AppendixThe Double Sine Function
The double sine function \(S_2(z):=S_2(z|\varvec)\), see [12] and references therein, is a meromorphic function that satisfies two functional relations
$$\begin \frac=2\sin \frac,\quad \frac=2\sin \frac \end$$
(A.1)
and inversion relation
$$\begin S_2(z)S_2(-z)=-4\sin \frac\sin \frac, \end$$
(A.2)
or equivalently
$$\begin S_2(z)S_2(\omega _1+\omega _2-z)=1. \end$$
(A.3)
The function \(S_2(z)\) has poles at the points
$$\begin z = m \omega _1 + k\omega _2, \quad m,k\ge 1 \end$$
(A.4)
and zeros at
$$\begin z=-m\omega _1-k\omega _2,\quad m,k\ge 0. \end$$
(A.5)
For \(\omega _1 / \omega _2 \not \in }\), all poles and zeros are simple. In the analytic region \( \textrm\,z \in ( 0, \textrm\,(\omega _1 + \omega _2) )\), we have the following integral representation for the logarithm of \(S_2(z)\)
$$\begin \ln S_2 (z) = \int _0^\infty \frac \left( \frac\left[ (2z - \omega _1 - \omega _2)t \right] }(\omega _1 t) \text (\omega _2 t) } - \frac \right) . \end$$
(A.6)
It is clear from this representation that the double sine function is homogeneous
$$\begin S_2( \gamma z | \gamma \omega _1, \gamma \omega _2 ) = S_2(z|\omega _1, \omega _2), \quad \gamma \in (0, \infty ) \end$$
(A.7)
and invariant under permutation of periods
$$\begin S_2(z| \omega _1, \omega _2) = S_2(z | \omega _2, \omega _1). \end$$
(A.8)
The double sine function can be expressed through the Barnes double Gamma function \(\Gamma _2(z|\varvec)\) [1],
$$\begin S_2(z|\varvec)=\Gamma _2(\omega _1+\omega _2-z|\varvec)\Gamma _2^(z|\varvec), \end$$
(A.9)
and its properties follow from the corresponding properties of the double Gamma function. It is also connected to the Ruijsenaars hyperbolic Gamma function \(G(z|\varvec)\) [16]
$$\begin G(z|\varvec) = S_2\Bigl (\imath z + \frac \,\Big |\, \varvec\Bigr ) \end$$
(A.10)
and to the Faddeev quantum dilogarithm \(\gamma (z|\varvec)\) [3]
$$\begin \gamma (z|\varvec) = S_2\Bigl (-\imath z + \frac\, \Big |\, \varvec\Bigr ) \exp \Bigl ( \frac \Bigl [z^2 + \frac \Bigr ]\Bigr ). \end$$
(A.11)
Both \(G(z|\varvec)\) and \(\gamma (z|\varvec)\) were investigated independently.
