2.3.1 Proof of Theorem 1.1. The domain of $f^\triangleleft$ .

In this section we show that $f^\triangleleft(x)$ is defined for all $x\in\mathbb{R}$ . By periodicity, we can assume $x\in (0,1)\smallsetminus\mathbb{Q}$ . We also recall that h is bounded on $[-1,1]$ by hypothesis. We let

(2.19) \begin{align} h_j(y) := h((-1)^j y), \quad w_j(y) := w(y, j, -k, h_j),\end{align}

for $y>0$ , and extend $h_j$ at $y=0$ by continuity, so that in particular

(2.20) \begin{equation} w_{r(x')}(x')=\operatorname{Den}(x') ^k h(0^\pm)\quad (x'\in\mathbb{Q},\ \pm = (-1)^{r(x')}).\end{equation}

Thus, with this notation, (2.1) reads

(2.21) \begin{equation} f(x') = \sum_{j\geqslant 0} w_j(x') + \operatorname{Den}(x') ^k(f(0)-h(0^\pm)). \end{equation}

It suffices to show that $\sup |f(x')-f(x'')|\to0$ as $\varepsilon\to0$ , where the sup is over all $x',x''\in\mathbb{Q}$ with $|x-x'|,|x-x''|<\varepsilon$ . Since $x\not\in\mathbb{Q}$ , there exists $m(x, \varepsilon) \to \infty$ as $\varepsilon \to 0$ such that, for all $x^* \in \mathbb{Q}$ , ${| {x-x^*} |}\leqslant \varepsilon$ , we have $r(x^*)>m(x, \varepsilon)$ and $b_j(x^*) = b_j(x)$ for $j\leqslant m(x, \varepsilon)$ . It follows that both $\operatorname{Den}(x^*) ^k$ and $\sum_{j\geqslant m(x, \varepsilon) } w_j(x^*)$ (by (2.16)) tend to 0 as $\varepsilon\to0^+$ . Thus, for any $\eta>0$ , we have

$$ {| {f(x') - f(x'')} |} \leqslant \eta + \sum_{j=1}^{m(x, \varepsilon)-1} {| {w_j(x') - w_j(x'')} |}, \quad \forall x',x''\in (x-\varepsilon,x+\varepsilon)\cap \mathbb{Q}$$

for $\varepsilon>0$ sufficiently small.

By Lemma 2.7, the function $y \to w_j(y)$ is continuous at x and thus for $x',x''\in (x-\varepsilon',x+\varepsilon')\cap \mathbb{Q}$ with $\varepsilon'\in(0,\varepsilon)$ sufficiently small we have ${| {f(x') - f(x'')} |} \leqslant2\eta$ . Since $\eta$ was arbitrary, this yields our claim.

In particular, approximating $x = [0; b_1, b_2, \dotsc] \in(0, 1)\smallsetminus\mathbb{Q}$ by the sequence $x_n = [0; b_1, \dotsc, b_n]$ , we get the normally converging series expression

(2.22) \begin{equation} f^\triangleleft(x) =\sum_{j\geqslant0} w_j(x) \end{equation}

for $x\notin\mathbb{Q}$ . Notice that the same expression holds also for $x\in\mathbb{Q}$ if $f(0)=h(0^\pm)$ , by (2.21).

2.3.2 Proof of Theorem 1.1. The continuity of $f^\triangleleft$ .

First, assume $x\in(0,1)\smallsetminus\mathbb{Q}$ . Then, using the expression (2.22) and an identical argument as above, the continuity of $f^\triangleleft$ at x follows immediately.

Assume, then, that $x\in[0, 1]\cap\mathbb{Q}$ . Let $r = r(x)$ and $q=\operatorname{Den}(x)$ . By Lemma 2.1 for all $y\neq x$ in a neighborhood of x, we have $r(y)\geqslant r + 1$ , and therefore, with $\pm = (-1)^{r}$ ,

\begin{align*} f^\triangleleft(y) - f(x) = {}& q^k(h(0^\pm)-f(0)) + O(\operatorname{Den}(y)^{\operatorname{Re}(k)}) + \sum_{j\geqslant 0}(w_j(y) - w_j(x))\end{align*}

where the term $O(\operatorname{Den}(y)^k)$ is to be ignored if $y\not\in\mathbb{Q}$ . We let $y\to x$ . First note that $\operatorname{Den}(y)^k \to 0$ . Also, for $j\not\in\{r, r+1\}$ , the function $w_j$ is continuous by Lemma 2.7, and so the jth summand in the sum tends to 0. By dominated convergence, we deduce

\begin{align*} f^\triangleleft(y) - f(x) & = {} o(1) + q^k(h(0^\pm) - f(0)) + w_r(y) - w_r(x) + w_{r+1}(y)\\ & = {} o(1) + w_r(y) + w_{r+1}(y)- q^k f(0),\end{align*}

by (2.20). Finally, by (2.13) and since $h_r(1) = h((-1)^r) = 0$ by (2.3), we have

(2.23) \begin{equation} \begin{aligned} w_r(y) = {}& o(1) + q^k \begin{cases} h_r(0) & (\textrm{sgn}(y-x) = (-1)^r), \\ 0 & (\textrm{sgn}(y-x) = (-1)^{r+1}), \end{cases} \\ w_{r+1}(y) = {}& o(1) + q^k \begin{cases} 0 & (\textrm{sgn}(y-x) = (-1)^r), \\ h_{r+1}(0) & (\textrm{sgn}(y-x) = (-1)^{r+1}). \end{cases} \end{aligned} \end{equation}

Since $h_r(0) = h(0^\pm)$ , $h_{r+1}(0) = h(0^\mp)$ , then splitting in cases depending on the sign $\pm$ and on whether $y\to0$ from the right or the left, we find

$$ f^\triangleleft(y) - f(x) \to \begin{cases} q^k(h(0^+) - f(0)) & (y\to x^+), \\ q^k(h(0^-) - f(0)) & (y \to x^-), \end{cases} $$

as claimed. We also notice for later use that if h is continuous at zero, then (2.23) gives

$$\begin{equation*} {\lim_{y\to x^+}} w_r(y)={\lim_{y\to x^-}} w_{r+1}(y),\quad{\lim_{y\to x^-}} w_r(y)={\lim_{y\to x^+}} w_{r+1}(y).\end{equation*}$$

2.3.3 Proof of Theorem 1.1. The differentiability of $f^\triangleleft$ .

We argue by induction, starting from the case $m=1$ . Assume that $h\in\mathcal{C}^1([-1, 1])$ and $\operatorname{Re}(k)<-2$ .

