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Existence and Uniqueness of Solutions to Y-Systems and TBA Equations


We consider Y-system functional equations of the form

$$\begin{aligned} \textstyle { Y_n(x+i)Y_n(x-i)=\prod _{m=1}^N \left( 1+Y_m(x)\right) ^{G_{nm}} } \end{aligned}$$

and the corresponding nonlinear integral equations of the thermodynamic Bethe ansatz. We prove an existence and uniqueness result for solutions of these equations, subject to appropriate conditions on the analytical properties of the \(Y_n\), in particular the absence of zeros in a strip around the real axis. The matrix \(G_{nm}\) must have non-negative real entries and be irreducible and diagonalisable over \(\mathbb {R}\) with spectral radius less than 2. This includes the adjacency matrices of finite Dynkin diagrams, but covers much more as we do not require \(G_{nm}\) to be integers. Our results specialise to the constant Y-system, proving existence and uniqueness of a strictly positive solution in that case.

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Fig. 1


  1. 1.

    We are grateful to Nathan Bowler for pointing this out to us.

  2. 2.

    Note that our results do not imply that the TBA integral operators are contracting, but only the operator to which the Banach theorem is directly applied, namely the right-hand side of (3.22) for \(\mathbf {C} = \nu \mathbf {G}\) with \(\nu \in [0,\frac{1}{2}]\) (which for \(\nu =0\) is the operator \(\mathbf {J}\)).

    However, our bound on the contraction constant \(\kappa \) is certainly not optimal (see Remark 3.12) and the TBA integral operator may well be contracting even if our bound yields \(\kappa >1\). The results of [6] simply show that sometimes it is not contracting.


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We would like to thank Nathan Bowler, Andrea Cavaglià, Patrick Dorey, Clare Dunning, Andreas Fring, Annina Iseli, Karol Kozlowski, Louis-Hadrien Robert, Roberto Tateo, Jörg Teschner, Stijn van Tongeren, Alessandro Torrielli, Benoît Vicedo and Gérard Watts for helpful discussions and comments. LH is supported by the SFB 676 “Particles, Strings, and the Early Universe” of the German Research Foundation (DFG).

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Correspondence to Lorenz Hilfiker.

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Communicated by Nikolai Kitanine.


Appendix A. Explicit Expression for the Green’s Function \(\phi _d(z)\)

In this section, we provide the proof of Lemma 2.15. The explicit solution (2.55) can also be found in tables, see, for example, [9, 1.9 (6)]. However, we are not aware of a more direct proof that does not require the analysis carried out in Sect. 2.1.

First, let us consider the case where \(\tfrac{\gamma }{\pi }\in (0,1)\) is rational. In this case, there exists \(P\in \mathbb {Z}_{\ge 0}\), such that \(P\gamma \in 2\pi \mathbb {Z}\). From Lemma 2.8 (iii) then follows the periodicity

$$\begin{aligned} \phi _d(z+imPs) ~=~ \phi _d(z) \quad \text {for all}\quad m\in \mathbb {Z}. \end{aligned}$$

Hence, there exists a meromorphic function \({\widetilde{\phi }}_d\) on \(\mathbb {C}^\times \), such that

$$\begin{aligned} \phi _d(z) ~=~ {\widetilde{\phi }}_d\left( e^{\frac{2\pi }{Ps}z}\right) . \end{aligned}$$

Due to Lemma 2.10, \({\widetilde{\phi }}_d(w)\rightarrow 0\) as \(w\rightarrow 0\) and \(w\rightarrow \infty \). In particular, \({\widetilde{\phi }}_d\) is analytic also in the origin.

Set \(\xi ~:=~ e^{\frac{2\pi i}{P}}\). According to Lemma 2.8 (i), the function \({\widetilde{\phi }}_d\) is analytic everywhere except in a subset of \(\lbrace \xi ^n \rbrace _{n\in \mathbb {Z}}\), where it may have poles of first order. The residues can be easily computed from the residues of \(\phi _d\) provided by Lemma 2.13:

$$\begin{aligned} \mathrm {Res}_{\xi ^n}\left( {\widetilde{\phi }}_d\right) ~=~ \frac{2\pi }{Ps}\,\xi ^n\,\mathrm {Res}_{isn}\left( \phi _d\right) ~=~ \frac{\xi ^n}{iPs} \frac{\sin (n\gamma )}{\sin (\gamma )}. \end{aligned}$$

We will now use this result in order to explicitly compute the expansion of \({\widetilde{\phi }}_d\) around the origin,

$$\begin{aligned} {\widetilde{\phi }}_d (w) ~=~ \sum _{m=1}^\infty \varphi _m w^m. \end{aligned}$$

This expansion converges in the open unit disc. Since \(\phi _d\) is an even function, we have \({\widetilde{\phi }}_d(w)={\widetilde{\phi }}_d(1/w)\) for all w. Thus, we will automatically also get an expression for \({\widetilde{\phi }}_d\) which converges in the whole complex plane without the closed unit disc.

