Asymptotics for Toeplitz determinants: perturbation of symbols with a gap

Theorem 1.1. Let $\theta_0 \in (0, \pi)$ and define $$x_c = -2\ln\tan\frac{\theta_0}{4}.\tag{1.7}$$ Let f be of the form (1.2) with W analytic in a neighbourhood of the unit circle. As $n \to \infty$ and simultaneously $s \to 0$ in such a way that $0 \le s \le e^{-x_c n}$ , we have $$\ln D_n(s, \theta_0, W) = \ln D_n(0, \theta_0, W) + o(1). \tag{1.8}$$ The error term o(1) is uniform for $0 \le s \le e^{-x_c n}$ and $\epsilon \le \theta_0 \le \pi - \epsilon$ , $\epsilon > 0$ , and can be specified as $$o(1) = \mathcal{O}(n^{-1/2}e^{x_c n}s). \tag{1.9}$$  Figure 1: As $n \to \infty$ , $\ln D_n(s, \theta_0, W)$ follows Widom asymptotics for s = 0, Fisher-Hartwig asymptotics for $s \in (0, 1)$ fixed, and Szegő asymptotics for s = 1. Theorem 1.1 implies that the Widom asymptotics remain valid as $n \to \infty$ with s = s(n) below the curve $s = e^{-x_c n}$ . As $n \to \infty$ with s = s(n) above the curve, a different type of asymptotic behavior is expected. In addition, the result extends to the case where $\theta_0$ approaches $\pi$ at a sufficiently slow rate: (1.8) holds for $\epsilon < \theta_0 < \pi - \frac{M}{n}$ with M sufficiently large and $0 \le s \le e^{-x_c n}$ , with the error term given by $$o(1) = \mathcal{O}((\pi - \theta_0)^{1/2} n^{-1/2} e^{nx_c} s). \tag{1.10}$$ Remark 1.2 We believe the bound $s \leq e^{-x_c n}$ is sharp: as $n \to \infty$ , $s \to 0$ with $s > e^{-x_c n}$ , the asymptotic behavior for $\ln D_n(s, \theta_0, W)$ is expected to be described in terms of elliptic $\theta$ -functions. For a heuristic discussion, see Section 5.4. Remark 1.3 If f is positive on $\gamma$ and even in $\theta$ , the asymptotics for $\ln D_n(0, \theta_0, W)$ in (1.8) are given by (1.3). The perturbative result (1.8) is valid for general analytic W, even if the positivity and symmetry conditions needed for the Widom asymptotics do not hold. Note that the error term $o(1) = \mathcal{O}((\pi - \theta_0)^{1/2} n^{-1/2} e^{nx_c} s)$ in (1.8) improves as $\theta_0$ approaches $\pi$ . This is reasonable since the perturbation of the arc-supported symbol then takes place on a shrinking arc near -1. For our proof, it is however crucial that $n(\pi - \theta_0)$ is sufficiently large.

Proposition 2.1. Let $D_n(s, \theta_0, 0)$ be the Toeplitz determinant with symbol (1.2) in the case where $W(e^{i\theta}) = 0$ . We have the following differential identity for $\ln D_n$ : $$\partial_s \ln D_n(s, \theta_0, 0) = -2n \frac{\partial_s \chi_n}{\chi_n} + \frac{2(1-s)}{\pi} \operatorname{Im} \left( \overline{\phi_n(e^{i\theta_0})} \partial_s \phi_n(e^{i\theta_0}) \right). \tag{2.7}$$ Proof. Taking the logarithm of both sides in (2.4), and then differentiating with respect to s, we get $$\partial_s \ln D_n(s, \theta_0, 0) = -2 \sum_{j=0}^{n-1} \frac{\partial_s \chi_j}{\chi_j}.