Proof. By the triangle inequality, definition of $\Xi _{\varepsilon }$ , non-negativity of the coefficients of $\Phi$ , and $\Phi (0)=1$ , we have

\begin{align*} |f_{x,z}(y)|&=|x\bigl ((\Phi (y)-1)-(\Phi (z)-1)\bigr )|\\ &\le (\rho +\varepsilon )(|\Phi (y)-1|+|\Phi (z)-1|)\\ &\le (\rho +\varepsilon )((\Phi (|{y}|)-1)+(\Phi (|z|)-1))\\ &\le (\rho +\varepsilon )(\Phi (\tau -\rho +\varepsilon )-1+\Phi (\varepsilon )-1). \end{align*}

For $\varepsilon \to 0$ , the upper bound converges to $\rho \Phi (\tau -\rho )-\rho$ because we are assuming that $\Phi (0)=1$ . As $\rho \Phi (\tau -\rho )-\rho \lt \rho \Phi (\tau )-\rho =\tau -\rho$ by (3), the assertion of the lemma holds for sufficiently small $\varepsilon \gt 0$ .

Proof. For any triple $(x, y, z)\in \Xi _{\varepsilon }$ ,

\begin{equation*} |f'_{x, z}(y)|=|x\Phi '(y)|\le (\rho +\varepsilon )\Phi '(\tau -\rho +\varepsilon ). \end{equation*}

For $\varepsilon \to 0$ , the upper bound converges to $\rho \Phi '(\tau -\rho )$ , which is less than $\rho \Phi '(\tau )=1$ (by (3)).

Proof. The fact that $f_{x,z}$ maps $\Xi _{\varepsilon }^{(2)}$ to itself for sufficiently small $\varepsilon \gt 0$ is a direct consequence of Lemma 3.1.

Making use of Lemma 3.2, the contraction property now follows by a standard argument: For $y_1$ , $y_2\in \Xi _{\varepsilon }^{(2)}$ , we have

\begin{equation*} |f_{x, z}(y_2)-f_{x, z}(y_1)|\le \int _{[y_1, y_2]}|f'_{x, z}(y)|\,|dy|\le \lambda |y_2-y_1|. \end{equation*}

Proof. Note that using Lemma 3.2, we have that $|\frac{\partial }{\partial y}(y-f_{x, z}(y))|=|1-f_{x,z}^{\prime}(y)| \geq 1 - |f_{x,z}^{\prime}(y)| \geq 1 - \lambda$ is bounded away from zero for sufficiently small $\varepsilon \gt 0$ and $(x, y, z)\in \Xi _{\varepsilon }$ . Thus the analytic implicit function theorem shows that $g$ as defined by (12) is analytic and has bounded partial derivative $\frac{\partial }{\partial z} g(x,z)$ on $\Xi _{\varepsilon }^{(1, 3)}$ for sufficiently small $\varepsilon \gt 0$ .

Proof. Since $g$ is defined as the fixed point of $f_{x, z}$ and $f_{x, z}$ is a contraction with Lipschitz constant $\lambda$ , we have

\begin{equation*} |\Lambda _h(x, y, z)| = |f_{x, z}^{(h-1)}(y) - f_{x, z}^{(h-1)}(g(x, z))| \le \lambda ^{h-1} |y-g(x, z)|=O(\lambda ^h) \end{equation*}

for $(x, y, z)\in \Xi _\varepsilon$ , so we have shown (16).

For $(x, y, z)\in \Xi _{\varepsilon/3}$ , Cauchy’s integral formula yields

\begin{equation*} \frac {\partial }{\partial z}\Lambda _h(x, y, z)=\frac {1}{2\pi i}\oint _{|\zeta -z|=\varepsilon/3}\frac {\Lambda _h(x, y, \zeta )}{(\zeta -z)^2}\,d\zeta. \end{equation*}

By (16), we can bound the integral by $O(\lambda ^h)$ . Thus replacing $\varepsilon$ by $\varepsilon/3$ yields (17).

Proof. To compute $\frac{\partial }{\partial z} g(x, z)$ , we differentiate (12) with respect to $z$ and obtain

\begin{equation*} \frac {\partial }{\partial z} g(x, z) = x\Phi '(g(x,z))\frac {\partial }{\partial z} g(x, z) - x\Phi '(z), \end{equation*}

which leads to

\begin{equation*} \frac {\partial }{\partial z} g(x, z) = - \frac {x\Phi '(z)}{1-x\Phi '(g(x, z))}. \end{equation*}

Note that the denominator is nonzero for $(x, z)\in \Xi _{\varepsilon }^{(1, 3)}$ by Lemma 3.2. We obtain

(19) \begin{equation} \left |\frac{\partial }{\partial z}(z-g(x, z))\right | = \left |{\frac{1+x(\Phi '(z)-\Phi '(g(x, z)))}{1-x\Phi '(g(x, z))}}\right |\ge \frac{1-(\rho +\varepsilon )\left |\Phi '(z)-\Phi '(g(x, z))\right |}{1+(\rho +\varepsilon )\left |\Phi '(g(x, z))\right |}. \end{equation}

By Lemma 3.4, $\frac{\partial g(x, z)}{\partial z}$ is analytic and bounded for $(x, z)\in \Xi _\varepsilon ^{(1, 3)}$ , and by (13), it follows that

\begin{equation*} g(x, z)=g(x, z)-g(x, 0)=\int _{[0, z]}\frac {\partial g(x, \zeta )}{\partial \zeta }\,d\zeta =O(|{z}|)=O(\varepsilon ) \end{equation*}

for $\varepsilon \to 0$ , uniformly in $x$ . Therefore, we have

\begin{equation*} \Phi '(z)-\Phi '(g(x, z)) = (\Phi '(z)-\Phi '(0))-(\Phi '(g(x,z))-\Phi '(0)) = O(\varepsilon ) \end{equation*}

and $|{\Phi '(g(x, z))}|= \Phi '(0)+O(\varepsilon )$ for $\varepsilon \to 0$ . So (19) yields

\begin{equation*} \left |{\frac {\partial }{\partial z}(z-g(x, z))}\right |\ge \frac {1-(\rho +\varepsilon )O(\varepsilon )}{1+(\rho +\varepsilon )(\Phi '(0)+O(\varepsilon ))} = \frac {1}{1+\rho \Phi '(0)}+O(\varepsilon ) \end{equation*}

for $\varepsilon \to 0$ . Setting $\delta \,{:\!=}\, \frac 12 \frac{1}{1+\rho \Phi '(0)}$ and choosing $\varepsilon$ small enough yields the result.

Proof. In view of (15) and (16), we have

(21) \begin{equation} g(x, z) - z = O(\lambda ^h). \end{equation}

By definition, $g(x, z)\in \Xi _{\varepsilon }^{(2)}$ . The implicit equation (12) for $g(x,z)$ and (21) imply

\begin{equation*} g(x, z) = x(\Phi (g(x, z)) - \Phi (z))=x \int _{[z, g(x, z)]}\Phi '(\zeta )\,d\zeta =O(|{g(x, z)-z}|)=O(\lambda ^h). \end{equation*}

Inserting this into (21) leads to (20).

Proof. We choose $h$ sufficiently large such that (17) implies

(23) \begin{equation} \left |{\frac{\partial }{\partial z}\Lambda _h(x, y, z)}\right |\le \frac{\delta }{2} \end{equation}

for $(x, y, z)\in \Xi _\varepsilon$ , where $\delta$ is taken as in Lemma 3.6, and such that (20) implies

(24) \begin{equation} |{z}|\le \frac{\varepsilon }{2} \end{equation}

for all $(x, y, z)\in \Xi _\varepsilon$ for which (15) holds.

By definition of $f$ , we have $f_{x, 0}(0) = 0$ and therefore $f_{x, 0}^{(h-1)}(0) = 0$ for every $x \in \Xi _\varepsilon ^{(1)}$ , so $z=0$ is a solution of (14) for $y=0$ . By (18) and (23), we have

(25) \begin{equation} \frac{\partial }{\partial z} (f_{x,z}^{(h-1)}(y) - z)\neq 0 \end{equation}

for $(x,y,z) \in \Xi _\varepsilon$ . The analytic implicit function theorem thus implies that, for every $x \in \Xi _\varepsilon ^{(1)}$ , there is an analytic function $q_h$ defined in a neighbourhood of $(x, 0)$ such that (22) holds there and such that $q_h(x, 0)=0$ . Next we show that this extends to the whole region $\Xi _\varepsilon ^{(1,2)}$ .