By Lemma 2.7 we have that $w_j$ is of $\mathcal{C}^1$ class on $[0, 1)\smallsetminus \{x, r(x)\in\{j-1,j\}\}$ (and thus on any neighborhood of irrational numbers). In particular, by (2.22), (2.16) and (2.17), it follows that $f^\triangleleft$ is differentiable on $\mathbb{R}\smallsetminus\mathbb{Q}$ . The same argument also gives that $\sum_{j\neq r,r+1} w_j(x)$ is differentiable at rationals. In particular, it suffices to show that the right and left derivatives of $w_r(y)+w_{r+1}(y)$ coincide at rationals. To show this, we start by observing that from the period relation (1.2), proceeding as in [Reference ZagierZag99, Equation (23)], we have

$$ h(x+1) - h(x) =-{| {1+x} |}^{-k}h(1-1/(x+1)) ,\quad x\in\mathbb{Q}\smallsetminus\{0,-1\}. $$

In particular, taking the limit as $x\to 0$ , we obtain $h(1)=0$ and $h'( 1) = k h(0)$ . Thus, since by (1.2) we have $h(-1/x)=-|x|^{-k}h(x)$ , we find $h'(- 1)=-kh(0)$ and thus $h_j'(1)=kh(0)$ for all j. Thus, recalling that $h_j(1)=0$ (and $h_j'(0)=(-1)^j h(0)$ ) and observing that

$$ -k\sum_{i=0}^{r(x)} \frac{(-1)^i}{u_i(x)u_{i+1}(x)} = -k\sum_{i=0}^{r(x)-1} \frac{(-1)^i}{u_i(x)u_{i+1}(x)} - (-1)^r k,\quad x\in\mathbb{Q}\cap(0,1), $$

we obtain from (2.15) that

\begin{equation*} w'_r(x^+)+w'_{r+1}(x^+)=w'_r(x^-) +w'_{r+1}(x^-), \quad x\in\mathbb{Q}\cap(0,1),\end{equation*}

thus giving that $f^\triangleleft$ is differentiable at any $x\in\mathbb{Q}\cap(0,1)$ . Finally, by (1.2), we see that $f^\triangleleft$ is differentiable at 0 also, with $(f^\triangleleft)'(0) = h'(0)$ . We conclude that $f^\triangleleft$ is differentiable on [0, 1), hence on $\mathbb{R}$ by periodicity. By (1.2), its derivative satisfies, for $x\neq 0$ ,

$$ h_1(x) = (f^\triangleleft)'(x) - {| {x} |}^{-k-2}(f^\triangleleft)'(-1/x), $$

where $h_1(x) := h'(x) - k\textrm{sgn}(x) {| {x} |}^{-k-1}f(-1/x)$ . The function $h_1$ , extended at 0 by $h_1(0)=h'(0)=(f^\triangleleft)'(0)$ , is continuous on $[-1, 1]$ . We can then apply the continuity property proven in § 2.3.2 to $(f^\triangleleft)'$ and deduce that $(f^\triangleleft)'$ is continuous on [0, 1], and hence on $\mathbb{R}$ .

If $k<-2m$ and $h\in\mathcal{C}^m$ , then this argument can be iterated m times, with functions $(h_n)_{n\leqslant m}$ defined inductively by

$$ h_0(x) = h(x), \quad h_{n+1}(x) = h_n'(x) - (k+2n) \textrm{sgn}(x) {| {x} |}^{-k-2n-1} (f^\triangleleft)^{(n)}(-1/x). $$

2.3.4 Proof of Theorem 1.2. Extending f when h is unbounded.

We now justify Theorem 1.2. We assume that h satisfies (1.5). First note that, due to the fact that periodicity is assumed separately on $\mathbb{R}_+$ and $\mathbb{R}_-$ , we have the following variant of (2.1),

$$ f(x) = \sum_{j=0}^{r-1} \bigg(\prod_{i=0}^{j-1} T^i(x)\bigg)^{\!{-}k} h((-1)^j T^j(x)) + q^k f((-1)^r). $$

Proof. Let $x = [b_0; b_1, b_2, \dotsc] \in \mathfrak{T}(m^{1+\delta})$ . By periodicity we can assume $x\in[0,1)$ , that is $b_0=0$ . If $U:[0,1)\to\mathbb{R}$ is the inverse branch of $T^m$ , given by $U(y) = [0; b_1, \dotsc, b_m+ y]$ , then any $x'\in V(m^{1+\delta}, m, x)$ is in the range of U. By [Reference Marmi, Moussa and YoccozMMY97, Proposition 1.14], we have

$$ \|(T^j\circ U)'\|_\infty = \sup_{y\in [0, 1]} \prod_{i=j}^{m-1} {| {T^i(U(y))} |}^2 \ll \varpi^{-2(m-j)}$$

uniformly for $0\leqslant j < m$ , and therefore, for any $x', x'' \in V(m^{1+\delta}, m, x)$ ,

(2.24) \begin{equation} {| {T^j(x') - T^j(x'')} |} \ll \varpi^{-2(m-j)} \end{equation}

for $j<m$ . We note also that $T^j(x') \asymp T^j(x)$ for $x' \in V(m^{1+\delta}, m, x)$ and $j<m$ , with a uniform constant, and that the condition $T^j(x') \gg \min(m^{-1-\delta}, j^{-1} (\log j)^{-2})$ holds uniformly over j. Finally, for any $x'\in V(m^{1+\delta}, m, x)$ , the condition $r(x')\geqslant m$ implies $\operatorname{Den}(x') \geqslant \varpi^m$ by virtue of (1.16) (with $j=r(x')$ ).

