Analysis of a tripartite entanglement distribution switch

Abstract

We study a quantum switch that distributes tripartite entangled states to sets of users. The entanglement switching process requires two steps: First, each user attempts to generate bipartite entanglement between itself and the switch, and second, the switch performs local operations and a measurement to create multipartite entanglement for a set of three users. In this work, we study a simple variant of this system, wherein the switch has infinite memory and the links that connect the users to the switch are identical. This problem formulation is of interest to several distributed quantum applications, while the technical aspects of this work result in new contributions within queueing theory. The state of the system is modeled as continuous-time Markov chain (CTMC), and performance metrics of interest (probability of an empty system, switch capacity, expectation, and variance of the number of qubit-pairs stored) are computed via the solution of a two-dimensional functional equation obtained by reducing it to a boundary value problem on a closed curve. This work is a follow-up of Nain et al. (Proc ACM Measure Anal Comput Syst(POMACS) 4, 2020) where a switch distributing entangled multipartite states to sets of users was studied, but only the switch capacity and the expected number of stored qubits were derived.

Appendices

A Zero of K(x, y) for $|x|=1$

Lemma 2

For $|x|=1$, the equation $K(x,y)=0$ has a unique solution $y=y(x)$ such that $|y(x)|\le 1$. Moreover, $y(1):=1$, $y^{\prime }(1)=-1$ and $y^{\prime \prime }(1)=\frac{2(k-2)}{k-3}$.

Proof

Recall that $k>3$.

Assume first that $x=1$. Since $K(1,y)=-(y-1)(y-(k-2))$, we see that K(1, y) has a unique zero in the unit disk $\{|y|\le 1\}$, located at $y=1$. Assume now that x is fixed with $|x|=1$ and $x\not =1$. Define $h(x,y)=x(k-x)y$. For $|y|=1$, we have

$$\begin{aligned} |K(x,y)-h(x,y)|=|xy^2 +k-2|\le k-1< |k-x|=|x(k-x)y|=|h(x,y)|, \end{aligned}$$

where the second inequality holds since²$|x|=1$ with $x\not = 1$. By Rouché’s theorem, we conclude that K(x, y) and h(x, y) have the same number of zeros in the unit disk $\{|y|<1\}$. Since h(x, y) has the only zero $y=0$ in $\{|y|<1\}$, K(x, y) has a unique zero in $\{|y|<1\}$. This proves the first part of the lemma.

The fact that y(x) has first and second derivatives at $x=1$ follows³ from the implicit function theorem [27] that applies here since K(x, y) is analytic in $\mathbb {C}\times \mathbb {C}$, $K(1,1)=1$, and $\frac{\partial }{\partial y}K(x,y)|_{(x,y)=(1,1)}=k-3\not =0$.

Differentiating K(x, y(x)) with respect to x and letting $x=1$ gives $y^{(1)}(1)=-1$ by using $y(1)=1$. Differentiating twice K(x, y(x)) with respect to x and letting $x=1$ gives $y^{\prime \prime }(1)=\frac{2(k-2)}{k-3}$ upon using $y(1)=1$ and $y^{\prime }(1)=-1$. This concludes the proof. $\square$

B Proof of Lemma 1

We may rewrite the set $\mathcal {L}$ in (34) as

$$\begin{aligned} \mathcal {L}:=\left\{ 0\le \rho \le 1, -\pi \le \theta \le \pi : 2\rho ^3 \cos \theta -k\rho ^2+k-2=0\right\} . \end{aligned}$$

Define $f(\theta ,\rho ):=2\rho ^3 \cos \theta -k\rho ^2+k-2$. Let us show that for any $\theta \in \mathbb {R}$, $f(\theta ,\rho )$ has a unique zero in (0, 1], denoted by $\rho =\rho (\theta )$.

Assume first that $\theta =2l\pi$ for $l\in \mathbb {Z}$. From

$$\begin{aligned} f(2l\pi , \rho )=(x-1)\left( x- \frac{k-2-\sqrt{k^2+4k-12}}{2}\right) \left( x- \frac{k-2+\sqrt{k^2+4k-12}}{2}\right) , \end{aligned}$$

we deduce that $f(2l\pi , \rho )$ has a single zero in (0, 1] given by $\rho (2l\pi )=1$.

Assume now that $\theta \in \mathbb {R}-\{2l\pi , l\in \mathbb {Z}\}$. From

$$\begin{aligned} f(\theta ,-1)=-2(\cos \theta +1)\le 0,\quad f(\theta ,0)=k-2>0, \quad f(\theta ,1)=2 (\cos \theta -1)<0 \end{aligned}$$

and

$$\begin{aligned}&\lim _{\rho \rightarrow -\infty } f(\theta ,\rho )=\left\{ \begin{array}{ll} -\infty , &{}\text{ if } \theta \in (2l\pi ,(2l+1)\pi ),\\ +\infty , &{}\text{ if } \theta \in ((2l+1)\pi ,2(l+1)\pi ), \end{array} \right. \\&\quad \lim _{\rho \rightarrow \infty } f(\theta ,\rho )=\left\{ \begin{array}{ll} +\infty , &{}\text{ if } \theta \in (2l\pi ,(2l+1)\pi ),\\ -\infty , &{}\text{ if } \theta \in (2l+1,2(l+1)), \end{array} \right. \end{aligned}$$

for $l\in \mathbb {Z}$, we readily deduce that the polynomial $f(\theta ,\rho )$ of degree 3 in the variable $\rho$ has a single zero $\rho =\rho (\theta )$ in (0, 1). In summary, for any $\theta \in \mathbb {R}$, $f(\theta ,\rho )$ has a unique zero $\rho =\rho (\theta )$ in (0, 1].