Let \(m\in \mathbb {Z}_{> 0}\). Denote by \(\Gamma \) a small contour inside the open unit disc, which circles once around the origin in anti-clockwise direction. The coefficient \(\varphi _m\) of the expansion is given by the Cauchy integral

$$\begin{aligned} \varphi _m ~=~ \oint _{\Gamma }\, w^{-m-1} {\widetilde{\phi }}_d(w)\, \frac{dw}{2\pi i}. \end{aligned}$$

Now denote by \(\Gamma _\infty \) a large contour outside the closed unit disc, which circles once around the origin in anti-clockwise direction, and by \(\Gamma _k\) (for \(k=0,1,2\dots , P-1\)) a small contour, which circles once around the pole \(\xi ^k\) in clockwise direction. We can deform the contour \(\Gamma \) in order to get

$$\begin{aligned} \varphi _m ~=~ \sum _{k=0}^{P-1}\oint _{\Gamma _k}\, w^{-m-1} {\widetilde{\phi }}_d(w)\, \frac{dw}{2\pi i} ~+~ \oint _{\Gamma _\infty }\, w^{-m-1} {\widetilde{\phi }}_d(w)\, \frac{dw}{2\pi i}. \end{aligned}$$

The last term vanishes because \({\widetilde{\phi }}_d\) vanishes at infinity. For the remaining integrals, we can compute the residues as follows:

$$\begin{aligned} \varphi _m ~&=~ - \sum _{k=0}^{P-1} \mathrm {Res}_{w=\xi ^k}\left( {\widetilde{\phi }}_d(w)w^{-m-1}\right) \overset{(A.3)}{=}~ \frac{i}{Ps}\frac{1}{\sin (\gamma )}\sum _{k=0}^{P-1} \xi ^{-km} \sin (k\gamma ) \nonumber \\ ~&=~ \frac{1}{2Ps\sin (\gamma )}\sum _{k=0}^{P-1} \left( e^{\frac{2\pi i}{P} k \left( -m+\frac{P\gamma }{2\pi }\right) } - e^{\frac{2\pi i}{P} k \left( -m-\frac{P\gamma }{2\pi }\right) } \right) \end{aligned}$$

For \(\alpha \in \mathbb {Z}\) we have \(\sum _{k=0}^{P-1} e^{\frac{2\pi i}{P} k \alpha } ~=~ P\cdot \delta _{\alpha \in P\mathbb {Z}}\). Inserting this identity in the expression for \(\varphi _m\) gives

$$\begin{aligned} \varphi _m ~=~ \frac{1}{2s\sin (\gamma )} \left( \delta _{m\in \frac{P\gamma }{2\pi } + P\mathbb {Z}} - \delta _{m\in -\frac{P\gamma }{2\pi } + P\mathbb {Z}} \right) . \end{aligned}$$

This coefficient can be plugged into the expansion (A.4),

$$\begin{aligned} {\widetilde{\phi }}_d(w) ~&=~ \frac{1}{2s\sin (\gamma )}\sum _{k=0}^{\infty } \left( w^{\left( \frac{\gamma }{2\pi } + k \right) P} - w^{\left( -\frac{\gamma }{2\pi } + k +1\right) P} \right) \nonumber \\ ~&=~ \frac{1}{2s\sin (\gamma )}\cdot \frac{ w^{\frac{\gamma P}{2\pi }} - w^{P-\frac{\gamma P}{2\pi }} }{1-w^P}. \end{aligned}$$

From this we can recover the function \(\phi _d\) according to (A.2), which yields precisely the expression (2.55).

Finally, the case where \(\frac{\gamma }{\pi }\) is irrational can be obtained from continuity of \(\phi _d(z)\) in d (Lemma 2.6). This completes the proof.

Appendix B. Fourier Transformation of \(\cosh (x)^{-m}\)

Lemma B.1

Let \(m\in \mathbb {Z}_{\ge 1}\), and \(f_m(x)=\cosh (x)^{-m}\). The Fourier transformation of \(f_m(x)\) is given by

$$\begin{aligned} \hat{f}_m(k)&~:=~\int _{-\infty }^\infty e^{-ikx}f_m(x) \; dx \nonumber \\&~=~ \frac{\pi }{(m-1)!}\left( \prod _{\begin{array}{c} l=m-2\\ \mathrm {step\, -2} \end{array}}^1 (k^2 + l^2)\right) \cdot {\left\{ \begin{array}{ll} \frac{1}{\cosh \left( \frac{\pi }{2}k\right) } &{} \mathrm {if} \; m \;\mathrm {odd} \\ \frac{k}{\sinh \left( \frac{\pi }{2}k\right) } &{}\mathrm {if} \; m \;\mathrm {even} \end{array}\right. }\, . \end{aligned}$$


It is convenient to first get rid of the infinite number of poles of the integrand. This is achieved by the variable transformation \(r=e^{2x}\), which results in

$$\begin{aligned} \hat{f}_m(k) = 2^{m-1}\int _0^\infty \frac{\left( r^{1/2}\right) ^{m-2-ik}}{(1+r)^m}\;\mathrm{d}r. \end{aligned}$$

This transformation comes at the expense of single-valuedness: the new integrand has only one simple pole at \(r=-1\), but also a branch cut connecting the origin and infinity via, say, the negative imaginary axis. Since the integrand decays fast enough for \(|r|\rightarrow \infty \), it is possible to revolve the integration contour once around the origin like the big hand of a watch. If we do this counter-clockwise, the whole integral picks up a residue, and the square root acquires a monodromy of \(-1\), which changes the numerator to

$$\begin{aligned} \left( -r^{1/2}\right) ^{m-2-ik} = (-1)^m e^{k\pi } \left( r^{1/2}\right) ^{m-2-ik}. \end{aligned}$$

This contour manipulation yields the equation

$$\begin{aligned} \int _0^\infty \frac{\left( r^{1/2}\right) ^{m-2-ik}}{(1+r)^m}\;\mathrm{d}r&= 2\pi i\,\mathrm {Res}_{z=-1}\left( \frac{\left( z^{1/2}\right) ^{m-2-ik}}{(1+z)^m}\right) \nonumber \\&\quad +\, (-1)^me^{k\pi } \int _0^\infty \frac{\left( r^{1/2}\right) ^{m-2-ik}}{(1+r)^m}\;\mathrm{d}r. \end{aligned}$$

