Asymptotics for Hankel Determinants Associated to a Hermite Weight with a Varying Discontinuity

Theorem 2.1. Let $\alpha \in (-1, \infty)$ and $t \in (-1, \infty)$ . As $n \to \infty$ , we have $$\log \frac{H_n(\sqrt{2n}t, 0, \alpha)}{H_n(0, 0, \alpha)} = C_1(t)n^2 + C_2(t, \alpha)n + C_3(t, \alpha) + \mathcal{O}(n^{-1}), \tag{2.1}$$ where the coefficients are given by $$C_1(t) = -\frac{2t^3}{27} \left( \sqrt{3+t^2} - t \right) - \left( \frac{4}{3}t^2 + \frac{5}{9}t\sqrt{3+t^2} \right) - \log\left( \frac{t + \sqrt{3+t^2}}{\sqrt{3}} \right), \tag{2.2}$$ $$C_2(t,\alpha) = \frac{\alpha t}{3} \left( t - \sqrt{3 + t^2} \right) - \alpha \log \left( \frac{t + \sqrt{3 + t^2}}{\sqrt{3}} \right), \tag{2.3}$$ $$C_3(t,\alpha) = \frac{1 - 3\alpha^2}{6} \log\left(\frac{t + \sqrt{3 + t^2}}{\sqrt{3}}\right) - \frac{1}{48} \log\left(\frac{3 + t^2}{3}\right) - \frac{1}{16} \left(1 - 4\alpha^2\right) \log\left(\frac{3 + 5t^2 + 4t\sqrt{3 + t^2}}{3}\right).$$ (2.4) Furthermore, the error term $\mathcal{O}(n^{-1})$ is uniform for t in a compact subset of $(-1, \infty)$ . Note that there is no critical transition in (2.1) as $t \to 1$ . Remark 2.2. The probability (1.7) with $v = \sqrt{2nt}$ can be rewritten as $$\mathbb{P}(x_{\min}^{(\sqrt{2n}t,\alpha)} \ge \sqrt{2n}t) = \frac{H_n(\sqrt{2n}t,0,\alpha)}{H_n(0,0,\alpha)} \frac{H_n(0,0,\alpha)}{H_n(0,1,\alpha)} \frac{H_n(0,1,\alpha)}{H_n(\sqrt{2n}t,1,\alpha)}.$$ Large n asymptotics of these three ratios are given up to the constant term by Theorem 2.1 (for t > -1), (1.8) and (1.6) (for $t \in (-1,1)$ ), respectively. Putting these asymptotics together, we obtain as $n \to \infty$ and for $t \in (-1,1)$ , $$\log \mathbb{P}\left(x_{\min}^{(\sqrt{2n}t,\alpha)} \ge \sqrt{2n}t\right) = \left(C_1(t) - \frac{\log 3}{2}\right)n^2 + \left(C_2(t,\alpha) - \frac{\alpha \log 3}{2} - \alpha t^2\right)n$$ $$+ \left(\frac{\alpha^2}{4} - \frac{1}{12}\right)\log n + C_3(t,\alpha) + c_0 - \frac{\alpha^2}{8}\log\left(1 - t^2\right) + \mathcal{O}\left(\frac{\log n}{n}\right). \tag{2.5}$$ It can be checked from (2.2) that $C_1(-1) = \frac{\log 3}{2}$ and $$C_1'(t) = -\frac{8}{27} (\sqrt{3+t^2} - t) (3+5t^2+4t\sqrt{3+t^2}) < 0,$$ for $t > -1$ , which shows that the leading term in (2.5) is negative. This implies that the above probability decays super exponentially fast as $n \to \infty$ for $t \in (-1,1)$ . To observe a transition in the large n asymptotics of $H_n(\sqrt{2nt}, s, \alpha)$ when s = 0 and when s is in a compact subset of (0, 1], we couple the parameter s with n in the form $$s = e^{-\lambda n}, \qquad \lambda \ge 0.$$ Large n asymptotics of $H_n(\sqrt{2n}t, e^{-\lambda n}, \alpha)$ will depend on whether $\lambda$ is greater or smaller than a critical value $\lambda_c(t)$ , which is explicit and given by $$\lambda_c(t) = \frac{2t}{\sqrt{3}} \sqrt{3 + t^2 + 2t\sqrt{3 + t^2}} + 2\log\left(2 + t^2 + t\sqrt{3 + t^2} + \frac{\sqrt{3 + t^2} + t}{\sqrt{3}}\sqrt{3 + t^2 + 2t\sqrt{3 + t^2}}\right).$$ (2.6)