Since the partial derivatives $\frac{d^m}{d\theta ^m}f(\theta ,\rho )$ and $\frac{d^m}{d\rho ^m}f(\theta ,\rho )$ exist for all $m=0,1,\ldots$ and $\frac{d}{d\rho }f(\theta ,\rho )= 2\rho (3\rho \cos \theta -k)$ does not vanish for any zero $(\theta ,\rho )$ of $f(\theta ,\rho )$ in $\mathbb {R}\times (0,1]$ as $k>3$, we conclude from the implicit function theorem [27] that $\rho (\theta )$ is infinitely differentiable in $\mathbb {R}$.

Because $\rho (\theta )=\rho (\theta +2\pi )$ for any $\theta \in \mathbb {R}$, the function $\rho (\theta )$ is periodic with period $2\pi$ and is infinitely differentiable in (in particular) $[-\pi , \pi ]$. The continuity of $\rho (\theta )$ in $[-\pi ,\pi ]$ together with $\rho (-\pi )=\rho (\pi )$ implies that the contour $\mathcal {L}$ defined by (35) is closed and the differentiability of $\rho (\theta )$ in $[-\pi ,\pi ]$ implies that it is smooth. The contour $\mathcal {L}$ is symmetric with respect to the real axis since $f(\theta ,\rho )=f(-\theta ,\rho )$ for all $\theta \in R$, $\rho \ge 0$.

Differentiating the equation $f(\theta ,\rho (\theta ))=0$ once gives $\rho ^\prime (0)=0$ by using $\rho (0)=1$, differentiating it twice gives $\rho ^{\prime \prime }(0)= \frac{1}{3-k}$ by using $\rho (0)=1$ and $\rho ^\prime (0)=0$, and differentiating it three times gives $\rho ^{\prime \prime \prime }(0)=0$ by using $\rho (0)=1$, $\rho ^\prime (0)=0$ and $\rho ^{\prime \prime }(0)= \frac{1}{3-k}$.

Last, if $\mathcal {L}$ intersects with itself, this means that there exist $\theta _1,\theta _2\in (0,\pi )$, $\theta _1\not =\theta _2$, such that $\rho (\theta _1)=\rho (\theta _2)$. (Because $\mathcal {L}$ is symmetric with respect to the real axis, we do not need to consider the case where $\theta _1,\theta _2\in (-\pi ,0)$.) Assume that $\rho (\theta _1)=\rho (\theta _2)=\rho _0$ for some $\theta _1,\theta _2\in (0,\pi )$. Note that $\rho _0\not =0$ as $\rho =0$ is not a zero of $f_\theta (\rho )$. The identities $f(\theta _i,\rho _0)=0$ for $i=1,2$ imply that $2\rho _0^3 (\cos (\theta _1)-\cos (\theta _2))=0$, which in turn implies that $\cos (\theta _1)=\cos (\theta _2)$ and that $\theta _1=\theta _2$ since $\theta _1,\theta _2\in (0,\pi )$. Therefore, the contour $\mathcal {L}$ does not intersect with itself. $\square$

C Proof that $I_2=1$

We have, by (55), (38), and (51),

$$\begin{aligned} \int _{\mathcal C} \frac{\log (u^{-1}G(u)}{u}du= & {} i\int _{-\pi }^\pi \log \left( e^{-i\theta }\frac{\rho (\phi (\theta ))^3 e^{i\phi (\theta )}}{1-\rho (\phi (\theta ))^3 e^{-i\phi (\theta )}}\right) d\theta \nonumber \\= & {} i\int _{-\pi }^0 \log \left( e^{-i\theta }\frac{\rho (\phi (\theta ))^3 e^{i\phi (\theta )}-1}{1-\rho (\phi (\theta ))^3 e^{-i\phi (\theta )}}\right) d\theta \nonumber \\&+ i\int _{-\pi }^0 \log \left( e^{i\theta }\frac{\rho (\phi (-\theta ))^3 e^{i\phi (-\theta )}-1}{1-\rho (\phi (-\theta ))^3 e^{-i\phi (-\theta )}}\right) d\theta \nonumber \\= & {} i\int _{-\pi }^0 \log \left( e^{-i\theta }\frac{\rho (\phi (\theta ))^3 e^{i\phi (\theta )}-1}{1-\rho (\phi (\theta ))^3 e^{-i\phi (\theta )}}\right) d\theta \nonumber \\&+ i\int _{-\pi }^0 \log \left( e^{i\theta }\frac{\rho (\phi (\theta ))^3 e^{-i\phi (\theta )}-1}{1-\rho (\phi (\theta ))^3 e^{i\phi (\theta )}}\right) d\theta \nonumber \\= & {} i\int _{-\pi }^0 \log \left( e^{-i\theta }\frac{\rho (\phi (\theta ))^3 e^{i\phi (\theta )}-1}{1-\rho (\phi (\theta ))^3 e^{-i\phi (\theta )}}\right) d\theta \nonumber \\&- i\int _{-\pi }^0 \log \left( e^{-i\theta }\frac{\rho (\phi (\theta ))^3 e^{i\phi (\theta )}-1}{1-\rho (\phi (\theta ))^3 e^{-i\phi (\theta )}}\right) d\theta \nonumber \\= & {} 0, \end{aligned}$$

(101)

where we have used (41) and the property that $\rho (\theta )=\rho (-\theta )$ for all $\theta$ (cf. Lemma 1) to establish (101). This concludes the proof from the definition of $I_2$ in (55).