This can now be solved for the integral:

$$\begin{aligned} \hat{f}_m(k) = \frac{1}{1-(-1)^me^{k\pi }} 2^m i\pi \; \mathrm {Res}_{z=-1}\left( \frac{\left( z^{1/2}\right) ^{m-2-ik}}{(1+z)^m}\right) . \end{aligned}$$

The residue can be computed as the coefficient of the Laurent expansion

$$\begin{aligned} \left( -z^{1/2}\right) ^a = \sum _{n=0}^\infty (-1)^n i^a (1+z)^n {{\frac{a}{2}}\atopwithdelims (){n}} \end{aligned}$$

for \(n=m-1\), namely

$$\begin{aligned} \mathrm {Res}_{z=-1}\left( \frac{\left( z^{1/2}\right) ^{m-2-ik}}{(1+z)^m}\right) = e^{\frac{\pi }{2}(k-im)}{{\frac{1}{2}(m-2-ik)}\atopwithdelims (){m-1}}. \end{aligned}$$

Plugging this into (B.5) gives

$$\begin{aligned} \hat{f}_m(k) = 2^{m-1}\pi e^{i\frac{3}{2}\pi (1-m)}{{\frac{1}{2}(m-2-ik)}\atopwithdelims (){m-1}} \cdot {\left\{ \begin{array}{ll} \frac{1}{\cosh \left( \frac{\pi }{2}k\right) } &{} \mathrm {if} \; m \;\mathrm {odd} \\ \frac{k}{\sinh \left( \frac{\pi }{2}k\right) } &{}\mathrm {if} \; m \;\mathrm {even} \end{array}\right. }.\nonumber \\ \end{aligned}$$

Finally, the following identity is straightforward to prove by induction \(m\rightarrow m+2\):

$$\begin{aligned} \prod _{j=1}^{m-1}\left( m-2j-ik\right) = e^{i\frac{3}{2}\pi (m-1)} \left( \prod _{\begin{array}{c} l=m-2\\ \mathrm {step\, -2} \end{array}}^1 (k^2 + l^2)\right) \cdot {\left\{ \begin{array}{ll} 1 &{} \mathrm {if} \; m \;\mathrm {odd} \\ k &{}\mathrm {if} \; m \;\mathrm {even} \end{array}\right. }\nonumber \\ \end{aligned}$$

Substituting (B.9) into (B.8) results in (B.1). \(\square \)

Appendix C. Sokhotski Integrals

The following proposition is the key ingredient in the proof of Proposition 2.23 in Appendix D. The proposition and its proof are adapted from [12, Ch. 1, §4], where a version of this theorem with contours of general shape but finite length is treated, and where the functions \(\varphi (z,t)\) below are constant in z.

Proposition C.1

Let \(D\subseteq \mathbb {C}\) be a complex domain such that \(\mathbb {S}_a\subseteq D\) for some \(a>0\). Let \(\varphi : D\times \mathbb {R} \rightarrow \mathbb {C}\) be a function with the following properties:

  1. 1.

    (Analyticity) For every \(t_0\in \mathbb {R}\), the function \(z\mapsto \varphi (z,t_0)\) is analytic in D.

  2. 2.

    (Hölder continuity) There exist \(0<\alpha \le 1\) and \(C>0\), such that for every \(z_0\in D\) the function \(t\mapsto \varphi (z_0,t)\) is \(\alpha \)-Hölder continuous with Hölder constant C.

  3. 3.

    (Decay) There exist \(\mu >0\) and \(T>0\), such that \(|\varphi (z,t)|\le |t-z|^{-\mu }\) for all \(z\in D\) and \(t\in \mathbb {R}\) with \(|t-z| \ge T\).

  4. 4.

    (Local majorisation) For every \(z_0\in D{\setminus } \mathbb {R}\) there exist a neighbourhood \(U\subseteq D{\setminus }\mathbb {R}\) and a function \(M\in L_1(\mathbb {R})\), such that \(|\varphi (z,t)|\le |z-t|M(t)\) for all \(z\in U\).

  5. 5.

    (Uniform convergence) The convergence \(\varphi (x\pm iy,t)\xrightarrow {y\searrow 0}\varphi (x,t)\) is uniform in \((x,t) \in \mathbb {R}^2\).

  6. 6.

    (Boundedness) \(\sup _{(x,t)\in \mathbb {R}^2}|\varphi (x,t)|<\infty \).

Then the function

$$\begin{aligned} F(z) = \int _{-\infty }^\infty \frac{\varphi (z,t)}{t-z}\;\mathrm{d}t \end{aligned}$$

is analytic in \(D{\setminus }\mathbb {R}\), and there exist limiting functions \(F^\pm :\mathbb {R}\rightarrow \mathbb {C}\) such that

$$\begin{aligned} F(x\pm iy)\rightarrow F^\pm (x) \end{aligned}$$

uniformly as \(y\searrow 0\). The functions \(F^\pm (x)\) are bounded and satisfy

$$\begin{aligned} F^+(x)-F^-(x) = 2i\pi \varphi (x,x) \qquad \text {for all} \quad x \in \mathbb {R}. \end{aligned}$$


  • F(z) is analytic on\(D {\setminus } \mathbb {R}\): Conditions 2 and 3 ensure that the integrand in (C.1) is always in \(L_1(\mathbb {R})\). Thus, F(z) is well defined. Analyticity of F(z) in \(D{\setminus }\mathbb {R}\) follows directly from lemma 2.19 together with condition 4.