$$ On the other hand, by (2.1), $$\frac{1}{2\pi} \int_0^{2\pi} \partial_s \left( \phi_j(e^{i\theta}) \overline{\phi_j(e^{i\theta})} \right) f(e^{i\theta}) d\theta = 2 \frac{\partial_s \chi_j}{\chi_j},$$ and this gives $$\partial_s \ln D_n(s, \theta_0, 0) = -\frac{1}{2\pi} \int_0^{2\pi} \partial_s \left[ \sum_{j=0}^{n-1} \phi_j(e^{i\theta}) \overline{\phi_j(e^{i\theta})} \right] f(e^{i\theta}) d\theta. \tag{2.8}$$ Here we can use the Christoffel-Darboux formula (see e.g. [21, 8] for a proof of it): $$\sum_{i=0}^{n-1} \phi_j(z)\overline{\phi_j}(z^{-1}) = -n\phi_n(z)\overline{\phi_n}(z^{-1}) + z\left(\overline{\phi_n}(z^{-1})\phi_n'(z) - \left(\overline{\phi_n}(z^{-1})\right)'\phi_n(z)\right).$$ Substituting this into (2.8), we obtain $$\partial_{s} \ln D_{n}(s, \theta_{0}, 0) = 2n \frac{\partial_{s} \chi_{n}}{\chi_{n}}$$ $$- \frac{1}{2\pi} \int_{0}^{2\pi} \partial_{s} \left[ e^{i\theta} \left( \overline{\phi_{n}}(e^{-i\theta}) \phi'_{n}(e^{i\theta}) - \left( \overline{\phi_{n}}(z^{-1}) \right)' \Big|_{z=e^{i\theta}} \phi_{n}(e^{i\theta}) \right) \right] f(e^{i\theta}) d\theta$$ $$= 2n \frac{\partial_{s} \chi_{n}}{\chi_{n}} + I_{1} + I_{2} + I_{3} + I_{4}, \tag{2.9}$$ where $$I_1 = -\frac{1}{2\pi} \int_0^{2\pi} e^{i\theta} \partial_s \left( \overline{\phi_n}(e^{-i\theta}) \right) \phi_n'(e^{i\theta}) f(e^{i\theta}) d\theta, \tag{2.10}$$ $$I_2 = -\frac{1}{2\pi} \int_0^{2\pi} e^{i\theta} \overline{\phi_n}(e^{-i\theta}) \partial_s \left( \phi_n'(e^{i\theta}) \right) f(e^{i\theta}) d\theta, \tag{2.11}$$ $$I_3 = \frac{1}{2\pi} \int_0^{2\pi} e^{i\theta} \partial_s \left( \left( \overline{\phi_n}(z^{-1}) \right)' \Big|_{z=e^{i\theta}} \right) \phi_n(e^{i\theta}) f(e^{i\theta}) d\theta, \tag{2.12}$$ $$I_4 = \frac{1}{2\pi} \int_0^{2\pi} e^{i\theta} \left( \overline{\phi_n}(z^{-1}) \right)' \Big|_{z=e^{i\theta}} \partial_s \phi_n(e^{i\theta}) f(e^{i\theta}) d\theta. \tag{2.13}$$ From the orthogonality relation (2.1), we easily get $$I_2 = I_3 = -n \frac{\partial_s \chi_n}{\chi_n}. (2.14)$$ The computation of $I_1$ and $I_4$ is slightly more involved. Using (1.2), we have $$I_{1} = -\frac{1}{2\pi i} \int_{0}^{2\pi} \partial_{s} \left( \overline{\phi_{n}}(e^{-i\theta}) \right) \frac{d}{d\theta} \left( \phi_{n}(e^{i\theta}) \right) d\theta + \frac{1-s}{2\pi i} \int_{\theta_{0}}^{2\pi-\theta_{0}} \partial_{s} \left( \overline{\phi_{n}}(e^{-i\theta}) \right) \frac{d}{d\theta} \left( \phi_{n}(e^{i\theta}) \right) d\theta. \quad (2.15)$$ Integrating by parts and then using orthogonality, we obtain $$I_{1} = \frac{1-s}{2\pi i} \left[ \partial_{s} \left( \overline{\phi_{n}}(e^{-i\theta}) \right) \phi_{n}(e^{i\theta}) \right]_{\theta_{0}}^{2\pi-\theta_{0}} + \frac{1}{2\pi i} \int_{0}^{2\pi} \phi_{n}(e^{i\theta}) \partial_{s} \frac{d}{d\theta} \left( \overline{\phi_{n}}(e^{-i\theta}) \right) f(e^{i\theta}) d\theta$$ $$= \frac{1-s}{2\pi i} \left[ \partial_{s} \left( \overline{\phi_{n}}(e^{-i\theta}) \right) \phi_{n}(e^{i\theta}) \right]_{\theta_{0}}^{2\pi-\theta_{0}} - n \frac{\partial_{s} \chi_{n}}{\chi_{n}}. \tag{2.16}$$ In the same way, we show that $$I_{4} = \frac{1}{2\pi} \int_{0}^{2\pi} e^{i\theta} \left( \overline{\phi_{n}}(z^{-1}) \right)' \Big|_{z=e^{i\theta}} \partial_{s} \phi_{n}(e^{i\theta}) f(e^{i\theta}) d\theta$$ $$= -\frac{1-s}{2\pi i} \left[ \overline{\phi_{n}}(e^{-i\theta}) \partial_{s} \phi_{n}(e^{i\theta}) \right]_{\theta_{0}}^{2\pi-\theta_{0}} - n \frac{\partial_{s} \chi_{n}}{\chi_{n}}.