D Properties of the mappings G and g

The following results will be used in the proofs of Lemmas 3 and 4. Since $\gamma _0$ is a one-to-one mapping from $\mathcal {L}$ onto $\mathcal {C}$, for any $t=e^{i\theta }$ there exists a unique $w\in \mathcal {L}$ such that $\gamma _0(w)=t$. Hence, cf. (51)–(52),

$$\begin{aligned} G(\gamma _0(w))= & {} -\frac{1-\rho (\theta )^3 e^{i \theta }}{1-\rho (\theta )^3 e^{-i\theta }}, \end{aligned}$$

(102)

$$\begin{aligned} g(\gamma _0(w))= & {} \frac{ 1-\rho (\theta )^3 \cos \theta }{1-\rho (\theta )^3 e^{-i\theta }}, \end{aligned}$$

(103)

for all $w\in \mathcal {L}$ by using $\gamma (\gamma _0(w))=w$ (in particular) for all $w\in \mathcal {L}$.

Introduce the mappings

$$\begin{aligned} \tilde{G}(\theta ):= & {} \frac{\rho (\theta )^3 e^{i\theta }-1}{1-\rho (\theta )^3 e^{-i\theta }}, \end{aligned}$$

(104)

$$\begin{aligned} \tilde{g}(\theta ):= & {} \frac{1-\rho (\theta )^3\cos \theta }{1-\rho (\theta )^3 e^{-i\theta }}. \end{aligned}$$

(105)

Observe from (51), (52) and (38) that, for $t=e^{i\theta }$,

$$\begin{aligned} G(t)=\tilde{G}(\phi (\theta )), \quad g(t)=\tilde{g}(\phi (\theta )). \end{aligned}$$

(106)

Lemma 3

(Properties of G and g) The function G(t) does not vanish on $\mathcal {C}$, and G(t), g(t), $\log (t^{-1}G(t))$ and $g(t)e^{-H(t)}$ are continuous on $\mathcal {C}$.

Also, $G(1)=1$ and $g(1)=0$.

Proof

Let $t_0=e^{i\theta _0}\in \mathcal {C}$. Assume first that $\theta _0\in [-\pi ,\pi ]-\{0\}$. Since $\phi (\theta )$ is continuous in $[-\pi ,\pi ]$, the functions $\tilde{G}(\phi (\theta ))$ and $\tilde{g}(\phi (\theta ))$ in (104)–(105) are continuous at $\theta _0$ if their common denominator $1-\rho (\phi (\theta ))^3 e^{-i\phi (\theta )}$ does not vanish at this point. This is true as $\rho (\theta )<1$ for all $\theta \in [-\pi ,\pi ]-\{0\}$ and that $\phi (\theta )=0$ iff $\theta =0$. The same argument shows that the numerator of $\tilde{G}(\phi (\theta ))$ does not vanish at $\theta _0$. Assume now that $\theta _0=0$. Applications of L’Hôpital’s rule show that $\tilde{G}(0)=1$ (or equivalently $G(1)=1$) and $\tilde{g}(0)=1$ (or equivalently $g(1)=0$), thereby proving all statements of the lemma (but the continuity of $g(t)e^{-H(t)}$) thanks to (106).

It remains to show that $g(t)e^{-H(t)}$ with H(t) defined in (60) is continuous on $\mathcal {C}$. We have just shown that the first term $\log (t^{-1}G(t))$ in (60) is continuous on $\mathcal {C}$. Rewriting the integral in (60) as $\int _{-\pi }^\pi \frac{\log (e^{-i\theta }G(e^{i\theta }))-\log (t^{-1}G(t))}{e^{i\theta }-t} \, e^{-i\theta } d\theta$, we observe that it is a continuous function of $t\in \mathcal {C}$ since, for each $\theta \in [-\pi ,\pi ]$, the mapping $t\rightarrow \frac{\log (e^{-i\theta }G(e^{i\theta }))-\log (t^{-1}G(t))}{e^{i\theta }-t} e^{-i\theta }$ is continuous on $\mathcal {C}$ and the range of integration (i.e., $[-\pi ,\pi ]$) is finite. Therefore, H(t) is continuous on $\mathcal {C}$ as the sum of two continuous functions on $\mathcal {C}$, and so is $g(t)e^{-H(t)}$ as the composition of continuous functions on $\mathcal {C}$. This ends the proof. $\square$

E Calculation of the index $\chi$

Lemma 4

(Index) The index $\chi$ [defined in (53)] is equal to one.