  • The auxiliary function\(\psi \): Below we make frequent use of the following simple integral. Let \(x,y,\eta ,L \in \mathbb {R}\), \(\eta \ge 0\), \(\eta \pm x \le L\), and suppose that \(y \ne 0\) in case \(\eta =0\). Denote \(B_\eta (x)=(x-\eta ,x+\eta )\). Then

    $$\begin{aligned} \int _{[-L,L]{\setminus } B_\eta (x)}\frac{1}{t-x-iy} \;\mathrm{d}t ~=~ \log \frac{L-x-iy}{L+x+iy} ~+~ {\left\{ \begin{array}{ll} \log \frac{\eta +iy}{\eta -iy}; &{} \eta >0 \\ i \pi \, \mathrm {sgn}(y); &{} \eta = 0 \end{array}\right. }.\nonumber \\ \end{aligned}$$

    Here, the branch cut of the logarithm is placed along the negative real axis. We now investigate the \(y\searrow 0\) limit of \(F(x \pm iy)\). To do so, we split F(z) into two integrals by adding and subtracting a term in the integrand. Namely, for \(z \in D{\setminus }\mathbb {R}\) we have

    $$\begin{aligned} F(z) =\lim _{L\rightarrow \infty }\int _{-L}^L \frac{\varphi (z,t)-\varphi (z,\mathrm {Re}(z))}{t-z}\;\mathrm{d}t +\varphi (z,\mathrm {Re}(z)) \lim _{L\rightarrow \infty }\int _{-L}^L\frac{1}{t-z} \;\mathrm{d}t.\nonumber \\ \end{aligned}$$

    The improper integral

    $$\begin{aligned} \psi (z):=\lim _{L\rightarrow \infty }\int _{-L}^L \frac{\varphi (z,t)-\varphi (z,\mathrm {Re}(z))}{t-z}\;\mathrm{d}t \end{aligned}$$

    exists because by (C.4) the limit in the second summand of (C.5) exists. We obtain, for \(z \in D{\setminus }\mathbb {R}\),

    $$\begin{aligned} F(z)=\psi (z) + i\pi \, \varphi (z,\mathrm {Re}(z)) \, \mathrm {sgn}(\mathrm {Im}(z)). \end{aligned}$$

    Since both F(z) and \(\varphi (z,\mathrm {Re}(z))\) are continuous in \(D{\setminus } \mathbb {R}\), \(\psi (z)\) is also continuous in \(D{\setminus } \mathbb {R}\). We now claim that the integral and limit defining \(\psi (z)\) in (C.6) also exist for \(z \in \mathbb {R}\). To see this, first note that due to the Hölder condition (condition 2)

    $$\begin{aligned} \left| \frac{\varphi (x,t)-\varphi (x,x)}{t-x}\right| \le \frac{C}{|t-x|^{1-\alpha }} \quad \forall x,t\in \mathbb {R}. \end{aligned}$$

    Hence, the integral

    $$\begin{aligned} \int _{x-1}^{x+1} \frac{\varphi (x,t)-\varphi (x,x)}{t-x}\;\mathrm{d}t \end{aligned}$$

    exists. On the other hand, by (C.4) and for \(|x|+1<L\),

    $$\begin{aligned}&\int _{[-L,L]{\setminus } B_1(x)} \frac{\varphi (x,t)-\varphi (x,x)}{t-x}\;\mathrm{d}t \nonumber \\&\qquad \qquad ~=~ \int _{[-L,L]{\setminus } B_1(x)} \frac{\varphi (x,t)}{t-x}\;\mathrm{d}t -\varphi (x,x) \log \frac{L-x}{L+x}. \end{aligned}$$

    The integral in the first summand has a well-defined \(L \rightarrow \infty \) limit by condition 3 Adding (C.9) and (C.10) shows that the limit and integral in (C.6) exist also for \(z \in \mathbb {\mathbb {R}}\), so that altogether \(\psi \) is defined on all of D.

  • Uniform convergence of\(\psi \): Next we study the continuity properties of \(\psi \) on D. Let us restrict \(\psi (z)\) to lines parallel to the real axis. Namely, for \(|y|<a\) we define \(\psi ^{[y]}:\mathbb {R}\rightarrow \mathbb {C}\) by \(\psi ^{[y]}(x):=\psi (x+iy)\). We will show that \(\psi ^{[y]}(x)\) converges to \({\psi ^{[0]}(x)}\) uniformly as \(y\rightarrow 0\) (from both sides). To do so, it is convenient to define the family of functions

    $$\begin{aligned} \Delta ^{[y]}(x,t):=\varphi (x+iy,t)-\varphi (x,t). \end{aligned}$$

    Due to the uniform convergence condition on \(\varphi (x+iy,t)\) (condition 5), \(\Delta ^{[y]}(x,t)\) converges to 0 uniformly in \((x,t) \in \mathbb {R}^2\) as \(y\rightarrow 0\). In particular, for |y| small enough, \(\Delta ^{[y]}(x,t)\) is bounded. Moreover, \(\Delta ^{[y]}(x,t)\) inherits the decay property (condition 3) from \(\varphi (z,t)\). These properties will be used later in the proof.