For $x_0 \in \Xi _\varepsilon ^{(1)}$ , let $r(x_0)$ be the supremum of all $r \lt \tau -\rho +\varepsilon$ for which there is an analytic extension of $y\mapsto q_h(x_0,y)$ from the open disk $D_r(0)$ to $\Xi _\varepsilon ^{(3)}$ . Suppose for contradiction that $r(x_0) \lt \tau - \rho + \varepsilon$ . Consider a point $y_0$ with $|{y_0}| = r(x_0)$ , and take a sequence $y_n \to y_0$ such that $|{y_n}| \lt r(x_0)$ . Note that $\left |{q_h(x_0,y_n)}\right | \leq \frac{\varepsilon }{2}$ by (24). Without loss of generality, we can assume that $q_h(x_0,y_n)$ converges to some $q_0$ with $|{q_0}| \leq \frac{\varepsilon }{2}$ as $n \to \infty$ (by compactness). By continuity, we have $q_0 = f_{x_0, q_0}^{(h-1)}(y_0)$ . Since $(x_0,y_0,q_0) \in \Xi _\varepsilon$ , we can still use the analytic implicit function theorem together with (25) to conclude that there is a neighbourhood of $(x_0,y_0,q_0)$ where the equation $f_{x, z}^{(h-1)}(y) = z$ has exactly one solution $z$ for every $x$ and $y$ , and an analytic function $\tilde{q}_h(x,y)$ such that $\tilde{q}_h(x,y) = f_{x, \tilde{q}_h(x,y)}^{(h-1)}(y)$ and $\tilde{q}_h(x_0,y_0) = q_0$ . We assume the neighbourhood to be chosen small enough such that $\tilde{q}_h(x, y)\in \Xi _\varepsilon ^{(3)}$ for all $(x, y)$ in the neighbourhood. For large enough $n$ , this neighbourhood contains $(x_0,y_n,q_h(x_0,y_n))$ , so we must have $q_h(x_0,y_n) = \tilde{q}_h(x_0,y_n)$ for all those $n$ . This implies that $\tilde{q}_h$ is an analytic continuation of $q_h$ in a neighbourhood of $(x_0, y_0)$ with values in $\Xi _\varepsilon ^{(3)}$ . Since $y_0$ was arbitrary, we have reached the desired contradiction.

So we conclude that there is indeed such an analytic function $q_h$ defined on all of $\Xi _\varepsilon ^{(1,2)}$ , with values in $\Xi _\varepsilon ^{(3)}$ . The fact that $q_h(x,y) = O(\lambda ^h)$ finally follows from Lemma 3.7.

Proof. Note first that $Y_{h,1}(x)$ is an increasing function of $x$ for positive real $x \lt \rho _h$ . Let $\tilde{\rho } = \min \!(\rho _h,\rho + \frac{\varepsilon }{2})$ . Suppose first that $\lim _{x \to \tilde{\rho }^-} Y_{h,1}(x) \geq \tau - \rho + \frac{\varepsilon }{2}$ . If $h$ is large enough, this implies together with (27) that

\begin{align*} \lim _{x \to \tilde{\rho }^-} \frac{\partial F_h}{\partial y} (x,Y_{h,1}(x)) &= \lim _{x \to \tilde{\rho }^-} \frac{\partial F_{\infty }}{\partial y} (x,Y_{h,1}(x)) + O(\lambda ^h) \\ &\geq \frac{\partial F_{\infty }}{\partial y} \Big (\rho,\tau -\rho + \frac{\varepsilon }{2}\Big ) + O(\lambda ^h) \\ &= \rho \Phi '\Big ( \tau + \frac{\varepsilon }{2}\Big ) + O(\lambda ^h) \gt \rho \Phi '(\tau ) = 1. \end{align*}

On the other hand, we also have

\begin{align*} \frac{\partial F_h}{\partial y} (\rho/2,Y_{h,1}(\rho/2)) &= \frac{\partial F_{\infty }}{\partial y} (\rho/2,Y_{h,1}(\rho/2)) + O(\lambda ^h) \\ &\leq \frac{\partial F_{\infty }}{\partial y} (\rho/2,Y(\rho/2)-\rho/2) + O(\lambda ^h) \\ &\lt \rho \Phi '(\tau ) = 1, \end{align*}

so by continuity there must exist some $x_0 \in (\rho/2, \tilde{\rho })$ such that

\begin{equation*} \frac {\partial F_h}{\partial y} (x_0,Y_{h,1}(x_0)) = 1. \end{equation*}

Moreover, if $h$ is large enough we have

\begin{equation*} \frac {\partial ^2 F_h}{\partial y^2} (x_0,Y_{h,1}(x_0)) = \frac {\partial ^2 F_{\infty }}{\partial y^2} (x_0,Y_{h,1}(x_0)) + O(\lambda ^h) \gt 0 \end{equation*}

as $x_0$ and thus also $Y_{h,1}(x_0)$ are bounded below by positive constants, and analogously $\frac{\partial F_h}{\partial x} (x_0,Y_{h,1}(x_0)) \gt 0$ . But this would mean that $Y_{h,1}$ has a square root singularity at $x_0 \lt \rho _h$ (compare the discussion in Section 3.3 later), and we reach a contradiction. Hence we can assume that

(28) \begin{equation} \lim _{x \to \tilde{\rho }^-} Y_{h,1}(x) \lt \tau - \rho + \frac{\varepsilon }{2}. \end{equation}

Assume next that $\rho _h \gt \rho + \lambda ^{h/2}$ . Now for $x_1 = \rho + \lambda ^{h/2} \lt \tilde{\rho }$ (the inequality holds if $h$ is large enough to make $\lambda ^{h/2} \lt \frac{\varepsilon }{2}$ ), $u_1 = Y_{h,1}(x_1) + x_1$ satisfies

(29) \begin{equation} u_1 = x_1 \Phi (u_1) + O(\lambda ^h), \end{equation}

since $F_h(x,y) = F_{\infty }(x,y) + O(\lambda ^h) = x (\Phi (y+x)-1) + O(\lambda ^h)$ . Note here that $u_1 \leq \tau + \frac{\varepsilon }{2} + \lambda ^{h/2}$ by (28), thus $u_1$ is in the region of analyticity of $\Phi$ (again assuming $h$ to be large enough). However, since $u \leq \rho \Phi (u)$ for all positive real $u$ for which $\Phi (u)$ is well-defined (the line $u \mapsto \frac{u}{\rho }$ is a tangent to the graph of the convex function $\Phi$ at $\tau$ ), for sufficiently large $h$ the right-hand side in (29) is necessarily greater than the left, and we reach another contradiction. So it follows that $\rho _h \leq \rho + \lambda ^{h/2}$ , and in particular $\tilde{\rho } = \rho _h \lt \rho + \frac{\varepsilon }{2}$ if $h$ is large enough. Since we know that $\lim _{x \to \tilde{\rho }^-} Y_{h,1}(x) \lt \tau - \rho + \frac{\varepsilon }{2}$ , we also have $\eta _{h,1} \,{:\!=}\, Y_{h,1}(\rho _h) \lt \tau -\rho + \frac{\varepsilon }{2}$ . We conclude that $(\rho _h,\eta _{h,1}) \in \Xi _{\varepsilon }^{(1,2)}$ , i.e., $(\rho _h,\eta _{h,1})$ lies within the region of analyticity of $F_h$ . So the singularity at $\rho _h$ must be due to the implicit function theorem failing at this point:

\begin{equation*} \eta _{h,1} = F_h(\rho _h,\eta _{h,1}) \text { and } 1 = \frac {\partial F_h}{\partial y} (\rho _h,\eta _{h,1}). \end{equation*}

The second equation in particular gives us

\begin{equation*} \rho _h \Phi '(\eta _{h,1} + \rho _h) = 1 + O(\lambda ^h) \end{equation*}

by (27). Since $\Phi '$ is increasing for positive real arguments and we know that $\rho \Phi '(\tau ) = 1$ and $\rho _h = \rho + O(\lambda ^{h/2})$ , we can conclude from this that $\eta _{h,1} = \tau - \rho + O(\lambda ^{h/2})$ .