Proof

Thanks to (102), the index $\chi$ in (53), is also given by

$$\begin{aligned} \chi =\frac{1}{2\pi }\arg [G(\gamma _0(w))]_{\mathcal {L}}. \end{aligned}$$

(107)

Take $w=\rho (\theta ) e^{i\theta } \in \mathcal {L}$. When $\theta =\pi$ (resp. $\theta =-\pi$), $w=-\rho (\pi )$ ($w=\rho (-\pi )=\rho (\pi )$) and $G(\gamma _0(w))= -\frac{1+\rho (\pi )^3}{1+\rho (\pi )^3 }=-1$. When $\theta =0$, $w=\rho (0)=1$ and

$$\begin{aligned} G(\gamma _0(1))=-\lim _{\theta \rightarrow 0} \frac{-3 \frac{d}{d\theta }\rho (\theta ) \rho (\theta )^2 e^{i\theta } - i \rho (\theta )^3 e^{i\theta }}{ -3 \frac{d}{d\theta }\rho (\theta ) \rho (\theta )^2 e^{-i\theta } +i \rho (\theta )^3 e^{-i\theta } }=1, \end{aligned}$$

(108)

since $\frac{d}{d\theta }\rho (\theta )|_{\theta =0}=0$ and $\rho (\theta )=1$ by (36), where the second equality holds by L’Hôpital’s rule.

The function $G(\gamma _0(w))= -\frac{1-\rho (\theta )^3 e^{i \theta }}{1-\rho (\theta )^3 e^{-i\theta }}$ in (102) is continuous in $[-\pi ,\pi ]$ since $\rho (\theta )$ is continuous in $[-\pi ,\pi ]$ (cf. Lemma 1), since its denominator does not vanish for $\theta \not =0$ as $\rho (\theta )<1$ for $\theta \not =0$ and since $\lim _{\theta \rightarrow 0} -\frac{1-\rho (\theta )^3 e^{i \theta }}{1-\rho (\theta )^3 e^{-i\theta }}=1$ as shown above.

Let us now show that $G(\gamma _0(w))$ crosses the real axis only at $\theta =0$ and at $\theta =\pi$ (resp. $\theta =-\pi$) when w describes once the contour $\mathcal {L}$ or, equivalently, that $\mathfrak {I}(G(\gamma _0(w)))\not =0$ when $\theta \in (-\pi ,\pi )$ with $\theta \not =0$. By (102),

$$\begin{aligned} \mathfrak {I}(G(\gamma _0(w)))=\frac{2(1-\rho (\theta )^3\cos \theta )\rho (\theta )^3 \sin \theta }{1- 2 \rho (\theta )^3\cos \theta + \rho (\theta )^6}. \end{aligned}$$

(109)

Fix $\theta _0\in (-\pi ,\pi )$ with $\theta _0\not =0$. The numerator of $\mathfrak {I}(G(\gamma _0(w)))$ vanishes if and only if $1=\rho (\theta _0)^3\cos (\theta _0)$. Since $\rho (\theta )<1$ for $\theta \not =0$, we see that $1>\rho (\theta _0)^3\cos (\theta _0)$. On the other hand, the denominator of $\mathfrak {I}(G(\gamma _0(w)))$ does not vanish since $1- 2 \rho (\theta _0)^3\cos (\theta _0) + \rho (\theta _0)^6> 1-2\rho (\theta _0)^3 + \rho (\theta _0)^6=(1-\rho (\theta _0)^3)^2 >0$ by using $\cos (\theta _0)<1$ and $\rho (\theta _0)<1$. Therefore, $\mathfrak {I}(G(\gamma _0(w)))>0$ for $\theta _0\in (0,\pi )$ and $\mathfrak {I}(G(\gamma _0(w)))<0$ for $\theta _0\in (-\pi , 0)$.

In summary, we have shown that as w describes once the contour $\mathcal {L}$ the mapping $G(\gamma _0(w))$ describes once a circuit around zero in the counter-clockwise direction, thereby proving that $\chi =1$ by (107). $\square$

F Hölder condition for first and second derivatives of $\alpha$ and $\beta$

Lemma 5

The mapping $\phi$ defined in (39) is twice differentiable in $[-\pi ,\pi ]$ and its second derivative is continuous in $[-\pi ,\pi ]$.

Proof

We recall that the stability condition $k\ge 4$ is enforced. By Lemma 3 in [7, p. 876], the conformal mapping $\gamma (z)$ from $\mathcal {C}$ onto $\mathcal {L}$ (see Section 5.1) is differentiable on $\mathcal {C}$. By (98) and the differentiability of $\gamma$ on $\mathcal {C}$, $\phi ^\prime (\theta )$ exists if (a) $\rho ^\prime (\phi (\theta ))+i\rho (\phi (\theta ))\not =0$ for $\theta \in [-\pi ,\pi ]$. The latter is always true since $\rho ^\prime (\theta )$ is a real number (cf. (96)) and $\rho (\theta )\not =0$ for all $\theta \in [-\pi ,\pi ]$. This proves the differentiability of $\phi$ on $[-\pi ,\pi ]$.

