Uniform asymptotics of Toeplitz determinants with Fisher–Hartwig singularities

Theorem 1.1. Assume that f is of the form (1), satisfying (a) – (c). Then as $n \to \infty$ , $$\log D_n(f) = nV_0 + \sum_{j=1}^m \left(\alpha_j^2 - \beta_j^2\right) \log n + \sum_{1 \le j < k \le m} 2(\alpha_j \alpha_k - \beta_j \beta_k) \log \left(\frac{1}{\sin \left|\frac{t_j - t_k}{2}\right| + n^{-1}}\right) + \mathcal{O}(1),$$ <span id="page-2-0"></span>where the error term is uniform for $0 \le t_1 < t_2 < \cdots < t_m < 2\pi$ .

Lemma 2.2. There is a constant 0 < c such that for all sufficiently small $\epsilon > 0$ , $$I_{\epsilon}(\alpha) \leq c\widehat{I}_{\epsilon}(\alpha).$$ Proof. Denote $U_j = [0, 2\pi(j-1)/(m+1)) \cup [2\pi j/(m+1), 2\pi)$ for j = 1, 2, ..., m+1. Since there are m points $t_1, ..., t_m$ and m+1 sets $U_j$ , it follows that there is always a j such that $\{t_1, ..., t_m\} \subset U_j$ , thus $$I_{\epsilon}(\alpha) \leq \sum_{j=1}^{m+1} \int_{U_j^m} \prod_{1 \leq j < k \leq m} \left( \sin \left| \frac{t_j - t_k}{2} \right| + \epsilon \right)^{-2\alpha^2} dt_1 \dots dt_m.$$ It follows that $$I_{\epsilon}(\alpha) \leq (m+1) \int_{[0,2\pi-2\pi/(m+1))^m} \prod_{1 \leq j \leq k \leq m} \left( \sin \left| \frac{t_j - t_k}{2} \right| + \epsilon \right)^{-2\alpha^2} dt_1 \dots dt_m.$$ Furthermore, for $0 \le x \le \pi - \pi/(m+1)$ , one has <span id="page-6-0"></span> $$\frac{x}{\pi}\sin\frac{\pi}{m+1} \le \sin x \le x,$$ and it follows that $$I_{\epsilon}(\alpha) \le \frac{(m+1)\pi^{2\alpha}}{\left(\sin\frac{\pi}{m+1}\right)^{2\alpha}} \int_{[0,2\pi-2\pi/(m+1))^m} \prod_{1 \le j \le k \le m} \left( \left| \frac{t_j - t_k}{2} \right| + \epsilon \right)^{-2\alpha^2} dt_1 \dots dt_m. \tag{18}$$ <span id="page-6-2"></span> The lemma follows easily from (18). We now take the change of variables $s_j = t_{j+1} - t_j$ for j = 1, ..., m-1 and find that $$\widehat{I}_{\epsilon}(\alpha) = \int_0^1 dt_1 \int \prod_{1 < j < k < m-1} \left( \sum_{i=j}^k s_i + \epsilon \right)^{-2\alpha^2} ds_1 \dots ds_{m-1},$$ with integration taken over $s_1, \ldots, s_{m-1} \ge 0$ such that $\sum_{j=1}^m s_j < 1 - t_1$ , from which it follows that $$\widehat{I}_{\epsilon}(\alpha) \le I_{\epsilon}^{(2)}(\alpha),$$ (19) where <span id="page-6-1"></span> $$I_{\epsilon}^{(2)}(\alpha) = \int_{[0,1)^m} \prod_{1 \le i \le k \le m-1} \left( \sum_{i=i}^k s_i + \epsilon \right)^{-2\alpha^2} ds_1 \dots ds_{m-1}.$$ (20) If $2\alpha^2 > 1$ , then $I_{\epsilon}^{(2)}(\alpha)$ is straightforward to evaluate – one simply notes that <span id="page-7-2"></span> $$\left(\sum_{i=j}^{k} s_i + \epsilon\right)^{-2\alpha^2} \le (s_i + \epsilon)^{-2\alpha^2}, \tag{21}$$ for any i = j, ..., k, and thus $$I_{\epsilon}^{(2)}(\alpha) \le \int_{[0,1)^m} \prod_{j=1}^{m-1} (s_j + \epsilon)^{-2j\alpha^2} ds_1 \dots ds_{m-1}.