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Rate of convergence in the central limit theorem for the determinantal point process with Bessel kernel S. M. Gorbunovabc a Landau Phystech School of Physics and Research, Moscow Institute of Physics and Technology, Dolgoprudny, Moscow Region, Russia b Ivannikov Institute for System Programming of the Russian Academy of Science, Moscow, Russia c Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, Russia
§ 1. IntroductionFor $f\in L_\infty(\mathbb{R}_+)\cap L_1(\mathbb{R}_+)$ consider the Bessel kernel
$$
\begin{equation}
B_f(x, y)=\int_{\mathbb{R}_+}t\sqrt{xy}J_\nu(xt)J_\nu(yt)f(t)\,dt,
\end{equation}
\tag{1.1}
$$
where $J_\nu$ is the Bessel function of order $\nu$. Fix $\nu> -1$. The kernel (1.1) induces a bounded linear operator on $L_2(\mathbb{R}_+)$, which we will also denote by $B_f$. We let $B_1$ be equal to the identity operator. We refer to $B_f$ as the Bessel operator. For any $h\in L_\infty(X)$ we also let $h$ denote the operator of pointwise multiplication by this function in $L_2(X)$. Let $\chi_A(x)$ be the characteristic function of the subset $A$. In this work we consider the Fredholm determinant $\det(I + \chi_{[0, R]}B_{f}\chi_{[0, R]})$ for $R>0$ and a sufficiently smooth function $f$. This determinant gives an exact expression for the Laplace transform of additive functionals in the determinantal point process with Bessel kernel (see [26]). We describe the relation between the determinantal point process and the operator in § 5. Briefly, the operator $B_{\chi_{[0, 1]}}$ induces a probability measure $\mathbb{P}_{J_\nu}$ on countable subsets without accumulation points of $\mathbb{R}_+$. These subsets, called configurations, are endowed with a certain $\sigma$-algebra (see [25]). For any measurable function $b$ on $\mathbb{R}_+$ an additive functional is defined as a measurable function on configurations by $S_b(X) = \sum_{x\in X}b(x)$. The probability measure on configurations induces a random variable $S_b$. For $b\in L_1(\mathbb{R}_+)\cap L_\infty(\mathbb{R}_+)$ the Laplace transform of $S_b$ is expressed by the formula
$$
\begin{equation*}
\mathbb{E}_{J_\nu}e^{\lambda S_b}= \det(\chi_{[0, 1]}B_{e^{\lambda b}}\chi_{[0, 1]}).
\end{equation*}
\notag
$$
We consider the additive functionals $S_b^R = S_{b(x/R)}$. We have
$$
\begin{equation*}
\det(\chi_{[0, 1]}B_{e^{\lambda b(x/R)}}\chi_{[0, 1]}) =\det(\chi_{[0, R]}B_{e^{\lambda b}}\chi_{[0, R]}).
\end{equation*}
\notag
$$
We conclude that the limit distribution of $S_b^R$ as $R\to\infty$ can be derived in terms of the asymptotic of the corresponding Fredholm determinant. Recall an analogous problem for the sine process $\mathbb{P}_{\mathcal{S}}$. Unlike the determinantal point process with Bessel kernel, it is a measure on configurations on the real line $\mathbb{R}$. It is induced by the integral kernel $K_{\mathcal{S}}(x, y)=(\sin \pi(x-y))/(\pi(x-y))$. For $b\in L_1(\mathbb{R})\cap L_\infty(\mathbb{R})$ the formula for the Laplace transform is in this case
$$
\begin{equation*}
\mathbb{E}_{\mathcal{S}}e^{\lambda S_b^R}= \det(\chi_{[0, 2\pi R]}W_{e^{\lambda b}}\chi_{[0, 2\pi R]}),
\end{equation*}
\notag
$$
where $W_{e^{\lambda b}}$ is the Wiener–Hopf operator with symbol $e^{\lambda b}$, which we define in § 2. For these determinants we recall two results. The first is the Kac–Akhiezer formula (see [10], § 10.13), which states that under certain conditions on $b$ we have
$$
\begin{equation*}
\det(\chi_{[0, 2\pi R]}W_{e^{\lambda b}}\chi_{[0, 2\pi R]})= \exp(\lambda R c_1^{\mathcal{S}}(b) + \lambda^2 c_2^{\mathcal{S}}(b))Q^{\mathcal{S}}_R(\lambda b),
\end{equation*}
\notag
$$
where $Q^{\mathcal{S}}_R(\lambda b)\to 1$ as $R\to\infty$; see Theorem 3 for the values of $c_1^{\mathcal{S}}(b)$ and $c_2^{\mathcal{S}}(b)$. It follows that the distribution of additive functionals approaches here the Gaussian distribution if $c^{\mathcal{S}}_1(b)\!=\!0$. The second result is a precise formula for the remainder term $Q^{\mathcal{S}}_R(\lambda b)$; see Theorem 3 again. These results are continuous analogues of the strong Szegö limit theorem and the Borodin–Okounkov identity for Toeplitz matrices (see [10], § 10.4, and [11], respectively). Borodin and Okounkov derived an explicit expression for the determinants of Toeplitz matrices from Gessel’s theorem and Schur measures. Several different proofs under less restrictive assertions were given later [6], [9]. In a similar way a precise formula for the Wiener–Hopf operators was obtained by Basor and Chen [2]. Under weaker assumptions about the symbol this result was deduced by Bufetov [12] directly from the Borodin–Okounkov formula. Having derived a formula for $Q^{\mathcal{S}}_R(\lambda b)$, it is possible to estimate the rate of convergence of additive functionals in the limit procedure under consideration. This was done by Bufetov in [13], Lemma 3.10. Theorem 2 gives a similar estimate for the Bessel kernel point process. An analogue of the Kac–Akhiezer formula in case of the Bessel kernel was derived by Basor and Ehrhardt [3]. We state it in Theorem 6. One consequence of Theorem 6 is the rate at which the distribution of additive functionals approaches the Gaussian one if $c_1^{\mathcal{B}}(b)=0$. In this paper we derive a precise expression for $Q^{\mathcal{B}}_R(\lambda b)$ and estimate the rate of its convergence to 1. This result has two special cases when $\nu=\pm 1/2$. In these cases the Bessel operator (1.1) is known to be the sum of the Wiener–Hopf and Hankel operators: where $H_f$ is the Hankel integral operator with kernel $\chi_{[0, \infty)^2}(x, y)\widehat{f}(x+y)$. In these cases an expression for $Q^{\mathcal{B}}_R(b)$ was derived by Basor, Ehrhardt and Widom [5].It is notable that in these cases $B_f$ has a discrete analogue, a sum of a Toeplitz and a Hankel matrix. An analogue of Szegő’s theorem for such matrices was proved by Johansson [18]. Furthermore, there is a counterpart of the Borodin–Okounkov formula for these matrices, a precise formula derived by Basor and Ehrhardt [4]. Another proof of this precise formula, which is similar to that of Borodin and Okounkov for Toeplitz matrices, was given by Betea [8], who used symplectic and orthogonal Schur measures. Lastly, recall some results on the rate of convergence of additive functionals for matrix ensembles. For example, classical compact groups were studied by Johansson [18]. In [19] authors considered the traces of random Haar-distributed matrices multiplied by deterministic ones. Lambert, Ledoux and Webb [20] studied the rate of convergence of additive functionals in $\beta$-ensembles. Gaussian Laguerre and Jacobi ensembles were considered by Bufetov and Berezin [7], who used the Deift–Zhou asymptotic analysis of the Riemann–Hilbert problem [14]. The authors of [1] studied convergence in Wigner ensembles. However, all ensembles mentioned above are finite-dimensional. On the other hand, the Bessel kernel point process is a measure on infinite configurations. Therefore, we pursue the operator-theoretic approach used in [2] and [3]. § 2. Statement of the resultFor a function $f\in L_1(\mathbb{R})\cap L_\infty(\mathbb{R})$ we define its Fourier transform by
$$
\begin{equation*}
\widehat{f}(\lambda)=\frac{1}{2\pi}\int_\mathbb{R} e^{-i\lambda x}f(x)\,dx.
\end{equation*}
\notag
$$
Here and below, if $f$ is defined on $\mathbb{R}_+$, then we extend it to an even function. In this case the Fourier transform coincides with the cosine transform. Recall that the Sobolev $p$-seminorm of a function $f$ on $\mathbb{R}_+$ is
$$
\begin{equation*}
\|f\|_{\dot{H}_{p}}^2=\int_{\mathbb{R}_+} |\widehat{f}(\lambda)|^2 |\lambda|^{2p}\,d\lambda.
\end{equation*}
\notag
$$
For an integer $p$ we have $\|f\|_{\dot{H}_{p}}=(2\pi)^{-1/2}\|f^{(p)}\|_{L_2(\mathbb{R}_+)}$ by Parseval’s theorem. We set $\|f\|_{H_p} = \|f\|_{L_2(\mathbb{R}_+)} \!+\! \|f\|_{\dot{H}_{p}}$ and define the Sobolev space $H_p(\mathbb{R}_+) \!=\! \{{f\!\in\! L_2(\mathbb{R}_+)}$: $\|f\|_{H_p}\!<\!+\infty\}$ to be a Banach space with the norm $\|\,{\cdot}\,\|_{H_p}$. The Sobolev space $H_p(\mathbb{R})$ on the real line is defined similarly, and we keep the same notation for the norm. We define the following seminorm of a function $f$ on $\mathbb{R}_+$:
$$
\begin{equation*}
\|f\|_{\dot{\mathcal B}}=\|f\|_{\dot{H}_{1}} + \|f\|_{\dot{H}_{3}} + \|xf(x)\|_{\dot{H}_{2}} + \|x^2f(x)\|_{\dot{H}_{3}}.
\end{equation*}
\notag
$$
Again, set $\|f\|_{\mathcal{B}}=\|f\|_{L_1(\mathbb{R}_+)} + \|f\|_{L_2(\mathbb{R}_+)}+\|xf'(x)\|_{L_\infty(\mathbb{R}_+)} +\|f\|_{\dot{\mathcal B}}$ and consider the space
$$
\begin{equation*}
\mathcal{B}=\{f\in L_2(\mathbb{R}_+)\colon \|f\|_{\mathcal{B}}<+\infty\}
\end{equation*}
\notag
$$
to be a Banach space with the norm $\|\,{\cdot}\,\|_{\mathcal{B}}$. For a function $f \in L_\infty(\mathbb{R})$ the Wiener–Hopf operator on $L_2(\mathbb{R}_+)$ is defined by
$$
\begin{equation*}
W_f=\chi_{\mathbb{R}_+}\mathcal{F}^{-1}f\mathcal{F}\chi_{\mathbb{R}_+},
\end{equation*}
\notag
$$
where $\mathcal{F}$ is the unitary Fourier transform on $L_2(\mathbb{R})$: $\mathcal{F}h = \sqrt{2\pi}\,\widehat{h}$, $h\in L_2(\mathbb{R})$. Again, if $f$ is defined on $\mathbb{R}_+$, then we extend it to an even function. Set
$$
\begin{equation*}
P_\pm=\mathcal{F}^{-1}\chi_{\mathbb{R}_\pm}\mathcal{F}\quad\text{and} \quad H_1^\pm(\mathbb{R})= P_\pm(H_1(\mathbb{R})).