In the paper, we deal only with ratios of double sine functions denoted by \(\mu (x)\) (1.6) and K(x) (1.11)
$$\begin \begin\mu (x)&=S_2(\imath x)S_2^ (\imath x+g),\\ K(x)&= S_2\left( \imath x+\frac+\frac\right) S_2^\left( \imath x+\frac-\frac\right) . \end \end$$
(A.12)
Now we will give the key asymptotic formulas and bounds for them, which were derived in [2, Appendices A, B] from the known results for the double Gamma function. In what follows we assume conditions (1.9), (1.10)
$$\begin \textrm\,\omega _j> 0, \quad 0< \textrm\,g < \textrm\,\omega _1 + \textrm\,\omega _2, \quad \textrm\,} > 0, \end$$
(A.13)
where we denoted
$$\begin } = \frac. \end$$
(A.14)
Let \(\sigma _i\) be the arguments of the periods \(\omega _i\), \(|\sigma _i|<\pi /2\). Because of the symmetry (A.8), we may assume that \(\sigma _1 \ge \sigma _2\). Let \(D_+\) and \(D_-\) be the cones of poles (A.4) and zeros (A.5) of \(S_2(z)\):
$$\begin \begin&D_+= \,\\ &D_-=\,\quad D= D_+\cup D_-. \end \end$$
Denote by d(z, D) the distance between a point z and the cones D. Using Barnes’ Stirling formula for the asymptotic of double Gamma function, for the ratio of double sines one obtains
$$\begin \frac = e^\pi \imath }\left( z-\frac\right) }\Bigl (1+\,O\Bigl (d^(z,D)\Bigr )\Bigr ), \end$$
(A.15)
where \(z \in } \setminus D\) and the sign − (or \(+\)) is taken for \(\textrm\,z > 0\) (or \(\textrm\,z < 0\)), see [2, eq.(A.18)]. Then from (A.15) for the functions \(\mu (x)\) and K(x) (A.12) with \(x \in }\) we have
$$\begin \mu (x) \sim e^} | x | \pm \imath \frac} g^*} }, \quad K(x) \sim e^} |x|}, \quad x\rightarrow \pm \infty . \end$$
(A.16)
Denote also
$$\begin \nu _g= \textrm\,}. \end$$
(A.17)
Under the conditions (1.9), (1.10) by using the same Stirling formula, we also have bounds
$$\begin |\mu (x)| \le C e^, \quad |K(x)| \le C e^, \quad x \in }\end$$
(A.18)
where C is a positive constant uniform for compact subsets of parameters \(\varvec, g\) preserving the mentioned conditions, see [2, eq.(B.3)].
Another key result that we need in the paper is the following Fourier transform formula given in [18, Proposition C.1], which we rewrite in terms of the double sine function using connection formula (A.10). This Fourier transform can be already found in [4, 14].
Proposition
[18] For real positive periods \(\omega _1, \omega _2\), we have
$$\begin \begin&\int _}} dx \, e^ y x} S_2\Bigl (\imath x - \imath \nu + \frac \Bigr ) S_2^ \Bigl ( \imath x - \imath \rho + \frac \Bigr ) \\&\quad = \sqrt \, e^ y (\nu + \rho ) } S_2(\imath \rho - \imath \nu ) \, S_2^\Bigl (\imath y + \frac \Bigr ) \, S_2^\Bigl (-\imath y + \frac \Bigr ), \end \nonumber \\ \end$$
(A.19)
while the parameters \(\nu , \rho , y\) satisfy the conditions
$$\begin -\frac< }\, \rho< }\, \nu< \frac, \quad |}\, y | < }\frac. \end$$
(A.20)
In the special case
$$\begin \nu = \frac, \quad \rho = -\frac \end$$
(A.21)
taking \(y = \omega _1 \omega _2 \lambda \) and using homogeneity of the double sine (A.7) (with \(\gamma = \omega _1 \omega _2\)), we arrive at the Fourier transform formula for the function K(x) (A.12)
$$\begin \int _}} dx \; e^ (x) = \sqrt \, S_2(g) \, }}(\lambda ), \end$$
(A.22)
where \(| }\, \lambda | < \textrm\,}/2\) and conditions (A.20) are satisfied due to the inequalities on the coupling constant g (1.9), (1.10). Here we recall the notations (1.32), (1.34)
$$\begin }(\lambda ) = K_}^*}(\lambda |}), \quad }^* = \frac, \quad }} = \Bigl ( \frac, \frac \Bigr ). \end$$
(A.23)
Note that the right-hand side of (A.22) is analytic function of \(\omega _1, \omega _2\) in the domain \(\textrm\,\omega _j > 0\). The integral from the left is also analytic with respect to periods. Indeed, due to the bound (A.18) it is absolutely convergent uniformly on compact sets of parameters \(\varvec, g\) preserving the conditions (1.9), (1.10). Hence, the formula (A.22) also holds for complex periods under the mentioned conditions.