Theorem 2.3. Let $s = e^{-\lambda n}$ with $\lambda \in [0, \infty)$ , we have the following asymptotic results. (1) If $t \in (-1,1)$ and $\lambda \geq \lambda_c(t)$ , then $$\log \frac{H_n(\sqrt{2nt}, e^{-\lambda n}, \alpha)}{H_n(\sqrt{2nt}, 0, \alpha)} = \mathcal{O}(n^{-1/2} e^{-n(\lambda - \lambda_c(t))}), \quad as \quad n \to \infty,$$ (2.7) and large n asymptotics for $\log H_n(\sqrt{2n}t, 0, \alpha)$ are given by Theorem 2.1. Furthermore, the $\mathcal{O}$ term in (2.7) is uniform for t in a compact subset of (-1, 1) and for $\lambda \geq \lambda_c(t)$ . (2) If $t \in (-1,1)$ and $0 \le \lambda \le \lambda_c(t)$ are fixed, then $$\lim_{n \to \infty} \frac{1}{n^2} \log \frac{H_n(\sqrt{2n}t, e^{-\lambda n}, \alpha)}{H_n(\sqrt{2n}t, 1, \alpha)} = -\int_0^\lambda \Omega(t, \tilde{\lambda}) d\tilde{\lambda}, \tag{2.8}$$ where $$\Omega(t,\lambda) = \int_a^b \rho(x;t,\lambda) dx, \qquad \rho(x;t,\lambda) = \frac{2}{\pi} \sqrt{c-x} \sqrt{\frac{x-b}{x-t}} \sqrt{x-a},$$ and a < b < t < c, with a, b and c depending on $\lambda$ and t, are uniquely determined by the following equations: $$t = a + b + c,$$ $$2 = a^2 + b^2 + c^2 - t^2,$$ $$\lambda = 4 \int_b^t \frac{\sqrt{c - x}}{\sqrt{t - x}} \sqrt{x - b} \sqrt{x - a} dx.$$ Remark 2.4. In Theorem 2.3, we restrict ourselves to the case $t \in (-1,1)$ . With increasing effort, this result can be extended for $t \in (-1,\infty)$ . If $t \ge 1$ , a new region appears in the $(t,\lambda)$ plane which deserves a separate analysis (which we expect to be straightforward but long). Therefore, we decided not to proceed in this direction. Remark 2.5. Note that the denominators on the left hand sides of (2.7) and (2.8) are different. We can use Theorem 2.3 to obtain information about the large deviation of the smallest thinned $\mathrm{GUE}(\sqrt{2n}t,\alpha)$ eigenvalue as follows. By (1.9) and (1.7) with $v=\sqrt{2n}t$ and $s=e^{-\lambda n}$ , we have $$\mathbb{P}\big(y_{\min}^{(\sqrt{2n}t,e^{-\lambda n},\alpha)} \ge \sqrt{2n}t\big) = \frac{H_n\big(\sqrt{2n}t,e^{-\lambda n},\alpha\big)}{H_n\big(\sqrt{2n}t,0,\alpha\big)} \mathbb{P}\big(x_{\min}^{(\sqrt{2n}t,\alpha)} \ge \sqrt{2n}t\big).$$ Therefore, for $t \in (-1,1)$ and $\lambda \geq \lambda_c(t)$ , by (2.7), as $n \to \infty$ we have $$\log \mathbb{P}\left(y_{\min}^{(\sqrt{2n}t,e^{-\lambda n},\alpha)} \ge \sqrt{2n}t\right) = \log \mathbb{P}\left(x_{\min}^{(\sqrt{2n}t,\alpha)} \ge \sqrt{2n}t\right) + \mathcal{O}\left(n^{-1/2}e^{-n(\lambda-\lambda_c(t))}\right),\tag{2.9}$$ and large n asymptotics of $\log \mathbb{P}(x_{\min}^{(\sqrt{2n}t,\alpha)} \geq \sqrt{2n}t)$ are given by (2.5). In the regime $t \in (-1,1)$ and $0 \leq \lambda \leq \lambda_c(t)$ , (2.8) implies $$\log \mathbb{P}\left(y_{\min}^{(\sqrt{2n}t,e^{-\lambda n},\alpha)} \ge \sqrt{2n}t\right) = \left(-\int_0^\lambda \Omega(t,\tilde{\lambda})d\tilde{\lambda}\right)n^2 + o(n^2). \tag{2.10}$$ Since (2.9) and (2.10) are both valid for $\lambda = \lambda_c(t)$ , by equalling the leading term, we have $$-\int_0^{\lambda_c(t)} \Omega(t,\lambda) d\lambda = C_1(t) - \frac{\log 3}{2}.$$ (2.11) We will give an independent and more direct proof of this formula at the end of Section 7. Remark 2.6. Note that the limit [\(2.8\)](#page-6-2) is independent of α. The subleading terms in the large n asymptotics of ^Hn( √ 2nt,e−λn,α) Hn( √ 2nt,1,α) are expected to depend on α and to be oscillatory and described in terms of elliptic θ-functions. These functions appear in our analysis (see, e.g., [\(6.12\)](#page-30-0)). This heuristic is also supported by the analogy of our situation with [\[6\]](#page-41-0), where the authors obtained θ-functions in the subleading terms.