\end{equation*}
\notag
$$
Then $H_1(\mathbb{R}) = H_1^+(\mathbb{R})\oplus H_1^-(\mathbb{R})$. For any function $b\in H_1(\mathbb{R})$ let $b_\pm = P_\pm b$; $b = b_+ + b_-$. The above space $\mathcal{B}$ is clearly a subspace of $H_1(\mathbb{R})$ with embedding given by even continuation, so the decomposition is well defined on $\mathcal{B}$, although the components can fail to remain in $\mathcal{B}$. We refer to this decomposition as Wiener–Hopf factorization. Our main result is then stated as follows. Theorem 1. Let $b \in \mathcal{B}$. Then for any $R>0$ Remark 1. Again, we mention the cases $\nu=\pm 1/2$. Substituting $B_b = W_b\mp H_b$ into the formula for $Q_R^{\mathcal{B}}(b)$ we obtain We now proceed to a corollary for the determinantal point process with Bessel kernel $\mathbb{P}_{J_\nu}$ (see § 5). Let $\overline{S_f^R} = S_f^R - \mathbb{E}_{J_\nu}S_f^R$. We denote by $F_{R, f}$ and $F_{\mathcal{N}}$ the cumulative distribution functions of $\overline{S_f^R}$ and the standard Gaussian, respectively. Theorem 2. Let $b\in\mathcal{B}$ be a real-valued function, satisfying $c_3^{\mathcal{B}}(b)=1/2$. Then there exists a constant $C=C(L(b), \|b_+\|_{L_\infty})$ providing the following estimate for the Kolmogorov–Smirnov distance for any $R\geqslant 1$ § 3. Structure of the paperFirst recall the result of Basor and Chen and the main steps of its proof. Theorem 3. For $f\in H_1(\mathbb{R})\cap L_1(\mathbb{R})$ we have Remark 2. We state Theorem 3 under assumptions different from [2] to outline clearer the proof of Theorem 1. The proof consists of three main steps. Step 1. The properties of Wiener–Hopf factorization give the following identity:
$$
\begin{equation}
\det(\chi_{[0, R]}W_{e^f}\chi_{[0, R]})= \exp(Rc_1^{\mathcal{S}}(f)) \det(\chi_{[0, R]}W_{e^{-f_+}}W_{e^f}W_{e^{-f_-}}\chi_{[0, R]}).
\end{equation}
\tag{3.1}
$$
Step 2. It is shown below that the operator $W_{e^{-f_+}}W_{e^f}W_{e^{-f_-}} - I$ is trace class. We can apply the Jacobi–Dodgson identity $\det(PAP) = \det(A)\det(QA^{-1}Q)$ for orthogonal projections $P$ and $Q$ satisfying $P+Q=I$ and a determinant-class invertible operator $A$. Then we have
$$
\begin{equation}
\frac{\det(\chi_{[0, R]}W_{e^{-f_+}}W_{e^f}W_{e^{-f_-}} \chi_{[0, R]})}{ \det(\chi_{[R, \infty)}W_{e^{f_-}} W_{e^{-f_+}}W_{e^{-f_-}}W_{e^{f_+}}\chi_{[R, \infty)})}= \det(W_{e^{-f_+}}W_{e^f}W_{e^{-f_-}}),
\end{equation}
\tag{3.2}
$$
where $W_{e^f}^{-1} = W_{e^{-f_+}}W_{e^{-f_-}}$, as follows from the properties of Wiener–Hopf factorization. Step 3. From the properties of the Fredholm determinant and Wiener–Hopf factorization we obtain $\det(W_{e^{-f_+}}W_{e^f}W_{e^{-f_-}}) = \det(W_{e^{f_-}}W_{e^{f_+}}W_{e^{-f_-}}W_{e^{-f_+}})$. Widom’s formula (see [27]) states that determinant on the right is well defined and its value is
$$
\begin{equation}
\det(W_{e^{f_-}}W_{e^{f_+}}W_{e^{-f_-}}W_{e^{-f_+}})=\exp(c_2^{\mathcal{S}}(f)).
\end{equation}
\tag{3.3}
$$
This finishes the proof. We now pass to the proof of Theorem 1. By contrast to the proof of Theorem 3, it is not known if $W_{e^{-b_+}}B_{e^{b}}W_{e^{-b_-}}-I$ is trace class, so the Jacobi–Dodgson identity cannot be applied directly. However, one can observe that equality (3.2) means that the ratio of the determinants in the left-hand side does not depend on $R$. In our case we can use an analogue of the Jacobi–Dodgson identity (Corollary 2) to show the independence of this ratio of $R$. Lemma 2. Let $b\in H_1(\mathbb{R}_+)\cap L_1(\mathbb{R}_+)$, and let $\chi_{[R,\infty)}(B_{e^{-b}} - W_{e^{-b}})\chi_{[R, \infty)}$ be a trace-class operator for each $R>0$. Then there exists $Z(b)$ such that for all $R>0$ we have Observe that the denominator in (3.5) equals $Q_R^{\mathcal{B}}(b)$. The central step in the proof of Theorem 1 is as follows. Lemma 3. For $b\in L_1(\mathbb{R}_+)\cap L_\infty(\mathbb{R}_+)$ such that $\|b\|_{\dot{\mathcal B}}< \infty$ and $z\in\mathbb{C}$ the operator $\chi_{[R, \infty)}(B_{b+z} - W_{b+z})\chi_{[R, \infty)}$ is trace class for any $R>0$. There exists a constant $C$ such that for any $R\geqslant1$, $z\in\mathbb{C}$ and $b$ satisfying the conditions above we have The operator $Q_R^{\mathcal{B}}(b)$ is related to the operator in Lemma 3 by formula (8.5) below. This allows us to obtain estimate (2.2). Further, the asymptotic for the numerator in (3.5) follows from Lemma 1 and an asymptotic result of Basor and Ehrhardt (see Theorem 6). This proves that $Z(b) = \exp(c_2^{\mathcal{B}}(b) + c_3^{\mathcal{B}}(b))$. The rest of the paper has the following structure. In § 4 we recall some facts and notation for trace-class and Hilbert–Schmidt operators. Then in § 5 we explain the relation between the Fredholm determinant in Theorem 1 and the determinantal point process with Bessel kernel. In § 6 we establish some properties of Wiener–Hopf factorization and deduce Lemma 1. Section 7 presents the proof of Lemma 3. We complete the proof of Theorems 1 and 2 in § 8. § 4. Trace class and Hilbert–Schmidt operatorsHere and below we denote the ideals of trace-class and Hilbert–Schmidt operators by $\mathcal{J}_1$ and $\mathcal{J}_2$, respectively. For more details of the theory, we refer the reader to [24] and [22]. We let $E$ denote $\mathbb{R}$ or $\mathbb{R}_+ = [0, \infty)$. Definition 1. The operator $K$ on $L_2(E)$ is locally trace class if for each bounded measurable subset $B$ of $ E$ the operator $\chi_B K \chi_B$ is trace class. Let $\mathcal{J}_1^{\mathrm{loc}}(L_2(E))$ denote the space of locally trace-class operators. Recall that for $K\in \mathcal{J}_1$ the Fredholm determinant $\det (I+K)$ is well defined, which is equal to a product, $\det(I+K)=\prod_{\lambda\in\sigma(K)}(1+\lambda)$, by Lidskii’s theorem. It is continuous with respect to the trace norm. We also define the regularized determinant by
$$
\begin{equation*}
\det{}_2 (I + K)=\exp(-\operatorname{Tr}(K))\det(I+K).
\end{equation*}
\notag
$$
It is continuous with respect to the Hilbert–Schmidt norm, so it can extended by continuity to all operators in $\mathcal{J}_2$. The following theorem implies that for a function $f\in L_1(\mathbb{R}_+)\cap L_\infty(\mathbb{R}_+)$ the determinants $\det (I + \chi_{[0, R]}B_f\chi_{[0, R]})$ and $\det(I + \chi_{[0, R]}W_f\chi_{[0, R]})$ are well defined. Theorem 4 ([22], Theorem 3.11.9). Let $K$ be a continuous kernel on $[a, b]^2$ inducing a positive self-adjoint operator $K$ on $L_2[a, b]$. Then the operator $K$ is trace class. In addition, for any trace-class $K$ with continuous kernel we have We refer to (4.1) as the calculation of the trace over the diagonal. Recall that the kernel of an integral operator is defined almost everywhere on $[a, b]^2$. Thus, for an operator satisfying the assumptions of Theorem 4, formula (4.1) can fail to hold after we change the kernel on the diagonal, although the induced operator remains the same. This explains the assumption of continuity. Let us now extend the class of kernels for which we can calculate the trace over the diagonal. Let $K$ be a self-adjoint trace-class operator. By the Hilbert–Schmidt theorem we have a spectral decomposition, so that its kernel can be chosen in the form
$$
\begin{equation}
K(x, y)=\sum_{i\in\mathbb{N}}\lambda_i\phi_i(x)\phi^*_i(y),
\end{equation}
\tag{4.2}
$$
where $\{\phi_i\}_{i\in\mathbb{N}}$ forms an orthonormal basis. Now, by definition the trace is
$$
\begin{equation*}
\operatorname{Tr}(K)=\int_a^b \sum_{i\in\mathbb{N}}\lambda_i\phi_i(x)\phi^*_i(x)\,dx,
\end{equation*}
\notag
$$
which is equal to $\displaystyle\int_a^b K(x, x)\,dx$ if we choose $K$ as in (4.2). Observe that this does not require the kernel to be continuous, but this choice is better defined: we can only change $K$ on subsets of $[a, b]^2$ having projections with zero measure. Choosing similarly the kernel of an operator $fK$:
$$
\begin{equation*}
(fK)(x, y)=\sum_{i\in\mathbb{N}}\lambda_i (f(x)\phi_i(x))\phi_i^*(y),
\end{equation*}
\notag
$$
we see that then $\displaystyle\operatorname{Tr}(fK)=\int_a^b f(x)K(x, x)\, dx$ in a similar way. We use these observations in § 5. We will also use the following theorem. Theorem 5 ([15], Theorem 2.2). For $A, B \in \mathfrak{B}(\mathcal{H})$ such that $[A,B]\in \mathcal{J}_1$ we have $e^Ae^Be^{-A-B}- I\in \mathcal{J}_1$. For the Fredholm determinant we have § 5. A link between the Bessel operator and the determinantal point process with Bessel kernelLet $E$ be $\mathbb{R}$ or $\mathbb{R}_+ = [0, \infty)$. Recall that a configuration on $E$ is a discrete subset of $E$. We denote the set of configurations by $\operatorname{Conf}(E)$. It can be endowed with a $\sigma$-algebra of measurable subsets $\mathfrak{X}$ (for instance, see [25]). A point process $\mathbb{P}$ is defined to be a probability measure on $(\operatorname{Conf}(E), \mathfrak{X})$. For a measurable function $f\colon E\to\mathbb{C}$ we define two measurable functions on configurations by the following formulae:
$$
\begin{equation*}
S_f(X)=\sum_{x\in X}f(x) \quad\text{and}\quad \Psi_{1+f}(X)=\prod_{x\in X}(1+f(x));
\end{equation*}
\notag
$$
they are called additive and multiplicative functionals, respectively. Definition 2. A point process $\mathbb{P}_K$ on $\operatorname{Conf}(E)$ is determinantal if there exists an operator $K\in \mathcal{J}_1^{\mathrm{loc}}(L_2(E))$ satisfying the following condition: for any bounded measurable function $f$ with compact support $B=\operatorname{supp} f$ We consider the limit distribution of $S_f^R = S_{f(x/R)}$ as $R\to\infty$ for the determinantal point process with Bessel kernel $\mathbb{P}_{J_\nu}$ on $\mathbb{R}_+$ [26] induced by the operator $B_{\chi_{[0, 1]}}$. Theorem 3 describes the limit under consideration for the sine process $\mathbb{P}_{\mathcal{S}}$ on $\mathbb{R}$ induced by the operator $\mathcal{F}^{-1}\chi_{[-\pi, \pi]} \mathcal{F}$ on $L_2(\mathbb{R})$. To be precise, we formulate the following statement. Proposition 1. For $f\in L_1(E)\cap L_\infty(E)$ we have and The Laplace transform of additive functionals is expressed in terms of multiplicative functionals by the formula
$$
\begin{equation*}
\mathbb{E}e^{\lambda S_b^R}=\mathbb{E}\Psi^R_{e^{\lambda b}}.