Inequalities with absolute value
First, we will prove a little lemma that will be crucial for all other estimates.
Lemma 1
For any \(\varepsilon \in [0, 2]\), \(y_1, y_2, y \in }\), we have
$$\begin |y_1 - y_2| - |y_1 - y| - |y_2 - y| \le \varepsilon \left( |y_1| + |y_2| - |y| \right) . \end$$
(B.24)
Proof 6
Consider two cases. First, assume that \(|y_1| + |y_2| \ge |y|\). Then using triangle inequality, we obtain
$$\begin |y_1 - y_2| - |y_1 - y| - |y_2 - y| \le 0 \le \varepsilon \left( |y_1| + |y_2| - |y| \right) \end$$
(B.25)
since \(\varepsilon \ge 0\).
Second, assume \(|y_1| + |y_2| \le |y|\). Now using \(|y_j - y| \ge |y| - |y_j|\) and \(|y_1 - y_2| \le |y_1| + |y_2|\), we arrive at
$$\begin |y_1 - y_2| - |y_1 - y| - |y_2 - y|\le & 2 \left( |y_1| + |y_2| - |y| \right) \nonumber \\\le & \varepsilon \left( |y_1| + |y_2| - |y| \right) \end$$
(B.26)
because \(\varepsilon \le 2\). \(\square \)
Remark 5
Note that Lemma 1 holds for any norm and points \(y_j\) in the corresponding normed space.
Denote by \(}_k\) a vector with k components \(}_k = \bigl ( y^_, \dots , y^_ \bigr )\), \(y_j^\in }\). Define a function \(S_n\) by a recurrence relation
$$\begin \begin S_n ( }_1, \dots , }_n )= \sum _ i, j = 1 \\ i \not = j \end}^n \bigl | y^_i - y^_j \bigr |&- \sum _^ \sum _^ \, \bigl | y^_i - y^_j \bigr | \\ &+S_ ( }_1, \dots , }_ ) \end \end$$
(B.27)
with \(S_1 = 0\). Also by \(\Vert }_k \Vert \) denote \(L^1\)-norm
$$\begin \Vert }_k \Vert = \sum _^k \bigl | y^_j \bigr |. \end$$
(B.28)
Lemma 2
The inequality
$$\begin S_n \le \frac\sum _ i, j = 1 \\ i \not = j \end}^n \bigl | y^_i - y^_j \bigr | + c_n \varepsilon \Vert }_n \Vert - \varepsilon \sum _^ \Vert }_k \Vert , \end$$
(B.29)
holds for any \(\varepsilon \in \left[ 0, \frac \right] \), where the number \(c_n\) is defined by recurrence relation \(c_n = (n - 1) (1 + c_)\) with \(c_1 = 0\).
Proof 7
The proof goes by induction. Consider the case \(n = 2\). Using Lemma 1, we have
$$\begin \begin S_2&= 2 \bigl | y_1^ - y_2^ \bigr | - \bigl | y_1^ - y_1^ \bigr | - \bigl | y_2^ - y_1^ \bigr | \\&\le \bigl | y_1^ - y_2^ \bigr | + \varepsilon \left( \bigl | y_1^ \bigr | + \bigl | y_2^ \bigr | \right) - \varepsilon \bigl | y_1^ \bigr | \end \end$$
(B.30)
for any \(\varepsilon \in [0, 2]\). Thus, we proved the base case.
