonymous, good, weak $S$ -predictor. Let $E: \mathbb {R} \to S$ be defined by $E(x) = [x]_\sim $ . We will get a contradiction by proving that $\mathcal {P}\Big ( E \setminus \{ x \} \Big ) \ne E(x)$ holds for positively many $x$ . It suffices to prove that

(23) $$ \begin{align} \forall x,y \in \mathbb{R} \ \ \mathcal{P}\Big( E \setminus \{ x \} \Big) = \mathcal{P} \Big( E \setminus \{ y \} \Big), \end{align} $$

because if the equivalence class $C$ were their constant value, then the equation

(24) $$ \begin{align} \mathcal{P}\Big( E \setminus \{ x \} \Big) = E(x) \ \Big( = [x]_\sim \Big) \end{align} $$

could hold only if $x \in C$ ; and since $C$ does not have full measure, this would yield positively many $x$ where equation (24) fails.

To prove (23), pick any $x,y \in \mathbb {R}$ . By assumption, there is a $\varphi \in U$ such that $\varphi (x)=y$ and $z \sim \varphi (z)$ whenever $z \ne x$ . It follows that $E \setminus \{x \} = E \setminus \{ y \} \circ \varphi \restriction \Big (\mathbb {R} \setminus \{ x \} \Big )$ . Then, by $U$ -anonymity of $\mathcal {P}$ , $\mathcal {P}\Big ( E \setminus \{ x \} \Big ) = \mathcal {P} \Big ( E \setminus \{ y \} \Big )$ .

Corollary 21 Suppose $U \subseteq \mathrm{Homeo}^+(\mathbb {R})$ and $\mathcal {F}$ is a family such that:

1. (1) $\mathcal {F}$ is a countable set of partial, injective functions from $\mathbb {R} \to \mathbb {R}$ ;

2. (2) for every $(x,y) \in \mathbb {R}^2$ , there is some $\varphi $ such that:

   1. (a) $\varphi \in U$ ;

   2. (b) $\varphi (x)=y$ ; and

   3. (c) if $z \ne x$ , then there is some $f \in \mathcal {F}$ such that $\varphi (z) = f(z)$ .

Let $\sim $ be the equivalence relation on $\mathbb {R}$ generated by the relation

$$\begin{align*}R:=\Big\{ (u,v) \ : \ \exists f \in \mathcal{F} \ \ f(u)=v \Big\}. \end{align*}$$

Then there is no good weak $U$ -anonymous $\mathbb {R}/\sim $ predictor.

We first sketch the outline of Bajpai–Velleman’s proof of Theorem 4. They first fix a single, increasing $C^\infty $ bijection

$$\begin{align*}s: [0,1] \to [0,1] \end{align*}$$

with vanishing derivatives at the endpoints (i.e., $s^{(k)}_-(0) = 0 = s^{(k)}_+(1)$ for all $k \in \mathbb {N}$ , where the minus and the plus denote left and right derivatives, respectively). A rational point is a point in the plane such that both coordinates are rational. By rational line, we will mean a line with positive slope that passes through two distinct rational points.Footnote ¹⁴ If $\Psi :\mathbb {R} \to \mathbb {R}$ is a rational line (i.e., an affine function whose graph is a rational line), we write $s \circ \Psi $ to denote the obvious partial function with domain $\Psi ^{-1} \Big ( [0,1] \Big )$ . They show that the countable collection

(25) $$ \begin{align} \begin{aligned} \mathcal{F}:= \Big\{ \Phi \circ s \circ \Psi \ :\ \Phi, \Psi \text{ are rational lines } \Big\} \end{aligned} \end{align} $$

witnesses the assumptions of Corollary 21. This basically amounts to showing that, given any $(x,y) \in \mathbb {R}^2$ , one can form an infinite concatenation of members of $\mathcal {F}$ to yield a strictly increasing $C^\infty $ function $h: (-\infty ,x) \to \mathbb {R}$ , whose approach (from the left) to the point $(x,y)$ is flat enough that $\hat {h}:=h ^\frown (x,y)$ will still be a $C^\infty $ function from $(-\infty ,x] \to \mathbb {R}$ , with $\hat {h}^{(k)}_-(x) = 0$ for all $k \in \mathbb {N}$ . Then extending $\hat {h}$ to all of $\mathbb {R}$ (in a $C^\infty $ manner) is easy.

We follow a similar strategy to prove Theorem 19, although we must use a different collection in order to avoid vanishing first derivatives. Fix any “bump” function

$$\begin{align*}b: [0,1] \to \mathbb{R} \end{align*}$$

with the following properties:

1. (i) $b(0) = 0 = b(1)$ ;

2. (ii) $b(x)> 0$ for all $x \in (0,1)$ ;

3. (iii) $b \in C^\infty $ (with one-sided derivatives taken at 0 and 1) and $b^{(k)}(0)=0=b^{(k)}(1)$ for all $k \in \mathbb {N}$ ; and

4. (iv) $\int _{[0,1]} b = 1$ .