Given $x\in \mathfrak{T}(m^{1+\delta})$ and $x'\in V(m^{1+\delta}, m, x)$ , with the same notation as in § 2.3.1 we have

\begin{equation*} f(x') = \sum_{0\leqslant j < m/2} w_j(x') + O\bigg(\operatorname{Den}(x)^{\operatorname{Re}(k)} + \sum_{j\geqslant m/2} \varpi^{\operatorname{Re}(k)j} \sup_{{| {y} |}\gg \min(m^{-1-\delta}, j^{-1}(\log j)^{-2})} {| {h(y)} |}\bigg). \end{equation*}

By our hypothesis (1.5), the second sum is $\ll \sum_{j\geqslant m/2} \varpi^{\operatorname{Re}(k) j}({e}^{j^{1-\delta}(\log j)^2} + {e}^{O(m^{1-\delta^2})}) \ll \varpi^{\operatorname{Re}(k)m/3}$ , and therefore

(2.25) \begin{equation} f(x') = \sum_{0\leqslant j < m/2} w_j(x') + O(\varpi^{\operatorname{Re}(k) m/3}). \end{equation}

We now let $x', x'' \in V(m^{1+\delta}, m, x)$ be given. Taking differences in the expansion (2.25), we obtain

$$ {| {f(x') - f(x'')} |} \ll \sum_{0\leqslant j < m/2} {| {w_j(x') - w_j(x'')} |} + \varpi^{\operatorname{Re}(k)m/3}. $$

Denote temporarily $ \Pi_j(y) := \prod_{0\leqslant i < j} T^i(y)$ , which is also $u_j(y) / u_0(y)$ in the notation of § 1.4. By the definition (2.12), (2.19) of $w_j$ , splitting the difference in $w_j(x') - w_j(x'')$ , we get

\begin{align*} \sum_{0\leqslant j < m/2} {| {w_j(x') - w_j(x'')} |} \leqslant \sum_{0\leqslant j < m/2} (&{| {(h_j(T^j(x')) - h_j(T^j(x''))) \Pi_j(x'')^{-k}} |} \\ & + {| {h_j(T^j(x'))(\Pi_j(x')^{-k} - \Pi_j(x'')^{-k})} |} ). \end{align*}

Regarding the first term inside the sum, the bound (2.24) and our hypothesis (1.5), applied with some value of $\varepsilon \asymp m^{-1-\delta}$ , give ${| {h_j(T^j(x')) - h_j(T^j(x''))} |} = o(1)$ as $m\to \infty$ uniformly for all $j<m/2$ . By the bound (1.16), we further have $\Pi_j(y) \ll \varpi^{-j}$ uniformly for $y\in[0, 1]$ and $j\geqslant 0$ , and therefore

$$ \sum_{0\leqslant j < m/2} {| {(h_j(T^j(x')) - h_j(T^j(x''))) \Pi_j(x'')^{-k}} |} \ll o(1) \times \sum_{j\geqslant 0} \varpi^{j \operatorname{Re}(k)} = o(1) $$

as $m \to \infty$ . Regarding the second term, we have, as above, $h_j(T^j(x')) \ll {e}^{m^{1-\delta^2}} + {e}^{j^{1-\delta} (\log j)^2}$ for all $j\geqslant 0$ . We use the inequality $u^{-k}-v^{-k} \ll_k {| {u-v} |}^{\min(1, -k)}$ , valid for $0\leqslant u, v \leqslant 1$ , which gives in our case

$$ \Pi_j(x')^{-k} - \Pi_j(x'')^{-k} \ll {| {\Pi_j(x') - \Pi_j(x'')} |}^{\min(-k, 1)}. $$

Splitting the difference, we have

\begin{align*} {| {\Pi_j(x') - \Pi_j(x'')} |} \leqslant {}& \sum_{0\leqslant i \leqslant j} \Pi_{i-1}(x') \Pi_{j-i-1}(T^{i+1}(x'')) {| {T^i(x') - T^i(x'')} |} \\ \ll {}& \sum_{0\leqslant i \leqslant j} \varpi^{-i} \times \varpi^{-(j-i)} \times \varpi^{-2(m - i)} \\ \ll {}& \varpi^{-2m+j} \end{align*}

using our bound (2.24) and thus we conclude that

\begin{align*} & \sum_{0\leqslant j < m/2} {| {h_j(T^j(x'))(\Pi_j(x')^{-k} - \Pi_j(x'')^{-k})} |} \\ &\quad \ll \varpi^{-2m \min(-k, 1)} \sum_{j\geqslant 0} \varpi^{j\min(-k, 1)}({e}^{O(m^{1-\delta^2})} + {e}^{O(j^{1-\delta} (\log j)^2)}) \\&\quad \ll \varpi^{-m \min(-k, 1)/2}, \end{align*}

which tends to 0 as $m\to \infty$ . Grouping our bounds, we deduce $\Delta_{1+\delta}(m) \to 0$ when $m\to\infty$ , as claimed.

We now turn to the proof of Theorem 1.2. Consider $X = (\mathbb{R}\smallsetminus\mathbb{Q}) \cap \mathfrak{S}$ , where $\mathfrak{S}$ is defined in Lemma 2.2, and let $x\in X$ . In particular, $x\in \mathfrak{T}(B)$ for some $B>0$ , and therefore so do the convergents $x_j$ of x. Summarizing, we have $x_j\in \mathbb{Q}\cap\mathfrak{T}(B)$ for all j, and $x_j \to x$ as $j\to\infty$ . Since f has the property $\mathscr{S}(1+\delta)$ by Lemma 2.9, the limit $f^\triangleleft(x) := f^*(x) = \lim_{j\to\infty} f(x_j)$ in (2.9) exists.

Moreover, for any $\varepsilon$ , by Lemma 2.2 we can find $B>0$ such that $X_\varepsilon :=(\mathbb{R}\smallsetminus\mathbb{Q}) \cap \mathfrak{T}(B) \subset X$ satisfies $\nu(X_\varepsilon\cap [0, 1])\geqslant 1-\varepsilon$ . This set is also trivially invariant by $x\mapsto x+1$ . Let $y\in X_\varepsilon$ and $\eta>0$ be arbitrary. We pick an integer $m\geqslant B^{1/(1+\delta)}$ such that the right-hand side of (2.10) has modulus at most $\delta$ . Having chosen m, and since $y\not\in\mathbb{Q}$ , we may find $\xi>0$ such that any number $x\in (\mathbb{R}\smallsetminus\mathbb{Q})\cap[y-\xi, y+\xi]$ satisfies $b_j(x) = b_j(y)$ for $j\leqslant m$ . Now let $x\in X_\varepsilon \cap[y-\xi, y+\xi]$ be arbitrary. Then $x\in \mathfrak{T}(B)$ by definition, and therefore the formula (2.10) gives ${| {f^\triangleleft(x) - f^\triangleleft(y)} |} \leqslant \eta$ . We have thus proven that $f^\triangleleft|_{X_\varepsilon}$ is continuous at y with the restricted topology, as claimed.