Now suppose we proved the statement for \(S_\). Both sides of the stated inequality are symmetric with respect to components of \(}_k\) for all \(k = 1, \dots , n\). Therefore, without loss of generality we can assume the ordering
$$\begin y^_1 \ge \cdots \ge y^_k, \quad k = 1, \dots , n. \end$$
(B.31)
For the vector \(}_n\) with ordered components, we can write
$$\begin \sum _ i, j = 1 \\ i \not = j \end}^n \bigl | y^_i - y^_j \bigr | = 2 \sum _^ (n - 2m + 1) \bigl | y_m^ - y_^ \bigr |. \end$$
(B.32)
Consequently, due to the recurrence relation (B.27) and induction assumption
$$\begin \begin S_n&\le 2 \sum _^ (n - 2m + 1) \bigl | y_m^ - y_^ \bigr | - \sum _^n \sum _^ \, \bigl | y^_i - y^_j \bigr | \\ &+ \sum _^ (n - 2m) \bigl | y_m^ - y_^ \bigr | + c_ \varepsilon \Vert }_ \Vert - \varepsilon \sum _^ \Vert }_k \Vert . \end \nonumber \\ \end$$
(B.33)
Next step is to regroup and estimate terms from the first three sums. For the term with \(m = 1\) from the first sum, we use the Lemma 1 and for all other terms we simply use triangle inequalities.
Consider term with \(m = 1\) from the first sum and terms with \(i = 1, n\) from the second double sum and write the estimate
$$\begin \begin&(n - 1) \bigl | y_1^ - y_^ \bigr | - \sum _^ \left( \bigl | y_1^ - y^_j \bigr | + \bigl | y_^ - y^_j \bigr | \right) \\&\quad \le (n - 1) \, \varepsilon ' \left( \bigl | y_1^ \bigr | + \bigl | y_n^ \bigr | \right) - \varepsilon ' \, \Vert }_ \Vert , \end \end$$
(B.34)
where we used Lemma 1 with parameter \(\varepsilon ' \in [0, 2]\) multiple times. Similarly let us estimate the term with \(m > 1\) from the first sum together with the corresponding terms from the second double sum
$$\begin \begin&(n - 2m + 1) \bigl | y_m^ - y_^ \bigr | \\&\quad - \sum _^ \left( \bigl | y_m^ - y^_j \bigr | + \bigl | y_^ - y^_j \bigr | \right) \le 0, \end \end$$
(B.35)
where we used triangle inequality multiple times. Remaining from the second double sum terms can be grouped with terms from the third sum
$$\begin \begin&(n - 2m) \bigl | y_m^ - y_^ \bigr | \\&\quad - \sum _^ \left( \bigl | y_i^ - y^_m \bigr | + \bigl | y_i^ - y^_ \bigr | \right) \le 0, \end \end$$
(B.36)
where we again used triangle inequalities.
So, starting from (B.33) and using inequalities (B.34), (B.35), (B.36) we obtain
$$\begin S_n\le & \sum _^ (n - 2m + 1) \bigl | y_m^ - y_^ \bigr | + (n - 1) \varepsilon ' \left( \bigl | y_1^ \bigr | + \bigl | y_n^ \bigr | \right) \nonumber \\ & \quad + (c_ \varepsilon - \varepsilon ') \Vert }_ \Vert - \varepsilon \sum _^ \Vert }_k \Vert . \end$$
(B.37)
Finally, for the first sum we again use equality (B.32). For the second sum, we have
$$\begin \bigl | y_1^ \bigr | + \bigl | y_^ \bigr | \le \Vert }_ \Vert . \end$$
(B.38)
Choosing \(\varepsilon ' = (1 + c_ )\varepsilon \), we get bound stated in the lemma. Note that since \(\varepsilon ' \le 2\) then \(\varepsilon \le 2/(1 + c_) = 2(n - 1)/c_n\). \(\square \)
Corollary 5
The inequality
$$\begin S_n \le (n - 1 + \varepsilon ) \Vert }_n \Vert - \frac \sum _^ \Vert }_k \Vert \end$$
(B.39)
holds for any \(\varepsilon \in \left[ 0, 2(n - 1) \right] \).