\end{equation*}
\notag
$$
By Proposition 1 it is equal to the Fredholm determinant $\det(\chi_{[0, R]}B_{e^{\lambda b}}\chi_{[0, R]})$ for $\mathbb{P}_{J_\nu}$ and to $\det(\chi_{[0, 2\pi R]} W_{e^{\lambda b}}\chi_{[0, 2\pi R]})$ for the sine process. To prove Proposition 1 we follow Bufetov (see [13], § 2.9). We extend the class of functions for which (5.1) holds. Let $K$ be a Hilbert–Schmidt self-adjoint operator on $L_2(E)$. We choose its kernel $K$ to coincide with its spectral decomposition (see § 4). We set
$$
\begin{equation}
\det (I+K)=\exp\biggl( \int_E K(x, x)\,dx \biggr)\det{}_2(I+K)
\end{equation}
\tag{5.4}
$$
if the integral of the diagonal exists. Observe, that this definition does not require $K$ to be trace class. However, if it is trace class, then we obtain the ordinary Fredholm determinant. Let $\Pi\in \mathcal{J}_1^{\mathrm{loc}}(L_2(E))$ be an orthogonal projection. By the Macchi–Soshnikov theorem (see [21] and [25]) $\Pi$ defines a point process $\mathbb{P}_\Pi$. We choose the kernel $\Pi(x, y)$ to coincide with the spectral decomposition of $\chi_B \Pi \chi_B$ for every compact subset $B$ of $ E$. Consider the measure $d\mu_\Pi(x)=\Pi(x, x)\,dx$ on $E$. Proposition 2. If $f\in L_1(E, d\mu_\Pi)\cap L_\infty(E, d\mu_\Pi)$, then $\Psi_{1+f}\in L_1(\operatorname{Conf}(E), \mathbb{P}_\Pi)$ and In other words, multiplicative functionals define a continuous mapping Proof. The formula holds for $f\in L_\infty(E)$ with compact support The exponential of the integral of the diagonal in (5.4) is continuous with respect to the norm in $L_1(E, d\mu_\Pi)$. The regularized determinant is continuous with respect to the Hilbert–Schmidt norm, which is equal to
$$
\begin{equation*}
\begin{aligned} \, \|f\Pi\|_{\mathcal{J}_2}^2 &=\int_E|f(x)\Pi(x, y)|^2\,dx\,dy =\int_E|f(x)|^2\Pi(x, y)\Pi(y, x)\,dx\,dy \\ &=\int_E|f(x)|^2\,d\mu_\Pi(x), \end{aligned}
\end{equation*}
\notag
$$
and therefore it is continuous with respect to the $L_2(E, d\mu_\Pi)$-norm. Thus, multiplicative functionals define a continuous mapping of a dense subset of $L_1(E, d\mu_\Pi)\cap L_\infty(E, d\mu_\Pi)$ and therefore of the whole space. This extension coincides by continuity with the expectation of the multiplicative functionals by Beppo Levi’s theorem, since one can approximate any multiplicative functional $\Psi_{1+|f|}$, $f\in L_1(E, d\mu_\Pi)\cap L_\infty(E, d\mu_\Pi)$ by the functionals $\Psi_{1+|f_n|}$, $|f_n|=\chi_{[-n, n]}|f|$. Hence (5.5) is an extension of (5.1) by continuity. The proposition is proved. Proof of Proposition 1. The proof for the sine process case can be found in [13], Lemma 3.3. We proceed differently by proving the equality between the extended determinants of $I + f\mathcal{F}^{-1}\chi_{[-\pi, \pi]}\mathcal{F}$ and $I + fB_{\chi_{[0, 1]}}$ and the Fredholm determinants of $I + \chi_{[0, 2\pi]}W_f\chi_{[0, 2\pi]}$ and $I + \chi_{[0, 1]}B_f\chi_{[0, 1]}$, respectively. Note that by Theorem 4 we can take the kernel of $B_{\chi_{[0, 1]}}$ to be equal to (1.1), and the kernel of $\mathcal{F}^{-1}\chi_{[-\pi,\pi]}\mathcal{F}$ to be $\sin(\pi(x-y))/(\pi(x-y))$. It is straightforward to see in both cases that the integrals over the diagonals of the kernels coincide.
Now we show that the regularized determinants coincide. This is a consequence of unitary invariance. Recall that $B_{\chi_{[0,1]}}$ is unitarily equivalent to $\chi_{[0, 1]}$ via the Hankel transform $H_\nu$ (see § 8 and formula (8.4)). Then we have the unitary equivalence
$$
\begin{equation*}
fH_\nu \chi_{[0, 1]}H_\nu \sim H_\nu f H_\nu\chi_{[0, 1]},
\end{equation*}
\notag
$$
where the last operator has the same regularized determinant as $\chi_{[0, 1]}B_f\chi_{[0, 1]}$. Notably, the operator obtained is trace class and its Fredholm determinant is well defined.
We turn to the proof for the sine kernel. We use the translation operator $T_tf(x) = f(x+t)$ to proceed as follows:
$$
\begin{equation*}
\mathcal{F}^*\chi_{[-\pi, \pi]}\mathcal{F}f= \mathcal{F}T_{-\pi}\chi_{[0, 2\pi]}T_{\pi} \mathcal{F}^*f=e^{-i\pi x}\mathcal{F} \chi_{[0, 2\pi]}\mathcal{F}^*fe^{i\pi x} \sim \mathcal{F} \chi_{[0, 2\pi]}\mathcal{F}^*f.
\end{equation*}
\notag
$$
Now the Fourier transform gives the following unitary equivalence:
$$
\begin{equation*}
\mathcal{F} \chi_{[0, 2\pi]}\mathcal{F}^*f \sim \chi_{[0, 2\pi]}\mathcal{F}^*f\mathcal{F},
\end{equation*}
\notag
$$
where the determinant of the obtained operator coincides, as above, with the one for $\chi_{[0, 2\pi]}W_f\chi_{[0, 2\pi]}$. Again, $\chi_{[0, 2\pi]}W_f\chi_{[0, 2\pi]}$ is trace class.
In both cases conjugation by the unitary operator $U_R f(x)=R^{-1/2}f(x/R)$ completes the proof. The proposition is proved. § 6. Wiener–Hopf factorization on $H_1( {\mathbb{R}})$Recall that the Wiener–Hopf operator as a map $\mathbf{W}\colon f\mapsto W_f$, $L_\infty(\mathbb{R})\to \mathfrak{B}(L_2(\mathbb{R}_+))$ is not a Banach algebra homomorphism. But it can be restricted to some subalgebras on which it preserves multiplication. Recall that $H_1(\mathbb{R})$ is a Banach algebra. The spaces
$$
\begin{equation*}
H_1^\pm(\mathbb{R})=\{f\in H_1(\mathbb{R})\colon \operatorname{supp}\widehat{f}\subset\mathbb{R}_\pm\}
\end{equation*}
\notag
$$
are Banach subalgebras since $\operatorname{supp}(\widehat{f}*\widehat{g}) \subset \operatorname{supp}\widehat{f} + \operatorname{supp}\widehat{g}$. The operators $P_\pm = \mathcal{F}^{-1} \chi_{\mathbb{R}_\pm}\mathcal{F}$ map $H_1(\mathbb{R})$ into itself and $P_\pm H_1(\mathbb{R})=H_1^\pm(\mathbb{R})$. Then $\mathbf{W}$, as restricted to $H_1^\pm(\mathbb{R})$, is a Banach algebra homomorphism. As before, we consider $H_1(\mathbb{R}_+)$ to be embedded in $H_1(\mathbb{R})$ by an even continuation, which defines a decomposition for functions on $\mathbb{R}_+$. We also let
$$
\begin{equation*}
H_1^\pm(\mathbb{R})\oplus \mathbb{C}=\{f+c\colon f\in H_1^\pm(\mathbb{R}),\, c\in\mathbb{C}\}
\end{equation*}
\notag
$$
be the Banach algebras $H_1^\pm(\mathbb{R})$ with adjoined unit. Then for $b\in H_1^\pm\oplus\mathbb{C}$ and an entire function $\Phi=\sum_{i=0}^\infty \Phi_iz^i$ the element
$$
\begin{equation*}
\Phi(b)=\sum_{i=0}^\infty \Phi_i b^i\in H_1^\pm(\mathbb{R})\oplus\mathbb{C}
\end{equation*}
\notag
$$
is well defined. Recall that for functions in the spaces under consideration the Wiener–Hopf operators have kernels. Lemma 4 ([3], Proposition 5.2). Let $a$ be a function on $\mathbb{R}$ such that $\|a\|_{H_{1/2}} + \|\widehat{a}\|_{L_1}<\infty$. Then $\mathcal{F}^{-1}a\mathcal{F}$ is an integral operator on $L_2(\mathbb{R})$ with kernel $k(x, y) = \widehat{a}(x-y)$. Proposition 3. 1. The map $\boldsymbol{W}\big|_{H_1^\pm(\mathbb{R})}\colon f \mapsto W_f$ defines Banach algebra homomorphisms $H_1^\pm(\mathbb{R})\oplus \mathbb{C}\to \mathfrak{B}(L_2(\mathbb{R}_+))$. 2. For $b_{\pm}\in H_1^\pm(\mathbb{R})\oplus \mathbb{C}$ we have
$$
\begin{equation*}
\begin{gathered} \, \chi_{[0, R]}W_{b_+}=\chi_{[0, R]}W_{b_+}\chi_{[0, R]}, \qquad W_{b_+}\chi_{[R, \infty)}=\chi_{[R, \infty)}W_{b_+}\chi_{[R, \infty)}, \\ W_{b_-}\chi_{[0, R]}=\chi_{[0, R]}W_{b_-}\chi_{[0, R]}, \qquad \chi_{[R, \infty)}W_{b_-}=\chi_{[R, \infty)}W_{b_-}\chi_{[R, \infty)}. \end{gathered}
\end{equation*}
\notag
$$
3. For $b_\pm\in H_1^\pm(\mathbb{R})\oplus\mathbb{C}$ we have $W_{b_-}W_{b_+}=W_{b_-b_+}$. In particular, $W_{e^{b_\pm}}=\exp(W_{b_\pm})$ and $W_{e^b}=\exp(W_{b_-})\exp(W_{b_+})$. Proof. It is sufficient to prove the statements for $b_\pm\in H_1^\pm(\mathbb{R})$. We give proofs for $H_1^+(\mathbb{R})$; the case of $H_1^-(\mathbb{R})$ is proved similarly.