    Choose \(\eta >0\) such that \(\eta <T\) and split the integration over the interval \([-L,L]\) (w.l.o.g. \(|x|+\eta <L\)) into the interval \(B_\eta (x)\) and its complement \([-L,L]{\setminus } B_\eta (x)\). A straightforward computation yields

    $$\begin{aligned} \psi ^{[y]}(x)-\psi ^{[0]}(x)&~=~ iy\int _{B_\eta (x)} \frac{\varphi (x,t)-\varphi (x,x)}{(t-x)(t-x-iy)}\;\mathrm{d}t \nonumber \\&\quad + iy\lim _{L\rightarrow \infty }\int _{[-L,L]{\setminus } B_\eta (x)} \frac{\varphi (x,t)-\varphi (x,x)}{(t-x)(t-x-iy)}\;\mathrm{d}t \nonumber \\&\quad + \int _{B_\eta (x)} \frac{\Delta ^{[y]}(x,t)-\Delta ^{[y]}(x,x)}{t-x-iy} \;\mathrm{d}t \nonumber \\&\quad + \lim _{L\rightarrow \infty }\int _{[-L,L]{\setminus } B_\eta (x)} \frac{\Delta ^{[y]}(x,t)-\Delta ^{[y]}(x,x)}{t-x-iy} \;\mathrm{d}t. \end{aligned}$$

    We will now show that all four integrals and limits exist and at the same time provide estimates for them. For the first three we compute, where (*) refers to the use of \(\alpha \)-Hölder continuity (condition 2) and (**) to boundedness (condition 6)—we set \(S := \sup _{(x,t)\in \mathbb {R}^2}|\varphi (x,t)|\),

    $$\begin{aligned}&\left| iy\int _{B_\eta (x)} \frac{\varphi (x,t)-\varphi (x,x)}{(t-x)(t-x-iy)}\;\mathrm{d}t\right| ~\le ~ \int _{B_\eta (x)} \left| \frac{\varphi (x,t)-\varphi (x,x)}{t-x}\right| \frac{|y|}{|t-x-iy|}\;\mathrm{d}t \nonumber \\&\quad \overset{(*)}{\le } ~C \int _{B_\eta (x)} |t-x|^{\alpha -1} \;\mathrm{d}t ~=~ 2C \int _0^\eta r^{\alpha -1} \;\mathrm{d}r ~=~ \frac{2C\eta ^\alpha }{\alpha }, \end{aligned}$$
    $$\begin{aligned}&\left| \int _{B_\eta (x)} \frac{\Delta ^{[y]}(x,t)-\Delta ^{[y]}(x,x)}{t-x-iy} \;\mathrm{d}t\right| \nonumber \\&\quad \le ~ \int _{B_\eta (x)} \frac{\left| \varphi (x+iy,t)-\varphi (x+iy,x)\right| +\left| \varphi (x,t)-\varphi (x,x)\right| }{\left| t-x\right| }\;\mathrm{d}t \nonumber \\&\quad \overset{(*)}{\le }~ 2C \int _{B_\eta (x)} |t-x|^{\alpha -1} \;\mathrm{d}t ~=~ 4C \int _0^\eta r^{\alpha -1} \;\mathrm{d}r = \frac{4C\eta ^\alpha }{\alpha }, \end{aligned}$$
    $$\begin{aligned}&\left| iy\lim _{L\rightarrow \infty }\int _{[-L,L]{\setminus } B_\eta (x)} \frac{\varphi (x,t)-\varphi (x,x)}{(t-x)(t-x-iy)}\;\mathrm{d}t \right| \nonumber \\&\quad ~\le ~ |y|\int _{\mathbb {R}{\setminus } B_\eta (x)} \frac{|\varphi (x,t)|+|\varphi (x,x)|}{\left| t-x\right| ^2}\;\mathrm{d}t \overset{(**)}{\le }~ 4 S |y| \eta ^{-1}. \end{aligned}$$

    Now let us turn to the fourth integral, which is slightly more involved. With the help of the decay condition on \(\Delta ^{[y]}(x,t)\) and (C.4), we can rewrite it as

    $$\begin{aligned}&\lim _{L\rightarrow \infty }\int _{[-L,L]{\setminus } B_\eta (x)} \frac{\Delta ^{[y]}(x,t)-\Delta ^{[y]}(x,x)}{t-x-iy} \;\mathrm{d}t \nonumber \\&\quad = \int _{\mathbb {R}{\setminus } B_\eta (x)} \frac{\Delta ^{[y]}(x,t)}{t-x-iy} \;\mathrm{d}t ~-~ \Delta ^{[y]}(x,x) \log \frac{\eta +iy}{\eta -iy}. \end{aligned}$$

    To estimate the integral over \(\mathbb {R}{\setminus } B_\eta (x)\), we split it as follows, for \(R > T\),

    $$\begin{aligned} \qquad \int _{\mathbb {R}{\setminus } B_\eta (x)} \frac{\Delta ^{[y]}(x,t)}{t-x-iy} \;\mathrm{d}t = \int _{\mathbb {R}{\setminus } B_R(x)} \frac{\Delta ^{[y]}(x,t)}{t-x-iy} \;\mathrm{d}t + \int _{B_R(x){\setminus } B_\eta (x)} \frac{\Delta ^{[y]}(x,t)}{t-x-iy} \;\mathrm{d}t.\nonumber \\ \end{aligned}$$

    We now estimate the two integrals separately:

    $$\begin{aligned}&\quad \left| \int _{\mathbb {R}{\setminus } B_R(x)} \frac{\Delta ^{[y]}(x,t)}{t-x-iy} \;\mathrm{d}t\right| \le \int _{\mathbb {R}{\setminus } B_R(x)} \left| \frac{\Delta ^{[y]}(x,t)}{t-x}\right| \;\mathrm{d}t \nonumber \\&\qquad \le \int _{\mathbb {R}{\setminus } B_R(x)} \frac{1}{|t-x|^{1+\mu }} \;\mathrm{d}t = 2\int _R^\infty \frac{1}{r^{1+\mu }} \;\mathrm{d}r = \frac{2}{\mu R^\mu } \end{aligned}$$
    $$\begin{aligned}&\quad \left| \int _{B_R(x){\setminus } B_\eta (x)} \frac{\Delta ^{[y]}(x,t)}{t-x-iy} \;\mathrm{d}t\right| \le \int _{B_R(x){\setminus } B_\eta (x)} \left| \frac{\Delta ^{[y]}(x,t)}{t-x}\right| \;\mathrm{d}t \nonumber \\&\qquad \le \frac{1}{\eta }\int _{x-R}^{x+R} |\Delta ^{[y]}(x,t)| \;\mathrm{d}t \le \frac{2R}{\eta } \sup _{t\in \mathbb {R}}|\Delta ^{[y]}(x,t)| \le \frac{2R}{\eta } \sup _{t,x\in \mathbb {R}}|\Delta ^{[y]}(x,t)| \end{aligned}$$