Proof 8
Use Lemma 2 rescaling the parameter \(\varepsilon \rightarrow \varepsilon /c_n\) and
$$\begin \frac \sum _ i, j = 1 \\ i \not = j \end}^n \bigl | y^_i - y^_j \bigr | \le (n - 1) \Vert }_n \Vert \end$$
(B.40)
to arrive at stated inequality. \(\square \)
In fact, Lemma 2 gives an estimate for the integrand in the integral representation of the wave function \(\Psi _}_n}(}_n)\) (1.24) and, as a consequence, the exponential bound for the wave function. Denote, as before (4.11),
$$\begin \mu '(}_n) = \prod _ i,j=1 \\ i<j \end}^n \mu (x_i - x_j). \end$$
(B.41)
Corollary 6
The wave function admits the bound
$$\begin \bigl | \mu '(}_n) \Psi _}_n}(}_n) \bigr | \le C \exp \pi \nu _g \left( \varepsilon \Vert }_n \Vert -\frac \underline}}_n }\,\lambda _n \right) \end$$
(B.42)
with some \(C(g, \varvec)\) for any \(\varepsilon \in (0, 2(n - 1)]\), assuming \(x_j \in }\) and
$$\begin | }(\lambda _k - \lambda _j) |< \frac \, \varepsilon , \quad k,j=1, \dots ,n. \end$$
(B.43)
Proof 9
The function \(\mu '(}_n)\) is estimated as
$$\begin |\mu '(}_n)| \le C_1 \exp \pi \nu _g \Biggl ( \; \sum _ i,j=1 \\ i<j \end}^n |x_i-x_j| \, \Biggr ) \end$$
(B.44)
with some \(C_1(g, \varvec)\), where we used bound from (A.18). The integral representation (1.24) in full form
$$\begin \begin \Psi _}_n} (}_n)= C_\Psi \int _}^ }&d}_ \dots d}_1 \; e^ \left[ \lambda _n \underline}}_n + (\lambda _ - \lambda _) \underline}}_ \right] } (}_n, }_) \\ &\times \prod _^ e^(\lambda _ - \lambda _) \underline}}_} \, \mu ( }_ ) \, (}_, }_ ), \end \end$$
(B.45)
where \(C_\Psi \) contains all constants \(d_k\). Denote the integrand by H. Assuming
$$\begin |}(\lambda _ - \lambda _)| \le \delta _\Lambda \frac \end$$
(B.46)
and using bounds (A.18), we arrive at
$$\begin \begin |H|&\le C_2 \exp \pi \nu _g \Biggl ( -\frac \underline}}_n\, }\,\lambda _n +\delta _\Lambda \sum _^\Vert }_k\Vert \\&\quad + S_n(}_1, \dots ,}_, }_n) - \sum _ i,j=1\\ i\not = j \end}^n|x_i - x_j| \Biggr ) \end \end$$
(B.47)
with some \(C_2(g,\varvec)\). Using Lemma 2 with rescaling \(\varepsilon \rightarrow \varepsilon /c_n\), we have
$$\begin |H|\le & C_2 \exp \pi \nu _g \Biggl ( -\frac \underline}}_n \,}\,\lambda _n + \Bigl (\delta _\Lambda - \frac \Bigr ) \sum _^\Vert }_k\Vert \nonumber \\ & + \varepsilon \Vert }_n\Vert - \frac \sum _ i,j=1\\ i\not = j \end}^n|x_i - x_j| \Biggr ) \end$$
(B.48)
for any \(\varepsilon \in [0, 2(n - 1)]\). For
$$\begin \delta _\Lambda< \frac < \frac, \end$$
(B.49)
see (3.35), the bound (B.48) represents integrable function. Combining it with (B.44), we arrive at the bound stated in the corollary. \(\square \)
For the next inequality, define another function that depends on vectors \(}_k\) and additional vector \(}_n = (t_1, \dots , t_n)\)
$$\begin T_n(}_1, \dots , }_n, }_n) = \sum _ i, j = 1 \\ i \not = j \end}^n | t_i - t_j | - \sum _^n \bigl | t_i - y_j^ \bigr | + S_n(}_1, \dots , }_n). \nonumber \\ \end$$
(B.50)
Lemma 3
The inequality
$$\begin \begin T_n \le ( n + r ) \Vert }_n \Vert - \frac \sum _^ \Vert }_k \Vert - r \, \biggl | \sum _^n \bigl ( t_j - y_j^ \bigr ) \biggr | \end \end$$
(B.51)
holds for any \(r \in [0, 1]\).