The proof that $\phi$ is twice differentiable is more tedious. Differentiating the equation $\gamma (\gamma _0(w))=w$ for $w\in \mathcal {L}$ (see Section 5.1) gives $\gamma _0^\prime (w)=\frac{1}{\gamma ^{\prime }(\gamma _0(w))}$. But Lemma 3 in [7, p. 876] also tells us that $\gamma ^\prime (z)\not =0$ on $\mathcal {C}-\{1\}$ for all $k\ge 4$, which allows us to conclude that

$$\begin{aligned} \gamma _0^\prime (w)\not =0, \quad \forall w\in \mathcal {L}-\{1\}. \end{aligned}$$

(110)

By (110), the denominator in (99) does not vanish for $\theta \in [-\pi ,\pi ]-\{0\}$ and it does not vanish either for $\theta =0$ since $\gamma _0^\prime (\rho (\phi (0))e^{i\phi (0)})=\gamma _0^\prime (1)\not =0$ by (136). (Here, we are using that $\rho ^\prime (\phi (\theta ))+i\rho (\phi (\theta ))\not =0$ for $\theta \in [-\pi ,\pi ]$ as already observed.) On the other hand, a glance at the r.h.s. of (99) indicates that the term within square brackets is well defined on $[-\pi ,\pi ]$ if (c) $\gamma _0^{\prime \prime }(w)$ is well defined on $\mathcal {L}$ which, by (131) with $i=2$, (133) and (135), amounts to showing that (c) $K-w\delta ^2\not =0$ for $w\in \mathcal {L}$ in the denominator of (135) (Hint: $w\not \in \mathcal {L}$ and $K-w$ does not vanish on $\mathcal {L}$ since $K=k-2\ge 2$ and $|w|\le 1$—see Figure 3). It can be checked that property (c) holds (a more direct argument is to notice that (c) holds as otherwise $\gamma _0$ in (44) would not be well defined on $\mathcal {L}$). This proves that $\phi$ is twice differentiable on $\theta \in [-\pi ,\pi ]$. Last, we observe from (99) that $\phi ^{\prime \prime }(\theta )$ is continuous on $[-\pi ,\pi ]$ as it is only composed of continuous mappings, which completes the proof. $\square$

Lemma 6

The mapping $\alpha$ defined in (82) is twice differentiable on $\mathcal {C}$, and its second derivative is continuous on $\mathcal {C}$. The same result holds for the mapping $\beta$ defined in (83).

Proof

Define

$$\begin{aligned} \tilde{G}(\theta )= & {} -\frac{1-\rho (\theta )^3e^{i\theta }}{1-\rho (\theta )^3 e^{-i\theta }}, \end{aligned}$$

(111)

$$\begin{aligned} \tilde{g}(\theta )= & {} \frac{1-\rho (\theta )^3\cos \theta }{1-\rho (\theta )^3 e^{-i\theta }}. \end{aligned}$$

(112)

Since $\rho (\theta )\in [0,1)$ for $\theta \in (0,1)$ and $\rho (0)=1$ (see Sect. 5.1), the denominator in the r.h.s. of (111) and (112) has a single zero of multiplicity one in $[-\pi ,\pi ]$ at $\theta =0$. Since both numerators in the r.h.s. of (111) and (112) vanish at $\theta =0$, this shows that the mappings $\tilde{G}(\theta )$ and $\tilde{g}(\theta )$ are continuous in $[-\pi ,\pi ]$.

Notice that (cf. (51), (52), (38))

$$\begin{aligned} G(e^{i\theta })=\tilde{G}(\phi (\theta )) \quad \hbox {and}\quad g(e^{i\theta })=\tilde{g}(\phi (\theta )). \end{aligned}$$

We have, for $u=e^{i\theta }$ (cf. (82))

$$\begin{aligned} \alpha ^\prime (u)= & {} \frac{d}{de^{i\theta }} \log (e^{-i\theta }G(e^{i\theta }))= -i e^{-i\theta } \frac{d}{d\theta }\log (e^{-i\theta }\tilde{G}(\phi (\theta )) \nonumber \\= & {} -i e^{-i\theta }\left( -i +\frac{\phi ^\prime (\theta )\tilde{G}^\prime (\phi (\theta ))}{\tilde{G}(\phi (\theta ))}\right) , \end{aligned}$$

(113)

and

$$\begin{aligned} \alpha ^{\prime \prime }(u)= & {} \frac{d}{du} \alpha ^\prime (u)=\frac{1}{ie^{i\theta }}\frac{d}{d\theta } \alpha ^\prime (u)\nonumber \\= & {} -ie^{-i\theta }\frac{d}{d\theta }\left\{ -i e^{-i\theta }\left( -i +\frac{\phi ^\prime (\theta )\tilde{G}^\prime (\phi (\theta ))}{\tilde{G}(\phi (\theta ))}\right) \right\} , \quad \hbox {by using} (113),\nonumber \\= & {} -ie^{-i\theta } \left( -e^{-i\theta }\left( -i +\frac{\phi ^\prime (\theta )\tilde{G}^\prime (\phi (\theta ))}{\tilde{G}(\phi (\theta ))}\right) -i e^{-i\theta } \frac{d}{d\theta }\frac{\phi ^\prime (\theta )\tilde{G}^\prime (\phi (\theta ))}{\tilde{G}(\phi (\theta ))}\right) \nonumber \\= & {} - e^{-i\theta } \alpha ^\prime (u)- e^{-2i\theta } \frac{d}{d\theta }\frac{\phi ^\prime (\theta )\tilde{G}^\prime (\phi (\theta ))}{\tilde{G}(\phi (\theta ))},\quad \hbox {by using again }(113) \nonumber \\= & {} - e^{-i\theta } \alpha ^\prime (u)- e^{-2i\theta } \left( \phi ^{\prime \prime }(\theta ) \frac{\tilde{G}^\prime (\phi (\theta ))}{\tilde{G}(\phi (\theta ))} +\phi ^\prime (\theta )^2 \frac{\tilde{G}^{\prime \prime }(\phi (\theta ))}{\tilde{G}(\phi (\theta ))} -\phi ^\prime (\theta )^2 \left( \frac{\tilde{G}^\prime (\phi (\theta ))}{\tilde{G}(\phi (\theta ))}\right) ^2 \right) , \nonumber \\ \end{aligned}$$