    Finally, we remark that with our choice of branch cut for the logarithm,

    $$\begin{aligned} \left| \log \tfrac{\eta +iy}{\eta -iy} \right| = 2 \left| \mathrm {arg}(\eta + iy) \right| \le \pi . \end{aligned}$$

    Assembling all of the above estimates, we obtain:

    $$\begin{aligned} \quad \left| \psi ^{[y]}(x)-\psi ^{[0]}(x)\right| \le \frac{6C\eta ^\alpha }{\alpha } + \frac{4 S |y|}{\eta } + \frac{2}{\mu R^\mu } + \left( \frac{2R}{\eta }+\pi \right) \sup _{x,t\in \mathbb {R}}|\Delta ^{[y]}(x,t)|\nonumber \\ \end{aligned}$$

    To establish uniform convergence, we need to show that for each \(\varepsilon >0\) there exists a \(\delta >0\) such that for all \(|y|<\delta \) and all \(x \in \mathbb {R}\) we have \(|\psi ^{[y]}(x)-\psi ^{[0]}(x)| \le \varepsilon \). To find \(\delta \), we choose \(\eta \) and R in the above estimate appropriately. Choose \(\eta \) such that the first term in (C.21) equals \(\varepsilon /4\):

    $$\begin{aligned} \eta =\left( \frac{\alpha \varepsilon }{24C}\right) ^{\frac{1}{\alpha }}. \end{aligned}$$

    The second term is smaller than \(\varepsilon /4\) provided \(|y|<\delta _1\), where

    $$\begin{aligned} \delta _1= \frac{\eta \, \varepsilon }{16 \, S} = \frac{1}{16 S} \left( \frac{\alpha }{24C}\right) ^{\frac{1}{\alpha }} \varepsilon ^{1+\frac{1}{\alpha }}. \end{aligned}$$

    The third term is smaller than \(\varepsilon /4\) if we set

    $$\begin{aligned} R=\left( \frac{8}{\mu \varepsilon }\right) ^{\frac{1}{\mu }}. \end{aligned}$$

    Finally, we remember that \(\Delta ^{[y]}(x,t)\xrightarrow {y\rightarrow 0}0\) uniformly in x and t. Hence, there exists a \(\delta _2>0\) such that for all \(|y|< \delta _2\),

    $$\begin{aligned} \qquad \sup _{x,t\in \mathbb {R}}|\Delta ^{[y]}(x,t)| \le \frac{\varepsilon }{4}\left( \frac{2R}{\eta }+\pi \right) ^{-1} = \frac{\varepsilon }{4}\left( 2\left( \frac{24C}{\alpha \varepsilon }\right) ^{\frac{1}{\alpha }}\left( \frac{8}{\mu \varepsilon }\right) ^{\frac{1}{\mu }}+\pi \right) ^{-1}.\nonumber \\ \end{aligned}$$

    This makes the last term smaller than \(\varepsilon /4\). Setting \(\delta := \min (\delta _1,\delta _2)\), this proves uniform convergence \(\psi ^{[y]} \xrightarrow {y \rightarrow 0} \psi ^{[0]}\).

  • Uniform convergence ofFand relation of\(F^\pm \): The claim of uniform convergence of (C.2) and formula (C.3) now both follow from (C.7).

  • Boundedness of\(F^\pm \): It is enough to provide a bound for \(\psi ^{[0]}(x)\). This can be achieved as follows. Split the integral:

    $$\begin{aligned} \int _{-L}^L \frac{\varphi (x,t)-\varphi (x,x)}{t-x}\;\mathrm{d}t&= \int _{B_1(x)} \frac{\varphi (x,t)-\varphi (x,x)}{t-x}\;\mathrm{d}t \nonumber \\&\quad + \int _{[-L,L]{\setminus } B_1(x)}\frac{\varphi (x,t)}{t-x}\; \mathrm{d}t \nonumber \\&\quad -~ \varphi (x,x)\int _{[-L,L]{\setminus } B_1(x)}\frac{1}{t-x}\; \mathrm{d}t \end{aligned}$$

    The first summand can be estimated using the Hölder inequality (condition 2)

    $$\begin{aligned} \left| \int _{B_1(x)} \frac{\varphi (x,t)-\varphi (x,x)}{t-x}\;\mathrm{d}t\right| \le C\int _{-1}^1 \left| r\right| ^{\alpha -1}\;\mathrm{d}r = \frac{2C}{\alpha } \end{aligned}$$

    For the second integral, we make use of both boundedness (condition 6, where as above we denote \(S:=\sup _{(x,t)\in \mathbb {R}^2}|\varphi (x,t)|\)) and the decay property (condition 3, w.l.o.g. \(1<T<L\)):

    $$\begin{aligned}&\left| \int _{[-L,L]{\setminus } B_1(x)} \frac{\varphi (x,t)}{t-x}\; \mathrm{d}t\right| \nonumber \\&\quad \le \int _{B_T(x){\setminus } B_1(x)} \left| \frac{\varphi (x,t)}{t-x}\right| \; \mathrm{d}t + \int _{[-L,L]{\setminus } B_T(x)} \left| \frac{\varphi (x,t)}{t-x}\right| \; \mathrm{d}t \nonumber \\&\quad \le \int _{B_T(x){\setminus } B_1(x)} \left| \varphi (x,t)\right| \; \mathrm{d}t + \int _{[-L,L]{\setminus } B_T(x)} \left| t-x\right| ^{-\mu -1}\; \mathrm{d}t \nonumber \\&\quad \le 2TS + 2\int _1^\infty r^{-\mu -1}\; \mathrm{d}r = 2TS+\frac{1}{\mu } \end{aligned}$$