Lemma 4.1. We have the following differential identities $$\partial_t \log H_n(\sqrt{2nt}, 0, \alpha) = 4nU_{1,11},\tag{4.5}$$ $$s\partial_s \log H_n(\sqrt{2n}t, s, \alpha) = \int_{-\infty}^t \frac{\widetilde{w}(x)}{2\pi i} [U^{-1}(x)U'(x)]_{21} dx.$$ (4.6) Proof. The differential identity (4.5) is obtained by substituting (4.4) and (4.3) into (3.6). Similarly, using (3.7) and (4.1), the differential identity (3.9) can be rewritten as $$s\partial_s \log H_n(\sqrt{2nt}, s, \alpha) = \int_{-\infty}^{\sqrt{2nt}} \frac{w(x)}{2\pi i} [Y^{-1}(x)Y'(x)]_{21} dx,$$ which gives (4.6) after using (4.4) and a change of variables. Remark 4.2. Note that $[Y^{-1}(z)Y'(z)]_{21}$ only involves the first column of Y, which is entire (see (4.1) or equivalently (4.2)). Thus $[Y_+^{-1}(x)Y_+'(x)]_{21} = [Y_-^{-1}(x)Y_-'(x)]_{21}$ for $x \in \mathbb{R}$ , and we simply denote it by $[Y^{-1}(x)Y'(x)]_{21}$ without ambiguity. The same remark holds for U.

Lemma 7.1. As $n \to \infty$ , we have $$U_{+}(x) \begin{pmatrix} 1 \\ 0 \end{pmatrix} = e^{\frac{n\ell}{2}} e^{ng_{+}(x)} e^{-\frac{n\ell}{2}\sigma_{3}} \begin{pmatrix} \mathcal{O}\left(n^{\frac{1}{2} + \max(\alpha, 0)}\right) \\ \mathcal{O}\left(n^{\frac{1}{2} + \max(\alpha, 0)}\right) \end{pmatrix}, \tag{7.7}$$ $$U'_{+}(x)\begin{pmatrix}1\\0\end{pmatrix} = e^{\frac{n\ell}{2}}e^{ng_{+}(x)}e^{-\frac{n\ell}{2}\sigma_{3}}\begin{pmatrix}\mathcal{O}\left(n^{\frac{5}{2}+\max(\alpha,0)}\right)\\\mathcal{O}\left(n^{\frac{5}{2}+\max(\alpha,0)}\right)\end{pmatrix},\tag{7.8}$$ uniformly for $x \in (-\infty, t) \cap D_t$ . Proof. For $z \in D_t$ , z outside the lenses, we have $$U(z)\begin{pmatrix} 1\\0 \end{pmatrix} = e^{\frac{n\ell}{2}} e^{ng(z)} e^{-\frac{n\ell}{2}\sigma_3} R(z) P(z) \begin{pmatrix} 1\\0 \end{pmatrix}. \tag{7.9}$$ If furthermore, $\Im z > 0$ , by (5.19) and (B.4) we have $$P(z)\begin{pmatrix} 1\\0 \end{pmatrix} = e^{\frac{\pi i \alpha}{2}} (z-t)^{-\frac{\alpha}{2}} e^{-n\xi(z)} E(z) \begin{pmatrix} I_{\alpha}(2n\sqrt{-f(z)})\\ -2\pi i n\sqrt{-f(z)} I'_{\alpha}(2n\sqrt{-f(z)}) \end{pmatrix}. \tag{7.10}$$ Let $x \in (-\infty, t) \cap D_t$ . Note that $\tilde{\xi}(x) > 0$ ( $\tilde{\xi}$ is defined in (5.16)) and thus $\sqrt{-f(x)}_+ = \frac{1}{2}\tilde{\xi}(x)$ . Inserting (7.10) into (7.9), we can take the limit $z \to x$ , this gives $$U_{+}(x) \begin{pmatrix} 1 \\ 0 \end{pmatrix} = (-1)^{n} e^{\frac{n\ell}{2}} e^{ng_{+}(x)} e^{-n\tilde{\xi}(x)} (t-x)^{-\frac{\alpha}{2}} e^{-\frac{n\ell}{2}\sigma_{3}}$$ $$\times R(x) E(x) \begin{pmatrix} I_{\alpha}(n\tilde{\xi}(x)) \\ -\pi i n\tilde{\xi}(x) I'_{\alpha}(n\tilde{\xi}(x)) \end{pmatrix}. \tag{7.11}$$ Since E is analytic in $D_t$ , one has from (5.21) that as $n \to \infty$ $$E(z) = \mathcal{O}(1)n^{\frac{\sigma_3}{2}}, \qquad E'(z) = \mathcal{O}(1)n^{\frac{\sigma_3}{2}}, \qquad \text{uniformly for } z \in D_t.