2.4.1 Proofs of Theorems 1.4 and 1.5. The domain and continuity of $f^\triangleright$ .

For the purpose of studying the example of the Kontsevich function (1.12), it will be convenient to generalize the period relations to

(2.26) \begin{equation} \begin{cases} f(x+1) = \vartheta f(x) & (x\in \mathbb{Q}), \\ h(x) = f(x) - \vartheta^{\pm 3} {| {x} |}^{-k} f(-1/x) & (\pm x \in \mathbb{Q}_{>0}), \end{cases}\end{equation}

where $\vartheta$ is some fixed root of unity. Define

(2.27) \begin{align} \vartheta_j = \vartheta_j(b_1, b_2, \dotsc) = \begin{cases} \vartheta^{\sum_{i=1}^j (-1)^i b_i} & (j\text{ even}), \\ \vartheta^{3+\sum_{i=1}^j (-1)^i b_i} & (j\text{ odd}), \end{cases}\end{align}

where $b_1, b_2, \dotsc$ are integers. By arguing in the same way as (2.2), we find for $x = [0; b_1, \dotsc, b_r]$ that

(2.28) \begin{equation} f(x) = \sum_{j=0}^{r-1} \vartheta_j\, \bigg(\frac{u_j}{u_0}\bigg)^{\!{-}k} h\bigg((-1)^j \frac{u_{j+1}}{u_j}\bigg) + \vartheta_r u_0^k f(0).\end{equation}

Note that by the period relation, we have

$$ h((-1)^r) = (\vartheta^{(-1)^r} - \vartheta^{3(-1)^r+(-1)^{r+1}}) f(0), $$

which confirms that the expression (2.28) holds for $x = [0; b_1, \dotsc, b_r]$ regardless of whether this expansion is canonical ( $b_r>1$ if $r>1$ ) or not. Similarly to (2.4), we may then work with r odd, for which we observe that

$$ \vartheta_{r-j}(b_1, \dotsc, b_r)\vartheta_{j}(b_r, \dotsc, b_1) = \vartheta_{r}(b_1, \dotsc, b_r). $$

Changing j to $r-j$ in the sum (2.28), and using the notation (1.17), we deduce that for $x = [0; b_1, \dotsc, b_r]$ , r odd, we have

(2.29) \begin{equation} \vartheta_r^{-1}(x) q^{-k} f(x) = \Psi(\overline{x}),\end{equation}

where now $\Psi(y)$ is defined for $y = [0; c_1, \dotsc, c_r]$ , r odd, as

(2.30) \begin{equation} \Psi(y) = \sum_{j=1}^r \vartheta_j^{-1}(y) v_j(y)^{-k} h\bigg((-1)^j \frac{v_{j-1}(y)}{v_j(y)}\bigg) + f(0),\end{equation}

where, whenever the expansion $x = [0;b_1, \dotsc, b_r]$ is clear from the context, we denote

(2.31) \begin{equation} \vartheta_j(x) = \vartheta_j(b_1, \dotsc),\end{equation}

but this quantity in fact depends on the tuple $(b_j)$ representing x.

In the case of $\operatorname{Re}(k)>0$ and $h(x)=O(|x|^{-\operatorname{Re}(k)})$ for $x\in[-1,1]\smallsetminus\{0\}$ , the existence of $f^\triangleright$ for all $x\in\mathbb{R}$ is straightforward. Indeed, by (2.28) we have

(2.32) \begin{align} f^\triangleright(x)=\begin{cases} \Psi( x) & \text{if }x\in\mathbb{Q}, \\ {\displaystyle \lim_{\mathbb{Q}\ni y\to x}\Psi( y)} & \text{if }x\notin\mathbb{Q}, \end{cases}\end{align}

where the limit exists by virtue of the fact that the sum in (2.28) is uniformly convergent and for each j, $v_j(y)$ depends only on finitely many of the functions $y \mapsto b_j(y) = {\lfloor {1/T^{j-1}(y)} \rfloor}$ which are locally constant at irrationals. Note that for irrational $x=[0;c_1,c_2,\ldots]$ we simply have

(2.33) \begin{equation} f^\triangleright(x)=\sum_{j=1}^{\infty}\vartheta_j^{-1}v_{j}^{-k} h\bigg((-1)^{j-1}\frac{v_{j-1}}{v_{j}}\bigg)+f(0).\end{equation}

The continuity of $f^\triangleright$ on $\mathbb{R}\smallsetminus\mathbb{Q}$ when $h(x)=O(|x|^{-\operatorname{Re}(k)})$ is immediate from (2.33). Suppose $x=[0;b_1,\ldots,b_r]\in\mathbb{Q}$ with r odd. If $x'\to x^-$ with x’ sufficiently close to x, then we have $x'=[0;b_1,\ldots,b_r,b',b_{r+2}',\ldots]$ with $b'\to\infty$ . It follows immediately from the definition that $f^\triangleright(x')\to f^\triangleright(x)$ if $h(x)=o(|x|^{-\operatorname{Re}(k)})$ as $x\to0$ , and so under this assumption $f^\triangleright$ is continuous at x from the left. Next, let $x'\to x^+$ , then if $b_r>1$ we have $x'=[0;b_1,\ldots,b_r-1,1,b',b'_{r+2},\ldots]$ with $b'\to\infty$ . Note that

$$ \frac{v_{r-1}(x)}{v_{r}(x)} = \overline x, \quad \frac{v_{r-1}(x')}{v_r(x')} = \frac{v_{r-1}(x)}{v_r(x)-v_{r-1}(x)} = \frac{\overline x}{1-\overline x}, \quad \frac{v_{r}(x')}{v_{r+1}(x')}=1-\overline x. $$

Then, under the hypothesis $h(x)=o(|x|^{-\operatorname{Re}(k)})$ , by (2.30) and the relations (2.26), we obtain

\begin{align*} f^\triangleright(x')-f^\triangleright(x) = {}& \vartheta_r^{-1} q^{-k} \bigg(\vartheta^{-1} (1-\overline x)^{-k} h\bigg(\frac{\overline x}{1-\overline x}\bigg) + \vartheta h(\overline x-1) - h(\overline x)\bigg) + o(1) \\ = {}& \vartheta_r^{-1} q^{-k} \bigg( \vartheta^{-1}(1-\overline x)^{-k}f\bigg(\frac{\overline x}{1-\overline x}\bigg) + \vartheta h(\overline x - 1) - f(\overline x) \bigg)+o(1) \\ = {}& o(1)\end{align*}

as $b'\to\infty$ . If $b_r=1$ , then $x'=[0;b_1,\ldots,b_{r-1}+1,b',b'_{r+1},\ldots]$ with $b'\to\infty$ . We then obtain in a similar way

\begin{align*} f^\triangleright(x')-f^\triangleright(x) = {}& \vartheta_r^{-1}q^{-k}\bigg( \vartheta h(\overline x - 1) - h(\overline x) - \vartheta^2 (\overline x)^{-k} h\bigg(\frac{\overline x - 1}{\overline x}\bigg) \bigg) + o(1) \\ = {}& o(1),\end{align*}

which goes to 0 as $x'\to x^+$ , and concludes the proof that $f^\triangleright$ is continuous on $\mathbb{R}$ . This proves the first part of Theorem 1.4.