Proof. Note that $(\rho,\eta _1)$ is a solution to the simultaneous equations $F_{\infty }(x,y) = x(\Phi (x+y)\,{-}\,1)\,{=}\,y$ and $\frac{\partial }{\partial y}F_{\infty } (x,y) = x \Phi '(x+y) = 1$ , and that there is no other solution with $|{x}| \leq \rho + \varepsilon$ and $|{y}| \leq \eta _1 + \varepsilon$ if $\varepsilon$ is chosen sufficiently small by our assumptions on the function $\Phi$ (see Section 2.1). We take $\Omega = \Xi _{\varepsilon }^{(1,2)}$ in Theorem 3.10 and set

\begin{equation*} u(x,y) = \Big (F_{\infty }(x,y) - y,\frac {\partial }{\partial y}F_{\infty }(x,y) - 1 \Big ). \end{equation*}

Moreover, take

\begin{equation*} v(x,y) = \Big (F_{h}(x,y) - F_{\infty }(x,y), \frac {\partial }{\partial y} F_{h}(x,y) - \frac {\partial }{\partial y} F_{\infty }(x,y) \Big ). \end{equation*}

Note that both coordinates of $v$ are $O(\lambda ^h)$ by (26) and (27). Since the boundary $\partial \Omega$ contains no zeros of $u$ , if we choose $h$ sufficiently large, then the conditions of Theorem 3.10 are satisfied. Consequently, $u$ and $u+v$ have the same number of zeros in $\Omega$ , namely $1$ . Solutions to the simultaneous equations $F_h(x,y) = y$ and $\frac{\partial }{\partial y} F_h(x,y) = 1$ are precisely zeros of $u+v$ , so this completes the proof.

Proof. Recall that we can approximate partial derivatives of $F_h$ by those of $F_{\infty }$ with an exponential error bound (as in (27)), giving us

\begin{align*} \frac{\partial }{\partial x}F_h(x,y)&=\frac{\partial }{\partial x}F_\infty (x,y) + O(\lambda ^h)\\&= \frac{\partial F_\infty }{\partial x}(\rho, \eta _1) + O(\lambda ^h) + O(x-\rho ) + O(y-\eta _1)\\&=\Phi (\tau ) + O(\lambda ^h)+O(x-\rho ) + O(y-\eta _1), \end{align*}

as well as

\begin{align*} \frac{\partial ^2}{\partial y^2}F_h(x,y)&=\frac{\partial ^2}{\partial y^2}F_\infty (x,y) + O(\lambda ^h)\\ &=\frac{\partial ^2 F_\infty }{\partial y^2}(\rho, \eta _1) + O(\lambda ^h)+ O(x-\rho ) + O(y-\eta _1)&\\ &=\rho \Phi ''(\tau ) + O(\lambda ^h)+ O(x-\rho ) + O(y-\eta _1) \end{align*}

for $(x, y)$ in a neighbourhood of $(\rho, \eta _1)$ contained in $\Xi _\varepsilon ^{(1, 2)}$ and $h\to \infty$ .

Using Lemma 3.9, we choose $\delta _1\gt 0$ and $\delta _2\gt 0$ small enough and $h_0$ large enough such that $|x-\rho _h|\le \delta _1$ , $|y-\eta _{h,1}|\le \delta _2$ , and $h\ge h_0$ imply that

(31) \begin{equation} \left |{\frac{\partial }{\partial x}F_h(x, y)-\Phi (\tau )}\right |\le \frac 12\Phi (\tau ) \end{equation}

and

(32) \begin{equation} \left |{\frac{\partial ^2}{\partial y^2}F_h(x,y)-\rho \Phi ''(\tau )}\right |\le \frac 12 \rho \Phi ''(\tau ), \end{equation}

and such that $\overline{D_{\delta _1}(\rho _h) \times D_{\delta _2}(\eta _{h,1})}\subseteq \Xi _\varepsilon ^{(1, 2)}$ . By Lemma 3.11, we have

(33) \begin{align} F_h(\rho _h, \eta _{h,1})&=\eta _{h,1}, \end{align}

(34) \begin{align} \frac{\partial F_h}{\partial y}(\rho _h, \eta _{h,1}) &=1. \end{align}

We now define

\begin{equation*} S_h(y)\,{:\!=}\, \frac {F_h(\rho _h, y)-y}{(y-\eta _{h,1})^2} \end{equation*}

for $y\in \overline{D_{\delta _2}(\eta _{h,1})} \setminus \{\eta _{h,1}\}$ . By (33) and (34), $S_h$ has a removable singularity at $\eta _{h,1}$ . Therefore it is analytic on $D_{\delta _2}(\eta _{h,1})$ . By (33), we have

\begin{align*} F_h(\rho _h, y)-y &= (F_h(\rho _h, y)-y) - (F_h(\rho _h, \eta _{h,1})-\eta _{h,1})\\ &=\int _{\eta _{h,1}}^{y} \Bigl (\frac{\partial }{\partial w}F_h(\rho _h, w)-1\Bigr )\,dw. \end{align*}

By (34), this can be rewritten as

\begin{align*} F_h(\rho _h, y)-y &=\int _{\eta _{h,1}}^{y} \Bigl (\Bigl (\frac{\partial F_h}{\partial y}(\rho _h, w)-1\Bigr ) - \Bigl (\frac{\partial F_h}{\partial y}(\rho _h, \eta _{h, 1})-1\Bigr )\Bigr )\,dw\\ &=\int _{\eta _{h,1}}^{y}\int _{\eta _{h, 1}}^w \frac{\partial ^2F_h}{\partial y^2}(\rho _h, v)\,dv\,dw\\ &=\int _{\eta _{h,1}}^{y}\int _{\eta _{h, 1}}^w \rho \Phi ''(\tau )\,dv\,dw+ \int _{\eta _{h,1}}^{y}\int _{\eta _{h, 1}}^w \Bigl (\frac{\partial ^2F_h}{\partial y^2}(\rho _h, v)-\rho \Phi ''(\tau )\Bigr )\,dv\,dw\\ &= \frac 12\rho \Phi ''(\tau )(y-\eta _{h, 1})^2+ \int _{\eta _{h,1}}^{y}\int _{\eta _{h, 1}}^w \Bigl (\frac{\partial ^2F_h}{\partial y^2}(\rho _h, v)-\rho \Phi ''(\tau )\Bigr )\,dv\,dw. \end{align*}

Rearranging and using the definition of $S_h(y)$ as well as (32) yields

\begin{equation*} \left |{S_h(y)-\frac 12\rho \Phi ''(\tau )}\right |\le \frac 14 \rho \Phi ''(\tau ) \end{equation*}

for all $y\in \overline{D_{\delta _2}(\eta _{h,1})}$ and $h\ge h_0$ . Thus $|{S_h(y)}|$ is bounded from below and above by positive constants for every such $y$ and $h$ .

We now define $R_h(x, y)$ such that (30) holds, which is equivalent to

\begin{equation*} R_h(x, y) \,{:\!=}\, \frac {F_h(x,y)-F_h(\rho _h, y)}{x-\rho _h} \end{equation*}

for $x\in \overline{D_{\delta _1}(\rho _h)}\setminus \{\rho _h\}$ and $y\in \overline{D_{\delta _2}(\eta _{h,1})}$ . We have

\begin{align*} F_h(\rho _h, y)-F_h(x,y)&=\int _x^{\rho _h}\frac{\partial F_h}{\partial x}(w, y)\,dw\\ &=\Phi (\tau )(\rho _h-x)+\int _x^{\rho _h}\Bigl (\frac{\partial F_h}{\partial x}(w, y)-\Phi (\tau )\Bigr )\,dw. \end{align*}

Rearranging and using the definition of $R_h(x, y)$ yields

\begin{equation*} \left |{R_h(x, y)-\Phi (\tau )}\right |\le \frac 12 \Phi (\tau ) \end{equation*}

by (31) for $x\in \overline{D_{\delta _1}(\rho _h)}\setminus \{\rho _h\}$ and $y\in \overline{D_{\delta _2}(\eta _{h,1})}$ and $h\ge h_0$ . In other words, $|{R_h(x,y)}|$ is bounded from below and above by positive constants for these $(x, y)$ and $h$ .