(114)

after easy algebra. Similarly, for $u=e^{i\theta }$ [cf. (83)]

$$\begin{aligned} \beta ^\prime (u)= & {} \frac{d}{de^{i\theta }} \left\{ g(e^{i\theta })e^{-H(e^{i\theta })} \right\} \nonumber \\= & {} -ie^{-i\theta }\frac{d}{d\theta } \left\{ \tilde{g}(\phi (\theta ))e^{-H(e^{i\theta })} \right\} \nonumber \\= & {} -ie^{-i\theta }\left( \phi ^\prime (\theta ) \tilde{g}^\prime (\phi (\theta ))- i e^{i\theta } \tilde{g}(\phi (\theta )) H^\prime (e^{i\theta }) \right) e^{-H(e^{i\theta })}, \end{aligned}$$

(115)

and

$$\begin{aligned} \beta ^{\prime \prime }(u)&= - e^{-2i\theta }\Biggl [-i\phi ^\prime (\theta ) \tilde{g}^\prime (\phi (\theta )) +\phi ^{\prime \prime }(\theta )\tilde{g}^\prime (\phi (\theta )) + \phi ^{\prime }(\theta )^2\tilde{g}^{\prime \prime }(\phi (\theta ))\nonumber \\&\qquad -i e^{i\theta }\phi ^\prime (\theta ) \tilde{g}^\prime (\phi (\theta )) H^\prime (e^{i\theta })\nonumber \\&\qquad +e^{2i\theta }\tilde{g}(\phi (\theta )) H^{\prime \prime }(e^{i\theta }) -e^{i\theta }\left( i\phi ^\prime (\theta ) \tilde{g}^\prime (\phi (\theta ))+ e^{i\theta } \tilde{g}(\phi (\theta )) H^\prime (e^{i\theta }) \right) H^\prime (e^{i\theta })\Biggr ]\nonumber \\&\qquad e^{-H(e^{i\theta })}, \end{aligned}$$

(116)

after lengthy but easy algebra. On the other hand,

$$\begin{aligned} \tilde{G}^\prime (\theta )= & {} -\frac{2i\rho (\theta )^2}{(1-\rho (\theta )^3e^{-i\theta })^2}\left( \rho (\theta )^4 -\rho (\theta ) \cos \theta -3\rho ^\prime (\theta ) \sin \theta \right) , \end{aligned}$$

(117)

$$\begin{aligned} \tilde{g}^\prime (\theta )= & {} \frac{i\rho (\theta )^2}{(1-\rho (\theta )^3e^{-i\theta })^2}\left( \rho (\theta )^4 -\rho (\theta ) \cos \theta -3\rho ^\prime (\theta ) \sin \theta \right) , \end{aligned}$$

(118)

so that

$$\begin{aligned} \tilde{G}^\prime (\theta )=-2\tilde{g}^\prime (\theta ), \end{aligned}$$

(119)

and

$$\begin{aligned} \tilde{G}^{\prime \prime }(\theta )= & {} \frac{\rho (\theta )}{ (1-\rho (\theta )^3e^{-i\theta })^3}\Biggl (-4 \rho (\theta )^8 e^{-i\theta }+\rho (\theta )^5(3+e^{-2i\theta })\nonumber \\&+3\rho (\theta )^4\left( 2i \rho ^\prime (\theta )(e^{-2i\theta }-3)- \rho ^{\prime \prime }(\theta )(1-e^{-2i\theta })\right) \nonumber \\&+ 12 \rho (\theta )^3 \rho ^\prime (\theta )^2(1-e^{-2i\theta })\nonumber \\&-2i\rho (\theta )^2 \sin \theta +6 i\rho (\theta )\left( 2 \rho ^\prime (\theta )\cos \theta + \rho ^{\prime \prime }(\theta )\sin \theta \right) +12 i \rho ^\prime (\theta )^2 \sin \theta \Biggr ), \nonumber \\ \end{aligned}$$

(120)

$$\begin{aligned} \tilde{g}^{\prime \prime }(\theta )= & {} -\frac{1}{2} \tilde{G}^{\prime \prime }(\theta ) \quad \hbox {from } (119). \end{aligned}$$

(121)

In particular,

$$\begin{aligned} \tilde{G}^\prime (0)=i\frac{k}{k-3}, \quad \tilde{G}^{\prime \prime }(0)=-\left( \frac{k}{k-3}\right) ^2, \quad \tilde{g}^\prime (0)= -\frac{i}{2}\cdot \frac{k}{k-3},\\\quad \tilde{g}^{\prime \prime }(0)=\frac{1}{2}\left( \frac{k}{k-3}\right) ^2. \end{aligned}$$

We are now in position to prove the lemma. We have shown in Lemma 5 above that $\phi (\theta )$ is twice differentiable in $[-\pi ,\pi ]$ and that its second derivative is continuous in $[-\pi ,\pi ]$. Also, recall that $\rho (\theta )$ is infinitely differentiable in $[-\pi ,\pi ]$ (cf. Lemma 1).