$$ Separating out the variables, it follows that <span id="page-7-0"></span> $$I_{\epsilon}^{(2)}(\alpha) = \mathcal{O}\left(\epsilon^{(m-1)(1-m\alpha^2)}\right),\tag{22}$$ as $\epsilon \to 0$ , for $2\alpha^2 > 1$ . However, if $\frac{1}{m} < \alpha^2 < \frac{1}{2}$ , this approach fails to yield (22), and in fact yields a worse error term¹. To achieve the optimal error term (22) also in the case $\frac{1}{m} < \alpha^2 < \frac{1}{2}$ , we need to consider ordered integrals. Since the integral (20) is taken over all possible orderings, it follows that $$I_{\epsilon}^{(2)}(\alpha) = \sum_{\sigma \in S_{m-1}} \int_{\mathcal{W}_{\sigma}} \prod_{1 \le j \le k \le m-1} \left( \sum_{i=j}^{k} s_i + \epsilon \right)^{-2\alpha^2} ds_1 \dots ds_{m-1},$$ where the integral is taken over $$W_{\sigma} = \{0 < s_{\sigma(m-1)} < s_{\sigma(m-2)} < \dots < s_{\sigma(1)} < 1\}.$$ Let $$V(\ell) = \{(j,k) : 1 \le j \le k \le m-1 \text{ and } j = \sigma(\ell) \text{ or } k = \sigma(\ell)\}.$$ Let $S(\ell) = V(\ell) \setminus \bigcup_{j=1}^{\ell-1} V(j)$ . Then $S(1), \ldots, S(m-1)$ are disjoint, and $$\prod_{1 \le j \le k \le m-1} \left( \sum_{i=j}^k s_i + \epsilon \right)^{-2\alpha^2} = \prod_{\ell=1}^{m-1} \prod_{(j,k) \in S(\ell)} \left( \sum_{i=j}^k s_i + \epsilon \right)^{-2\alpha^2}.$$ By (21) it follows that $$\prod_{1 \le j \le k \le m-1} \left( \sum_{i=j}^k s_i + \epsilon \right)^{-2\alpha^2} \le \prod_{\ell=1}^{m-1} \prod_{(j,k) \in S(\ell)} \left( s_{\sigma(\ell)} + \epsilon \right)^{-2\alpha^2}.$$ $$I_{\epsilon}^{(\alpha)} \le \int_{0}^{1} ds_1 \int_{0}^{1} ds_2 (s_1 + \epsilon)^{-4/5} (s_2 + \epsilon)^{-8/5} = \mathcal{O}\left(\epsilon^{-3/5}\right),$$ as $\epsilon \to 0$ . However the optimal bound we are looking to obtain is of order $\epsilon^{(m-1)(1-m\alpha^2)} = \epsilon^{-2/5}$ . <span id="page-7-1"></span>^&lt;sup>1</sup>As an example where the approach fails to provide optimal error terms, consider m=3 and $\alpha^2=2/5$ . Then we obtain It is easily seen that $|S(\ell)| = m - \ell$ , and it follows that <span id="page-8-1"></span><span id="page-8-0"></span> $$\prod_{1 \le j \le k \le m-1} \left( \sum_{i=j}^k s_i + \epsilon \right)^{-2\alpha^2} \le \prod_{\ell=1}^{m-1} \left( s_{\sigma(\ell)} + \epsilon \right)^{-2\alpha^2(m-\ell)}. \tag{23}$$ By (23), it follows that $$\int_{\mathcal{W}_{\sigma}} \prod_{1 \le i \le k \le m-1} \left( \sum_{i=i}^{k} s_i + \epsilon \right)^{-2\alpha^2} ds_1 \dots ds_{m-1} \le \int_{\mathcal{W}_{\sigma}} \prod_{\ell=1}^{m-1} \left( s_{\sigma(\ell)} + \epsilon \right)^{-2\alpha^2(m-\ell)} ds_{\sigma(m-1)} \dots ds_{\sigma(1)}.