To get such a $b$ , one could start, for example, with the common bump function

$$\begin{align*}\Psi(x)= \begin{cases} \exp \left( \frac{-1}{1-x^2}\right), & \text{ if } x \in (-1,1), \\ 0, & \text{ otherwise,} \end{cases} \end{align*}$$

let $\Phi (x)=\Psi (2x-1)$ , and then define $b:[0,1] \to \mathbb {R}$ by $b(x) = \frac {\Phi (x)}{\int _{[0,1]} \Phi }$ .

Definition 22 Given any point $A=(p,q)$ , any $\Delta>0$ , and any $\gamma> 0$ , define

$$\begin{align*}F_{A,\Delta,\gamma}: [p,p + \Delta] \to [q, q + \Delta(1 + \gamma) ] \end{align*}$$

by

$$\begin{align*}F_{A,\Delta,\gamma}(x)=q + \Delta \int_{\left[ 0, \frac{x-p}{\Delta} \right]} (1 + \gamma b). \end{align*}$$

$F_{A,\Delta ,\gamma }$ will serve as a smooth transition function from the point $A=(p,q)$ to the point $\big (p+\Delta , q + \Delta (1 + \gamma ) \big )$ , with derivative 1 (and vanishing higher derivatives) at those endpoints (see Figure 2). The $\Delta $ and $\gamma $ parameters are used to control the norm of the higher derivatives in our subsequent construction.

Figure 2: The function $F_{A,\Delta ,\gamma }$ .

The key features are:

1. (I) $F_{A,\Delta ,\gamma }(p) = q$ and $F_{A,\Delta ,\gamma }(p+\Delta ) = q + \Delta (1+\gamma )$ ;

2. (II) The Fundamental Theorem of Calculus and Chain Rule yield, for all $x \in [p,p + \Delta ]$ :

   $$\begin{align*}F_{A,\Delta,\gamma}'(x) = \Delta \left( 1 + \gamma b\left( \frac{x-p}{\Delta} \right) \right)\left( \frac{1}{\Delta}\right) =1 + \gamma b\left( \frac{x-p}{\Delta} \right). \end{align*}$$
   In particular, since $\gamma> 0$ and $b> 0$ on $(0,1)$ , $F_{A,\Delta ,\gamma }$ is increasing. Together with (I), it follows that $F_{A,\Delta ,\gamma }$ is an increasing bijection from $[p,p+\Delta ]$ to $[q,q+\Delta (1+\gamma )]$ . Moreover, notice that for all $x \in [p,p + \Delta ]$ ,
   $$\begin{align*}\lvert F^{\prime}_{A,\Delta,\gamma}(x)-1 \rvert \le \gamma \lVert b \rVert, \end{align*}$$
   where $\lVert \cdot \rVert $ denotes the sup-norm.

3. (III) Since $b(0)=0=b(1)$ , $F_{A,\Delta ,\gamma }'(p) = 1 = F_{A,\Delta ,\gamma }'(p + \Delta )$ .

4. (IV) Following part (II), we see that

   $$\begin{align*}\forall k \ge 2 \ \ F^{(k)}_{A,\Delta, \gamma}(x)= \frac{\gamma}{\Delta^{k-1}} b^{(k-1)}\left( \frac{x-p}{\Delta} \right), \end{align*}$$
   so, in particular,
   $$ \begin{align*} \forall k \ge 2 \ \ \lVert F^{(k)}_{A,\Delta,\gamma} \rVert \le \frac{\gamma}{\Delta^{k-1}} \lVert b^{(k-1)} \rVert. \end{align*} $$

Let $\mathcal {F}$ denote the collection of all rational lines of positive slope, together with all functions of the form $F_{A,\Delta ,\gamma }$ where $A$ is a rational point and $\Delta $ and $\gamma $ are positive rational numbers. We will show that $\mathcal {F}$ satisfies the assumptions of Corollary 21, with $U:= D^\infty _{\text {Lipschitz}}$ . Clearly, $\mathcal {F}$ is countable, and consists of injective functions. It remains to verify clause (2) of Corollary 21.

Proof Fix an increasing sequence $(p_n)_n$ of rational numbers converging to $w$ such that $\Delta _n := p_{n+1}-p_n < 1$ for all $n$ . Let $L$ be the line of slope 1 that passes through $P$ .Footnote ¹⁵ For any point $B$ , let $\text {vDist}(L,B)$ denote the vertical distance from $L$ to $B$ .