We start with $\alpha$. The first derivative of $\alpha (u)$ in (113) is continuous on $\mathcal {C}$ if the numerator in the ratio⁴

$$\begin{aligned} \frac{\tilde{G}^\prime (\theta )}{\tilde{G}(\theta )}=\frac{2i\rho (\theta )^2}{(1-\rho (\theta )^3e^{-i\theta }) (1-\rho (\theta )^3e^{i\theta }) }\left( \rho (\theta )^4 -\rho (\theta ) \cos \theta -3\rho ^\prime (\theta ) \sin \theta \right) \end{aligned}$$

(122)

vanishes when the denominator vanishes. Since $\rho (\theta )\in [0,1)$ for $\theta \not =0$ and $\rho (0)=1$ (see Lemma 1), the denominator in (122) has a single zero in $[-\pi ,\pi ]$ at $\theta =0$ and this zero has multiplicity two. It is easily checked that $\rho (\theta )^4 -\rho (\theta ) \cos \theta -3\rho ^\prime (\theta ) \sin \theta$ has a zero of order two at $\theta =0$. This shows that $\alpha ^\prime$ is well defined (actually continuous) on $\mathcal {C}$.

Now, since $\alpha ^\prime$ is continuous on $\mathcal {C}$, we see from (114) and from the fact that the ratio in (117) is continuous for $\theta \in [-\pi ,\pi ]$, that $\alpha ^{\prime \prime }$ is continuous on $\mathcal {C}$ if the numerator in the ratio

$$\begin{aligned} \frac{\tilde{G}^{\prime \prime }(\theta )}{\tilde{G}(\theta )}= & {} -\frac{\rho (\theta )}{ (1-\rho (\theta )^3e^{-i\theta })^2(1-\rho (\theta )^3 e^{i\theta })}\nonumber \\&\quad \Biggl (-4 \rho (\theta )^8 e^{-i\theta }+\rho (\theta )^5(3+e^{-2i\theta })\ \nonumber \\&+3\rho (\theta )^4\left( 2i \rho ^\prime (\theta )(e^{-2i\theta }-3)- \rho ^{\prime \prime }(\theta )(1-e^{-2i\theta })\right) \nonumber \\&+ 12 \rho (\theta )^3 \rho ^\prime (\theta )^2(1-e^{-2i\theta })-2i\rho (\theta )^2 \sin \theta \nonumber \\&+6 i\rho (\theta )\left( 2 \rho ^\prime (\theta )\cos \theta + \rho ^{\prime \prime }(\theta )\sin \theta \right) +12 i \rho ^\prime (\theta )^2 \sin \theta \Biggr ) \end{aligned}$$

(123)

vanishes when the denominator vanishes. As observed above, the denominator in (123) has a single zero in $[-\pi ,\pi ]$ at $\theta =0$ and this zero has multiplicity three. One can check that the term within the parentheses in (123) has a zero of order three at $\theta =0$ (Hint: $\rho ^\prime (0)=\rho ^{\prime \prime \prime }(0)=0$ and $\rho ^{\prime \prime }(0)=1/(3-k)$ from Lemma 1), which proves that $\alpha$ is twice differentiable on $\mathcal {C}$ and that $\alpha ^{\prime \prime }$ is continuous on $\mathcal {C}$.

We now turn our attention to $\beta$. It is easy to show that $\tilde{g}$ is twice differentiable in $[-\pi ,\pi ]$ and that its second derivative is continuous (Hint: check that the numerator in (118) [resp. (121)] vanishes at $\theta =0$ as many times as the denominator does). We are therefore left with proving that the mapping H is twice differentiable on $\mathcal {C}$ and that its second derivative is continuous on $\mathcal {C}$. With (82), H(t) in (60) may be written

$$\begin{aligned} H(t)= \frac{1}{2}\alpha (t)+\frac{1}{2\pi i} \displaystyle \int _{\mathcal {C}} \frac{\alpha (u)}{u-t} du. \end{aligned}$$

(124)

Thanks to (74), H(t) in (124) is also given by

$$\begin{aligned} H(t)=\alpha (t) +\frac{1}{2\pi i} \displaystyle \int _{\mathcal {C}} \frac{\alpha (u) -\alpha (t)}{u-t} du,\quad t\in \mathcal {C}. \end{aligned}$$

(125)

For every $t\in \mathcal {C}$, notice that the integrand in (125) is continuous on $\mathcal {C}$. This is clearly true for $u\not =t$. When $u=t$, it is equal to $\alpha ^\prime (t)$ by L’Hôpital’s rule. Differentiating (125) gives, for $t\in \mathcal {C}$,

$$\begin{aligned} H^\prime (t)= & {} \alpha ^\prime (t) +\frac{1}{2\pi i} \displaystyle \int _{\mathcal {C}} \frac{\alpha (u) -\alpha (t) -\alpha ^\prime (t)(u-t)}{(u-t)^2} du, \end{aligned}$$