$$ (24) If m=2 and $m\alpha^2 > 1$ , then the right hand side is of order $e^{-2\alpha^2+1}$ , and we are done. We assume that m>2, and integrate in $s_{\sigma(m-1)}$ on the right hand side of (24). The power of $s_{\sigma(m-1)}$ is $-2\alpha^2$ , which could very well be equal to -1, so we need to take this into account. Clearly $$\int_0^{s_{\sigma(m-2)}} (s_{\sigma(m-1)} + \epsilon)^x ds_{\sigma(m-1)} = \mathcal{O}\left(\log(s_{\sigma(m-2)}/\epsilon + 3)(\epsilon^{x+1} + (s_{\sigma(m-2)} + \epsilon)^{x+1})\right)$$ for any fixed x as $\epsilon \to 0$ . Since $$\log(s_{\sigma(m-2)}/\epsilon + 3) \le \log(s_{\sigma(1)}/\epsilon + 3),$$ it follows that <span id="page-8-2"></span> $$\int_{\mathcal{W}_{\sigma}} \prod_{\ell=1}^{m-1} \left( s_{\sigma(\ell)} + \epsilon \right)^{-2\alpha^{2}(m-\ell)} ds_{\sigma(m-1)} \dots ds_{\sigma(1)} = \mathcal{O}\left( \int \log(s_{\sigma(1)}/\epsilon + 3) \right) \\ \times \prod_{\ell=1}^{m-2} \left( s_{\sigma(\ell)} + \epsilon \right)^{-2\alpha^{2}(m-\ell)} \left( \epsilon^{-2\alpha^{2}+1} + \left( s_{\sigma(m-2)} + \epsilon \right)^{-2\alpha^{2}+1} \right) ds_{\sigma(m-2)} \dots ds_{\sigma(1)} , (25)$$ as $\epsilon \to 0$ where integration on the right hand side is taken over $0 < s_{\sigma(m-2)} < \cdots < s_{\sigma(1)} < 1$ . We will next integrate out $s_{\sigma(m-2)}$ , then $s_{\sigma(m-3)}$ , etc. To do this, we introduce the following notation for $v = 1, 2, \ldots, m-2$ : $$J_{\epsilon}(v) = \int \left[ \log(s_{\sigma(1)}/\epsilon + 3) \right]^{v} \prod_{\ell=1}^{m-v-1} \left( s_{\sigma(\ell)} + \epsilon \right)^{-2\alpha^{2}(m-\ell)}$$ $$\times \left[ \sum_{r=0}^{v} \epsilon^{r-2\alpha^{2} \sum_{j=1}^{r} j} \left( s_{\sigma(m-v-1)} + \epsilon \right)^{v-r-2\alpha^{2} \sum_{j=r+1}^{v} j} \right] ds_{\sigma(m-v-1)} \dots ds_{\sigma(1)},$$ with integration taken over $0 < s_{\sigma(m-v-1)} < \cdots < s_{\sigma(1)} < 1$ , and where we interpret $\sum_{j=1}^{0} j = \sum_{j=v+1}^{v} j = 0$ . We observe that the error term on the right hand side of (25) is equal to $J_{\epsilon}(1)$ . It is easily verified that $$J_{\epsilon}(v) = \mathcal{O}\left(J_{\epsilon}(v+1)\right),\,$$ as $\epsilon \to 0$ , for $v = 1, 2, \dots, m - 3$ . Iterating, we obtain $$\int_{\mathcal{W}_{\sigma}} \prod_{\ell=1}^{m-1} \left( s_{\sigma(\ell)} + \epsilon \right)^{-2\alpha^{2}(m-\ell)} ds_{\sigma(m-1)} \dots ds_{\sigma(1)} = \mathcal{O}(J_{\epsilon}(m-2)) =$$ $$\mathcal{O}\left( \int_{0}^{1} \left[ \log(s/\epsilon + 3) \right]^{m-2} (s+\epsilon)^{-2\alpha^{2}(m-1)} \left[ \sum_{r=0}^{m-2} \epsilon^{r-2\alpha^{2} \sum_{j=1}^{r} j} (s+\epsilon)^{m-2-r-2\alpha^{2} \sum_{r+1}^{m-2} j} \right] ds \right),$$ as $\epsilon \to 0$ , where we interpret $\sum_{j=1}^{0} j = \sum_{j=m-1}^{m-2} j = 0$ . Since $m\alpha^2 > 1$ , it follows that the power of $s + \epsilon$ is smaller than -1, namely: $$-2\alpha^{2}(m-1) + m - 2 - r - 2\alpha^{2} \sum_{r+1}^{m-2} j < -1,$$ for any r = 0, 1, 2, ..., m - 2. If x < -1, then $$\int_0^1 (s+\epsilon)^x \log(s/\epsilon+3)^{m-2} ds = \mathcal{O}(\epsilon^{x+1}),$$ as $\epsilon \to 0$ , and thus it follows that $$\int_{\mathcal{W}_{\sigma}} \prod_{\ell=1}^{m-1} \left( s_{\sigma(\ell)} + \epsilon \right)^{-2\alpha^2(m-\ell)} ds_{\sigma(m-1)} \dots ds_{\sigma(1)} = \mathcal{O}\left( \epsilon^{(m-1)(1-m\alpha^2)} \right),$$ as $\epsilon \to 0$ . Thus by (19) and (24), $I_{\epsilon}(\alpha) = \mathcal{O}\left(\epsilon^{(m-1)(1-m\alpha^2)}\right)$ , as $\epsilon \to 0$ , which combined with the lower bound (17) and Theorem 1.1 proves Corollary 2.1.

Proposition 4.1. (a) As $n \to \infty$ , $$\log \chi_n = -V_0/2 + \mathcal{O}(1/n),$$ uniformly for $t_1, \ldots, t_m \in \mathcal{S}_t$ . (b) There exists $U_1 > U_0 > 0$ , C > 0 and $n_0 > 0$ , such that if the parameters $t_1, \ldots, t_m$ satisfy condition $(U_0, U_1, n)$ and $n > n_0$ , then $$\left|\log \chi_n + V_0/2 + H_n(\alpha_j, \beta_j, t_j)_{j=1}^m\right| < C\left(\frac{1}{n^2} + \frac{1}{ne^{\widehat{u}_n}} + \frac{\epsilon_n}{n}\right),$$ where $$H_n(\alpha_j, \beta_j, t_j)_{j=1}^m = \frac{1}{2n} \sum_{j=1}^m (\alpha_j^2 - \beta_j^2) + \frac{1}{n} \sum_{1 \le j < k \le m} (\alpha_j \alpha_k - \beta_j \beta_k) \mathbb{1}_{U_0/n}(|t_k - t_j|),$$ where <span id="page-15-0"></span> $$\mathbb{1}_{U_0/n}(x) = \begin{cases} 1 & 0 < x < U_0/n, \\ 0 & U_0/n < x. \end{cases}$$ Using Proposition 4.1, we now compute the asymptotics of $D_n(f)$ as $n \to \infty$ , for a specific configuration $0 \le t_1 < t_2 < \cdots < t_m < 2\pi - \pi/m$ , but with error terms which are uniform over all configurations. Let $n_0$ be a fixed positive integer such that the asymptotics of Proposition 4.1 are valid for $n \ge n_0$ . Then $D_{n_0}(f)$ is a continuous function in terms of $t_j$ on the compact set $t_1, \ldots, t_m \in [0, 2\pi]$ , and is thus uniformly bounded as $t_j$ vary. Thus by (35) $$\log D_n = -2\sum_{N=n_0}^{n-1} \log \chi_N + \mathcal{O}(1), \tag{38}$$ as $n \to \infty$ , uniformly over $\mathcal{S}_t$ . Denote $\mathbb{N}_0 = \mathbb{N} \setminus \{0, 1, 2, \dots, n_0\}$ , and let $$J_t = \{ N \in \mathbb{N}_0 : t_1, \dots, t_m \text{ satisfy condition } (U_0, U_1, N) \},$$ with complement $J_t^c = \mathbb{N}_0 \setminus J_t$ . Then $$J_t^c = \bigcup_{j=1}^{m-1} I_{j,t}, \qquad I_{j,t} = \{ N \in \mathbb{N}_0 : (t_{j+1} - t_j) N \in [U_0, U_1) \}.