Recursively define the $q_n$ , $A_n$ , and $\gamma _n$ , together with the auxiliary variable

$$\begin{align*}v_n:= \text{vDist}(L,A_n), \end{align*}$$

as follows: fix a rational $q_1$ such that $A_1:=(p_1,q_1)$ lies strictly below the line $L$ , and

(26) $$ \begin{align} v_1 < \Delta_1. \end{align} $$

Given that $q_n$ is defined (and hence so are $A_n=(p_n,q_n)$ and $v_n = \text {vDist}(L,A_n)$ ), let $\gamma _n$ be a positive rational number such that

(27) $$ \begin{align} v_n - \frac{\left( \Delta_{n+1}\right)^{n+1}}{n+1} < \gamma_n \Delta_n < v_n; \end{align} $$

this is possible because the $\Delta $ ’s are positive. Define $q_{n+1}:=q_n + \Delta _n(1 + \gamma _n)$ , $A_{n+1} := (p_{n+1},q_{n+1})$ , and $v_{n+1}:= \text {vDist}(L,A_{n+1})$ . Note that $q_{n+1}$ is rational, provided that $q_n$ was rational.

We inductively verify that

(IH _n ) $$\begin{align} v_n < \frac{(\Delta_n)^n}{n} \end{align}$$

holds for all $n$ . It holds for $n=1$ by (26). Now, suppose $\text {IH}_n$ holds. Since $L$ has slope 1, it follows from the definition of $A_{n+1}$ that $v_{n+1} = v_n - \gamma _n \Delta _n$ , which by (27) is less than $\frac {(\Delta _{n+1})^{n+1}}{n+1}$ ; so (IH $_{n+1}$ ) holds.

Inequalities (IH $_n$ ) and (27) yield

$$\begin{align*}\gamma_n \Delta_n < v_n < \frac{\left( \Delta_n \right)^n}{n}, \end{align*}$$

and hence

$$\begin{align*}\frac{\gamma_n}{\left( \Delta_n \right)^{n-1}} < \frac{1}{n} \end{align*}$$

for all $n$ . This ensures part (4) of the lemma.

Fix any point $P=(w,z)$ in the plane, and let $(A_n)_n=(p_n,q_n)_n$ , $\Delta _n=p_{n+1}-p_n$ , and $\gamma _n$ be rational numbers as given by Lemma 23. For each $n$ , let

$$\begin{align*}F_n:=F_{A_n, \Delta_n, \gamma_n}: [p_n, p_{n+1}] \to [q_n, \underbrace{q_n + \Delta_n(1+\gamma_n)}_{q_{n+1}}] \end{align*}$$

using notation from Definition 22. See Figure 3.

Figure 3: The $F_n$ functions.

Let $F:= \bigcup _n F_n$ , which maps from $[p_1,w) \to \mathbb {R}$ . By (III) on page 19, $F_n$ and $F_{n+1}$ meet at the point $A_{n+1}$ , both with first derivative 1 and vanishing higher derivatives; in particular,

(28) $$ \begin{align} F \text{ is a } C^\infty \text{ function on } [p_1,w) \text{ and } \lim_{x \nearrow w} F(x) = z. \end{align} $$

By (II), we have the following for all $n \in \mathbb {N}$ :

(29) $$ \begin{align} \forall x \in [p_n,\underbrace{p_n + \Delta_n}_{p_{n+1}} ] \ \lvert F^{\prime}_{n}(x)-1 \rvert \le \gamma_n \lVert b \rVert. \end{align} $$

Since $\gamma _n \to 0$ as $n \to \infty $ (by requirements (3) and (4) of Lemma 23), it follows that

(30) $$ \begin{align} \lim_{x \nearrow w} F'(x) = 1. \end{align} $$

By (IV), the following also holds for all $n \in \mathbb {N}$ , where $\alpha _{k,n}$ is defined as indicated:

(31) $$ \begin{align} \forall k \ge 2 \ \ \lVert F^{(k)}_{n} \rVert \le \underbrace{ \frac{\gamma_n}{(\Delta_n)^{k-1}} }_{\alpha_{k,n}} \lVert b^{(k-1)} \rVert. \end{align} $$

Requirement (3) of Lemma 23 (that $0 < \Delta _n < 1$ ) ensures that

$$\begin{align*}n \ge k \ \implies \ 0 < \underbrace{\frac{\gamma_n}{(\Delta_n)^{k-1}}}_{\alpha_{k,n}} \le \frac{\gamma_n}{(\Delta_n)^{n-1}}, \end{align*}$$

the last term of which goes to 0 as $n \to \infty $ by requirement (4) of Lemma 23. So

$$\begin{align*}\lim_{n \to \infty} \alpha_{n,k} = 0, \end{align*}$$

and together with (31), this yields

(32) $$ \begin{align} \forall k \ge 2 \ \ \lim_{n \to \infty} \lVert F^{(k)}_n \rVert =0. \end{align} $$