Assume now that f is not (twisted) periodic, but instead satisfies $f(x) = \vartheta f(x+1)$ only for $x \in \mathbb{Q}\smallsetminus[-1, 0]$ , and that h satisfies $h(x)=O({e}^{|x|^{-1+\delta}})$ for some $\delta>0$ . In this situation, we use Proposition 2.5 to extend $\Psi$ .

Lemma 2.10. Let $\delta>0$ , and assume that f satisfies

$$\begin{equation*} \begin{cases} f(x+1) = \vartheta f(x) & (x\in \mathbb{Q}\smallsetminus[-1, 0]), \\ h(x) = f(x) - \vartheta^{\pm 3} {| {x} |}^{-k} f(-1/x) & (\pm x \in \mathbb{Q}_{>0}), \end{cases} \end{equation*}$$

and that the map h defined through (1.2) satisfies $h(x) = O({e}^{|x|^{-1+\delta}})$ . Then the formula (2.29) holds with

$$ \Psi(y) := \sum_{j=1}^r \vartheta_j^{-1}(y) v_j(y)^{-k} h\bigg((-1)^j \frac{v_{j-1}(y)}{v_j(y)}\bigg) + \vartheta^{-1} f(-1) \quad (y=[0;c_1, \dotsc, c_r]; r\text{ odd}), $$

and the map $\Psi$ has the property $\mathscr{S}(1+\delta)$ .

Proof. The verification that Equations (2.29) and (2.30) hold, with f(0) replaced with $\vartheta^{-1}f(-1)$ , is straightforward. Let $B = m^{1+\delta}$ , $x\in \mathfrak{T}(B)$ and $x', x'' \in V(B, m, x)$ . Note, in particular, that $b_j(x') = b_j(x'')$ for all $j\leqslant m$ , and therefore also $v_j(x') = v_j(x'')$ and $\vartheta_j(x') = \vartheta_j(x'')$ . We deduce

$$ {| {\Psi(x') - \Psi(x'')} |} \leqslant \sum_{j=m+1}^{r(x')} v_j(x')^{-\operatorname{Re}(k)} {\bigg| {h\bigg(\!(-1)^j \frac{v_{j-1}(x')}{v_j(x')}\bigg)} \bigg|} + \sum_{j=m+1}^{r(x'')} v_j(x'')^{-\operatorname{Re}(k)} {\bigg| {h\bigg(\!(-1)^j \frac{v_{j-1}(x'')}{v_j(x'')}\bigg)}\! \bigg|}. $$

Since $x'\in\mathfrak{T}(B)$ , we also have $v_{j-1}(x') / v_j(x') \gg 1/b_j(x') \gg j^{-1}(\log j)^{-2} + B^{-1}$ . The same holds for x”. By our hypothesis on h, and combining this with (1.18), we deduce

$$ {| {\Psi(x') - \Psi(x'')} |} \ll \sum_{j\geqslant m+1} \varpi^{-j \operatorname{Re}(k)} ({e}^{O(j^{1-\delta}(\log j)^2)} + {e}^{O(m^{1-\delta^2})}) \ll \varpi^{-m \operatorname{Re}(k)/2}, $$

which tends to 0 as $m\to \infty$ , uniformly in x, x’, x”. This shows the property $\mathscr{S}(1+\delta)$ for $\Psi$ .

Letting $f^\triangleright(x) := \Psi^*(x)$ with the notation of Proposition 2.5, the deduction Theorem 1.5 follows by arguments identical to those of § 2.3.4, which were used to prove Theorem 1.2.

2.4.2 Proof of Theorem 1.4. The differentiability of $f^\triangleright$ .

We assume here the hypotheses of Theorem 1.4, in particular, that $h(x)=O(|x|^{-\operatorname{Re}(k)})$ for $0<|x|<1$ . Let $X = \mathfrak{S}$ be as in Lemma 2.2, let $\varepsilon>0$ and suppose $x=[0;b_1,b_2,\ldots] \in \mathfrak{S}$ . Let $m := m(x,\varepsilon)$ be as in § 2.3.1. Let $x'\in(0,1)$ be such that $|x-x'|<\varepsilon$ , so that $x'=[b_1,\ldots,b_m,b_{m+1}',\ldots]$ . Let $n\geqslant m$ be the least integer such that $b_{n+1}\neq b_{n+1}'$ . We write $x=[0;b_1,\ldots,b_n,z]$ and $x'=[0;b_1,\ldots,b_n,z']$ with $z=[b_{n+1};b_{n+2},\ldots]$ , $z'=[b'_{n+1};b'_{n+2},\ldots]$ (and $b_{n+1}\neq b_{n+1}'$ by hypothesis). By standard properties of continued fractions, we have

\begin{align*} |x-x'|=\frac{|z-z'|}{(z v_n+v_{n-1})(z' v_n+v_{n-1})}.\end{align*}

This is $\gg b_{n+1}^{-2}v_n^{-2}$ , unless $b_{n+1}-b_{n+1}'$ is equal to 1 or $-1$ . In the first case we have $z-z'=1+1/({b_{n+2}+y})-1/({b_{n+2}'+y'})\geqslant b_{n+2}^{-1}$ for some $0\leqslant y,y'\leqslant1$ . In the second case we have $z'-z=1-1/({b_{n+2}+y})+1/({b_{n+2}'+y'})\geqslant y/2\geqslant -(b_{n+3}+1)^{-1}$ . Since $x\in \mathfrak{S}$ , in all cases we have $\varepsilon>|x-x'|\gg_x v_{n}^{-2} n^{-3}(\log n)^{-6} \gg v_n^{-2} (\log v_n)^{-4}$ .