1. By Lemma 4 Wiener–Hopf operators have kernels. Let $a$ and $b$ be functions in $H_1^+(\mathbb{R})$. Since $\widehat{b}_\pm(x) = \chi_{\mathbb{R}_\pm}(x)\widehat{b}_\pm(x)$, we have
$$
\begin{equation*}
\begin{aligned} \, W_{ab}(x, y) &=\chi_{\mathbb{R}_+^2}(x, y)\int_\mathbb{R} \widehat{a}(x-y-t)\chi_{\mathbb{R}_+}(t)\widehat{b}(t)\chi_{\mathbb{R}_\pm}(t)\,dt \\ &=\chi_{\mathbb{R}_+^2}(x, y)\int_\mathbb{R}\widehat{a}(x-t) \chi_{\mathbb{R}_+}(t-y)\widehat{b}(t-y)\,dt \\ &=\chi_{\mathbb{R}_+^2}(x, y)\int_\mathbb{R}\widehat{a}(x-t) \chi_{\mathbb{R}_+}(t)\widehat{b}(t-y)\,dt =(W_a W_b)(x, y), \end{aligned}
\end{equation*}
\notag
$$
where the identity $\chi_{\mathbb{R}_+}(y)\chi_{\mathbb{R}_+}(t-y) = \chi_{\mathbb{R}_+}(y)\chi_{\mathbb{R}_+}(t-y)\chi_{\mathbb{R}_+}(t)$ has been used.
2. Since $\chi_{[0, R]}(x)\chi_{\mathbb{R}_+}(x-y) = \chi_{[0, R]}(x)\chi_{\mathbb{R}_+}(x-y)\chi_{[0, R]}(y)$ for all $(x, y) \in \mathbb{R}_+$, for the kernel of $(\chi_{[0, R]}W_{b_+})(x, y)$ we can write
$$
\begin{equation*}
\begin{aligned} \, (\chi_{[0, R]}W_{b_+})(x, y) &=\chi_{[0, R]}(x) \widehat{b}_+(x-y)\chi_{\mathbb{R}_+}(y) \\ &=\chi_{[0, R]^2}(x, y)\widehat{b}_+(x-y) =(\chi_{[0, R]}W_{b_+}\chi_{[0, R]})(x, y). \end{aligned}
\end{equation*}
\notag
$$
The proof for $W_{b_+}\chi_{[R, +\infty)}$ follows from the equality
$$
\begin{equation*}
\chi_{[R, \infty)}(y)\chi_{\mathbb{R}_+}(x-y)\chi_{\mathbb{R}_+}(x)= \chi_{[R, \infty)}(y)\chi_{\mathbb{R}_+}(x-y)\chi_{[R, \infty)}(x).
\end{equation*}
\notag
$$
3. This is similar to the first part and follows from the equality
$$
\begin{equation*}
W_{b_-b_+}(x, y)=\chi_{\mathbb{R}_+^2}(x, y) \int_\mathbb{R}\widehat{b}_-(x - y - t)\widehat{b}_+(t)\chi_{\mathbb{R}_+}(t)\,dt
\end{equation*}
\notag
$$
after the substitution $\chi_{\mathbb{R}_+}(y)\chi_{\mathbb{R}_+}(t-y) = \chi_{\mathbb{R}_+}(y) \chi_{\mathbb{R}_+}(t-y)\chi_{\mathbb{R}_+}(t)$.
The proposition is proved. Proof of Lemma 1. The properties of Fredholm determinants and the first and second parts of Proposition 3 yield the following expression:
$$
\begin{equation*}
\begin{aligned} \, \det (\chi_{[0, R]}B_{e^b}\chi_{[0, R]}) &= \det(\chi_{[0, R]}W_{e^{b_+}}W_{e^{-b_+}}B_{e^b} W_{e^{-b_-}}W_{e^{b_-}}\chi_{[0, R]}) \\ &=\det (\chi_{[0, R]}W_{e^{-b_+}}B_{e^b}W_{e^{-b_-}}\chi_{[0, R]}) \det (\exp(\chi_{[0, R]}W_{b_-}\chi_{[0, R]}) \\ &\qquad\times \exp(\chi_{[0, R]}W_{b_+}\chi_{[0, R]})). \end{aligned}
\end{equation*}
\notag
$$
It remains to prove that
$$
\begin{equation}
\det (\exp(\chi_{[0, R]}W_{b_-}\chi_{[0, R]}) \exp(\chi_{[0, R]}W_{b_+}\chi_{[0, R]}))=\exp(R\widehat{b}(0)).
\end{equation}
\tag{6.1}
$$
By Theorem 4 the operator $\chi_{[0, R]}W_b\chi_{[0, R]}$ is trace class for $b\in L_1(\mathbb{R}_+)\cap L_\infty(\mathbb{R}_+)$, so the left-hand side of (6.1) is equal to
$$
\begin{equation*}
\begin{aligned} \, &\det (\exp(\chi_{[0, R]}W_{b_-}\chi_{[0, R]}) \exp(\chi_{[0, R]}W_{b_+}\chi_{[0, R]})\exp(-\chi_{[0, R]} W_b\chi_{[0, R]})) \\ &\qquad\times\exp(\operatorname{Tr}(\chi_{[0, R]}W_b\chi_{[0, R]})), \end{aligned}
\end{equation*}
\notag
$$
where $\operatorname{Tr}(\chi_{[0, R]}W_b\chi_{[0, R]}) = R\widehat{b}(0)$ is calculated over the diagonal. A direct calculation shows that the operator $\chi_{[0, R]}W_f\chi_{[0, R]}$ is Hilbert–Schmidt for $f\in H_1(\mathbb{R}_+)$. Hence the commutator $[\chi_{[0, R]} W_{b_-}\chi_{[0, R]}, \chi_{[0,R]}W_{b_+}\chi_{[0, R]}]$ is trace class, and its trace is zero. Theorem 5 gives
$$
\begin{equation*}
\det (\exp(\chi_{[0, R]}W_{b_-}\chi_{[0, R]}) \exp(\chi_{[0, R]}W_{b_+}\chi_{[0, R]}) \exp(-\chi_{[0, R]}W_b\chi_{[0, R]}))=1.
\end{equation*}
\notag
$$
This finishes the proof of (6.1). Lemma 1 is proved. § 7. Proof of Lemma 3In this section we establish Lemma 3. Our calculations follow Basor and Ehrhardt (see [3], Lemmas 2.6–2.8). The section is structured as follows. In Propositions 4 and 5 and Corollary 1 we establish estimates for the trace norms of integral operators with kernels of certain form. Then we recall the necessary properties of Bessel functions and their derivatives. We end the preparations to the proof by Lemma 5, establishing some properties of the function spaces in question. The rest of the section is devoted to the proof of Lemma 3 itself. Proposition 4. Let $K$ be an integral operator on $L_2[R, \infty)$ with kernel where $h_1$, $h_2$ and $a$ are some measurable functions. Then the estimate holds, provided that the right-hand side is finite. Proof. Let $f, g\in L_2[R, \infty)$. Set $h^t_i(x) = h_i(x, t)$, $i=1, 2$. Then
$$
\begin{equation}
\begin{aligned} \, \notag \langle f, Kg\rangle_{L_2} &=\int_R^\infty\int_R^\infty \biggl (\int_0^\infty a(t)h_1(x, t)h_2(y, t)\,dt\biggr )g(y)\,dy f^*(x)\,dx \\ \notag &=\int_0^\infty a(t)\biggl ( \int_R^\infty h_1(x, t)f^*(x)\,dx\biggr ) \biggl ( \int_R^\infty h_2(y, t)g(y)\,dy\biggr) dt \\ &=\int_0^\infty a(t)\langle f, h^t_1\rangle_{L_2} \langle (h^t_2)^*, g\rangle_{L_2} \,dt, \end{aligned}
\end{equation}
\tag{7.2}
$$
where Fubini’s theorem can be applied since the function under the integral is absolutely integrable by the assumption that the right-hand side of (7.1) is finite. Recall that $\mathfrak{B}(L_2[R, \infty)) \simeq \mathcal{J}_1^*(L_2[R, \infty))$, where $A\in\mathfrak{B}(L_2[R, \infty)$ acts on $\mathcal{J}_1^*(L_2[R, \infty))$ by the rule $X\mapsto \operatorname{Tr}(AX)$. For the norm of $K$, considering it as a functional on $\mathfrak{B}(L_2[R, \infty))$ and using the definition of a trace and the expression (7.2), we have
$$
\begin{equation*}
\begin{aligned} \, \|K\|_{\mathcal{J}_1} &=\sup_{B\in\mathfrak{B}(L_2), \|B\|=1}|{\operatorname{Tr}(BK)}| \\ &=\sup_{B\in\mathfrak{B}(L_2), \|B\|=1}\biggl|\sum_{i\in \mathbb{N}}\int_0^\infty a(t)\langle f_i, B h_1^t\rangle_{L_2}\langle (h_2^t)^*, f_i\rangle_{L_2}\,dt\biggr|, \end{aligned}
\end{equation*}
\notag
$$
where $\{f_i\}_{i\in\mathbb{N}}$ is an arbitrary orthonormal basis in $L_2[R, \infty)$. Next we use the Cauchy–Bunyakovsky–Schwarz inequality to obtain
$$
\begin{equation*}
\begin{aligned} \, &\sum_{i\in \mathbb{N}}|\langle f_i, B h_1^t\rangle_{L_2} \langle (h_2^t)^*, f_i\rangle_{L_2}| \\ &\qquad\leqslant \biggl (\sum_{i\in\mathbb{N}}|\langle f_i, B h_1^t\rangle_{L_2}|^2\biggr)^{1/2} \biggl (\sum_{i\in\mathbb{N}}|\langle (h_2^t)^*, f_i\rangle_{L_2}|^2\biggr)^{1/2} \leqslant \|h_1^t\|_{L_2}\,\|h_2^t\|_{L_2}, \end{aligned}
\end{equation*}
\notag
$$
which finishes the proof. Proposition 5. Let the kernel $K(x, y)$ of an integral operator $K$ be absolutely continuous with respect to $y$ on $[a, b]$. Assume that $K = K\chi_{[a, b]}$. Consider the operator $\partial_yK$ with kernel ${\partial K(x, y)}/{\partial y}$ and assume that $\partial_y K\in\mathcal{J}_2$. Then Proof. Let $Pf(x)=(1/(b-a))\chi_{[a, b]}(x)\langle \chi_{[a, b]}, f\rangle_{L_2}$ be a one-dimensional projector onto $\chi_{[a, b]}$, and let $Q=\chi_{[a, b]}-P$. We decompose the operator: $K = KP + KQ$. The operator $KP$ is also a one-dimensional projector with trace norm of at most
Recall that the Volterra operator $V$ on $L_2([a, b])$ is defined by the formula $\displaystyle Vf(x)=\int_a^x f(t)\,dt$. It is a Hilbert–Schmidt operator with kernel $V(x, y) = \chi_{[a, x]}(y)$, and its Hilbert–Schmidt norm is
$$
\begin{equation*}
\|V\|_{\mathcal{J}_2}^2=\int_a^b\,dx\int_a^x\,dy=\frac{(b-a)^2}{2}.