$$ (2.17) Summing up (2.14), (2.16), and (2.17), and using the fact that $\overline{\phi_n} = \phi_n$ if W = 0, we get the result. As a consequence of Proposition 2.1 and (2.5), we can express the right hand side of (2.7) in terms of $Y = Y^{(n)}$ given by (2.5):

Proposition 3.1. (a) For $x = +\infty$ , the support of the equilibrium measure and its density are given by $$J = J^{(\infty)} = \gamma, \qquad u(e^{i\theta}) = u^{(\infty)}(e^{i\theta}) = \frac{1}{2\pi} \sqrt{\frac{\cos \theta + 1}{\cos \theta - \cos \theta_0}}.$$ (3.6) The constant $\ell = \ell^{(\infty)}$ in the variational conditions (3.4)-(3.5) is given by $$\ell^{(\infty)} = -2\ln\sin\frac{\theta_0}{2},\tag{3.7}$$ and the variational inequality (3.5) is strict for $z \in S_1 \setminus \gamma$ . (b) Let $x_c$ be given by (1.7). For $x \ge x_c$ , we have the same result as for $x = +\infty$ : $$J = \gamma$$ , $u(e^{i\theta}) = u^{(\infty)}(e^{i\theta})$ , $\ell = \ell^{(\infty)}$ . (3.8) Moreover, the variational inequality (3.5) is strict for $z \in S_1 \setminus \gamma$ if $x \neq x_c$ ; if $x = x_c$ , it is strict for $z \in S_1 \setminus (\gamma \cup \{-1\})$ , and there is equality for z = -1. (c) For $0 < x < x_c$ , the support of $\mu$ consists of two disjoint arcs: we have $$J = \{e^{i\theta} : \theta \in [-\theta_0, \theta_0] \cup [\pi - \theta_1, \pi + \theta_1]\}, \tag{3.9}$$ where $\theta_1 = \theta_1(x) \in (0, \pi - \theta_0)$ is the unique solution of $$2\int_{[-\theta_0,\theta_0]\cup[\pi-\theta_1,\pi+\theta_1]} \log \left| \frac{1+e^{i\theta}}{1-e^{i\theta}} \right| u(e^{i\theta})d\theta = x, \tag{3.10}$$ and the density is given by $$u(e^{i\theta}) = \frac{1}{2\pi} \sqrt{\frac{\cos\theta + \cos\theta_1}{\cos\theta - \cos\theta_0}}.$$ (3.11) The constant $\ell$ is given by $$\ell = -\int_0^1 \frac{1}{s} \left( 1 - \sqrt{\frac{s^2 + 2\cos(\theta_1)s + 1}{s^2 - 2\cos(\theta_0)s + 1}} \right) ds, \tag{3.12}$$ and the variational inequality (3.5) is strict for $z \in S_1 \setminus J$ . Remark 3.2 Although part (c) of the proposition is not needed for the proof of Theorem 1.1, we present it here for completeness and to support the heuristic arguments in Section 5.4, where we will discuss asymptotics for $D_n$ if $x < x_c$ . Proof. Note first that the equation (3.10) has indeed a unique solution $\theta_1$ for $x < x_c$ , since the function $$\theta_1 \mapsto 2 \int_{[-\theta_0,\theta_0] \cup [\pi-\theta_1,\pi+\theta_1]} \log \left| \frac{1+e^{i\theta}}{1-e^{i\theta}} \right| u(e^{i\theta}) d\theta$$ decreases as a function of $\theta_1 \in (0, \pi - \theta_0)$ , and is bijective from $(0, \pi - \theta_0)$ to $(0, x_c)$ . The equilibrium measure $\mu$ is uniquely characterized by the conditions (3.4)-(3.5), which means that it is sufficient for us to show that the measure $\mu$ defined in cases (a), (b), and (c) satisfy these variational conditions. To do so, consider the function $$f(z) = 2 \int \log|z - e^{i\theta}| d\mu^{(x)}(e^{i\theta}).