$$ (7.12) To obtain a uniform bound from (7.11), we distinguish three cases. Let M > 0 be an arbitrary large but fixed constant and let m > 0 be an arbitrary small but fixed constant. Case (a): $n\tilde{\xi}(x) \geq M$ as $n \to \infty$ . In this case we need large $\zeta$ asymptotics for $I_{\alpha}(\zeta)$ and $I'_{\alpha}(\zeta)$ . From (B.2), we have $$I_{\alpha}(\zeta) = \frac{e^{\zeta}}{\sqrt{2\pi\zeta}} \left( 1 + \mathcal{O}(\zeta^{-1}) \right), \qquad I_{\alpha}'(\zeta) = \frac{e^{\zeta}}{\sqrt{2\pi\zeta}} \left( 1 + \mathcal{O}(\zeta^{-1}) \right), \qquad \text{as} \quad \zeta \to \infty.$$ (7.13) If we insert (7.13) into (7.11), the result follows for Case (a) from (5.26), (7.12) and from the fact that $(t-x)^{-\frac{\alpha}{2}} = \mathcal{O}(n^{\max(\alpha,0)})$ . Case (b): $m \leq n\tilde{\xi}(x) \leq M$ as $n \to \infty$ . In this case we have $I_{\alpha}(n\tilde{\xi}(x)) = \mathcal{O}(1)$ , $n\tilde{\xi}(x)I'_{\alpha}(n\tilde{\xi}(x)) = \mathcal{O}(1)$ , $e^{-n\tilde{\xi}(x)} = \mathcal{O}(1)$ , $(t-x)^{-\frac{\alpha}{2}} = \mathcal{O}(n^{\alpha})$ . Again from (5.26) and (7.12), we obtain $$U_{+}(x)\begin{pmatrix} 1\\0 \end{pmatrix} = e^{\frac{n\ell}{2}} e^{ng_{+}(x)} e^{-\frac{n\ell}{2}\sigma_{3}} \begin{pmatrix} \mathcal{O}\left(n^{\frac{1}{2}+\alpha}\right)\\ \mathcal{O}\left(n^{-\frac{1}{2}+\alpha}\right) \end{pmatrix}, \tag{7.14}$$ which is even slightly better than (7.7). Case (c): $n\xi(x) \leq m$ as $n \to \infty$ . From [36, formula (10.25.2)], we have $$I_{\alpha}(\zeta) = \left(\frac{\zeta}{2}\right)^{\alpha} \left(\frac{1}{\Gamma(1+\alpha)} + \mathcal{O}(\zeta^2)\right),$$ $$I'_{\alpha}(\zeta) = \left(\frac{\zeta}{2}\right)^{\alpha-1} \left(\frac{\alpha}{\Gamma(1+\alpha)} + \mathcal{O}(\zeta^2)\right), \quad \text{as} \quad \zeta \to 0.$$ From the above expansion, we have for Case (c) that $$\frac{I_{\alpha}(n\tilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}} = \mathcal{O}(n^{\alpha}), \qquad \frac{n\tilde{\xi}(x)I_{\alpha}'(n\tilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}} = \mathcal{O}(n^{\alpha})$$ and $e^{-n\tilde{\xi}(x)} = \mathcal{O}(1)$ . Thus, from (5.26) and (7.12) we obtain again (7.14), which finishes the proof of (7.7). We now turn to the proof of (7.8). From (7.11), we have $U'_{+}(x) \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \tilde{U}_{1}(x) + \tilde{U}_{2}(x)$ , where $$\widetilde{U}_{1}(x) = n(g'_{+}(x) - \widetilde{\xi}'(x))U_{+}(x)\begin{pmatrix} 1\\0 \end{pmatrix} + (-1)^{n}e^{\frac{n\ell}{2}}e^{ng_{+}(x)}e^{-n\widetilde{\xi}(x)}(t-x)^{-\frac{\alpha}{2}}e^{-\frac{n\ell}{2}\sigma_{3}} \\ \times (R'(x)E(x) + R(x)E'(x))\begin{pmatrix} I_{\alpha}(n\widetilde{\xi}(x))\\ -\pi i n\widetilde{\xi}(x)I'_{\alpha}(n\widetilde{\xi}(x)) \end{pmatrix}, \\ \widetilde{U}_{2}(x) = (-1)^{n}e^{\frac{n\ell}{2}}e^{ng_{+}(x)}e^{-n\widetilde{\xi}(x)}e^{-\frac{n\ell}{2}\sigma_{3}}R(x)E(x)\begin{pmatrix} \left(\frac{I_{\alpha}(n\widetilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}}\right)'\\ \left(\frac{-\pi i n\widetilde{\xi}(x)I'_{\alpha}(n\widetilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}}\right)' \end{pmatrix}.