\end{equation*}
\notag
$$
Integrating $KQ$ by parts we obtain where by the definition of $V$ and $Q$. The inequality $\|\partial_yKVQ\|_{\mathcal{J}_1}\leqslant \|\partial_yK\|_{\mathcal{J}_2}\,\|V\|_{\mathcal{J}_2}$ finishes the proof.Corollary 1. There exists a constant $C$ such that for any kernel $K$ of an integral operator on $L_2(\mathbb{R}_+)$ with the estimates for some constant $A$, for any $R>0$ we have Proof. By Proposition 5 we have
$$
\begin{equation*}
\begin{aligned} \, &\|\chi_{[R, \infty)}K\chi_{[R, \infty)}\|_{\mathcal{J}_1} \leqslant \sum_{l=0}^\infty \|\chi_{[R, \infty)}K\chi_{[R + l,R + l+1]}\|_{\mathcal{J}_1} \\ &\qquad \leqslant \sum_{l=0}^\infty(\|\chi_{[R, \infty)} K\chi_{[R+l, R+l+1]}\|_{\mathcal{J}_2} +\|\chi_{[R, \infty)} \partial_y K\chi_{[R+l, R+l+1]}\|_{\mathcal{J}_2}), \end{aligned}
\end{equation*}
\notag
$$
where for each term we have by assumption
$$
\begin{equation*}
\begin{aligned} \, &\|\chi_{[R, \infty)} K \chi_{[R+l,R+l+1]})\|_{\mathcal{J}_2}^2 + \|\chi_{[R, \infty)} \partial_y K\chi_{[R+l, R+l+1]})\|_{\mathcal{J}_2}^2 \\ &\qquad \leqslant 2A^2\int_R^\infty dx \int_{R+l}^{R+l+1} dy\,\frac{1}{(x+y)^4} =3 A^2\frac{4R + 2l + 1}{(2R+l)^2(2R+l+1)^2} \\ &\qquad\leqslant \frac{6A^2}{(2R+l)^2(2R+l+1)}. \end{aligned}
\end{equation*}
\notag
$$
Therefore, the trace norm satisfies Finally, we have which finishes the proof. Recall several properties of Bessel functions. Set $\mathfrak{J}(x) = \sqrt{x}\,J_\nu(x)$ and $\mathfrak{D}(x) = \mathfrak{J}_\nu(x)-\sqrt{2/\pi}\cos(x-\phi_\nu)$, where $\phi_\nu={\pi}/{4} + ({\pi}/{2})\nu$. We have the following asymptotics for the Bessel function and its derivative as $x$ approaches infinity:
$$
\begin{equation}
\mathfrak{D}(x)=-\sqrt{\frac{2}{\pi}} \sin(x-\phi_\nu)\frac{\nu^2-1/4}{x} + O(x^{-2})
\end{equation}
\tag{7.5}
$$
and
$$
\begin{equation}
\mathfrak{D}'(x)=A_\nu\frac{\cos(x-\phi_\nu)}{x} + O(x^{-2}),
\end{equation}
\tag{7.6}
$$
where $A_\nu$ is some constant. These imply uniform estimates on $\mathbb{R}_+$ for $\nu> -1$ and some constant $C_\nu$:
$$
\begin{equation}
|\mathfrak{D}(x)| \leqslant \frac{C_\nu}{\sqrt{x}(1+\sqrt{x})}
\end{equation}
\tag{7.7}
$$
and
$$
\begin{equation}
\biggl|\mathfrak{D}'(x) - A_\nu\frac{\cos(x-\phi_\nu)}{x}\biggr| \leqslant \frac{C_\nu}{x^{3/2}(1+\sqrt{x})}.
\end{equation}
\tag{7.8}
$$
Recall the following improper integral. Proposition 6 ([3], Lemma 2.6). We have Lastly, we establish some properties of the function spaces we have introduced. Lemma 5. There exists a constant $C$ such that for any $a\in L_1(\mathbb{R}_+)\cap L_\infty(\mathbb{R}_+)$ satisfying $\|a\|_{\dot{\mathcal B}}<\infty$ we have (1) $\|a''\|_{L_1} \leqslant C(\|a\|_{\dot{H}_{1}}+\|a\|_{\dot{H}_{2}} + \|ta(t)\|_{\dot{H}_{2}})$; (2) $\|ta'''(t)\|_{L_1} \leqslant C(\|a\|_{\dot{H}_{3}} + \|a\|_{\dot{H}_{1}} + \|ta(t)\|_{\dot{H}_{2}} + \|t^2a(t)\|_{\dot{H}_{3}})$; (3) $\lim_{t\to+\infty}a(t)=0$, $\lim_{t\to+\infty}a'(t)=0$ and $\lim_{t\to+\infty}ta''(t)=0$; (4) $|a'(0)| \leqslant \|a\|_{\dot{H}_{1}} + \|a\|_{\dot{H}_{2}}$. In particular, the above estimates are bounded by $C\|a\|_{\dot{\mathcal B}}$. Proof. (1) Using the Cauchy–Bunyakovsky–Schwarz inequality, for the $L_1$-norm we write Since $ta''(t) = (at)'' - 2a'$, the second term is estimated by $2\|a\|_{\dot{H}_{1}} + \|ta(t)\|_{\dot{H}_{2}}$.
(2) The proof is quite similar to the previous one. (3) The statement on $a$ follows from $\widehat{a}\in L_1(\mathbb{R})$ and the Riemann–Lebesgue lemma. Since $a'\in H_1(\mathbb{R}_+)$ we see that $\widehat{a'}$ is absolutely integrable and the statement on $a'$ follows again from the Riemann–Lebesgue lemma. By the first and second assertions $(ta''(t))' = a''(t) + ta'''(t)$ is absolutely integrable, so $ta''(t)$ tends to a finite limit as $t$ approaches infinity. Since $(ta(t))''$ and $a'(t)$ are square integrable, so is $ta''(t)$, which yields that the limit is zero. (4) The expression of $a'(0)$ in terms of the cosine transform yields
$$
\begin{equation*}
|b'(0)| \leqslant \frac{1}{\pi}\int_0^\infty \lambda|\widehat{b}(\lambda)|\,d\lambda,
\end{equation*}
\notag
$$
and by the Cauchy–Bunyakovsky–Schwarz inequality and Parseval’s theorem we have
$$
\begin{equation*}
\int_0^1\lambda|\widehat{b}(\lambda)|\,d\lambda \leqslant \|b\|_{\dot{H}_{1}}\quad\text{and}\quad \int_1^\infty \frac{\lambda^2}{\lambda}|\widehat{b}(\lambda)|\,d\lambda \leqslant \|b\|_{\dot{H}_{2}}.
\end{equation*}
\notag
$$
Lemma 5 is proved. We devote the rest of this section to the proof of Lemma 3. Proof of Lemma 3. Since $B_1 - W_1 = 0$, it is sufficient to prove the statement for $b\in L_1(\mathbb{R}_+)\cap L_\infty(\mathbb{R}_+)$ satisfying $\|b\|_{\dot{\mathcal B}}<\infty$.
Recall the formula for the kernel of the difference $\mathcal{R}_b(x, y) = B_b(x, y) - W_b(x, y)$:
$$
\begin{equation*}
\mathcal{R}_b(x, y)=\int_0^\infty \biggl( \mathfrak{J}(xt)\mathfrak{J}(yt) - \frac{1}{\pi}\cos( (x-y)t) \biggr)b(t)\,dt.
\end{equation*}
\notag
$$
Let us outline the plan of the proof. One can observe from the asymptotics (7.5) and (7.6) that the above integral contains an asymptotically small function for large $x$ and $y$. Thus, we substitute $\mathfrak{J}(x) = \sqrt{x}\,J_\nu(x) = \mathfrak{D}(x) + \sqrt{2/\pi}\cos(x-\phi_\nu)$ into the expression for $\mathcal{R}_b(x, y)$. This substitution and several integrations by parts represent the kernel as a sum of other kernels, to which Corollary 1 and Proposition 4 can be applied.
To be precise, we perform a sequence of decompositions of our kernel, for which we use the following notation:
$$
\begin{equation}
\mathcal{R}_b(x, y)=\mathcal{R}_1(x, y) - \mathcal{R}_2(x, y),
\end{equation}
\tag{7.10}
$$
and Explicit formulae are presented below (see (7.13)–(7.15)). We prove an estimate for the trace norm for each of the above kernels separately. In particular, the estimate of the trace norm for $\chi_{[R, \infty)} \mathcal{R}_1\chi_{[R, \infty)}$ follows from Corollary 1. Estimates for the trace norms of the operators with the kernels
$$
\begin{equation*}
\begin{gathered} \, \chi_{[R, \infty)^2}(x, y)S(x, y), \qquad \chi_{[R, \infty)^2}(x, y)T_0(x, y), \\ \chi_{[R, \infty)^2}(x, y)T_1(x, y), \qquad \chi_{[R, \infty)^2}(x, y)(Z(x, y) + Z(y, x)) \end{gathered}
\end{equation*}
\notag
$$
follow from Proposition 4.
The sequence of decompositions1. The decomposition (7.10). First we substitute the following formula into the kernel $\mathcal{R}_b(x, y)$:
$$
\begin{equation*}
\frac{1}{\pi}\cos( (x-y)t)=\frac{2}{\pi} \cos(xt-\phi_\nu)\cos(yt-\phi_\nu) - \frac{1}{\pi}\cos( (x+y)t-2\phi_\nu).