Finally, we have

\begin{align*} f^\triangleright(x)-f^\triangleright(x')\ll \sum_{j\geqslant {n+1}}\frac{1}{v_{j-1}^{\operatorname{Re}(k)}}+\sum_{j\geqslant {n+1}}\frac{1}{v_{j-1}^{\prime\,\operatorname{Re}(k)}} \ll v_n^{-\operatorname{Re}(k)} \ll_x \varepsilon^{\operatorname{Re}(k)/2}|\log \varepsilon|^{\operatorname{Re}(k)}\end{align*}

and so it follows that $f^\triangleright$ is locally $\alpha$ -Hölder continuous at x, for any $\alpha<\frac12\operatorname{Re}(k)$ . This completes the proof of Theorem 1.4.

3.1.1 Almost sure stability of the CF expansion.

First we need the following lemmas.

Lemma 3.1. Let $q\geqslant2$ and let

$$ \mathfrak{A}_q:=\{0\leqslant a<q\mid (a,q)=1: a/q \in \mathfrak{T}((\log q)(\log\log q)^2)\}. $$

Then $|\mathfrak{A}_q|=\varphi(q)(1+o(1))$ as $q\to\infty$ , where $\varphi$ denotes Euler’s totient function.

Lemma 3.2. Let $q\geqslant2$ and let

$$ \mathfrak{T}_{q}:= ([0,1)\smallsetminus\mathbb{Q}) \cap \mathfrak{T}((\log q)(\log\log q)^2).$$

Then $\nu(\mathfrak{T}_q)=1+o(1)$ as $q\to\infty$ .

We can now deduce the following two lemmas, which show that $f^\triangleright$ and $f^\triangleleft$ can be well approximated by their values at suitable close-by rationals. Here again, we will obtain this as a consequence of the property $\mathscr{S}(\lambda)$ for some $\lambda>1$ .

Lemma 3.3. Suppose that there exists $\lambda>1$ such that $f:\mathbb{Q}\to\mathbb{C}$ has the property $\mathscr{S}(\lambda)$ . Then the map $f^*$ defined in Proposition 2.5 satisfies

(3.1) \begin{equation} f^*(a/q) = f^*(x) + o(1) \quad \forall a\in \mathfrak{A}_q, \quad \forall x\in \bigg(\frac {a-q^{1/4}}q, \frac {a+q^{1/4}}q\bigg) \cap \bigg(\mathfrak{T}_q \cup \frac1q \mathfrak{A}_q\bigg) \end{equation}

as $q\to\infty$ , uniformly for a and x within the stated sets.

Proof. Let $B := (\log q)(\log\log q)^2$ . We assume first $x\in(a/q, (a+q^{1/4})/q)$ , and we let $m:=2\lceil B^{1/\lambda}\rceil+1$ , so that, in particular, $m=(\log q)^{1/\lambda+o(1)}$ as $q\to\infty$ . Let $a\in \mathfrak{A}_q$ , denote $x' := a/q$ and let $x\in(a/q, (a+q^{1/4})/q)$ with either $x \cap \mathfrak{T}_q$ or $x=b/q$ , where $b\in \mathfrak{A}_q$ . We wish to prove that

$$ f^*(x) - f^*(x') = o(1) $$

as $q\to\infty$ , with a rate of decay depending at most on h. This follows from Proposition 2.5 if we can prove that $x, x'\in \mathfrak{T}(B)$ , $r(x) \geqslant m$ and $b_j(x) = b_j(x')$ for $j\leqslant m$ . The fact that $x, x' \in \mathfrak{T}(B)$ follows by definition. Write $x' = a/q = [0; b_1, \dotsc, b_r]$ . Note that $r\ll \log q$ by Euclid’s algorithm, so that by definition of $\mathfrak{T}(B)$ and our choice of B, we have $b_j \ll (\log q)(\log\log q)^2$ . By induction (see the proof of Theorem 31 in [Reference KhintchineKhi63]), we have $q^{1/r} \leqslant 2 (b_1 \dotsc b_r)^{1/r} \ll (\log q)(\log \log q)^2$ , and therefore $r \gg (\log q)/\log\log q$ . We deduce that $r(x') \geqslant m$ for q large enough. Finally, let $a^* / q^* := [0, b_1, \dotsc, b_m]$ be the mth convergent of $a/q$ . Then we have $(q^*)^{1/m} \ll (\log q)(\log\log q)^2$ as above, and therefore $\log(q^*) \ll m \log\log q = o(\log q)$ as $q\to\infty$ . In particular, $q^*<q^{1/3}$ for q large enough. Since m is odd, we deduce

$$ \frac{a^*}{q^*} - \frac{a}{q} > \frac1{(q^*)^2(b_{m+1}+2)} \gg \frac1{q^{2/3}(\log q)^2} \geqslant \frac{q^{1/4}}q $$

for q large enough. In particular, any $x\in (a/q, ({a+q^{1/4}})/q)$ satisfies $a/q<x<a^*/q^*$ , and it follows that $b_j(x) = b_j(a/q)$ for $j\leqslant m$ . This concludes the proof.

The remaining case $x\in((a-q^{1/4})/q, a/q)$ follows by an identical argument taking m even instead of odd.

3.1.2 Proof of Theorems 1.3 and 1.7: The convergence in distribution.

Let $\delta>0$ . We now assume that either $\operatorname{Re}(k)<0$ and h satisfies (1.5), or $\operatorname{Re}(k)>0$ and h satisfies $h(x) = O({e}^{{| {x} |}^{-1+\delta}})$ . Let $f^*$ be either $f^\triangleleft$ or $f^\triangleright$ , depending on the sign of $\operatorname{Re}(k)$ . By Lemmas 2.9 and 2.10, the map $f^*$ satisfies the conclusion of Lemma 3.3.

We will show that for any function $G\in\mathcal C_c^\infty(\mathbb{C})$ , we have

\begin{equation*} \mathcal{R} := \frac1{\varphi(q)} \sum_{\substack{a=0 \\ (a, q)=1}}^{q-1} G(f^*(a/q))=\int_0^1 G(f^*(x))\,\mathop{}\!{d} \nu+o(1)\end{equation*}

as $q\to\infty$ .