Proof 10
Both sides of the stated inequality are symmetric with respect to components of the vectors \(}_k, }_n\) (separately). Therefore, without loss of generality we assume ordering
$$\begin y^_1 \ge \cdots \ge y_k^, \quad t_1 \ge \cdots \ge t_n, \quad k = 1, \dots , n. \end$$
(B.52)
Then, as in the proof of the previous lemma, we can write
$$\begin \sum _ i, j = 1 \\ i \not = j \end}^n | t_i - t_j | = 2 \sum _^ (n - 2m + 1) | t_m - t_ |. \end$$
(B.53)
Next, take arbitrary \(r \in [0, 1]\) and use triangle inequalities for the second double sum in (B.50) to obtain
$$\begin - \sum _^n \bigl | t_i - y_j^ \bigr | \le - \left( 1 - \frac \right) \sum _^n \bigl | t_i - y_j^ \bigr | - r \, \biggl | \sum _^n \bigl ( t_j - y_j^ \bigr ) \biggr |. \nonumber \\ \end$$
(B.54)
Combining two previous formulas (B.53), (B.54) and Lemma 2 with the parameter
$$\begin \varepsilon = \frac \, \in \, \left[ 0, \frac \right] \end$$
(B.55)
we arrive at inequality
$$\begin \begin T_n&\le \sum _^ (n - 2m + 1) \left( 2 \, | t_m - t_ | + \bigl | y_m^ - y^_ \bigr | \right) \\&\quad - \left( 1 - \frac \right) \sum _^n \bigl | t_i - y_j^ \bigr | + \frac \Vert }_n \Vert \\&\quad - \frac \sum _^ \Vert }_k \Vert - r \, \biggl | \sum _^n \bigl ( t_j - y_j^ \bigr ) \biggr |. \end \nonumber \\ \end$$
(B.56)
Notice that two last sums from the right coincide with the corresponding terms in the stated inequality (B.51), so it is left to bound the rest ones. Denote them by \(R_n\)
$$\begin \begin R_n&= \sum _^ (n - 2m + 1) \left( 2 \, | t_m - t_ | + \bigl | y_m^ - y^_ \bigr | \right) \\&\quad - \left( 1 - \frac \right) \sum _^n \bigl | t_i - y_j^ \bigr | + \frac \Vert }_n \Vert . \end\nonumber \\ \end$$
(B.57)
We wish to prove the inequality
$$\begin R_n \le (n + r) \Vert }_n \Vert - \frac \Vert }_n \Vert . \end$$
(B.58)
Now fix \(m \in \\) and from all of the remaining terms \(R_n\) consider the following ones
$$\begin \begin R_&= 2(n - 2m + 1) |t_m - t_| + (n - 2m + 1) \bigl | y_m^ - y^_ \bigr | \\ &-\left( 1 - \frac \right) \sum _^ \left( \bigl | t_i - y_m^ \bigr | + \bigl | t_i - y_^ \bigr | \right) \\ &- \left( 1 - \frac \right) \sum _^ \left( \bigl | t_m - y_j^ \bigr | + \bigl | t_ - y_j^ \bigr | \right) \\ &+ \frac \left( \bigl | y_m^ \bigr | + \bigl | y_^ \bigr | \right) . \end\nonumber \\ \end$$
(B.59)
Then for even n, we have
$$\begin R_n = \sum _^ R_ \end$$
(B.60)
and for odd n
$$\begin R_n = \sum _^ R_ - \left( 1 - \frac \right) \bigl | t_} - y_}^ \bigr | + \frac \, \bigl | y_}^ \bigr |, \end$$
(B.61)
since in this case two variables \(t_\), \(y_^\) don’t enter the first sum in (B.57).