\end{equation*}
\notag
$$
By part (3) of Lemma 5 the following integral can twice be integrated by parts and thus expressed as follows:
$$
\begin{equation*}
\begin{aligned} \, &\frac{1}{\pi}\int_0^\infty \cos( (x+y)t - 2\phi_\nu)b(t)\,dt \\ &\qquad =\frac{\sin(2\phi_\nu)b(0)}{\pi(x+y)} - \frac{b'(0)\cos(2\phi_\nu)}{\pi(x+y)^2} - \frac{1}{\pi(x+y)^2}\int_0^\infty \cos( (x+y)t -2\phi_\nu)b''(t)\,dt. \end{aligned}
\end{equation*}
\notag
$$
Next we substitute the expression (7.9) for $\sin(2\phi_\nu)/(\pi(x+y))$ into the above identity. These calculations prove the decomposition (7.10) for the following kernels $\mathcal{R}_1$ and $\mathcal{R}_2$:
$$
\begin{equation}
\begin{aligned} \, \mathcal{R}_1(x, y) &=\frac{1}{\pi(x+y)^2} \biggl(\cos(2\phi_\nu)b'(0) + \int_0^\infty\cos( (x+y)t-2\phi_\nu)b''(t)\,dt\biggr), \\ \mathcal{R}_2(x, y) &=\int_0^\infty\biggl( \mathfrak{J}(xt)\mathfrak{J}(yt) - \frac{2}{\pi}\cos(xt-\phi_\nu)\cos(yt-\phi_\nu)\biggr)b_0(t)\,dt, \end{aligned}
\end{equation}
\tag{7.13}
$$
where $b_0(x) = b(x) - b(0)$. 2. The decomposition (7.11). It can be verified directly that (7.11) holds if we take $S$ and $T$ to be
$$
\begin{equation}
\begin{aligned} \, S(x, y)&=\int_0^\infty \mathfrak{D}(xt)\mathfrak{D}(yt)b_0(t)\,dt, \\ T(x, y)&=\int_0^\infty \mathfrak{D}(xt) \sqrt{\frac{2}{\pi}}\cos(yt-\phi_\nu)b_0(t)\,dt. \end{aligned}
\end{equation}
\tag{7.14}
$$
3. The decomposition (7.12). We integrate the expression for $T(x, y)$ by parts:
$$
\begin{equation*}
\begin{aligned} \, &\int_0^\infty \mathfrak{D}(xt)\sqrt{\frac{2}{\pi}}\, \frac{1}{y}\, b_0(t)\,d(\sin(yt-\phi_\nu)) =\mathfrak{D}(xt) \sqrt{\frac{2}{\pi}}\, \frac{1}{y}\, b_0(t)\sin(yt-\phi_\nu)\bigg|_0^{\infty} \\ &\qquad\qquad - \int_0^\infty \mathfrak{D}(xt)\sqrt{\frac{2}{\pi}}\, \frac{1}{y}\, \sin(yt-\phi_\nu)b'(t)\,dt \\ &\qquad\qquad-\int_0^\infty x\mathfrak{D}'(xt)\sqrt{\frac{2}{\pi}}\, \frac{1}{y}\, \sin(yt-\phi_\nu)b_0(t)\,dt, \end{aligned}
\end{equation*}
\notag
$$
where the first term is zero by Lemma 5 and (7.7). Next we take the following kernels for the decomposition (7.12):
$$
\begin{equation}
\begin{aligned} \, T_0(x, y) &=- \int_0^\infty \mathfrak{D}(xt) \sqrt{\frac{2}{\pi}}\, \frac{1}{y}\, \sin(yt-\phi_\nu)b'(t)\,dt, \\ T_1(x, y) &=- \int_0^\infty x\biggl(\mathfrak{D}'(xt) - A_\nu\frac{\cos(xt-\phi_\nu)}{xt}\biggr) \sqrt{\frac{2}{\pi}}\, \frac{1}{y}\, \sin(yt-\phi_\nu)b_0(t)\,dt, \\ Z(x, y)&=- \int_0^\infty x A_\nu\frac{\cos(xt-\phi_\nu)}{xt}\, b_0(t)\sqrt{\frac{2}{\pi}}\, \frac{1}{y}\, \sin(yt-\phi_\nu)\,dt. \end{aligned}
\end{equation}
\tag{7.15}
$$
Trace norm estimatesBefore diving into calculations we present several inequalities we need. Firstly, the Cauchy–Bunyakovsky–Schwarz inequality implies the following:
$$
\begin{equation}
|b_0(t)|=\biggl|\int_0^tb'(x)\,dx\biggr| \leqslant \sqrt{t}\, \|b\|_{\dot{H}_{1}}.
\end{equation}
\tag{7.16}
$$
The same argument for the derivative gives
$$
\begin{equation}
|b'(t)| \leqslant \sqrt{t}\, (|b'(0)| + \|b\|_{\dot{H}_{2}}).
\end{equation}
\tag{7.17}
$$
Further, using (7.17) we have
$$
\begin{equation}
|b_0(t)|=\biggl|\int_0^tb'(x)\,dx\biggr| \leqslant (|b'(0)| + \|b\|_{\dot{H}_{2}})t^{3/2}.
\end{equation}
\tag{7.18}
$$
Lastly, observe the identity
$$
\begin{equation*}
\frac{d}{dt}\biggl (\frac{b_0(t)}{t}\biggr )= \frac{1}{t^2}\biggl(\int_0^tb'(x)\,dx - b_0(t) + \int_0^txb''(x)\,dx\biggr) =\frac{1}{t^2}\int_0^txb''(x)\,dx.
\end{equation*}
\notag
$$
The Cauchy–Bunyakovsky–Schwarz inequality implies that
$$
\begin{equation*}
\biggl|\int_0^txb''(x)\,du\biggr| \leqslant t^{3/2}\|b\|_{\dot{H}_{2}}.
\end{equation*}
\notag
$$
Substituting this in yields
$$
\begin{equation}
\biggl|\frac{d}{dt}\biggl (\frac{b_0(t)}{t}\biggr )\biggr| \leqslant \frac{\|b\|_{\dot{H}_{2}}}{\sqrt{t}}.
\end{equation}
\tag{7.19}
$$
1. An estimate for $\mathcal{R}_1$. We have
$$
\begin{equation*}
|\mathcal{R}_1(x, y)| \leqslant \frac{1}{(x+y)^2}(|b'(0)| + \|b''\|_{L_1}).
\end{equation*}
\notag
$$
By Lemma 5 $b''$ is absolutely integrable, so we have the following expression for the derivative:
$$
\begin{equation*}
\begin{aligned} \, \partial_y\mathcal{R}_1(x, y) &=-\frac{2}{\pi(x+y)^3} \biggl(\cos(2\phi_\nu)b'(0) + \int_0^\infty\cos( (x+y)t-2\phi_\nu) b''(t)\,dt\biggr) \\ &\qquad- \frac{1}{\pi(x+y)^2}\int_0^\infty\sin( (x+y)t-2\phi_\nu)tb''(t)\,dt. \end{aligned}
\end{equation*}
\notag
$$
Next we integrate the second term by parts:
$$
\begin{equation*}
\begin{aligned} \, &\frac{1}{(x+y)}\int_0^\infty d(\cos((x+y)t-2\phi_\nu))tb''(t)\,dt =\cos((x+y)t - 2\phi_\nu)tb''(t)\big|_0^{\infty} \\ &\qquad\qquad -\frac{1}{(x+y)}\int_0^\infty \cos( (x+y)t-2\phi_\nu)(tb'''(t) + b''(t))\,dt, \end{aligned}
\end{equation*}
\notag
$$
where the limit at infinity is zero by Lemma 5. These calculations imply the following estimate for $x, y\geqslant 1$:
$$
\begin{equation*}
|\partial_y\mathcal{R}_1(x, y)| \leqslant \frac{2}{(x+y)^2} (|b'(0)| + \|b''\|_{L_1} + \|tb'''(t)\|_{L_1}).
\end{equation*}
\notag
$$
Finally, applying Corollary 1 and Lemma 5 we obtain the required estimate for some constant $C$ and $R\geqslant 1$:
$$
\begin{equation*}
\|\chi_{[R, \infty)}\mathcal{R}_1\chi_{[R, \infty)}\|_{\mathcal{J}_1} \leqslant \frac{C\|b\|_{\dot{\mathcal B}}}{\sqrt{R}}.
\end{equation*}
\notag
$$
2. An estimate for $S$. Using (7.7) we obtain
$$
\begin{equation*}
\int_R^\infty |\mathfrak{D}(xt)|^2 \,dx \leqslant C_\nu^2\int_R^\infty \frac{1}{xt(1+\sqrt{xt})^2}\,dx=\frac{2C_\nu^2}{t}\biggl(\ln \biggl(1+\frac{1}{\sqrt{tR}}\biggr)-\frac{1}{1+\sqrt{tR}}\biggr).
\end{equation*}
\notag
$$
Set $G(x) = \ln(1+1/\sqrt{x})-1/(1+\sqrt{x})$. Using Proposition 4 we have
$$
\begin{equation*}
\|\chi_{[R, \infty)}S\chi_{[R, \infty)}\|_{\mathcal{J}_1} \leqslant 2C_\nu^2 \int_0^\infty \frac{|b_0(t)|}{t}G(tR)\,dt.
\end{equation*}
\notag
$$
Next we substitute (7.16) in to obtain
$$
\begin{equation*}
\|\chi_{[R, \infty)}S\chi_{[R, \infty)}\|_{\mathcal{J}_1} \leqslant 2C_\nu^2\|b\|_{\dot{H}_{1}}\int_0^\infty \frac{G(tR)}{\sqrt{t}}\,dt =\frac{4C_\nu^2}{\sqrt{R}}\|b\|_{\dot{H}_{1}}\leqslant \frac{4C_\nu^2}{\sqrt{R}}\|b\|_{\dot{\mathcal B}}.
\end{equation*}
\notag
$$
3. An estimate for $T_0$. Again, Proposition 4 in combination with (7.7) gives
$$
\begin{equation*}
\|\chi_{[R, \infty)}T_0\chi_{[R, \infty)}\|_{\mathcal{J}_1} \leqslant \frac{C_\nu}{\sqrt{R}}\int_0^\infty \frac{|b'(t)|}{\sqrt{t}}\sqrt{G(tR)}\,dt.
\end{equation*}
\notag
$$
For the integral over $[0, 1]$ using inequality (7.17), for some constant $C$ we can write
$$
\begin{equation*}
\int_0^1\frac{|b'(t)|}{\sqrt{t}}G^{1/2}(tR)\,dt \leqslant (|b'(0)| + \|b\|_{\dot{H}_{2}})\int_0^1\sqrt{G(tR)}\,dt \leqslant C(|b'(0)| + \|b\|_{\dot{H}_{2}}).
\end{equation*}
\notag
$$
And for the integral over $[1, \infty)$ we use the Cauchy–Bunyakovsky–Schwarz inequality:
$$
\begin{equation*}
\int_1^\infty\frac{|b'(t)|}{\sqrt{t}}\sqrt{G(tR)}\,dt \leqslant \|b\|_{\dot{H}_{1}}\sqrt{\int_1^\infty \frac{G(xR)}{x}\,dx} \leqslant \sqrt{2}\,\|b\|_{\dot{H}_{1}}.
\end{equation*}
\notag
$$
These calculations yield the following estimate by Lemma 5:
$$
\begin{equation*}
\|\chi_{[R, \infty)}T_0\chi_{[R, \infty)}\|_{\mathcal{J}_1} \leqslant \frac{2CC_\nu}{\sqrt{R}}(|b'(0)| + \|b\|_{\dot{H}_{1}} + \|b\|_{\dot{H}_{2}}) \leqslant \frac{4CC_\nu}{\sqrt{R}}\|b\|_{\dot{\mathcal B}}.