$$ Written differently, we have $$I_{j,t} = \left\{ N \in \mathbb{N}_0 : \frac{U_0}{t_{j+1} - t_j} \le N < \frac{U_1}{t_{j+1} - t_j} \right\},$$ and it follows that $$\sum_{N \in I_{i,t}} \frac{1}{N} = \log(U_1/U_0) + \mathcal{O}(1),$$ uniformly for $t_{j+1} - t_j > 0$ . Since $U_0$ and $U_1$ are fixed, it follows that the right hand side is uniformly bounded, and by Proposition 4.1 (a) and the fact that $H_N = \mathcal{O}(1/N)$ we have <span id="page-15-1"></span> $$\sum_{N \in J_s^c} \left( \log \chi_n + V_0 / 2 + H_N(\alpha_j, \beta_j, t_j)_{j=1}^m \right) = \mathcal{O}(1), \tag{39}$$ uniformly for $t_1, \ldots, t_m \in [0, 2\pi - \pi/m)$ . Suppose that $t_1, \ldots, t_m$ satisfy condition $(U_0, U_1, N)$ for N in an interval $N_1, N_1 + 1, \ldots, N_2$ . By Proposition 4.1 (b), <span id="page-16-0"></span> $$\sum_{N_1}^{N_2} \left( \log \chi_n + V_0 / 2 + H_N(\alpha_j, \beta_j, t_j)_{j=1}^m \right) < C \sum_{N=N_1}^{N_2} \left( \frac{1}{N^2} + \frac{1}{Ne^{\widehat{u}_N}} + \frac{\epsilon_N}{N} \right), \tag{40}$$ Since $\hat{u}_N = N(t_i - t_{i-1})$ for some $i \in \{2, 3, ..., m\}$ (where i is fixed for $N \in [N_1, N_2]$ ), it follows that $\frac{\hat{u}_N}{N} = \frac{\hat{u}_{N_1}}{N_1}$ , and as a consequence (bearing in mind that $\frac{\hat{u}_{N_1}}{N_1} > U_1$ ) we have $\hat{u}_N > NU_1$ . Thus, bounding the sum by a suitable integral, $$\sum_{N=N_1}^{N_2} \frac{1}{Ne^{\hat{u}_N}} \le \frac{e^{-U_1}}{U_1}. \tag{41}$$ Similarly, $\epsilon_N/N = \epsilon_{N_2}/N_2$ , and thus <span id="page-16-1"></span> $$\sum_{N=N_1}^{N_2} \frac{\epsilon_N}{N} \le \epsilon_{N_2} \le U_0. \tag{42}$$ Since $J_t^c$ is composed of at most m-1 disjoint intervals, it follows that $J_t$ is composed of at most m disjoint intervals, and it follows by (40)-(42) that <span id="page-16-2"></span> $$\sum_{\substack{N \in J_t \\ N \le n}} \left( \log \chi_n + V_0 / 2 + H_N(\alpha_j, \beta_j, t_j)_{j=1}^m \right) \le Cm \left( \frac{1}{n_0} + \frac{e^{-U_1}}{U_1} + U_0 \right). \tag{43}$$ Since $U_0, U_1, n_0$ are just arbitrary constants, the right hand side is bounded uniformly over $S_t$ . Thus, by (38), (39), (43), it follows that $$\log D_n(f) = nV_0 + 2\sum_{N=1}^n H_N(\alpha_j, \beta_j, t_j)_{j=1}^m + \mathcal{O}(1),$$ uniformly over $\mathcal{S}_t$ . Since $$\sum_{N=1}^{n} \frac{1}{N} \mathbb{1}_{U_0/N}(t_j - t_i) = \log \frac{1}{t_j - t_i + 1/n} + \mathcal{O}(1),$$ as $n \to \infty$ , with the implicit constant depending only on $U_0$ which is fixed, it follows that $$\log D_n(f) = nV_0 + \sum_{j=1}^m (\alpha_j^2 - \beta_j^2) \log n + \sum_{1 \le j < k \le m} 2(\alpha_j \alpha_k - \beta_j \beta_k) \log \left(\frac{1}{|t_j - t_k| + n^{-1}}\right) + \mathcal{O}(1),$$ as $n \to \infty$ , uniformly over $S_t$ . For such $t_1, \ldots, t_m$ , we have $$\log\left(\frac{1}{|t_j - t_k| + n^{-1}}\right) = \log\left(\frac{1}{\sin\frac{|t_j - t_k|}{2} + n^{-1}}\right) + \mathcal{O}(1)$$ (44) with uniform error terms, which yields Theorem 1.1 for $t_1, \ldots, t_m \in \mathcal{S}_t$ , and the full theorem follows from the aforementioned rotational invariance of the Toeplitz determinant.