To prove analyticity of $R_h$ , we use Cauchy’s formula to rewrite it as

\begin{equation*} R_h(x, y) = \frac {1}{2\pi i}\oint _{|{\zeta -\rho _h}|=\delta _1} \frac {F_h(\zeta, y)-F_h(\rho _h, y)}{\zeta -\rho _h}\,\frac {d\zeta }{\zeta -x} \end{equation*}

for $x \neq \rho _h$ (note that the integrand has a removable singularity at $\zeta =\rho _h$ in this case). The integral is also defined for $x=\rho _h$ and clearly defines an analytic function on $D_{\delta _1}(\rho _h)\times D_{\delta _2}(\eta _{h,1})$ whose absolute value is bounded from above and below by a constant.

To see uniform convergence of $R_h$ , we use Cauchy’s formula once more and get

(35) \begin{equation} R_h(x, y) = \frac{1}{(2\pi i)^2}\oint _{|{\zeta -\rho _h}|=\delta _1}\oint _{|{\eta -\eta _{h,1}}|=\delta _2} \frac{F_h(\zeta, \eta )-F_h(\rho _h, \eta )}{\zeta -\rho _h}\,\frac{d\eta }{\eta -y}\,\frac{d\zeta }{\zeta -x} \end{equation}

for $x\in D_{\delta _1}(\rho _h)$ and $y\in D_{\delta _2}(\eta _{h,1})$ . Without loss of generality, $h_0$ is large enough such that $|{\rho _h-\rho }|\lt \delta _1/4$ and $|{\eta _{h,1}-\eta _1}|\lt \delta _2/4$ . By Cauchy’s theorem, we can change the contour of integration such that (35) implies

\begin{equation*} R_h(x, y) = \frac {1}{(2\pi i)^2}\oint _{|{\zeta -\rho }|=\delta _1/2}\oint _{|{\eta -\eta _{1}}|=\delta _2/2} \frac {F_h(\zeta, \eta )-F_h(\rho _h, \eta )}{\zeta -\rho _h}\,\frac {d\eta }{\eta -y}\,\frac {d\zeta }{\zeta -x} \end{equation*}

for $x\in D_{\delta _1/4}(\rho )$ and $y\in D_{\delta _2/4}(\eta _{1})$ , as the deformation is happening within the region of analyticity of the integrand. Using (26) and the fact that the denominator of the integrand is bounded away from zero shows that

\begin{equation*} R_h(x, y) = \frac {1}{(2\pi i)^2}\oint _{|{\zeta -\rho }|=\delta _1/2}\oint _{|{\eta -\eta _{1}}|=\delta _2/2} \frac {F_\infty (\zeta, \eta )-F_\infty (\rho _h, \eta )}{\zeta -\rho _h}\,\frac {d\eta }{\eta -y}\,\frac {d\zeta }{\zeta -x} + O(\lambda ^h) \end{equation*}

for $x\in D_{\delta _1/4}(\rho )$ and $y\in D_{\delta _2/4}(\eta _{1})$ . By Lemma 3.9, replacing the remaining occurrences of $\rho _h$ by $\rho$ induces another error term of $O(\lambda ^{h/2})$ , so that we get

\begin{equation*} R_h(x, y) = R(x, y) + O(\lambda ^{h/2}) \end{equation*}

with

\begin{equation*} R(x, y) \,{:\!=}\, \frac {1}{(2\pi i)^2}\oint _{|{\zeta -\rho }|=\delta _1/2}\oint _{|{\eta -\eta _{1}}|=\delta _2/2} \frac {F_\infty (\zeta, \eta )-F_\infty (\rho, \eta )}{\zeta -\rho }\,\frac {d\eta }{\eta -y}\,\frac {d\zeta }{\zeta -x} \end{equation*}

for $x\in D_{\delta _1/4}(\rho )$ and $y\in D_{\delta _2/4}(\eta _{1})$ . Of course, the $O$ constants do not depend on $x$ and $y$ ; therefore, we have uniform convergence. Analogously, we get

(36) \begin{align} S_h(y) &=\frac{1}{2\pi i} \oint _{|{\eta -\eta _{h,1}}|=\delta _2} \frac{F_h(\rho _h, \eta )-\eta }{(\eta -\eta _{h,1})^2}\,\frac{d\eta }{\eta -y} \end{align}

(37) \begin{align} &=S(y) + O(\lambda ^{h/2}) \end{align}

with

\begin{equation*} S(y)\,{:\!=}\, \frac {1}{2\pi i} \oint _{|{\eta -\eta _{1}}|=\delta _2/2} \frac {F_h(\rho, \eta )-\eta }{(\eta -\eta _{1})^2}\,\frac {d\eta }{\eta -y}, \end{equation*}

for $y\in D_{\delta _2/4}(\eta _{1})$ . Analogous results hold for partial derivatives.

We replace $\delta _1$ by $\delta _1/4$ and $\delta _2$ by $\delta _2/4$ to get the result as stated in the lemma.

Proof. We first choose $\delta _1$ and $\delta _2$ as in Lemma 3.12. Then $y=F_h(x,y)$ and Lemma 3.12 imply that

\begin{equation*} |{y-\eta _{h,1}}|=\sqrt {|{x-\rho _h}|\left |{\frac {R_h(x, y)}{S_h(y)}}\right |}. \end{equation*}

The fraction on the right-hand side is bounded by some absolute constant according to Lemma 3.12. So by decreasing $\delta _1$ if necessary, the right-hand side is at most $\delta _2/2$ .

Proof. By (30), the function $Y_{h,1}$ is determined by the implicit equation

(39) \begin{equation} 0 = F_h(x,Y_{h,1}(x)) - Y_{h,1}(x) = (x-\rho _h)R_h(x,Y_{h,1}(x)) + (Y_{h,1}(x)-\eta _{h,1})^2 S_h(Y_{h,1}(x)). \end{equation}

For $r\gt 0$ , set $C(r)\,{:\!=}\, \{x\in D_r(\rho _h)\colon |{\mathrm{Arg}(x-\rho _h)}|\ge \pi/4\}$ and $\widetilde{C}(r)\,{:\!=}\, \{x\in{\mathbb{C}}\colon |{x-\rho _h}|\,{=}\,r \text{ and } |{\mathrm{Arg}(x-\rho _h)}|\ge \pi/4\}$ . Choose $\delta _1$ , $\delta _2$ , $h_0$ as in Lemma 3.13. For some $h\ge h_0$ , let $r_h$ be the supremum of all $r\le \delta _1$ such that $Y_{h,1}$ can be continued analytically to $C(r)$ with values in $D_{\delta _2/2}(\eta _{h,1})$ . We claim that $r_h=\delta _1$ .

Suppose for contradiction that $r_h\lt \delta _1$ and let $x_{\infty }\in \widetilde C(r_h)$ . Choose a sequence of elements $x_n\in C(r_h)$ converging to $x_\infty$ for $n\to \infty$ and set $y_n\,{:\!=}\, Y_{h, 1}(x_n)$ for all $n$ . By assumption, we have $|{y_n-\eta _{h,1}}|\le \delta _2/2$ . By replacing the sequence $x_n$ by a subsequence if necessary, we may assume that the sequence $y_n$ is convergent to some limit $y_{\infty }$ . Note that $|{y_\infty -\eta _{h,1}}|\le \delta _2/2$ . By continuity of $F_h$ , we also have $y_\infty =F_h(x_\infty, y_\infty )$ . As $(x_\infty, y_\infty )\in \Xi _\varepsilon ^{(1, 2)}$ with $x_\infty \neq \rho _h$ , Lemma 3.11 and the analytic implicit function theorem imply that $Y_{h,1}$ can be continued analytically in a suitable open neighbourhood of $x_\infty$ . This neighbourhood can be chosen small enough such that the inequality $|{Y_{h,1}(x)-\eta _{h,1}}|\le \delta _2$ holds for all $x$ in this neighbourhood. However, Lemma 3.13 implies that we then actually have $|{Y_{h,1}(x)-\eta _{h,1}}|\le \delta _2/2$ for all such $x$ .