$$ The analysis of $\widetilde{U}_1(x)$ and $\widetilde{U}_2(x)$ can be done very similarly to the first part of the proof and we do not provide here all the details. From (5.4), one has $g'_+(x) - \widetilde{\xi}'(x) = 2x$ and thus by (5.26), (7.7) and (7.12), $$\widetilde{U}_1(x) = e^{\frac{n\ell}{2}} e^{ng_+(x)} e^{-\frac{n\ell}{2}\sigma_3} \begin{pmatrix} \mathcal{O}(n^{\frac{3}{2} + \max(\alpha, 0)}) \\ \mathcal{O}(n^{\frac{3}{2} + \max(\alpha, 0)}) \end{pmatrix}.$$ Again, by splitting the analysis into the same three cases as in the first part of the proof, we obtain the estimates $$\left(\frac{I_{\alpha}(n\tilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}}\right)' = \mathcal{O}\left(n^2 \left(\frac{I_{\alpha}(n\tilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}}\right)\right),$$ $$\left(\frac{-i\pi n\tilde{\xi}(x)I'_{\alpha}(n\xi(x))}{(t-x)^{\frac{\alpha}{2}}}\right)' = \mathcal{O}\left(n^2 \left(\frac{-i\pi n\tilde{\xi}(x)I'_{\alpha}(n\xi(x))}{(t-x)^{\frac{\alpha}{2}}}\right)\right),$$ which yields $$\widetilde{U}_2(x) = e^{\frac{n\ell}{2}} e^{ng_+(x)} e^{-\frac{n\ell}{2}\sigma_3} \begin{pmatrix} \mathcal{O}(n^{\frac{5}{2} + \max(\alpha, 0)}) \\ \mathcal{O}(n^{\frac{5}{2} + \max(\alpha, 0)}) \end{pmatrix}$$ and finishes the proof. Note that $g_+(x) + g_-(x) - 2x^2 + \ell$ is continuous on $\mathbb{R}$ and equal to 0 at x = t by (4.7). Thus, from (4.24) and (4.8) and the fact that V(x) has a jump discontinuity at x = t, we have $$\lim_{\substack{x \to t \\ r < t}} g_{+}(x) + g_{-}(x) - V(x) + \ell = -\lambda < -\lambda_{c} < 0.$$ (7.15) Therefore, by using first Lemma 7.1 and then (7.15), there exists $c \in (0, \lambda_c)$ such that $$\frac{\tilde{w}(x)}{2\pi i} \left[ U^{-1}(x)U'(x) \right]_{21} = |x - t|^{\alpha} e^{n(g_{+}(x) + g_{-}(x) - V(x) + \ell)} \mathcal{O}\left(n^{3 + 2\max(\alpha, 0)}\right) = |x - t|^{\alpha} \mathcal{O}\left(e^{-(\lambda - c)n}\right),$$ (7.16) as n → ∞ uniformly for x ∈ D^t ∩ (−∞, t). Now, we will split the integral of the dif ferential identity [\(4.6\)](#page-10-3) into two parts: $$s\partial_s \log H_n(\sqrt{2n}t, s) = I_1(s) + I_2(s),$$ $$I_1(s) = \int_{(-\infty, t) \setminus D_t} \frac{\widetilde{w}(x)}{2\pi i} [U^{-1}(x)U'(x)]_{21} dx,$$ $$I_2(s) = \int_{(-\infty, t) \cap D_t} \frac{\widetilde{w}(x)}{2\pi i} [U^{-1}(x)U'(x)]_{21} dx.$$ The first integral can be evaluated using [\(7.6\)](#page-32-4). By [\(5.1\)](#page-15-3), [\(5.2\)](#page-16-5), [\(5.4\)](#page-16-0) and [\(5.7\)](#page-16-1) (see also the comment just after), we have g+(b) + g−(b) + ` − V (b) = −(λ − λc) and $$(g_{+}(x) + g_{-}(x) + \ell - V(x))'\big|_{x=\bar{b}} = 0, \qquad (g_{+}(x) + g_{-}(x) + \ell - V(x)''\big|_{x=\bar{b}} < 0.