\end{equation*}
\notag
$$
4. An estimate for $T_1$. Using estimate (7.8) we obtain
$$
\begin{equation*}
\int_R^\infty x^2\biggl (\mathfrak{D}'(xt) - A_\nu\frac{\cos(xt - \phi_\nu)}{xt}\biggr)^2\,dx \leqslant \frac{C_\nu^2}{t^3}\int_{R}^\infty \frac{1}{x(1+\sqrt{xt})^2}\,dx= \frac{2C_\nu^2}{t^3}G(tR).
\end{equation*}
\notag
$$
In combination with Proposition 4, this implies that
$$
\begin{equation*}
\|\chi_{[R, \infty)}T_1\chi_{[R, \infty)}\|_{\mathcal{J}_1} \leqslant \frac{\sqrt{2}C_\nu}{\sqrt{R}}\int_0^\infty \frac{|b_0(t)|}{t^{3/2}}\sqrt{G(tR)}\,dt.
\end{equation*}
\notag
$$
For the integral over $[0, 1]$ we use inequality (7.18) to obtain
$$
\begin{equation*}
\int_0^1 \frac{|b_0(t)|}{t^{3/2}}\sqrt{G(tR)}\,dt \leqslant (|b'(0)| + \|b\|_{\dot{H}_{2}})\int_0^1\sqrt{G(tR)}\,dt \leqslant C(|b'(0)| + \|b\|_{\dot{H}_{2}})
\end{equation*}
\notag
$$
for some constant $C$. For the integral over $[1, \infty)$, using (7.16), for some constant $\widetilde{C}$ we have
$$
\begin{equation*}
\int_1^\infty \frac{|b_0(t)|}{t^{3/2}}\sqrt{G(tR)}\,dt \leqslant \|b\|_{\dot{H}_{1}}\int_1^\infty\frac{\sqrt{G(tR)}}{t}\,dt \leqslant \widetilde{C}\|b\|_{\dot{H}_{1}}.
\end{equation*}
\notag
$$
Therefore, we conclude that
$$
\begin{equation*}
\begin{aligned} \, \|\chi_{[R, \infty)}T_1\chi_{[R, \infty)}\|_{\mathcal{J}_1} &\leqslant \frac{\sqrt{2}(C+\widetilde{C})C_\nu}{\sqrt{R}}(\|b\|_{\dot{H}_{1}} + \|b\|_{\dot{H}_{2}} + |b'(0)|) \\ &\leqslant \frac{2\sqrt{2}(C+\widetilde{C})C_\nu}{\sqrt{R}}\|b\|_{\dot{\mathcal B}}. \end{aligned}
\end{equation*}
\notag
$$
5. An estimate for $Z(x, y) + Z(y, x)$. We denote the operator induced by this kernel by $\widetilde{Z}$. Integration of the kernel by parts yields
$$
\begin{equation*}
\begin{aligned} \, \widetilde{Z}(x, y) &=Z(x, y) + Z(y, x)=A_\nu\sqrt{\frac{2}{\pi}}\int_0^\infty \frac{b_0(t)}{xyt}\,\frac{d}{dt}(\sin(xt-\phi_\nu)\sin(yt-\phi_\nu))\,dt \\ &=-A_\nu\sqrt{\frac{2}{\pi}}\frac{b'(0)}{xy}\sin^2(\phi_\nu) \\ &\qquad - A_\nu\sqrt{\frac{2}{\pi}}\frac{1}{xy}\int_0^\infty\frac{d}{dt} \biggl (\frac{b_0(t)}{t}\biggr)\sin(xt - \phi_\nu) \sin(yt-\phi_\nu)\,dt. \end{aligned}
\end{equation*}
\notag
$$
The first term induces the one-dimensional projector $h_1(x)\langle h_2(y), -\rangle_{L_2}$ with trace norm bounded by $\|h_1\|_{L_2}\|h_2\|_{L_2}$. By Lemma 5, for the trace norm of the corresponding operator on $L_2[R, \infty)$ we have $|A_\nu b'(0)|/R \leqslant |A_\nu|\|b\|_{\dot{\mathcal B}}/R$. We apply Proposition 4 to the second term. This results in the estimate
$$
\begin{equation*}
\|\chi_{[R, \infty)}\widetilde{Z}\chi_{[R, \infty)}\|_{\mathcal{J}_1} \leqslant \frac{|A_\nu|\|b\|_{\dot{\mathcal B}}}{R} + \frac{|A_\nu|}{R}\int_0^\infty \biggl|\frac{d}{dt}\biggl (\frac{b_0(t)}{t}\biggr)\biggr|\,dt.
\end{equation*}
\notag
$$
The integral over $[0, 1]$ is estimated as follows in view of (7.19):
$$
\begin{equation*}
\int_0^1\biggl|\frac{d}{dt}\biggl (\frac{b_0(t)}{t}\biggr)\biggr|\,dt \leqslant \|b\|_{\dot{H}_{2}}.
\end{equation*}
\notag
$$
The integral over $[1, \infty)$ is estimated using (7.16) and the Cauchy–Bunyakovsky–Schwarz inequality:
$$
\begin{equation*}
\int_1^\infty\biggl|\frac{d}{dt}\biggl (\frac{b_0(t)}{t}\biggr)\biggr|\,dt \leqslant \int_1^\infty\biggl (\frac{|b'(t)|}{t} + \frac{\|b\|_{\dot{H}_{1}}}{t^{3/2}}\biggr )dt \leqslant 3\|b\|_{\dot{H}_{1}}.
\end{equation*}
\notag
$$
Therefore, we conclude that
$$
\begin{equation*}
\|\chi_{[R, \infty)}\widetilde{Z}\chi_{[R, \infty)}\|_{\mathcal{J}_1} \leqslant \frac{5|A_\nu|\|b\|_{\dot{\mathcal B}}}{R}.
\end{equation*}
\notag
$$
This finishes the proof of Lemma 3.
§ 8. Proof of Theorems 1 and 2Recall the Jacobi–Dodgson identity. Proposition 7 ([23], Proposition 6.2.9). Let $A$ be an invertible operator on a separable Hilbert space $\mathcal{H}$ and $A-I\in \mathcal{J}_1(\mathcal{H})$. Let $P$ be an orthogonal projection in $\mathcal{H}$ and $Q=I-P$. Then We use the following variation of this identity. Corollary 2. Let $P_1$ and $P_2$ be two commuting orthogonal projections in a separable Hilbert space $\mathcal{H}$. Let $Q_i=I-P_i$, $i=1, 2$. Then if $A$ is invertible, $A-I\in\mathcal{J}_2(\mathcal{H})$ and all Fredholm determinants are well defined. Proof of Corollary 2. Let $A = I + K$, where $K\in\mathcal{J}_2(\mathcal{H})$. We choose a joint orthogonal basis $\{e_i\}_{i\in\mathbb{N}}$ of eigenvectors of $P_1$ and $P_2$. Let $R_n$ be the orthogonal projection onto $\operatorname{span}(\{e_i\}_{i\in 1,\dots,n})$. We regard $R_n\mathcal{H}$ as a separable Hilbert space. Since the $P_i$ and $Q_i$ commute with $R_n$, they remain orthogonal projections in $R_n\mathcal{H}$.
Since $K$ is compact, $R_nKR_n \to K$ in the operator norm. This implies that for all sufficiently large $n$ the operator $I + R_n K R_n$ is invertible on $R_n\mathcal{H}$. Applying Proposition 7 we have
$$
\begin{equation}
\frac{\det(R_nP_1AP_1R_n)}{\det(Q_1(R_nAR_n)^{-1}Q_1)}= \frac{\det(R_nP_2AP_2R_n)}{\det(Q_2(R_nAR_n)^{-1}Q_2)}=\det(R_nAR_n).
\end{equation}
\tag{8.3}
$$
It remains to show that $(R_n A R_n)^{-1}-R_nA^{-1}R_n\xrightarrow{\mathcal{J}_1(\mathcal{H})} 0$ as $n\to \infty$. For the above difference we have
$$
\begin{equation*}
\begin{aligned} \, & (R_n+R_nK R_n)^{-1}-R_n(I+K)^{-1}R_n \\ &\qquad =(R_nKR_n)^2(R_n + R_nKR_n)^{-1}-R_nK^2(I+K)^{-1}R_n, \end{aligned}
\end{equation*}
\notag
$$
where $(R_n + R_nKR_n)^{-1}\to (I+K)^{-1}$ and $R_n\to I$ strongly, and $(R_nKR_n)^2\to K$ and $R_nK^2\to K^2$ in the trace norm since $K\in\mathcal{J}_2(\mathcal{H})$. Therefore, we have
$$
\begin{equation*}
\lim_{n\to\infty}\det(Q_i(R_nAR_n)^{-1}Q_i)= \lim_{n\to\infty}\det(R_nQ_iA^{-1}Q_iR_n)= \det (Q_i A^{-1}Q_i).
\end{equation*}
\notag
$$
Equality (8.2) follows by taking limits in (8.3) as $n\to\infty$.
The corollary is proved. Let us extend the definition (1.1) to all functions in $L_\infty (\mathbb{R}_+)$. Recall that the Hankel transform $H_\nu$ on $L_2(\mathbb{R}_+)$ is defined by
$$
\begin{equation*}
H_\nu f(\lambda)=\int_0^\infty \sqrt{\lambda x}\, J_\nu (\lambda x) f(x)\,dx.
\end{equation*}
\notag
$$
Similarly to the Fourier transform, it is an isometry of $L_1(\mathbb{R}_+)\cap L_2(\mathbb{R}_+)$ and extends to $L_2(\mathbb{R}_+)$ by continuity. Also recall that $H_\nu^* = H_\nu^{-1} = H_\nu$. Now for $b\in L_\infty(\mathbb{R}_+)$ we define the Bessel operator by This coincides with the definition (1.1) for $b\in L_1(\mathbb{R}_+)\cap L_\infty(\mathbb{R}_+)$, but from (8.4) it is clear that $\mathbf{B}\colon f\to B_f$, $L_\infty(\mathbb{R}_+)\to \mathfrak{B}(L_2(\mathbb{R}_+))$ is a Banach algebra homomorphism and, in particular, $B_{e^b}^{-1} = B_{e^{-b}}$. Next we want to show that for $b\in \mathcal{B}$ the operator $B_b - W_b$ is Hilbert–Schmidt. By Lemma 3 we have $\chi_{[1, \infty)}(B_b - W_b)\chi_{[1, \infty)}\in \mathcal J_1$ and $\chi_{[0, 1]}(B_b - W_b)\chi_{[0, 1]}\in \mathcal J_1$ by Theorem 4. The following statement, completing the argument, can be verified directly. Lemma 6 (see [3], Lemmas 3.1 and 3.2). For $b\in H_1(\mathbb{R}_+)\cap L_1(\mathbb{R}_+)$ the operators $\chi_{[0, 1]}B_b$ and $\chi_{[0, 1]}W_b$ are Hilbert–Schmidt. Proof of Lemma 2. For arbitrary $R_1, R_2 >0$ we set $P_1 = \chi_{[0, R_1]}$, $P_2 = \chi_{[0, R_2]}$ and $A = W_{e^{-b_+}}B_{e^b}W_{e^{-b_-}}$. By the first statement of Proposition 3 and the definition (8.4) $A$ is invertible and its inverse is $W_{e^{b_-}}B_{e^{-b}}W_{e^{b_+}}$. In the notation of Corollary 2 we also have $Q_1 = \chi_{[R_1, \infty)}$ and $Q_2 = \chi_{[R_2, \infty)}$. Using the second and third assertions of Proposition 3 we can write
$$
\begin{equation}
Q_i A^{-1} Q_i=\chi_{[R_i, \infty)} + \chi_{[R_i, \infty)}W_{e^{b_-}}\chi_{[R_i, \infty)}\mathcal{R}_{e^{-b}} \chi_{[R_i, \infty)}W_{e^{b_+}}\chi_{[R_i, \infty)},
\end{equation}
\tag{8.5}
$$
where $\mathcal{R}_b = B_b - W_b$. Hence by the assumptions and since $\mathcal{R}_{e^{-b}}$ is Hilbert–Schmidt, we can apply Corollary 2 to obtain
$$
\begin{equation}
\frac{\det(\chi_{[0, R_1]}W_{e^{-b_+}}B_{e^{b}}W_{e^{-b_-}} \chi_{[0, R_1]})}{\det(\chi_{[R_1, \infty)}W_{e^{b_-}}B_{e^{-b}} W_{e^{b_+}}\chi_{[R_1, \infty)})}\,{=}\, \frac{\det(\chi_{[0, R_2]}W_{e^{-b_+}}B_{e^{b}}W_{e^{-b_-}} \chi_{[0, R_2]})}{\det(\chi_{[R_2, \infty)}W_{e^{b_-}} B_{e^{-b}}W_{e^{b_+}}\chi_{[R_2, \infty)})}=Z(b),
\end{equation}
\tag{8.6}
$$
where all Fredholm determinants are well defined.