$$ (3.13) If we define $\theta_1 > 0$ as the unique solution of (3.10) if $x < x_c$ , and if we let $\theta_1 = 0$ for $x \ge x_c$ , we have $$f(e^{i\alpha}) = \frac{1}{\pi} \int_{[-\theta_0, \theta_0] \cup [\pi - \theta_1, \pi + \theta_1]} \log |e^{i\alpha} - e^{i\theta}| \sqrt{\frac{\cos \theta + \cos \theta_1}{\cos \theta - \cos \theta_0}} d\theta.$$ (3.14) The derivative $\frac{d}{d\alpha}f(e^{i\alpha})$ can be written as a contour integral $$\frac{d}{d\alpha}f(e^{i\alpha}) = -\frac{1}{2\pi i} \int_{\Sigma} \frac{1}{\xi} \frac{\xi + e^{i\alpha}}{\xi - e^{i\alpha}} \left( \frac{(\xi - z_1)(\xi - \overline{z_1})}{(\xi - z_0)(\xi - \overline{z_0})} \right)^{1/2} d\xi, \qquad z_1 = e^{i\theta_1}, \qquad (3.15)$$ where the square root is analytic off J and tends to 1 as $\xi \to \infty$ , and where the contour $\Sigma$ consists of one (if $x \geq x_c$ ) or two (if $x < x_c$ ) counterclockwise oriented circles around the arc(s) of J. If $e^{i\alpha} \in S_1 \setminus J$ , the contour has to be chosen sufficiently small such that $e^{i\alpha}$ lies in the exterior of $\Sigma$ . If $e^{i\alpha} \in J$ , a residue calculation shows that $\frac{d}{d\alpha}f(e^{i\alpha}) = 0$ . If $e^{i\alpha} \in S_1 \setminus J$ on the other hand, we have $$\frac{d}{d\alpha}f(e^{i\alpha}) = \begin{cases} \sqrt{\frac{\cos\theta_1 + \cos\alpha}{\cos\theta_0 - \cos\alpha}}, & \text{if } \theta_0 < \alpha < \pi - \theta_1, \\ -\sqrt{\frac{\cos\theta_1 + \cos\alpha}{\cos\theta_0 - \cos\alpha}}, & \text{if } \pi + \theta_1 < \alpha < 2\pi - \theta_0. \end{cases}$$ (3.16) It follows that $f(e^{i\alpha})$ is constant on $[-\theta_0, \theta_0] \cup [\pi - \theta_1, \pi + \theta_1]$ , and that it achieves its maximum on $[\pi - \theta_1, \pi + \theta_1]$ (i.e. at $\pi$ if $x \geq x_c$ ). We have $$f(e^{i\alpha}) = f(1),$$ for $-\theta_0 \le \alpha \le \theta_0,$ (3.17) $$f(e^{i\alpha}) = f(-1),$$ for $\pi - \theta_1 \le \alpha \le \pi + \theta_1,$ (3.18) $$f(e^{i\alpha}) < f(-1),$$ for $\alpha \notin [\pi - \theta_1, \pi + \theta_1].$ (3.19) If we show that $$f(1) = -\ell, \ f(-1) < f(1) + x,$$ for $x > x_c,$ (3.20) $$f(1) = -\ell, \ f(-1) = f(1) + x,$$ for $x \le x_c,$ (3.21) then (3.17)-(3.19) imply the Euler-Lagrange conditions (3.4)-(3.5). For the case x ≥ xc, using (3.16), we obtain after a straightforward calculation, $$f(-1) - f(1) = f(e^{i\pi}) - f(e^{i\theta_0}) = \int_{\theta_0}^{\pi} \sqrt{\frac{1 + \cos \theta}{\cos \theta_0 - \cos \theta}} d\theta = -2 \ln \tan \frac{\theta_0}{4} = x_c \le x,$$ which proves the inequality in (3.20), and the fact that there is equality at −1 if x = xc. Moreover, using an other residue calculation, we get $$f(1) = \int_0^1 f'(s)ds = \int_0^1 \frac{1}{s} \left( 1 - \frac{1+s}{\sqrt{s^2 - 2\cos(\theta_0)s + 1}} \right) ds = 2\ln\sin\frac{\theta_0}{2} = -\ell^{(\infty)}.