$$ Therefore, we obtain $$|I_1(s = e^{-\lambda n})| = \mathcal{O}(n^{-1/2}e^{-n(\lambda - \lambda_c)}), \quad \text{as} \quad n \to \infty.$$ On the other hand, from [\(7.16\)](#page-34-1), it immediately follows that $$|I_2(s=e^{-\lambda n})| = \mathcal{O}(e^{-(\lambda-c)n}), \quad \text{as} \quad n \to \infty,$$ where c ∈ (0, λc). Therefore, the dif ferential identity becomes $$\partial_s \log H_n(v, s, \alpha) \big|_{s=e^{-\lambda n}} = \mathcal{O}(n^{-1/2}e^{n\lambda_c}), \quad \text{as} \quad n \to \infty,$$ where in the above expression the O term is uniform for t in a compact subset of (−1, ∞) and for λ ≥ λc(t). Thus, we can integrate it from s = 0 to s = e ⁻λn, and it gives $$\log H_n(\sqrt{2n}t, e^{-\lambda n}, \alpha) = \log H_n(\sqrt{2n}t, 0, \alpha) + \mathcal{O}(n^{-1/2}e^{-n(\lambda - \lambda_c(t))}), \quad \text{as} \quad n \to \infty,$$ which is the claim [\(2.7\)](#page-6-0).

Proposition 7.2. Let W ⊂ R be an arbitrary small but fixed neighbourhood of the four points {a, b, t, c}. We have as n → ∞ $$\frac{\tilde{w}(x)}{2\pi i} \left[ U^{-1}(x)U'(x) \right]_{21} - n\rho(x)\chi_{\mathcal{S}}(x) = e^{n(g_{+}(x) + g_{-}(x) + \ell - V(x))} \mathcal{O}(1),$$ uniformly for x ∈ R \ W. Proof . We can assume without loss of generality that the disks of the local parametrices are suf ficiently small such that D ⊂ W. Let z be outside the lenses and outside the disks. In this region, by inverting the transformations R 7→ S 7→ T 7→ U, we have $$U(z) = e^{-\frac{n\ell}{2}\sigma_3} R(z) P^{(\infty)}(z) e^{ng(z)\sigma_3} e^{\frac{n\ell}{2}\sigma_3}.$$ Since the dependence in n of the global parametrix [\(6.12\)](#page-30-0) appears only in the form nΩ ∈ R, and as an argument of the θ-function, by the periodicity property [\(6.10\)](#page-29-0), as n → ∞ we have $$P^{(\infty)}(z) = \mathcal{O}(1), \qquad P^{(\infty)}(z)' = \mathcal{O}(1), \quad \text{uniformly for } z \text{ outside the disks.}$$ Therefore, using also the large n asymptotics for R (6.13), we have $$[U(z)^{-1}U'(z)]_{21} = e^{n\ell}e^{2ng(z)}\mathcal{O}(1), \quad \text{as} \quad n \to \infty,$$ (7.17) uniformly for z outside the lenses and outside the disks. For $x \in \mathbb{R} \setminus (S \cup W)$ , we can take the limit $z \to x$ in (7.17). As $n \to \infty$ , we have $$\frac{\tilde{w}(x)}{2\pi i} \left[ U^{-1}(x)U'(x) \right]_{21} = |x - t|^{\alpha} e^{n(g_{+}(x) + g_{-}(x) + \ell - V(x))} \mathcal{O}(1),$$ uniformly for $x \in \mathbb{R} \setminus (\mathcal{S} \cup \mathcal{W})$ . Now, we consider the case when z is still outside the disks but inside $\mathcal{I}_1$ , see (6.9) and Fig. 3. Inverting the transformations in this region, we get $$U(z) = e^{-\frac{n\ell}{2}\sigma_3} R(z) P^{(\infty)}(z) \begin{pmatrix} 1 & 0 \\ (z-t)^{-\alpha} e^{-2n\xi_1(z)} & 1 \end{pmatrix} e^{ng(z)\sigma_3} e^{\frac{n\ell}{2}\sigma_3}.