(126)

$$\begin{aligned} H^{\prime \prime }(t)= & {} \alpha ^{\prime \prime } (t)+\frac{1}{2\pi i} \displaystyle \int _{\mathcal {C}} \frac{2(\alpha (u)-\alpha (t))-2\alpha ^\prime (t)(u-t)- \alpha ^{\prime \prime }(t)(u-t)^2}{(u-t)^3} du. \nonumber \\ \end{aligned}$$

(127)

For every $t\in \mathcal {C}$, notice that the integrand in (126) [resp. (127)] is continuous on $\mathcal {C}$. This is clearly true for $u\not =t$. When $u=t$, it is equal to $\frac{1}{2}\alpha ^{\prime \prime }(t)$ (resp. $\frac{1}{3}\alpha ^{\prime \prime \prime }(t)$) by L’Hôpital’s rule. The third derivative $\alpha ^{\prime \prime \prime }(t)$ can be calculated similarly to the calculation of $\alpha ^{\prime \prime }$ in (114); this calculation is omitted. (For the sake of simplicity, we recommend discarding the point $u=t$ when evaluating the integral in (127).) Since we have shown that $\alpha$, $\alpha ^\prime$, and $\alpha ^{\prime \prime }$ are all continuous on $\mathcal {C}$, the mappings H, $H^\prime$ and $H^{\prime \prime }$ are all continuous on $\mathcal {C}$ as the range of integration in (125)–(127) is bounded. This concludes the proof. $\square$

G First and second derivatives of $\gamma _0$

Set $K=k-2$. First, rewrite $\gamma _0(w)$ in (44) as

$$\begin{aligned} \gamma _0(w)= 1-\frac{2}{(1-\delta )^2 (K-\delta )}\left( \delta \gamma _{0,1}(w) +\gamma _{0,2}(w)\right) , \end{aligned}$$

(128)

with

$$\begin{aligned} \gamma _{0,1}(w)= & {} \frac{(1-w)^2(K-w)}{w}, \end{aligned}$$

(129)

$$\begin{aligned} \gamma _{0,2}(w)= & {} \frac{(w-\delta )(1-w)}{w} \sqrt{(K-w\delta ^2)(K-w)}. \end{aligned}$$

(130)

For $i=1,2$, we find

$$\begin{aligned} \frac{d^i}{dw^i}\gamma _0(w)= -\frac{2}{(1-\delta )^2 (K-\delta )}\left( \delta \frac{d^i}{dw^i}\gamma _{0,1}(w)+ \frac{d^i}{dw^i}\gamma _{0,2}(w)\right) , \end{aligned}$$

(131)

with

$$\begin{aligned} \frac{d}{dw}\gamma _{0,1}(w)= & {} \frac{(w-1) (Kw-2w^2+K)}{w^2}, \end{aligned}$$

(132)

$$\begin{aligned} \frac{d^2}{dw^2}\gamma _{0,1}(w)= & {} \frac{2(K-w^3)}{w^3}, \end{aligned}$$

(133)

and

$$\begin{aligned} \frac{d}{dw}\gamma _{0,2}(w)= & {} \frac{1}{2w^2\sqrt{(K-w\delta ^2) (K-w)}} \nonumber \\&\Biggl (-K\delta ^3w^2+3K\delta ^2 w^3+2\delta ^3w^3-4\delta ^2w^4-K\delta ^3 w-K\delta ^2 w^2 \nonumber \\&+2\delta ^2 w^3-2 K^2 w^2-K\delta w^2+3Kw^3+2K^2\delta -K\delta w-Kw^2 \Biggr ),\nonumber \\ \end{aligned}$$

(134)

$$\begin{aligned} \frac{d^2}{dw^2}\gamma _{0,2}(w)= & {} -\frac{1}{4 w^3 \left( (K-w\delta ^2 )(K-w)\right) ^{3/2}} \Biggl ( K^2\delta ^5 w^3+3 K^2\delta ^4 w^4\nonumber \\&-12 K\delta ^4 w^5+8\delta ^4 w^6+3 K^2\delta ^5 w^2\nonumber \\&+K^2\delta ^4 w^3-4 K\delta ^5 w^3-4 K^3\delta ^2 w^3-2 K^2\delta ^3 w^3\nonumber \\&+18 K^2\delta ^2 w^4-12 K\delta ^2 w^5-12 K^3\delta ^3 w\nonumber \\&+18 K^2\delta ^3 w^2-2 K^2\delta ^2 w^3-4 K\delta ^3 w^3-4 K^3 w^3+K^2\delta w^3\nonumber \\&+3 K^2 w^4+8 K^4\delta -12 K^3\delta w+3 K^2\delta w^2+K^2 w^3 \Biggr ). \end{aligned}$$

(135)

It is easy to see that

$$\begin{aligned} \frac{d}{dw}\gamma _0(w)|_{w=1} = \frac{8\sqrt{2(k-3)}}{\sqrt{6-k+\sqrt{(k+6)(k-2)}} \cdot (3\sqrt{k-2}+\sqrt{k+6}))}. \end{aligned}$$

(136)