For $0\leqslant j < q^{3/4}$ , let $I_j:=[j q^{1/4},(j+1)q^{1/4}]\cap [0,q].$ If $j<q^{3/4}-1$ , by a standard estimate [Reference Fouvry and TenenbaumFT91, Equations (1.13), (4.11)], we have

(3.2) \begin{equation} \sum_{a\in \mathfrak{A}_q\cap I_j}1\leqslant \sum_{\substack{(a, q)=1 \\ a\in I_j}} 1 = \frac{\varphi(q)}{q}q^{1/4}+ O(q^{1/10 }).\end{equation}

Given $\varepsilon\in(0, 1)$ , we let $E_{\pm}=\{j \mid \pm ( \# I_{j}\cap \mathfrak{A}_q-\varphi(q)q^{-3/4}) > \varepsilon \varphi(q)q^{-3/4}\}$ and $E=E_{+}\cup E_-$ . By (3.2) it is clear that $E_+$ is empty if we assume q is large enough. Also, summing (3.2) over $j\notin E_{-}$ , we deduce from Lemma 3.1 that $\# E=\# E_{-}=O(\varepsilon q^{3/4})$ . Now, for each $j\notin E$ we fix any $y_j\in I_{j}\cap \mathfrak{A}_q$ . Since $\|G\|_\infty, \|G'\|_\infty<\infty$ , by Lemma 3.1 and Equations (3.2) and (3.1), we have

\begin{equation*} \begin{split} \mathcal{R} {}& = \frac1{\varphi(q)} \sum_{a\in \mathfrak{A}_q} G(f^*(a/q)) + o(1) \\ &{}= \frac1{\varphi(q)} \sum_{\substack{0\leqslant j \leqslant q^{3/4}\\ j\notin E}} \sum_{a\in I_j\cap \mathfrak{A}_q} G(f^*(a/q)) + o(1)+O(\varepsilon)\\ &{}= \frac1{\varphi(q)} \sum_{\substack{0\leqslant j \leqslant q^{3/4}\\ j\notin E}} G(f^*(y_j)) \sum_{a\in I_j\cap \mathfrak{A}_q} 1+ o(1)+O(\varepsilon). \end{split}\end{equation*}

For $j\notin E$ the inner sum is $({\varphi(q)}/{q})q^{1/4}(1+O(\varepsilon))$ and thus by (3.1) and Lemma 3.2 we deduce

\begin{align*} \mathcal{R} {} &{}= \frac1{q^{3/4}} \sum_{\substack{0\leqslant j \leqslant q^{3/4}\\ j\notin E}} G(f^*(y_j)) + o(1)+O(\varepsilon)\\ &{}= \sum_{\substack{0\leqslant j \leqslant q^{3/4}\\ j\notin E}} \int_{(\frac1qI_j)\cap \mathfrak{T}_q}G(f^*(x)) \mathop{}\!{d} x + o(1)+O(\varepsilon)\\ &{}= \int_{ \mathfrak{T}_q}G(f^*(x)) \mathop{}\!{d} x + o(1)+O(\varepsilon)\\ &{}= \int_{ [0,1]}G(f^*(x)) \mathop{}\!{d} x + o(1)+O(\varepsilon),\end{align*}

since $\nu(\cup_{j\in E}((1/q)I_j))=O(\varepsilon)$ . Letting $\varepsilon\to0$ sufficiently slowly, we obtain the claimed result.

This proves the first parts of Theorems 1.3 and 1.7.

3.3.1 The case of bounded h.

Let $N\in\mathbb{Z}_{>0}$ be the order of $\vartheta$ as a root of unity, so that

$$ \vartheta^N = 1, $$

where $\vartheta$ is the automorphy factor in the generalized period relation (2.26). We recall that $\vartheta_j$ was defined in (2.27). Recall also from (2.31) that the notation $\vartheta_j(x)$ for a number $x\in\mathbb{Q}$ depends on how we expand x in CF when x is rational. In what follows, we will work with the expansion of odd length. For each $g\geqslant 1$ , we define $e_g(x) \in \mathbb{Z}/N\mathbb{Z}$ through $ \vartheta_g(x) = \vartheta^{e_g(x)}$ and notice that by definition we have

(3.9) \begin{equation} e_g(x)\equiv e_{g-1}(x)+(-1)^g b_g+3(-1)^{g+1}\pmod{N}.\end{equation}

Given $c_1, \dotsc, c_\ell\geqslant 1$ , we denote

(3.10) \begin{equation} \begin{split} J(c_1, \dotsc, c_\ell) := {}& \{x\in[0, 1], b_i(x) = c_i \text{ for } 1\leqslant i \leqslant \ell \}, \\ \mu(c_1, \dotsc, c_\ell) := {}& \nu(J(c_1, \dotsc, c_\ell)). \end{split}\end{equation}

By [Reference KhintchineKhi63, p. 57], we have

(3.11) \begin{equation} \mu(c_1, \dotsc, c_\ell) = \frac1{v_\ell(x)(v_\ell(x) + v_{\ell-1}(x))} \asymp \frac1{v_\ell(x)^2} \quad (x = [0; c_1, \dotsc, c_\ell]).\end{equation}

Lemma 3.9. Let $m\in \mathbb{N}$ , $K>1$ and $(L_1,\ldots,L_m)\in\mathbb{N}^m$ be fixed. Let $V\geqslant1$ , and for each $e\in\mathbb{Z}/N\mathbb{Z}$ and $\omega>0$ , let $\mathscr{B}(e, \omega) \subset \mathbb{Z}\cap[\sqrt{V}, K\sqrt{V}]$ be given, and

$$ T := \inf_{e\in\mathbb{Z}/N\mathbb{Z},\, \omega>0} \# \mathscr{B}(e, \omega). $$

Also, for $x\in(0, 1)\smallsetminus\mathbb{Q}$ , let

$$ G_V(x):=\{g\in [V,2 V] \mid b_g(x) \in \mathscr{B}(e_{g-1}(x), v_{g-1}(x)),\, b_{g+i}(x) =L_i\ \forall i=1,\ldots,m\} $$

and let $X_{V, \mathscr{B}} = X_{V,\mathscr{B}}(K, L_1, \dotsc, L_m)$ be the subset of $[0,1]\smallsetminus\mathbb{Q}$ such that $G_V(x)$ contains at least an even and an odd integer. Then $\nu(X_{V,\mathscr{B}})=1+o(1)$ as $V, T \to \infty$ , where the rate of decay of o(1) depends at most on $K, L_1, \dotsc, L_m$ .