Let us prove the following bound
$$\begin R_ \le ( n + r ) \left( |t_m| + |t_| \right) - \frac \left( \bigl | y_m^ \bigr | + \bigl | y_^ \bigr | \right) . \end$$
(B.62)
Note that for even n the inequality we wish to prove (B.58) directly follows from it due to (B.60). For odd n, we have two more terms in (B.61), but they can be easily bounded using triangle inequality
$$\begin \begin&-\! \left( 1\! -\! \frac \right) \bigl | t_} \!-\! y_}^ \bigr | \!+\! \frac \, \!\bigl | y_}^ \bigr | \le \left( 1 \!-\! \frac \right) \bigl | t_} \bigr | \!+\! \frac \, \bigl | y_}^ \bigr | \\&\quad \le (n + r ) \bigl | t_} \bigr | - \frac \, \bigl | y_}^ \bigr |. \end \nonumber \\ \end$$
(B.63)
The proof of (B.62) requires several steps. First, using triangle inequalities we can obtain
$$\begin \begin&(n - 2m + 1) \bigl | y_m^ - y^_ \bigr | \\&\quad - \frac \sum _^ \left( \bigl | t_i - y_m^ \bigr | + \bigl | t_i - y_^ \bigr | \right) \le 0. \end \nonumber \\ \end$$
(B.64)
For the rest of the first sum in (B.59), we use obvious inequality
$$\begin \begin&-\biggl ( \frac - \frac \biggr ) \sum _^ \left( \bigl | t_i - y_m^ \bigr | + \bigl | t_i - y_^ \bigr | \right) \\&\quad \le -\frac \, \Bigl ( \bigl | t_m - y_m^ \bigr | + \bigl | t_m - y_^ \bigr | \\&\qquad + \bigl | t_ - y_m^ \bigr | + \bigl | t_ - y_^ \bigr | \Bigr ). \end \nonumber \\ \end$$
(B.65)
Then, we take the right-hand side of this formula and estimate it together with a part of \(|t_m - t_|\) term using Lemma 1 with the parameter \(\varepsilon \) = 1
$$\begin \begin&\frac \, |t_m - t_| - \frac \, \Bigl ( \bigl | t_m - y_m^ \bigr | \\&\quad + \bigl | t_m - y_^ \bigr | + \bigl | t_ - y_m^ \bigr | + \bigl | t_ - y_^ \bigr | \Bigr ) \\&\qquad \le \frac \left( 2 |t_m| + 2 |t_| - \bigl | y_^ \bigr | - \bigl | y_^ \bigr | \right) . \end\nonumber \\ \end$$
(B.66)
To bound the second sum in (B.59), we again use triangle inequalities
$$\begin \begin&(n - 2m) \biggl (1 - \frac \biggr ) |t_m - t_| \\&\quad - \left( 1 - \frac \right) \sum _^ \left( \bigl | t_m - y_j^ \bigr | + \bigl | t_ - y_j^ \bigr | \right) \le 0. \end \nonumber \\ \end$$
(B.67)
Collecting results of the four inequalities (B.64), (B.65), (B.66), (B.67), we obtain the bound
$$\begin \begin R_&\le \biggl ( n + r + 2 - 2m - \frac - \frac \biggr ) | t_m - t_ | \\&\quad + \frac \left( |t_m| + |t_| \right) - \frac \left( \bigl | y_^ \bigr | + \bigl | y_^ \bigr | \right) . \end \nonumber \\ \end$$
(B.68)
Finally, to arrive at the inequality (B.62) we use \(|t_m - t_| \le |t_m| + |t_|\) together with
$$\begin 2 - 2m - \frac \le 0, \hspace \frac \ge \frac \,. \end$$
(B.69)
Thus, we proved the key bound (B.62) and, consequently, the lemma. \(\square \)
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