Lemma 2 is proved. Recall an asymptotic result due to Basor and Ehrhardt. Theorem 6 ([3], Theorem 1.1). Suppose the function $b\in L_1(\mathbb{R}_+)\cap L_\infty(\mathbb{R}_+)$ satisfies the following conditions: By Lemma 5 the functions in $\mathcal{B}$ satisfy the assumptions of Theorem 5. The following lemma shows that for each $b\in \mathcal{B}$ the function $e^{-b}$ satisfies the assumption of Lemma 3. Lemma 7. Let $b\in\mathcal{B}$. Then $e^b - 1\in L_1\cap L_\infty$. There exists a constant $C$ such that for any $b\in\mathcal{B}$ we have Proof. The first assertion follows directly from $L_\infty$ being a Banach algebra with pointwise multiplication and the inequality
For an estimate of the $\dot{\mathcal{B}}$-seminorm, we have by definition
$$
\begin{equation*}
\|e^b\|_{\dot{H}_{1}} \leqslant \exp(\|b\|_{L_\infty})\|b\|_{\dot{H}_{1}}.
\end{equation*}
\notag
$$
The third derivative of $e^b$ is Observe that
$$
\begin{equation*}
\|b'\|_{L_\infty} \leqslant \|\lambda \widehat{b}(\lambda)\|_{L_1} \leqslant \|b\|_{\dot{H}_{1}} + \|b\|_{\dot{H}_{2}}.
\end{equation*}
\notag
$$
This yields the estimate
$$
\begin{equation*}
\|e^b\|_{\dot{H}_{3}} \leqslant \exp(\|b\|_{L_\infty}) (\|b\|_{\dot{H}_{3}} + \|b\|_{\dot{\mathcal B}}^2\|b\|_{\dot{H}_{1}} + 3\|b\|_{\dot{\mathcal B}}\|b\|_{\dot{H}_{2}}).
\end{equation*}
\notag
$$
Next we write out the second derivative of $xe^{b(x)}$: Therefore,
$$
\begin{equation*}
\|xe^{b(x)}\|_{\dot{H}_{2}} \leqslant \exp(\|b\|_{L_\infty}) (\|xb(x)\|_{\dot{H}_{2}} + \|xb'(x)\|_{L_\infty}\|b\|_{\dot{H}_{1}}).
\end{equation*}
\notag
$$
Finally, the third derivative of $x^2e^{b(x)}$ is as follows:
$$
\begin{equation*}
(x^2e^{b(x)})'''=e^{b(x)}(6x(b'(x))^2 +3x^2b'(x)b''(x) + (x^2b(x))''' + x^2(b'(x))^3).
\end{equation*}
\notag
$$
This implies that
$$
\begin{equation*}
\begin{aligned} \, \|x^2e^{b(x)}\|_{\dot{H}_{3}} &\leqslant \exp(\|b\|_{L_\infty}) (3\|xb'(x)\|_{L_\infty}\|xb(x)\|_{\dot{H}_{2}} \\ &\qquad+\|x^2b(x)\|_{\dot{H}_{3}}+\|xb'(x)\|_{L_\infty}^2\|b\|_{\dot{H}_{1}}). \end{aligned}
\end{equation*}
\notag
$$
The lemma is proved. Proof of Theorem 1. Recall the inequality
$$
\begin{equation*}
|\det(I+K) - 1| \leqslant \|K\|_{\mathcal{J}_1} \exp(\|K\|_{\mathcal{J}_1}).
\end{equation*}
\notag
$$
Also recall that for any trace-class operator $A$ and bounded $B$ we have $\|AB\|_{\mathcal{J}_1} \leqslant \|A\|_{\mathcal{J}_1}\|B\|$. By the definition of Wiener–Hopf operators $\|W_b\| = \|b\|_{L_\infty}$. These facts and Lemmas 3 and 7, in combination with the expression (8.5), prove estimate (2.2).
By the above argument the denominator in (3.5) approaches $1$. From Lemma 1 and Theorem 6, for the numerator in (3.5) we have It is now clear that $Z(b) = \exp(c_2^{\mathcal{B}}(b) + c_3^{\mathcal{B}}(b))$. This finishes the proof.We now proceed to the proof of Theorem 2. First we establish an estimate for the rate of convergence of $\mathbb{E}_{J_\nu}S_f^R$. The argument is based on the proof of Proposition 5.1 in [3]. Proof. Recall the expression for the expectation of an additive functional:
$$
\begin{equation*}
\mathbb{E}_{J_\nu}S_f^R=\int_{\mathbb{R}_+}f\biggl(\frac xR\biggr) B_{\chi_{[0, 1]}}(x, x)\,dx.
\end{equation*}
\notag
$$
The Bessel kernel $B_{\chi_{[0, 1]}}(x, y)$ is defined on the diagonal by
$$
\begin{equation*}
B_{\chi_{[0, 1]}}(x, x)=\frac{x}{2}(J_\nu^2(x) - J_{\nu+1}(x)J_{\nu-1}(x)).
\end{equation*}
\notag
$$
Its asymptotic as $x\to\infty$ is
$$
\begin{equation*}
B_{\chi_{[0, 1]}}(x, x)=\frac{1}{\pi} + \frac{\sin(2(x-\phi_\nu))}{x} + O(x^{-2}).
\end{equation*}
\notag
$$
Consider the function
$$
\begin{equation*}
F(\xi)=-\int_\xi^\infty\biggl (\frac{t}{2}(J_\nu^2(t) - J_{\nu+1}(t)J_{\nu-1}(t)) - \frac{1}{\pi}\biggr)dt.
\end{equation*}
\notag
$$
It is clear that $F(\xi) = O(\xi^{-1})$ as $\xi\to\infty$. Therefore, we have $|F(\xi)| \leqslant C(1+\xi)^{-1}$ for some constant $C$. Integrating by parts in the formula for the expectation we obtain
$$
\begin{equation}
\begin{aligned} \, \notag &\int_{\mathbb{R}_+}f\biggl(\frac xR\biggr)B_{\chi_{[0, 1]}}(x, x)\,dx - \frac{1}{\pi}\int_{\mathbb{R}_+}f\biggl(\frac xR\biggr)\,dx \\ &\qquad=f\biggl(\frac xR\biggr)F(x)\bigg|_0^{\infty} - \frac{1}{R}\int_{\mathbb{R}_+}f'\biggl(\frac xR\biggr)F(x)\,dx, \end{aligned}
\end{equation}
\tag{8.8}
$$
where $f(\infty)F(\infty)=0$ by Lemma 5. For the value at zero, using the equality $J_{\nu-1}(t) + J_{\nu+1}(t) =(2\nu/t)J_\nu(t)$ we write
$$
\begin{equation*}
\begin{aligned} \, F(0)&=\int_0^\infty \biggl (\frac{t}{2}(J_\nu^2(t) - J_{\nu+1}(t)J_{\nu-1}(t)) - \frac{1}{\pi}\biggr)dt \\ &=\int_0^\infty\biggl(\frac{t}{2}(J_\nu^2(t) + J_{\nu+1}^2(t)) - \frac{1}{\pi}\biggr)dt - \nu\int_0^\infty J_{\nu+1}(t)J_\nu(t)\,dt. \end{aligned}
\end{equation*}
\notag
$$
The second integral is equal to $1/2$ (see [17], § 6.512(3)). The first term is asymptotically equal to Recall that $\phi_{\nu+1}=\phi_{\nu}+\pi/{2}$. Therefore, the first term is zero. We conclude that $F(0)b(0)=({\nu}/{2})b(0)$.
What is left to do is to estimate the second term in (8.8). Using $F(\xi)\leqslant C(1+\xi)^{-1}$ we can write which is at most ${C\|b\|_{\dot{H}_{1}}}/{\sqrt{R}}$. This completes the proof. Proof of Theorem 2. Recall that we denote by $F_{R, b}$ and $F_{\mathcal{N}}$ the cumulative distribution functions of $\overline{S_b^R}$ and the standard Gaussian distribution $\mathcal{N}(0, 1)$, respectively. Using Feller’s smoothing estimate (see [16], Ch. XVI, § 3), Theorem 1 and Proposition 1, for any $T>0$ we have
$$
\begin{equation}
\begin{aligned} \, \notag &\sup_{x}|F_{R, b}(x) - F_{\mathcal{N}}(x)| \leqslant \frac{24}{\sqrt{2\pi^3}\,T} \\ &\qquad\qquad +\frac{1}{\pi}\int_{-T}^T \frac{1}{|k|}\biggl|\exp\biggl(ikR\widehat{b}(0) - ik\, \frac{\nu}{2}\, b(0)-ik \mathbb{E}_{J_\nu}S^R_b\biggr)Q_R^{\mathcal{B}}(kb) - 1\biggr|\,dk. \end{aligned}
\end{equation}
\tag{8.9}
$$
By the second statement of Theorem 1 and Lemma 8 the integrand can be estimated by the following expression for some $C>0$ depending only on $\|b_+\|_{L_\infty}$ and $L(b)$:
$$
\begin{equation*}
\frac{C}{\sqrt{R}}(1 + |k|^2)e^{C|k|}\exp\biggl(\frac{C|k|}{\sqrt{R}}(1 + |k|^2)e^{C|k|}\biggr).
\end{equation*}
\notag
$$
Observe that if $|k| \leqslant C_1\ln R$, where $C_1C <1/2$, then there exists a constant $\widetilde{C}$ such that for any $R\geqslant 1$ the above expression is at most Therefore, if we take $T = C_1\ln R$, then the integral in (8.9) is at most and the statement of the theorem follows.
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