$$ This proves the equality in (3.20). For 0 < x < xc, by (3.16), $$f(-1) - f(1) = f(e^{i(\pi - \theta_1)}) - f(e^{i\theta_0}) = \int_{\theta_0}^{\pi - \theta_1} \sqrt{\frac{\cos \theta_1 + \cos \theta}{\cos \theta_0 - \cos \theta}} d\theta = x,$$ by definition of f and θ1, and $$f(1) = \int_0^1 \frac{1}{s} \left( 1 - \sqrt{\frac{s^2 + 2\cos(\theta_1)s + 1}{s^2 - 2\cos(\theta_0)s + 1}} \right) ds = -\ell.$$ This completes the proof. ✷

Proposition 5.1. Let W = 0. As $n \to \infty$ with $x > x_c$ , we have $$Y_{12}(0) = e^{-n\ell} \left[ \sin \frac{\theta_0}{2} + \mathcal{O}(n^{-1/2}) \right], \tag{5.9}$$ $$\partial_s \ln Y_{12}(0) = \mathcal{O}(n^{-1/2}e^{nx_c}),$$ (5.10) $$Y_{11}(z_0) = e^{-\frac{n\ell}{2}} \mathcal{O}(n^{1/2}). \tag{5.11}$$ Proof. Using (5.2) and the expressions $g(0) = \pi i$ and $P_{12}^{(\infty)}(0) = \sin \frac{\theta_0}{2}$ (if W = 0), we get the result for $Y_{12}(0)$ . For $\partial_s \ln Y_{12}(0)$ , we have $$\partial_s \ln Y_{12}(0) = \frac{\partial_s \left( R_{11}(0) P_{12}^{(\infty)}(0) + R_{12}(0) P_{22}^{(\infty)}(0) \right)}{R_{11}(0) P_{12}^{(\infty)}(0) + R_{12}(0) P_{22}^{(\infty)}(0)}.$$ (5.12) By (4.39), this yields (5.10). For the rest of this proof, we assume that |z| < 1 and that z lies outside of the lenses and in $D(z_0, r)$ . Then we have by (5.1), $$Y_{11}(z) = e^{ng(z)} \left[ R_{11}(z) P_{11}(z) + R_{12}(z) P_{21}(z) \right]. \tag{5.13}$$ By (4.1) and (4.2), we can show that $$g(z_0) = -\frac{\ell}{2} + i\frac{\theta_0 + \pi}{2}.$$ On the other hand, by (4.25), as $z \to z_0$ for fixed n, we have $$\Psi_{11}(n^2\zeta(z)) = 1 + \mathcal{O}(z - z_0), \quad \Psi_{21}(n^2\zeta(z)) = \mathcal{O}(z - z_0).$$ This implies, by (4.26), that $$P_{i1}(z_0) = E_{i1}(z_0)(1 + \mathcal{O}(e^{-nx}))e^{-\frac{n}{2}\phi(z_0)},$$ as $n \to \infty$ . Since $\phi(z_0) = 0$ , we have $P_{i1}(z_0) = \mathcal{O}(\sqrt{n})$ , j = 1, 2, and $$Y_{11}(z_0) = e^{n\left(-\frac{\ell}{2} + i\frac{\theta_0 + \pi}{2}\right)} \left(P_{11}(z_0) + \mathcal{O}(1)\right) = e^{-\frac{n\ell}{2}} \mathcal{O}(n^{1/2}),$$ as $$n \to \infty$$ . By Proposition 5.1 and (2.18), we have $$n\partial_s \ln Y_{12}(0) = \mathcal{O}(n^{1/2}e^{nx_c}),$$ (5.14) $$\frac{2(1-s)}{\pi} \operatorname{Im} \left( \frac{\overline{Y_{11}(e^{i\theta_0})}}{\sqrt{Y_{12}(0)}} \partial_s \left( \frac{Y_{11}(e^{i\theta_0})}{\sqrt{Y_{12}(0)}} \right) \right) = \mathcal{O}(n), \tag{5.15}$$ as $n \to \infty$ . Therefore, using (2.19), we get $$\ln D_n(s, \theta_0, 0) = \ln D_n(0, \theta_0, 0) + \int_0^s \mathcal{O}(n^{1/2} e^{nx_c}) ds'$$ = \ln D\_n(0, \theta\_0, 0) + \mathcal{O}(n^{1/2} e^{-n(x-x\_c)}), and we rederive Theorem 1.1 with a slightly worse error term which is only small if $s = o(n^{1/2}e^{-nx_c})$ .