Proposition 5.1. Let $\alpha_j \geq 0$ and Re $\beta_j = 0$ for $j = 1, 2, ... \mu$ . There exists a unique solution to the Riemann–Hilbert problem for $\Phi$ .

Lemma 5.2. Let $\alpha_j \geq 0$ and Re $\beta_j = 0$ for $j = 1, 2, ..., \mu$ . Then the following two statements hold. (a) Given u > 0, <span id="page-21-1"></span> $$\Phi(\zeta)e^{\frac{\zeta}{2}\sigma_3}\prod_{j=1}^{\mu}(\zeta-iw_j)^{\beta_j\sigma_3}\exp\left[\pi i(-\beta_j+\alpha_j)\chi_{w_j}(\zeta)\sigma_3\right] = I + \mathcal{O}\left(\frac{1}{\zeta}\right),\tag{48}$$ as $\zeta \to \infty$ , uniformly for $-u/2 \le w_{\mu} < \cdots < w_1 \le u/2$ . (b) As $w_1 - w_{\mu} \to 0$ , <span id="page-21-2"></span> $$\Phi_1\left(\mu; (w_j, \alpha_j, \beta_j)_{j=1}^{\mu}\right) = \begin{pmatrix} \mathcal{A}^2 - \mathcal{B}^2 & -e^{-\pi i(\mathcal{A} + \mathcal{B})} \frac{\Gamma(1 + \mathcal{A} - \mathcal{B})}{\Gamma(\mathcal{A} + \mathcal{B})} \\ e^{\pi i(\mathcal{A} + \mathcal{B})} \frac{\Gamma(1 + \mathcal{A} + \mathcal{B})}{\Gamma(\mathcal{A} - \mathcal{B})} & \mathcal{B}^2 - \mathcal{A}^2 \end{pmatrix} + \mathcal{O}(w_1 - w_\mu), \tag{49}$$ where $\mathcal{A} = \sum_{j=1}^{\mu} \alpha_j$ and $\mathcal{B} = \sum_{j=1}^{\mu} \beta_j$ . The first step in the proof of Lemma 5.2 is to transform the RH problem for $\Phi$ to a RH problem for $\widehat{\Phi}$ which is analytic except on the imaginary axis Re z=0, and in particular the jump contour is independent of the locations of the singularities $w_j$ (though the jumps themselves will vary with the location of the singularities).

Lemma 6.1. Let $\alpha_j \geq 0$ and $\operatorname{Re} \beta_j = 0$ for $j = 1, \ldots, m$ . $R(z) = I + \mathcal{O}(\widehat{u}_n^{-1})$ , as $n \to \infty$ , uniformly for $z \in \mathbb{C}$ and $t_1, \ldots, t_m$ satisfying condition $(u, \widetilde{U}, n)$ , for some $\widetilde{U} > 0$ .

Lemma 6.2. Let $\alpha_j \geq 0$ and Re $\beta_j = 0$ . Then the following two statements hold. (a) As $n \to \infty$ , $$R(0) = I + \mathcal{O}(1/n),$$ uniformly for $t_1, \ldots, t_m$ satisfying condition $(u, \widetilde{U}, n)$ . (b) As $n \to \infty$ $$R(0) = I + \sum_{j=1}^{r} \int_{\partial U_j} \frac{\Delta_1(s)}{s} \frac{ds}{2\pi i} + \mathcal{O}\left(\frac{1}{n\widehat{u}_n}\right),\,$$ uniformly for $t_1, \ldots, t_m$ satisfying condition $(u, \widetilde{U}, n)$ , where $$\Delta_{1,22}(z) = \frac{\Phi_{1,11}\left(\mu_j; \left(w_{\nu}^{(j)}, \alpha_{\nu}^{(j)}, \beta_{\nu}^{(j)}\right)_{\nu=1}^{\mu_j}\right)}{\zeta_j(z)},$$ for $z \in \partial U_i$ .