By [Reference KhintchineKhi63, Equation (57)] we have uniformly for $\ell\geqslant 1$ and $c_1, \dotsc, c_{\ell+1}\geqslant 1$ ,

(3.12) \begin{align} \tfrac13 c_{\ell+1}^{-2}\mu(c_1,\ldots,c_{\ell})\leqslant \mu(c_1,\ldots,c_{\ell+1})\leqslant 2 c_{\ell+1}^{-2}\mu(c_1,\ldots,c_{\ell}). \end{align}

In particular, writing $L:=\max(L_1,\ldots, L_m)$ , we have for any $c\in\mathbb{Z}_{>0}$ ,

$$ \mu(c_1,\ldots,c_\ell,c,L_1,\ldots,L_m)\geqslant 3^{-m-1} c^{-2}L^{-2m}\mu(c_1,\ldots,c_{\ell}).$$

Since $\sum_{c\in\mathscr{B}(e_\ell, v_\ell)}c^{-2}\geqslant (1/{K^2V})\#{\mathscr{B}(e_\ell, v_\ell)}$ , where $e_\ell = e_\ell([0;c_1, \dotsc, c_\ell])$ , it then follows that the measure of $x\in [0,1]\smallsetminus\mathbb{Q}$ such that none of the integers $g\in\{2{\lfloor {V/2} \rfloor} + \ell (2m+2) \mid 1\leqslant \ell \leqslant V /(2m+2)\}$ satisfy $b_g(x) \in \mathscr{B}$ is

$$ \ll (1- K^{-2}(3L)^{-3m}TV^{-1})^{{\lfloor {V/(2m+2)} \rfloor}} = o(1) $$

as $V, T\to\infty$ with m, L fixed. The lemma then follows.

We now prove Theorem 3.5. To illustrate Remark 3.6, we assume only that h is bounded on $[-1, 1]$ , not necessarily continuous, and that it has finite left or right limit at 0. We assume it is the former, the latter case being analogous, as we will comment after the proof. If $\operatorname{Im}(k)=0$ , we suppose that $\phi(\vartheta^p h(0^-))\neq0$ for some $p\in\mathbb{Z}$ , whereas if $\operatorname{Im}(k)\neq0$ , we assume that $h(0^-)\neq0$ and set $p:=0$ . Moreover, we let

(3.13) \begin{equation} \kappa\in\mathbb{R}, \quad \phi({e}(-\kappa) \vartheta^p h(0^-)) \text{ is maximal.}\end{equation}

This means, in other terms, that for any $z\in\mathbb{C}$ ,

(3.14) \begin{equation} \phi(\vartheta^p h(0^-) z) = {| {h(0^-)} |} \operatorname{Re}(e(\kappa) z).\end{equation}

If $\operatorname{Im}(k)=0$ , then setting $z=1$ we see that $\kappa\not\equiv\pm\frac12\ (\textrm{mod}\,1)$ by hypothesis.

We define our choice of sets $\mathscr{B} = \mathscr{B}(e, \omega)$ . We let $\xi\in(0,1)$ be a parameter, on which these sets will depend. For $e\in\mathbb{Z}/N\mathbb{Z}$ , $\omega>0$ , let $K=2$ if $\operatorname{Im}(k)=0$ and $K=\max(2,e^{3\pi/{| {\operatorname{Im}(k)} |}})$ otherwise. Also, let $ \mathscr{B}(e, \omega) = \mathscr{B}_{V, \xi, N, k}(e, \omega)$ be defined by

(3.15) \begin{equation} \begin{aligned} \mathscr{B}(e, \omega) := {}& \Bigg\{b\in\mathbb{Z}\cap [\sqrt V, K\sqrt V] \colon b\equiv 3 - (p+e)\pmod{N} \\ & \quad \text{and }\left. \begin{cases} -\xi \leqslant \bigg\{\dfrac{\operatorname{Im} (k)}{2\pi}\log (\omega\sqrt V)-\kappa\bigg\} + \dfrac{\operatorname{Im} (k)}{2\pi}\log (b/\sqrt V) \leqslant \xi & \text{if } k\notin\mathbb{R} \\[2pt] 0 \leqslant \log (b/\sqrt V) \leqslant \xi & \text{if } k\in\mathbb{R} \end{cases}\right\}, \end{aligned}\end{equation}

where $\{\cdot\}$ denotes the fractional part, and $\kappa$ was defined at (3.13). Here our choice of K ensures that the set $\mathscr{B}$ is not void for V large enough, and in fact

(3.16) \begin{equation} \#\mathscr{B} \gg_{k, N} \xi \sqrt{V}\end{equation}

for V large enough in terms of k, N and $\xi$ . Moreover, on the one hand, by (3.9) the congruence condition ensures that whenever a number $x\in[0,1]\smallsetminus\mathbb{Q}$ satisfies $b_g(x) \in \mathscr{B}(e_{g-1}(x), v_{g-1}(x))$ with g even, we have $e_g(x) \equiv -p\ (\textrm{mod}\,N)$ , and therefore we deduce

(3.17) \begin{equation} \vartheta_g(x) = \vartheta^{-p} \quad (g\in G_V(x)\cap 2\mathbb{Z}).\end{equation}

On the other hand, the condition involving $\|\cdot\|_{\mathbb{R}/\mathbb{Z}}$ ensures that

(3.18) \begin{equation} \operatorname{Re}({e}(\kappa) (v_{g-1}(x) b_g(x))^{-i\operatorname{Im} k}) = 1 + O(\xi^2) \quad (\operatorname{Im} k \neq 0, g\in G_V(x)) \end{equation}

by the Taylor expansion of the cosine, with an absolute implicit constant.

We let $L_1,\ldots,L_m\in\mathbb{N}$ be chosen later and take $X_{V,\mathscr{B}}$ , $G_V(x)$ as in Lemma 3.9. For all even g and all $c_1, \dotsc, c_{g-1}\in \mathbb{N}$ , we let

(3.19) \begin{equation} S_{g, (c_i), \mathscr{B}} := \{x \in (0,1)\smallsetminus\mathbb{Q}: \forall i<g, b_i(x) = c_i; g\in G_V(x)\cap2\mathbb{Z}; \forall j\geqslant 1, g-2j \not\in G_V(x)\},\end{equation}

that is, $S_{g, (c_i), \mathscr{B}}$ is the subset of $ J(c_1,\ldots,c_{g-1})$ such that g is the smallest even element of $G_V(x)$ . We will drop the subscript $\mathscr{B}$ from the notation.