The set of these open neighbourhoods associated with all $x_\infty \in \widetilde{C}(r_h)$ covers the compact set $\widetilde{C}(r_h)$ , so a finite subset of these open neighbourhoods can be selected. Thus we find an analytic continuation of $Y_{h,1}$ to $C(\widetilde{r}_h)$ for some $\widetilde{r}_h\in (r_h, \delta _1)$ with values still in $D_{\delta _2/2}(\eta _{h,1})$ , which is a contradiction to the choice of $r_h$ .

Thus we have $r_h=\delta _1$ . In particular, choosing $h$ large enough that $|{\eta _{h, 1} - \eta _1}| \lt (\varepsilon - \delta _2)/2$ gives $|{Y_{h,1}(x)-\eta _1}| \le |{Y_{h,1}(x) - \eta _{h,1}}| + |{\eta _{h,1}-\eta _1}| \lt \delta _2/2 + (\varepsilon -\delta _2)/2 = \varepsilon/2$ for all $x\in C(\delta _1)$ .

Rearranging (39) yields

(40) \begin{equation} (\eta _{h, 1} - Y_{h,1}(x))^2 = \Bigl (\rho _h\frac{R_h(x, Y_{h,1}(x))}{S_h(Y_{h, 1}(x))}\Bigr ) \Bigl (1-\frac{x}{\rho _h}\Bigr ). \end{equation}

We know from Lemma 3.12 that $R_h$ is bounded above and $S_h$ is bounded below on $D_{\delta _1}(\rho _h)\times D_{\delta _2}(\eta _{h,1})$ and $D_{\delta _2}(\eta _{h,1})$ , respectively. Therefore, the absolute value of the first factor on the right-hand side of (40) is bounded above and below by positive constants for $x\in D_{\delta _1}(\rho _h)$ . For $x\lt \rho _h$ , we have that the factor $(1-x/\rho _h)$ is trivially positive and that $\eta _{h,1}\gt Y_{h,1}(x)$ because $Y_{h,1}$ is strictly increasing on $(0, \rho _h)$ , so the first factor on the right-hand side of (40) must be positive. Thus we may take the principal value of the square root to rewrite (40) as

(41) \begin{equation} \eta _{h, 1} - Y_{h,1}(x) = \sqrt{\rho _h\frac{R_h(x, Y_{h,1}(x))}{S_h(Y_{h, 1}(x))}} \Bigl (1-\frac{x}{\rho _h}\Bigr )^{1/2} \end{equation}

for $x\in C(\delta _1)$ . The above considerations also show that the radicand in (41) remains positive in the limit $x\to \rho _h^{-}$ (i.e., as $x$ approaches $\rho _h$ from the left) and then for $h\to \infty$ .

As we just observed that the first factor on the right-hand side of (41) is bounded, (41) implies

(42) \begin{equation} Y_{h,1}(x)-\eta _{h,1}=O\bigl ((x-\rho _h)^{1/2}\bigr ), \end{equation}

with an $O$ -constant that is independent of $h$ . We can now iterate this argument: using Taylor expansion along with the fact that partial derivatives of $R_h$ and $S_h$ are uniformly bounded above while $S_h$ is also uniformly bounded below, we obtain

\begin{align*} \frac{R_h(x, Y_{h,1}(x))}{S_h(Y_{h, 1}(x))} &= \frac{R_h(\rho _h,\eta _{h,1}) + O(x-\rho _h) + O(Y_{h,1}(x)-\eta _{h,1})}{S_h(\eta _{h,1}) + O(Y_{h,1}(x)-\eta _{h,1})} \\&= \frac{R_h(\rho _h,\eta _{h,1})}{S_h(\eta _{h,1})} + O\bigl ((x-\rho _h)^{1/2}\bigr ). \end{align*}

Plugging this into (41) yields

\begin{equation*} \eta _{h, 1} - Y_{h,1}(x) = \sqrt {\rho _h\frac {R_h(\rho _h, \eta _{h,1})}{S_h(\eta _{h,1})}} \Bigl (1-\frac {x}{\rho _h}\Bigr )^{1/2} + O( x - \rho _h), \end{equation*}

still with an $O$ -constant that is independent of $h$ . This can be continued arbitrarily often to obtain further terms of the expansion and an improved error term (for our purposes, it is enough to stop at $O((x - \rho _h)^2)$ ). Indeed it is well known (cf. [Reference Flajolet and Sedgewick10, Lemma VII.3]) that an implicit equation of the form (39) has a solution as a power series in $(1-x/\rho _h)^{1/2}$ . In particular, (38) follows with an error term that is uniform in $h$ . The coefficients $a_h,b_h,c_h$ can be expressed in terms of $R_h$ , $S_h$ and their partial derivatives evaluated at $(\rho _h,\eta _{h,1})$ : specifically,

\begin{align*} a_h &= - \sqrt{\rho _h\frac{R_h(\rho _h, \eta _{h,1})}{S_h(\eta _{h,1})}},\\ b_h &= \frac{\rho _h S_h(\eta _{h,1}) \frac{\partial R_h}{\partial y}(\rho _h,\eta _{h,1})-\rho _h S_h^{\prime}(\eta _{h,1}) R_h(\rho _h,\eta _{h,1})}{2 S_h(\eta _{h,1})^2},\\ c_h &=\frac{\rho _h^{3/2}N}{8\sqrt{R_h(\rho _h, \eta _{h,1})S_h(\eta _{h,1})^7}}, \end{align*}

where the numerator $N$ is a polynomial in $R_h(\rho _h,\eta _{h,1})$ , $S_h(\eta _{h,1})$ and their derivatives. By Lemma 3.12, $R_h$ and $S_h$ as well as their partial derivatives converge uniformly to $R$ and $S$ as well as their partial derivatives, respectively, with an error bound of $O(\lambda ^{h/2})$ . We also know that $\rho _h$ and $\eta _{h,1}$ converge exponentially to $\rho$ and $\eta _1$ , respectively, see Lemma 3.9. This means that first replacing all occurrences of $R_h$ and $S_h$ by $R$ and $S$ , respectively, and then replacing all occurrences of $\rho _h$ and $\eta _{h,1}$ by $\rho$ and $\eta _1$ , respectively, shows that $a_h=a+O(\lambda ^{h/2})$ , $b_h=b+O(\lambda ^{h/2})$ , and $c_h=c+O(\lambda ^{h/2})$ where $a$ , $b$ , and $c$ are the results of these replacements. Taking the limit for $h\to \infty$ in (30) shows that $R$ and $S$ and therefore $a$ , $b$ , and $c$ play the same role with respect to $F_\infty$ as $R_h$ , $S_h$ , $a_h$ , $b_h$ , and $c_h$ play with respect to $F_h$ , which implies that $a$ , $b$ , and $c$ are indeed the constants from (4).

Proof. Let us define $r_h = \sup \mathcal{R}_h$ , where

\begin{equation*} \mathcal {R}_h = \Big \{ r \,{:}\, Y_{h,1} \text { extends analytically to } D_r(0) \setminus D_{\delta _0}(\rho _h) \text { and satisfies } |Y_{h,1}(x)| \lt \eta _1 + \frac {\varepsilon }{2} \text { there} \Big \}, \end{equation*}

with $\delta _0$ as in the previous lemma. Note that trivially, $r_h \geq \rho$ . If $\liminf _{h \to \infty } r_h \gt \rho$ , we are done: in this case, there is some $\delta \gt 0$ such that $Y_{h,1}$ extends analytically to $D_{\rho (1+\delta )}(0) \setminus D_{\delta _0}(\rho _h)$ and satisfies $|Y_{h,1}(x)| \lt \eta _1 + \frac{\varepsilon }{2}$ there. As the previous lemma covers $D_{\delta _0}(\rho _h)$ , this already completes the proof.