$$ Since $P^{(\infty)}(z) = \mathcal{O}(1)$ as $n \to \infty$ uniformly for z in this region, we have $$\left[U^{-1}(z)U'(z)\right]_{21} = (z-t)^{-\alpha}e^{n(2g(z)+\ell)}\left(-2n\xi_1'(z)e^{-2n\xi_1(z)} + \mathcal{O}(1)\right), \quad \text{as} \quad n \to \infty, (7.18)$$ where we have also used (6.7) and $\Re \xi_{1,+}(x) = 0$ for $x \in (t,c)$ . Note that from (6.1), we have $\xi'_{1,+}(x) = -\pi i \rho(x)$ for $x \in (t,c)$ . Thus, if we let $z \to x \in (t,c) \setminus \mathcal{W}$ in (7.18), from (4.7) and (6.3), we have $$\frac{\tilde{w}(x)}{2\pi i} \left[ U^{-1}(x)U'(x) \right]_{21} = n\rho(x) \left( 1 + \mathcal{O}(n^{-1}) \right), \quad \text{as} \quad n \to \infty,$$ (7.19) where the $\mathcal{O}$ term in the above expression is uniform for $x \in (t,c) \setminus \mathcal{W}$ . For $x \in (a,b) \setminus \mathcal{W}$ , we can invert the transformations for $z \in \widetilde{\mathcal{I}}_1$ and then take the limit $z \to x$ . The computations are similar and we obtain the same asymptotics as (7.19). By (4.1) and (4.4), note that (3.10) can be rewritten as $\int_{\mathbb{R}} \frac{\tilde{w}(x)}{2\pi i} \left[ U^{-1}(x)U'(x) \right]_{21} dx = n$ . Thus, a consequence of Proposition 7.2 (by taking $\mathcal{W}$ arbitrarily small) and (4.6) is that for fixed $t \in (-1,1)$ and fixed $\lambda \in (0,\lambda_c(t))$ , we have $$\lim_{n \to \infty} \frac{s}{n} \partial_s \log H_n(\sqrt{2n}t, s, \alpha) \Big|_{s=e^{-\lambda n}} = \lim_{n \to \infty} \int_{-\infty}^t \frac{\widetilde{w}(x)}{2\pi i n} \big[ U^{-1}(x) U'(x) \big]_{21} dx = \Omega(t, \lambda).$$ (7.20) A simple change of variables shows that $$\frac{s}{n}\partial_s \log H_n(\sqrt{2nt}, s, \alpha)\big|_{s=e^{-\lambda n}} = -\frac{1}{n^2}\partial_\lambda \log H_n(\sqrt{2nt}, e^{-\lambda n}, \alpha).$$ By (7.20), for every $(t, \lambda)$ such that $t \in (-1, 1)$ and $\lambda \in (0, \lambda_c(t))$ , the right-hand side of the above expression converges to $\Omega(t, \lambda)$ as $n \to \infty$ . Also, by (3.8) and (3.9), we have $$\frac{s}{n}\partial_s \log H_n(v, s, \alpha) = \frac{\mathcal{E}_n(v, s, \alpha)}{n} \le 1.$$ Since the constant function 1 is integrable on any bounded interval, we can apply Lebesgue's dominated convergence theorem, and we have $$\lim_{n \to \infty} \frac{-1}{n^2} \int_0^{\lambda} \partial_{\tilde{\lambda}} \log H_n(\sqrt{2nt}, e^{-\tilde{\lambda}n}, \alpha) d\tilde{\lambda} = \int_0^{\lambda} \Omega(t, \tilde{\lambda}) d\tilde{\lambda},$$ which finishes the proof. Remark 7.3. As mentioned in Remark 4.5, we have indeed only used pointwise convergence for $\lambda \in (0, \lambda_c(t))$ of the quantity $\frac{s}{n} \partial_s \log H_n(\sqrt{2n}t, s, \alpha)\big|_{s=e^{-\lambda n}}$ to $\Omega(t, \lambda)$ as $n \to \infty$ and Lebesgue's theorem. The technical RH analysis as $\lambda \to 0$ or $\lambda \to \lambda_c(t)$ was thus not needed.