    By (C.4), the third integral is simply bounded as follows:

    $$\begin{aligned} \left| -\varphi (x,x)\int _{[-L,L]{\setminus } B_1(x)} \frac{1}{t-x}\; \mathrm{d}t\right| \le S\left| \log \left( \frac{L-x}{L+x}\right) \right| \end{aligned}$$

    Altogether, we obtain the bound

    $$\begin{aligned} \left| \int _{-L}^L \frac{\varphi (x,t)-\varphi (x,x)}{t-x}\;\mathrm{d}t\right| \le \frac{2C}{\alpha } +\left( 2T+\left| \log \left( \frac{L-x}{L+x}\right) \right| \right) S+\frac{1}{\mu }.\nonumber \\ \end{aligned}$$

    Since this bound itself converges as \(L\rightarrow \infty \), we obtain a bound for \(\left| \psi ^{[0]}(x)\right| \). \(\square \)

Write \(D^+ := \{z\in D | \mathrm {Im}(z) > 0 \}\), \(D^- := \{ z\in D | \mathrm {Im}(z) < 0 \}\) and \({\tilde{D}}^\pm := D^\pm \cup \mathbb {R}\). Consider the functions

$$\begin{aligned} {\tilde{F}}^\pm : {\tilde{D}}^\pm \rightarrow \mathbb {C}, \quad {\tilde{F}}^\pm (z) := {\left\{ \begin{array}{ll} F(z); &{}~ \mathrm {Im}(z) \ne 0\\ F^\pm (z); &{}~ \mathrm {Im}(z) = 0 \end{array}\right. }. \end{aligned}$$

Corollary 5.1

\({\tilde{F}}^\pm \) is a continuous extension of F from \(D^\pm \) to \({\tilde{D}}^\pm \).


It suffices to show that \(\psi (z)\) is continuous in D. We know already that it is continuous in \(D{\setminus } \mathbb {R}\). Now let us show that it is continuous in \(x_0\in \mathbb {R}\).

Let \(\varepsilon >0\). By uniform convergence \(\psi ^{[y]}(x)\rightarrow \psi ^{[0]}(x)\), there is a \(\delta _1>0\) such that for all \(|y|<\delta _1\) and all \(x \in \mathbb {R}\) we have \(|\psi ^{[y]}(x)-\psi ^{[0]}(x)|<\varepsilon /2\). By continuity of \(F^\pm \) on \(\mathbb {R}\), there is \(\delta _2>0\) such that for all x with \(|x-x_0| < \delta _2\) we have \(|\psi ^{[0]}(x)-\psi ^{[0]}(x_0)|<\varepsilon /2\).

Take \(\delta := \min (\delta _1,\delta _2)\). For \(z = x + iy \in D\) with \(|z-x_0|<\delta \), we have

$$\begin{aligned} \left| \psi (z)-\psi (x_0)\right| =\left| \psi ^{[y]}(x)-\psi ^{[0]}(x_0)\right| \le |\psi ^{[y]}(x)-\psi ^{[0]}(x)|+|\psi ^{[0]}(x)-\psi ^{[0]}(x_0)|\le \varepsilon . \end{aligned}$$

\(\square \)

Appendix D. Proof of Proposition 2.23

Here we prove Proposition 2.23 as a special case of Proposition C.1 and Corollary 5.1. Recall the setting of Proposition 2.23:

  • \(D=\mathbb {S}_a\) for some \(a>0\).

  • \(\varphi (z,t)=h(z-t)g(t)\) where \(h:\mathbb {S}_a\rightarrow \mathbb {C}\) is analytic and \(g:\mathbb {R}\rightarrow \mathbb {C}\) is bounded and Hölder continuous.

  • zh(z) and \(\frac{d}{dz}h(z)\) are bounded in \(\mathbb {S}_a\).

Let us check the conditions on \(\varphi (z,t)\) one by one. Denote by G the bound of g(t) and by H the bound of zh(z).

Condition 1 This is obvious since h(z) is analytic.

Condition 2 Let \(C,\alpha \) be constants expressing the Hölder continuity of g:

$$\begin{aligned} \forall t,t' \in \mathbb {R}~:~ |g(t')-g(t)| \le C |t-t'|^\alpha . \end{aligned}$$

Boundedness of \(\frac{d}{dz}h(z)\) implies that, in particular, the derivative \(\frac{d}{dx}h(x+iy)\) in the real direction is bounded for every \(y<(-a,a)\). As a consequence, for any given \(z_0\in \mathbb {S}_a\) the function \(t\mapsto h(z_0-t)\) is Lipschitz continuous with Lipschitz constant \(L := \mathrm {sup}_{z \in \mathbb {S}_a}(\frac{d}{dz}h(z))\). Since zh(z) is bounded, we also have boundedness of h on \(\mathbb {S}_a\): \(|h(z)|\le B\) for some \(B\ge 0\). Accordingly,

$$\begin{aligned}&|\varphi (z_0,t)-\varphi (z_0,t')| = |h(z_0-t)g(t)-h(z_0-t')g(t')| \nonumber \\&\qquad \qquad \le |h(z_0-t)-h(z_0-t')|\,|g(t)| + |h(z_0-t')|\,|g(t)-g(t')| \nonumber \\&\qquad \qquad \le BC\, |t-t'|^\alpha + {\left\{ \begin{array}{ll} GL \, |t-t'|; &{}~ |t-t'| \le 1\\ 2BG; &{} ~ |t-t'|>1 \end{array}\right. } \quad . \end{aligned}$$