So let us assume that $\liminf _{h \to \infty } r_h = \rho$ and derive a contradiction. The assumption implies that there is an increasing sequence of positive integers $h_j$ such that $\lim _{j\to \infty }r_{h_j}=\rho$ . Without loss of generality, we may assume that $r_{h_j} \leq \rho + \frac{\varepsilon }{2}$ for all $j$ . Pick (for each sufficiently large $j$ ) a point $x_{h_j}$ with $|x_{h_j}| = r_{h_j}$ and $|Y_{{h_j},1}(x_{h_j})| = \eta _1 + \frac{\varepsilon }{2}$ . If this were not possible, we could analytically continue $Y_{{h_j},1}$ at every point $x$ with $|{x}| = r_{h_j}$ and $x \notin D_{\delta _0}(\rho _{h_j})$ to a disk where $Y_{h_j,1}$ is still bounded by $\eta _1 + \frac{\varepsilon }{2}$ . This analytic continuation is possible, since by Lemma 3.11 the pair $(\rho _{h_j},\eta _{{h_j},1})$ is the only solution to the simultaneous equations $F_{h_j}(x,y) = y$ and $\frac{\partial }{\partial y} F_{h_j}(x,y) = 1$ with $(x,y) \in \Xi _{\varepsilon }^{(1,2)}$ , so the analytic implicit function theorem becomes applicable (compare e.g. the analytic continuation of $q_h$ in Lemma 3.8). By compactness, this would allow us to extend $Y_{{h_j},1}$ to $D_r(0) \setminus D_{\delta _0}(\rho _{h_j})$ for some $r \gt r_{h_j}$ while still maintaining the inequality $|Y_{{h_j},1}(x)| \lt \eta _1 + \frac{\varepsilon }{2}$ , contradicting the choice of $r_{h_j}$ .

Without loss of generality (choosing a subsequence if necessary), we can assume that $x_{h_j}$ and $Y_{h_j,1}(x_{h_j})$ have limits $x_{\infty }$ and $y_{\infty }$ , respectively. By construction, $|x_{\infty }| = \rho$ and $|y_{\infty }| = \eta _1 + \frac{\varepsilon }{2}$ .

Since $x_{h_j} \notin D_{\delta _0}(\rho )$ for all $j$ , $\mathrm{Arg}\, x_{h_j}$ is bounded away from $0$ . Thus we can find $\alpha \gt 0$ such that $|\mathrm{Arg}\, x_{h_j}| \geq 2\alpha$ for all $j$ . Define the region $A$ by

\begin{equation*} A = \Bigl \{z \in {\mathbb {C}}\,:\, |z| \lt \frac 12 \text { or } (|z| \lt 1 \text { and } |\mathrm {Arg}\, z| \lt \alpha ) \Bigr \}. \end{equation*}

Figure 5. Illustration of the domain $x_{h_j}A$ .

Note that $x_{h_j}A$ avoids the part of the real axis that includes $\rho _{h_j}$ (see Fig. 5), so the function $Y_{h_j,1}(x)$ is analytic in this region for all $j$ by construction since $(x, Y_{h_j,1}(x))\in \Xi _\varepsilon ^{(1,2)}$ whenever $x \in x_{h_j}A$ . So we have a sequence of functions $W_j(z) \,{:\!=}\, Y_{h_j,1}(x_{h_j}z)$ that are all analytic on $A$ and are uniformly bounded above by $\eta _1 + \frac{\varepsilon }{2}$ by our choice of $x_{h_j}$ . By Montel’s theorem, there is a subsequence of these functions (without loss of generality the sequence itself) that converges locally uniformly and thus to an analytic function $W_\infty$ on $A$ . This function needs to satisfy the following:

• • $W_{\infty }(0) = 0$ , since $W_j(0) = 0$ for all $j$ ,

• • $W_{\infty }(z) = F_{\infty }(x_{\infty }z, W_{\infty }(z)) = x_{\infty }z(\Phi (x_{\infty }z + W_{\infty }(z))-1)$ for $z\in A$ , since we have the uniform estimate

  \begin{equation*}W_j(z)=Y_{h_j,1}(x_{h_j}z)=F_{h_j}(x_{h_j}z, Y_{h_j,1}(x_{h_j}z))=F_{\infty }(x_{h_j}z, Y_{h_j,1}(x_{h_j}z)) + O(\lambda ^h).\end{equation*}
  This is also equivalent to
  \begin{equation*} x_{\infty }z + W_{\infty }(z) = x_{\infty }z\Phi (x_{\infty }z + W_{\infty }(z)). \end{equation*}

These two properties imply that $W_{\infty }(z) =Y(x_{\infty }z) - x_{\infty }z$ , since $Y$ is the unique function that is analytic at $0$ and satisfies the implicit equation $Y(x) = x \Phi (Y(x))$ . Implicit differentiation of $Y_{h_j,1}(x)=F_{h_j}(x, Y_{h_j, 1}(x))$ for $x \in x_{h_j}A$ yields

(43) \begin{equation} Y_{h_j,1}^{\prime}(x) = \frac{\frac{\partial F_{h_j}}{\partial x} (x, Y_{h_j, 1}(x))}{1 - \frac{\partial F_{h_j}}{\partial y} (x, Y_{h_j, 1}(x))} = \frac{\frac{\partial F_{\infty }}{\partial x} (x, Y_{h_j, 1}(x)) + O(\lambda ^{h_j})}{1 - \frac{\partial F_{\infty }}{\partial y} (x, Y_{h_j, 1}(x)) + O(\lambda ^{h_j})}. \end{equation}

Note that the numerator is uniformly bounded. Moreover, we recall again that the only solution to the simultaneous equations $F_{\infty }(x,y) = x(\Phi (y+x) - 1) = y$ and $\frac{\partial }{\partial y}F_{\infty }(x,y) = x \Phi '(y+x) = 1$ with $|{x}| \leq \rho + \varepsilon$ and $|x+y| \leq \tau + \varepsilon$ is $(x,y) = (\rho,\tau -\rho ) = (\rho,\eta _1)$ by our assumptions on $\Phi$ . By construction, there is a constant $\varepsilon _A \gt 0$ such that $|x - \rho | \geq \varepsilon _A$ whenever $x \in x_{h_j}A$ for some $j$ . The map $(x, y)\mapsto \| (F_{\infty }(x,y) - y, \frac{\partial }{\partial y}F_{\infty }(x,y) - 1)\|$ is continuous on the compact set

\begin{equation*}\mathcal {K}\,{:\!=}\, \{(x,y)\,{:}\, |{x}| \leq \rho + \varepsilon \text { and } |x+y| \leq \tau + \varepsilon \text { and } |x-\rho | \ge \varepsilon _A \}\end{equation*}

and has no zero there (using the Euclidean norm on ${\mathbb{C}}^2$ ). Therefore, it attains a minimum $\delta _A\gt 0$ on $\mathcal{K}$ .

Now for $x \in x_{h_j}A$ , $|{x}| \leq \rho + \varepsilon$ holds by assumption, as does $|x + Y_{h_j, 1}(x)| \leq \tau + \varepsilon$ . Moreover, $|x - \rho | \geq \varepsilon _A$ . Thus we can conclude that $(x, Y_{h_j,1}(x)) \in \mathcal{K}$ and therefore $\| (F_{\infty }(x,Y_{h_j, 1}(x)) - Y_{h_j, 1}(x), \frac{\partial F_{\infty }}{\partial y}(x,Y_{h_j, 1}(x)) - 1)\| \geq \delta _A$ for all such $x$ . Since

\begin{equation*} F_{\infty }(x,Y_{h_j, 1}(x)) - Y_{h_j, 1}(x) = F_{h_j}(x,Y_{h_j, 1}(x)) - Y_{h_j, 1}(x) + O(\lambda ^{h_j}) = O(\lambda ^{h_j}), \end{equation*}

this means that $\left |{1 - \frac{\partial F_{\infty }}{\partial y} (x, Y_{h_j, 1}(x))}\right | \geq \delta _A - O(\lambda ^{h_j})$ , so that the denominator in (43) is bounded below by a positive constant for sufficiently large $j$ .