This entire construction, starting from Lemma 23, has a straightforward analogue “from the right” of the point $P=(w,z)$ , yielding a decreasing sequence $(p^R_{n})_n$ of rationals converging to $w$ , and functions

$$\begin{align*}F^R_n: [p^R_{n+1}, p^R_n] \to \mathbb{R} \end{align*}$$

such that $F^R_n \in \mathcal {F}$ , and $F^R:= \bigcup _n F^R_n$ is an increasing $C^\infty $ function from $(w,p_1^R] \to \mathbb {R}$ with nonvanishing first derivative that has the following analogues of the properties above:

$$ \begin{align*} \lim_{x \searrow w} F^R(x) = z, \ \lim_{x \searrow w} (F^R)'(x) = 1, \text{ and } \forall k \ge 2 \ \ \lim_{n \to \infty} \lVert (F^R_n)^{(k)} \rVert =0. \end{align*} $$

Let $\Psi ^L$ be the affine function with slope 1 passing through the rational point $(p_1,q_1)$ , $\Psi ^R$ be the affine function with slope 1 passing through the rational point $(p^R_1, q^R_1)$ , and define $G:\mathbb {R} \to \mathbb {R}$ by

$$\begin{align*}G(x)= \begin{cases} \Psi^L(x), & \text{ if } x < p_1, \\ F(x), & \text{ if } x \in [p_1,w), \\ z, & \text{ if } x=w, \\ F^R(x), & \text{ if } x \in (w,p_1^R], \\ \Psi^R(x), & \text{ if } x> p_1^R. \end{cases} \end{align*}$$

Then $G$ is continuous, invertible, and has the property that for every $x \ne w$ , there is some $f \in \mathcal {F}$ such that $G(x) = f(x)$ .Footnote ¹⁶

It remains to prove that $G \in D^\infty _{\text {Lipschitz}}$ . Now, $G$ is affine on $(-\infty ,p_1]$ and on $[p_1^R,\infty )$ . In particular,

(33) $$ \begin{align} G |_{(-\infty,p_1] \cup [p_1^R,\infty)} \text{ has bounded derivatives of all orders.} \end{align} $$

So, to prove that $G \in D^\infty _{\text {Lipschitz}}$ , it suffices to show that $G \in D^\infty $ ; i.e., that both $G$ and $G^{-1}$ are $C^\infty $ functions.Footnote ¹⁷ Furthermore, by the Inverse Function Theorem, to prove that the invertible function $G$ is in $D^\infty $ , it suffices to show that $G \in C^\infty $ and that $G'$ is never zero. Since $F_n$ agrees with $F_{n+1}$ (and $F^R_n$ agrees with $F^R_{n+1}$ ) at all derivatives at their meeting points, and their first derivatives are always positive, the only nontrivial point to deal with is $w$ ; we will show that:

• • $G'(w) = 1$ ; and

• • $G^{(k)}(w) = 0$ , for $k \ge 2$ .

We will prove these from the left of $w$ ; the proof from the right is similar. The following lemma is a minor variant [Reference Bajpai and Velleman1, Lemma 7].

Lemma 24 Suppose $f$ is continuous on $(a,b]$ , differentiable on $(a,b)$ , and

$$\begin{align*}\lim_{x \nearrow b} f'(x) = L. \end{align*}$$

Then $f$ is left differentiable at $b$ , and $f^{\prime }_-(b) = L$ .

By (28), $G$ is $C^\infty $ on $(-\infty ,w)$ , and we already noted that $G$ is continuous on $\mathbb {R}$ . By (30), continuity of $G$ , and Lemma 24, $G^{\prime }_-(w) = 1 = \lim _{x \nearrow w} G'(x)$ and hence $G'$ is continuous on $(-\infty ,w]$ . Then, using (32) (with $k=2$ ) and applying Lemma 24 to $G'$ , we get that $G^{(2)}_-(w) = 0 = \lim _{x \nearrow w} G^{(2)}(x)$ and $G^{(2)}$ is continuous on $(-\infty ,w]$ . Continuing by induction, we get that $G^{(k)}_-(w) = 0 = \lim _{x \nearrow w} G^{(k)}(x)$ for all $k \ge 2$ .

6 Concluding remarks

Hardin–Taylor’s Question 2 (page 2) is still very much open, as there is still a big gap between the “lower frontier” of our Theorem 5 and the “upper frontier” of our Theorem 8.

Our Theorem 5 isolated some (topological) group-theoretic properties that guarantee existence of good anonymous predictors. It is natural to pursue this further. In particular, Theorem 11 implies that any group $U \le \mathrm{Homeo}^+(I)$ satisfying the assumptions of Theorem 5 is a metabelian group, meaning that its commutator subgroup is abelian. And Theorem 5 encompasses all known examples of groups for which good, anonymous predictors exist (for every set $S$ ). This suggests the following question.