Note that BGL are all independent of \(z_0\). Hence, there is an \(M>0\), independent of \(z_0\), such that for all \(t,t' \in \mathbb {R}\):

$$\begin{aligned} |\varphi (z_0,t)-\varphi (z_0,t')| \le M |t-t'|^\alpha . \end{aligned}$$

Condition 3 For all \(z_0\in \mathbb {S}_a\), we have the inequality

$$\begin{aligned} |\varphi (z_0,t)|\le G|h(z_0-t)| \le GH\,|z_0-t|^{-1}. \end{aligned}$$

Fix \(0<\mu <1\). There exists a \(T>0\), independent of \(z_0\), such that \(GH\,|z_0-t|^{-1}\le |z_0-t|^{-\mu }\) whenever \(|z_0-t|>T\).

Condition 4 For given \(z_0\in D{\setminus } \mathbb {R}\), set \(\rho :=\frac{1}{2}\mathrm {Im}(z_0)\) and \(U := \mathbb {S}_a \cap B_\rho (z_0)\), where \(B_\rho (z_0)\) is the open ball of radius \(\rho \) with centre \(z_0\). Then for all \(z\in U\), we have

$$\begin{aligned} |\varphi (z,t)| \le G|h(z-t)| \le GH\frac{1}{|z-t|} = |z-t|\frac{GH}{|z-t|^2}. \end{aligned}$$

Now set

$$\begin{aligned} M(t):={\left\{ \begin{array}{ll} \frac{GH}{|\mathrm {Re}(z_0)-\rho -t|^2} &{} \mathrm {if}\; t<\mathrm {Re}(z_0)-2\rho \\ \frac{GH}{\rho ^2}&{} \mathrm {if}\; t\in [\mathrm {Re}(z_0)-2\rho ,\mathrm {Re}(z_0)+2\rho ]\\ \frac{GH}{|\mathrm {Re}(z_0)+\rho -t|^2} &{} \mathrm {if}\; t>\mathrm {Re}(z_0)+2\rho \end{array}\right. }. \end{aligned}$$

Then \(M\in L_1(\mathbb {R})\) and one quickly checks that

$$\begin{aligned} \frac{GH}{|z-t|^2}\le M(t). \end{aligned}$$

Condition 5 Since \(\varphi (x\pm iy,t)=h(x\pm iy -t)g(t)\), the condition is satisfied if \(h(x+iy)\xrightarrow {y\rightarrow 0}h(x)\) uniformly. This is easily established with the following lemma.

Lemma D.1

Let \(a>0\), \(b\in \mathbb {R}\), and set \(D:=\mathbb {S}_a\cap \lbrace z\in \mathbb {C}|\mathrm {Re}(z)> b\rbrace \). Let \(f:D\rightarrow \mathbb {C}\) be an analytic function such that zf(z) and \(\frac{d}{dz}f(z)\) are bounded. Then \(f(x+iy)\xrightarrow {y\rightarrow Y}f(x+iY)\) uniformly on \([t,\infty )\) for any \(Y\in (-a,a)\) and any \(t>b\).


Pointwise convergence is clear by continuity. Now we claim that the convergence is uniform. Let \(\varepsilon >0\). Set \(x_0:=\frac{2B}{\varepsilon }\), where \(B>0\) is the bound of zf(z). Without loss of generality, assume \(b<t<x_0\). Then for all \(x\ge x_0\) one has

$$\begin{aligned} \left| f(x+iy)-f(x+iY)\right|&\le \left| f(x+iy)\right| +\left| f(x+iY)\right| \\&\le \frac{B}{|x+iy|}+\frac{B}{|x+iY|} \le \frac{2B}{|x|} \le \frac{2B}{x_0} =\varepsilon \end{aligned}$$

for all \(|y|<a\).

Thus, it remains to be shown that convergence on the compact interval \([t,x_0]\) is uniform. Since f(z) is bounded, the family of functions \(f^{[y]}(x):=f(x+iy)\) on the interval \([t,x_0]\) is uniformly bounded. Moreover, boundedness of \(\frac{d}{dz}f(z)\) means that \(\frac{d}{dx}f^{[y]}(x)\) are uniformly bounded. But this implies that \(f^{[y]}(x)\) are equicontinuous. Thus, we can apply the Arzela–Ascoli theorem: for every sequence \(y_n\rightarrow Y\), the sequence of functions \(f^{[y_n]}(x)\) has a uniformly convergent subsequence. Now assume that \(f^{[y]}(x)\) do not converge uniformly on \([t,x_0]\). Then there exists a sequence \(u_n\rightarrow Y\) and a sequence \(x_n\) of points in \([t,x_0]\) such that \(\left| f^{[u_n]}(x_n)-f(x_n)\right| \ge \varepsilon \) for all n. But then \(f^{[u_n]}(x)\) has no uniformly convergent subsequence, which is a contradiction. \(\square \)

Condition 6 This is again obvious, since both h(z) and g(t) are bounded.

This completes the proof of Proposition 2.23.

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Hilfiker, L., Runkel, I. Existence and Uniqueness of Solutions to Y-Systems and TBA Equations. Ann. Henri Poincaré 21, 941–991 (2020). https://doi.org/10.1007/s00023-019-00866-4

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