So we can conclude that $Y_{h_j,1}^{\prime}(x)$ is uniformly bounded by a constant for $x \in x_{h_j}A$ , implying that $W_j'(z)$ is uniformly bounded (for all $z \in A$ and all sufficiently large $j$ ) by a constant that is independent of $j$ . Therefore, $W_j(z)$ is a uniformly equicontinuous sequence of functions on $\overline{A}$ , the closure of $A$ . By the Arzelà–Ascoli theorem, this implies that $W_j(z) \to W_{\infty }(z) = Y(x_{\infty }z) - x_{\infty }z$ holds even for all $z \in \overline{A}$ , not only on $A$ . In particular, $y_{\infty } = W_{\infty }(1) = Y(x_{\infty }) - x_{\infty }$ . Here, we have $|{x_{\infty }}| \leq \rho$ and $|{y_{\infty }}| = \eta _1 + \frac{\varepsilon }{2}$ by assumption. However,

\begin{equation*} |{Y(x) - x}| \leq |{Y(\rho ) - \rho }| =\eta _1 \end{equation*}

holds for all $|{x}| \leq \rho$ by the triangle inequality, so we finally reach a contradiction.

Proof. Since we already know that $\eta _{h,1}$ converges to $\tau - \rho$ and that $\rho _h$ converges to $\rho$ , $\eta _{h,0}$ converges to $\tau$ by (45). By the monotonicity property (44), all $\eta _{h,k}$ must therefore be bounded by a single constant $M$ for sufficiently large $h$ . Since $\eta _{h,1}$ converges to $\tau - \rho$ , we must have that $\rho _h\Phi '(\eta _{h, 1})$ converges to $\rho \Phi '(\tau -\rho )$ . Therefore, $\rho _h\Phi '(\eta _{h, 1}) \le \rho \Phi '(\tau - \rho/2)$ for sufficiently large $h$ . It follows that $\rho _h \Phi '(\eta _{h,1}) \leq \rho \Phi '(\tau - \rho/2) \lt \rho \Phi '(\tau ) = 1$ . For all $1 \leq j \leq h$ , we now have

\begin{equation*} \eta _{h,j} = \rho _h \Phi (\eta _{h,j-1}) - \rho _h \Phi (\eta _{h,h}) \leq \rho _h \Phi '(\eta _{h,j-1}) (\eta _{h,j-1} - \eta _{h,h}) \leq \rho _h \Phi '(\eta _{h,1}) \eta _{h,j-1}. \end{equation*}

Thus by induction

\begin{equation*} \eta _{h,k} \leq \eta _{h,1} \big ( \rho _h \Phi '(\eta _{h,1}) \big )^{k-1} \leq M \big ( \rho \Phi '(\tau - \rho/2) \big )^{k-1}. \end{equation*}

This proves the desired inequality for sufficiently large $h$ and $1 \leq k \leq h$ with $B_1 = \rho \Phi '(\tau - \rho/2)\,{\lt}\,1$ , and we are done.

Proof. For a suitable choice of $B_2$ , the estimate for $\rho _h$ has been established by Lemma 3.9, as has the estimate for $\eta _{h,k}$ in the cases where $k = 0$ and $k = 1$ . Set $\delta _{h, k} = \eta _{h, k} - \eta _k$ . Since $\eta _{h,h} \leq C B_1^h$ by Lemma 4.1, we have $\Phi (\eta _{h,h}) = \Phi (0) + O(B_1^h) = 1 + O(B_1^h)$ . Without loss of generality, suppose that $B_2 \geq B_1$ . Then, using (46), we obtain

\begin{align*} \eta _{h, k} & = \rho _h\Phi (\eta _{h, k-1}) - \rho _h\Phi (\eta _{h, h}) \\ & = (\rho + O(B_2^h))\Phi (\eta _{k-1}+ \delta _{h, k-1}) - (\rho + O(B_2^h))(1 + O(B_1^h))\\ & = \rho (\Phi (\eta _{k-1}) + \Phi '(\xi _{h, k-1})\delta _{h, k-1}) - \rho + O(B_2^h) \\ & = \eta _k + \rho \Phi '(\xi _{h,k-1}) \delta _{h,k-1} + O(B_2^h) \end{align*}

where $\xi _{h, k-1}$ is between $\eta _{k-1}$ and $\eta _{h, k-1}$ (by the mean value theorem) and the $O$ -constant is independent of $k$ . Let $M$ be this $O$ -constant. We already know (compare the proof of Lemma 4.1) that $\eta _{h,k-1} \leq \eta _{h,1} \leq \tau - \rho/2$ for every $k \geq 2$ if $h$ is sufficiently large. Likewise, it is easy to see that $\eta _k$ is decreasing in $k$ , hence $\eta _{k-1} \leq \eta _1 = \tau - \rho$ . Thus, $\xi _{h,k-1} \leq \tau - \rho/2$ and $\rho \Phi '(\xi _{h, k-1}) \leq \rho \Phi '(\tau - \rho/2) = B_1 \lt 1$ . So we have, for every $k \gt 1$ ,

\begin{equation*} |{\delta _{h,k}}| = |{\eta _{h,k} - \eta _k}| \leq B_1 |{\delta _{h,k-1}}| + MB_2^h. \end{equation*}

Iterating this inequality yields

\begin{equation*} |{\delta _{h,k}}| \leq B_1^{k-1} |{\delta _{h,1}}| + (1 + B_1 + \cdots + B_1^{k-2}) MB_2^h \leq |{\delta _{h,1}}| + \frac {MB_2^h}{1-B_1}, \end{equation*}

and the desired statement follows.

Proof. Recall that we defined the sequence $(\eta _k)_{k \geq 0}$ by $\eta _0 = \tau$ and $\eta _k = \rho \Phi (\eta _{k-1}) - \rho$ for $k\ge 1$ . Using Taylor expansion, we obtain

\begin{equation*} \eta _k = \rho \Phi '(0) \eta _{k-1} (1 + O(\eta _{k-1})) = \zeta \eta _{k-1} (1 + O(\eta _{k-1})). \end{equation*}

Since we already know that $\eta _{k-1} \leq C \cdot B_1^{k-1}$ , this implies that

\begin{equation*} \eta _k = \zeta \eta _{k-1} (1 + O(B_1^k)). \end{equation*}

Now it follows that the infinite product

\begin{equation*} \lambda _1 = \eta _0 \prod _{j \geq 1} \frac {\eta _j}{\zeta \eta _{j-1}} = \lim _{k \to \infty } \eta _0 \prod _{j = 1}^k \frac {\eta _j}{\zeta \eta _{j-1}} = \lim _{k \to \infty } \zeta ^{-k} \eta _k \end{equation*}

converges. The error bound follows from noting that

\begin{equation*} \zeta ^{-k} \eta _k = \lambda _1 \prod _{j \geq k+1} \frac {\zeta \eta _{j-1}}{\eta _j} = \lambda _1 \prod _{j \geq k+1} (1 + O(B_1^j)). \end{equation*}

Proof. Note that

\begin{equation*} \prod _{j = k}^{h} (\rho _h \Phi '(\eta _{h, j})) = \rho _h^{h-k+1} \prod _{j = k}^{h} \Phi '(\eta _{h,j}). \end{equation*}

In view of Lemma 4.2, we have $\rho _h^{h-k+1} = \rho ^{h-k+1} (1+O(B_2^h))^{h-k+1} = \rho ^{h-k+1}(1+ O(hB_2^h))$ , uniformly in $k$ . Moreover, Lemma 4.1 yields $\Phi '(\eta _{h,j}) = \Phi '(0) + O(\eta _{h,j}) = \Phi '(0) + O(B_1^j)$ , uniformly in $h$ . Thus

\begin{equation*} \prod _{j = k}^{h} \Phi '(\eta _{h,j}) = \Phi '(0)^{h-k+1} \prod _{j=k}^h (1 + O(B_1^j)) = \Phi '(0)^{h-k+1} (1 + O(B_1^k)). \end{equation*}

Hence the expression simplifies to

\begin{equation*} 1 + \sum _{k = 2}^{h}\prod _{j = k}^{h} (\rho _h \Phi '(\eta _{h, j})) = 1 + (1 + O(hB_2^h)) \sum _{k = 2}^h \zeta ^{h-k+1}(1 + O(B_1^k)). \end{equation*}