Lemma 7.4. For t ∈ (−1, 1) and λ ∈ (0, λc(t)), there holds a relation between Ω(t, λ), the density ρ(x;t, λ) given by [\(4.11\)](#page-12-4), and the Euler–Lagrange constant `(t, λ) given by [\(4.15\)](#page-12-5): $$\Omega(t,\lambda) = \partial_{\lambda}\ell(t,\lambda) + \partial_{\lambda} \int_{\mathcal{S}} 2x^{2} \rho(x;t,\lambda) dx + \lambda \partial_{\lambda} \Omega(t,\lambda).$$ (7.21) Proof . Consider the function $$H(x;t,\lambda) = -2\int_{\mathcal{S}} \log|x-y|\rho(y;t,\lambda)dy.$$ By the Euler–Lagrange equality [\(4.7\)](#page-11-1), we have $$H(x;t,\lambda) = \ell(t,\lambda) - 2x^2, \qquad x \in (t,c), \tag{7.22}$$ $$H(x;t,\lambda) = \ell(t,\lambda) - 2x^2 - \lambda, \qquad x \in (a,b). \tag{7.23}$$ Thus, by integrating it with respect to ρ(x;t, λ)dx, we obtain $$\int_{\mathcal{S}} H(x;t,\lambda)\rho(x;t,\lambda)\mathrm{d}x = \ell(t,\lambda) - \lambda\Omega(t,\lambda) - \int_{\mathcal{S}} 2x^2\rho(x;t,\lambda)\mathrm{d}x.$$ We will evaluate ∂^λ R ^S H(x;t, λ)ρ(x;t, λ)dx in two dif ferent ways. From the above expression, it gives $$\partial_{\lambda} \int_{\mathcal{S}} H(x; t, \lambda) \rho(x; t, \lambda) dx = \partial_{\lambda} \ell(t, \lambda) - \Omega(t, \lambda) - \lambda \partial_{\lambda} \Omega(t, \lambda) - \partial_{\lambda} \int_{\mathcal{S}} 2x^{2} \rho(x; t, \lambda) dx. (7.24)$$ On the other hand, by Lebesgue's dominated convergence theorem, and by the symmetry in x and y, we have $$\partial_{\lambda} \int_{\mathcal{S}} \int_{\mathcal{S}} \log|x - y| \rho(y; t, \lambda) \rho(x; t, \lambda) dy dx = 2 \int_{\mathcal{S}} \int_{\mathcal{S}} \log|x - y| \partial_{\lambda} (\rho(y; t, \lambda)) \rho(x; t, \lambda) dy dx.$$ Therefore, by dif ferentiating [\(7.22\)](#page-37-0) and [\(7.23\)](#page-37-1) with respect to λ, we obtain $$\partial_{\lambda} \int_{\mathcal{S}} H(x; t, \lambda) \rho(x; t, \lambda) dx = 2 \int_{\mathcal{S}} \partial_{\lambda} \left( H(x; t, \lambda) \right) \rho(x; t, \lambda) dx = 2 \partial_{\lambda} \ell(t, \lambda) - 2\Omega(t, \lambda). \tag{7.25}$$ Putting [\(7.24\)](#page-37-2) and [\(7.25\)](#page-37-3) together, we obtain [\(7.21\)](#page-37-4). Let us consider the function $$F(t,\lambda) = \ell(t,\lambda) + \int_{\mathcal{S}} 2x^2 \rho(x;t,\lambda) dx,$$

Lemma 7.5. For t ∈ (−1, 1) and λ ∈ (0, λc(t)), we have the following relation $$\Omega(t,\lambda) = \partial_{\lambda} \left[ \frac{F(t,\lambda_{c}(t))}{2} \left( \frac{\lambda}{\lambda_{c}(t)} \right)^{2} + \int_{\frac{\lambda}{\lambda_{c}(t)}}^{1} \xi F(t,\frac{\lambda}{\xi}) d\xi \right].$$ (7.26) Proof . From a direct calculation and a change of variables, the right-hand side of [\(7.26\)](#page-37-5) is equal to $$\int_{\frac{\lambda}{\lambda_c(t)}}^{1} \xi \partial_{\lambda} F(t, \frac{\lambda}{\xi}) d\xi = \int_{\frac{\lambda}{\lambda_c(t)}}^{1} \partial_{u} F(t, u) \Big|_{u = \frac{\lambda}{\xi}} d\xi = \lambda \int_{\lambda}^{\lambda_c(t)} \frac{\partial_{u} F(t, u)}{u^2} du.$$ (7.27) By using [\(7.21\)](#page-37-4), which can be rewritten as ∂λF(t, λ) = Ω(t, λ) − λ∂λΩ(t, λ), the right-hand side of [\(7.27\)](#page-38-1) becomes $$\lambda \int_{\lambda}^{\lambda_c(t)} \frac{\Omega(t, u) - u \partial_u \Omega(t, u)}{u^2} du = \Omega(t, \lambda),$$ where the last equality is obtained via an integration by parts, and using the identity Ω(t, λc(t)) = 0.