Since $\zeta \lt 1$ and $B_1 \lt 1$ , we can simply evaluate the geometric series, and the expression further simplifies to

\begin{equation*} 1 + \sum _{k = 2}^h \zeta ^{h-k+1} + O(B_3^h) = \frac {1-\zeta ^h}{1-\zeta } + O(B_3^h) = \frac {1}{1-\zeta } + O(B_3^h) \end{equation*}

for an appropriately chosen $B_3 \lt 1$ . This proves the first statement. For the second statement, we also use Lemma 4.2, along with the monotonicity of $\Phi '$ and the assumption that $\Phi '(0) \neq 0$ , which implies that $\Phi '(\eta _j)$ is bounded away from $0$ . This yields

\begin{align*} \prod _{j=1}^{h}\rho _h\Phi '(\eta _{h, j}) = \prod _{j=1}^{h}(\rho +O(B_2^h))(\Phi '(\eta _j)+O(B_2^h)) = \rho ^h (1 + O(hB_2^h))\prod _{j=1}^h\Phi '(\eta _j). \end{align*}

Since $\Phi '(\eta _j) = \Phi '(0) + O(\zeta ^j)$ (by Lemma 4.3), the product that defines $\lambda _2$ converges. So we can rewrite the product term as

\begin{equation*} \prod _{j=1}^h\Phi '(\eta _j) = \Phi '(0)^h \prod _{j=1}^h\frac {\Phi '(\eta _j)}{\Phi '(0)} = \lambda _2 \Phi '(0)^h \prod _{j\geq h+1}\frac {\Phi '(0)}{\Phi '(\eta _j)}, \end{equation*}

and thus, using again the estimate $\Phi '(\eta _j) = \Phi '(0) + O(\zeta ^j)$ on the remaining product,

\begin{align*} \prod _{j=1}^{h}\rho _h\Phi '(\eta _{h, j}) & = \lambda _2 \zeta ^h (1 + O(hB_2^h))(1+O(\zeta ^h)). \end{align*}

This proves the desired formula for a suitable choice of $B_3$ .

Proof. Taking the asymptotic formulas from the statement of Lemma 4.4 and applying them to (47) we obtain the formula after solving for $\rho _h\Phi '(\eta _{h, 0})$ .

Proof. Pick some $\alpha \in (0,1)$ in such a way that $\zeta ^{\alpha } \gt B_2$ , with $B_2$ as in Lemma 4.2, and set $m = \lfloor \alpha h \rfloor$ . From Lemma 4.3, we know that $\eta _m = \Theta ( \zeta ^{\alpha h})$ . By Lemma 4.2, $\eta _{h,m} = \eta _m + O(B_2^h)$ , so by our choice of $\alpha$ there is some $B_4 \lt 1$ such that $\eta _{h,m} = \eta _m (1 + O(B_4^h))$ for sufficiently large $h$ .

Next, recall from (46) that

\begin{equation*} \eta _{h,k} = \rho _h \big ( \Phi (\eta _{h,k-1}) - \Phi (\eta _{h,h}) \big ). \end{equation*}

By the mean value theorem, there is some $\xi _{h,k} \in (\eta _{h,h},\eta _{h,k-1})$ such that

\begin{equation*} \eta _{h,k} = \rho _h (\eta _{h,k-1} - \eta _{h,h}) \Phi '(\xi _{h,k}) = \rho _h (\eta _{h,k-1} - \eta _{h,h}) (\Phi '(0) + O(\eta _{h,k-1})). \end{equation*}

Assume now that $k \geq m$ , so that $\eta _{h,k-1} = O(B_1^{\alpha h})$ by Lemma 4.1. Moreover, $\rho _h = \rho + O(B_2^h)$ by Lemma 4.2. So with $B = \max\! (B_2,B_1^{\alpha })$ , it follows that

\begin{equation*} \eta _{h,k} = \zeta (\eta _{h,k-1} - \eta _{h,h}) (1 + O(B^h)), \end{equation*}

uniformly for all $k \geq m$ . Rewrite this as

\begin{equation*} \eta _{h,k-1} = \eta _{h,h} + \frac {\eta _{h,k}}{\zeta } (1 + O(B^h)). \end{equation*}

Iterate this $h-m$ times to obtain

\begin{align*} \eta _{h,m} &= \sum _{j=0}^{h-m} \frac{\eta _{h,h}}{\zeta ^j} (1 + O(B^h))^j \\ &= \eta _{h,h} \zeta ^{-(h-m)} \frac{1 - \zeta ^{h-m+1}}{1-\zeta } (1 + O(h B^h)). \end{align*}

Now recall that $\eta _{h,m} = \eta _m (1 + O(B_4^h))$ , and that $\eta _m = \lambda _1 \zeta ^m (1 + O(B_1^{\alpha h}))$ by Lemma 4.3. Plugging all this in and solving for $\eta _{h,h}$ , we obtain (49), provided that $B_4$ was also chosen to be greater than $B$ and $\zeta ^{1-\alpha }$ .

Proof. From (45) and (46) with $k = 1$ , we have

(52) \begin{equation} \eta _{h,0} = \rho _h \big ( \Phi (\eta _{h,0}) - \Phi (\eta _{h, h}) + 1 \big ). \end{equation}

By means of Taylor expansion and Lemma 4.6, we get

\begin{equation*} \eta _{h, 0} = \rho _h \big ( \Phi (\eta _{h, 0}) - \Phi '(0)\eta _{h, h} + O(\eta _{h, h}^2) \big ). \end{equation*}

We multiply this by (48) and divide through by $\rho _h$ to obtain

(53) \begin{equation} \eta _{h, 0} \Phi '(\eta _{h, 0}) = \big ( \Phi (\eta _{h, 0}) - \Phi '(0)\eta _{h, h} + O(\eta _{h, h}^2) \big ) \big (1 + \lambda _2 (1 - \zeta ) \zeta ^{h} (1+ O(B_3^{h})) \big ) \end{equation}

or, with $H(x) = x\Phi '(x) - \Phi (x)$ ,

\begin{equation*} H(\eta _{h,0}) = \big ({-} \Phi '(0)\eta _{h, h} + O(\eta _{h, h}^2) \big ) (1 + O(\zeta ^{h})) + \Phi (\eta _{h,0}) \lambda _2 (1 - \zeta ) \zeta ^h (1+ O(B_3^{h})). \end{equation*}

We plug in the asymptotic formula for $\eta _{h,h}$ from Lemma 4.6 and also note that $\Phi (\eta _{h,0}) = \Phi (\tau + O(B_2^h)) = \Phi (\tau ) + O(B_2^h)$ by Lemma 4.2. This gives us

(54) \begin{equation} H(\eta _{h,0}) = (\Phi (\tau ) \lambda _2 - \Phi '(0) \lambda _1 )(1 - \zeta ) \zeta ^h + O((\zeta B_5)^h), \end{equation}

where $B_5 = \max\! (\zeta,B_2,B_3,B_4)$ . Now note that the function $H$ is increasing (on the positive real numbers within the radius of convergence of $\Phi$ ) with derivative $H'(x) = x \Phi ''(x)$ and a unique zero at $\tau$ . So by inverting (54), we finally end up with

\begin{equation*} \eta _{h,0} = \tau + \frac {1}{H'(\tau )} (\Phi (\tau ) \lambda _2 - \Phi '(0) \lambda _1 )(1 - \zeta ) \zeta ^h + O((\zeta B_5)^h), \end{equation*}

completing the proof of the first formula. Now we return to (48), which gives us

\begin{equation*} \rho _h = \frac {1 + \lambda _2 (1 - \zeta )\zeta ^h (1+ O(B_3^{h}))}{\Phi '(\eta _{h, 0})} \\ = \frac {1 + \lambda _2 (1 - \zeta )\zeta ^h (1+ O(B_3^{h}))}{\Phi '(\tau ) + \Phi ''(\tau )(\eta _{h,0}-\tau ) + O((\eta _{h,0}-\tau )^2)}. \end{equation*}

Plugging in (50) and simplifying by means of the identities $\rho \Phi (\tau ) = \tau$ and $\rho \Phi '(\tau ) = 1$ now yields (51).