Asymptotic Analysis of von Neumann Entropy in Conformal Field Theory

Abstract

Given a QFT net ${{\mathcal {A}}}$ of local von Neumann algebras ${{\mathcal {A}}}(O)$, we consider the von Neumann entropy $S_{{\mathcal {A}}}(O, {\widetilde{O}})$ of the restriction of the vacuum state to the canonical intermediate type I factor for the inclusion of von Neumann algebras ${{\mathcal {A}}}(O)\subset {{\mathcal {A}}}({\widetilde{O}})$ (split property). This canonical entanglement entropy $S_{{\mathcal {A}}}(O, {\widetilde{O}})$ is finite for the chiral conformal net on the circle generated by finitely many free Fermions (here double cones are intervals). The finiteness property is derived by an explicit formula of entropy and an observation that the operators in the definition are closely related to Hankel operators. In this paper we give further analysis of this entropy using a variety of techniques that have been developed in different context, and in particular we show that there is an upper bound given by a positive constant multiply by $|\ln \eta |$, where $\eta$ is the cross ratio of the underlying system, when $\eta \rightarrow 0$.

1 Introduction

von Neumann entropy is the basic concept in quantum information and extends the classical Shannon’s information entropy notion to the non commutative setting. The role of entanglement in Quantum Field Theory is more recent and increasingly important; it represents a piece of the quantum information framework in this subject. It appears in relation with several primary research topics in theoretical physics as area theorems, c-theorems, quantum null energy inequality, etc. (see for instance [5, 6, 40] and refs. therein).

Despite the rich physical literature on the subject, the rigorous definition of entanglement entropy in QFT is however not obvious. The point is that the von Neumann algebra ${{\mathcal {A}}}(O)$ associated with a double cone spacetime region O is typically a factor of type III, so no trace exists on ${{\mathcal {A}}}(O)$ and one cannot naively extends the definition of entropy as one would do with $A = {{\mathcal {A}}}(O)$, $B = {{\mathcal {A}}}(O')$, where O is a double cone and $O'$ is its causal complement and $\omega$ the vacuum state. Due to ultraviolet divergence, such a measure of the vacuum entanglement would always result to be infinite. By Haag duality, that holds in much generality, ${{\mathcal {A}}}(O')$ is the commutant ${{\mathcal {A}}}(O)'$ of ${{\mathcal {A}}}(O)$ on the vacuum Hilbert space ${{\mathcal {H}}}$, so the von Neumann algebra ${{\mathcal {A}}}(O)\vee {{\mathcal {A}}}(O')$ generated by ${{\mathcal {A}}}(O)$ and ${{\mathcal {A}}}(O')$ is equal to $B({{\mathcal {H}}})$, a type I factor, and cannot be naturally isomorphic to the von Neumann tensor product ${{\mathcal {A}}}(O)\otimes {{\mathcal {A}}}(O')$ which is type III.

To get rid of short distance divergences, one may however consider a slightly larger double cone $O\subset {\widetilde{O}}$, namely the closure of O is contained in the interior of ${\widetilde{O}}$. The split property states that there is a natural isomorphism of von Neumann algebras

$$\begin{aligned} {{\mathcal {A}}}(O)\vee {{\mathcal {A}}}({\widetilde{O}}') \simeq {{\mathcal {A}}}(O)\otimes {{\mathcal {A}}}({\widetilde{O}}')\, , \end{aligned}$$

that identifies ${{\mathcal {A}}}(O)$ with ${{\mathcal {A}}}(O)\otimes 1$ and ${{\mathcal {A}}}(O')$ with $1\otimes {{\mathcal {A}}}(O')$.

The split property expresses the statistical independence of ${{\mathcal {A}}}(O)$ and ${{\mathcal {A}}}({\widetilde{O}}')$; it was verified for the free, neutral Boson QFT case in [2]. It was studied in [11] and led to important structural features both in Mathematics and in Physics. It follows under natural, general physical requirements [4]. It holds automatically in chiral conformal QFT [28]. (See [16] for a discussion of its validity in topologically non trivial spacetimes).

Approaches to the entanglement entropy by means of the split property are studied in [8, 12, 17, 27, 29, 40].

The split property is a local property, in fact it is equivalent to the existence of an intermediate type I factor ${{\mathcal {F}}}$ between ${{\mathcal {A}}}(O)$ and ${{\mathcal {A}}}({\widetilde{O}})$

$$\begin{aligned} {{\mathcal {A}}}(O) \subset {{\mathcal {F}}}\subset {{\mathcal {A}}}({\widetilde{O}}) \ . \end{aligned}$$

(1)

A type I factor ${{\mathcal {F}}}$ is a von Neumann algebra isomorphic to $B({{\mathcal {K}}})$, the algebra of all bounded linear operators on some Hilbert space ${{\mathcal {K}}}$.

We may then define the entanglement entropy of the net ${{\mathcal {A}}}$ associated with the double cones $O \subset {\widetilde{O}}$ as the vacuum von Neumann entropy associated with the ${{\mathcal {F}}}$ where the global systems is $B({{\mathcal {H}}})$, the factorization is given by ${{\mathcal {F}}}$, namely $A = {{\mathcal {F}}},$ $B = {{\mathcal {F}}}'$ with a tensor product decomposition

$$\begin{aligned} {{\mathcal {H}}}= {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\, , \quad A \simeq B({{\mathcal {H}}}_A)\otimes 1, \quad B \simeq 1 \otimes B({{\mathcal {H}}}_B)\, , \end{aligned}$$

and the pure state is the vacuum state.

This definition however depends on the choice of ${{\mathcal {F}}}$. Actually, if the split property holds, there are infinitely many intermediate type I factors ${{\mathcal {F}}}$ in (1). Yet, as shown in [11], there is a canonical intermediate type I factor ${{\mathcal {F}}}$, associated with the $O, {\widetilde{O}}$ and the vacuum vector $\text{\O}mega$, given by the formula

$$\begin{aligned} {{\mathcal {F}}}= {{\mathcal {A}}}(O)\vee J{{\mathcal {A}}}(O)J = {{\mathcal {B}}}({\widetilde{O}})\cap J{{\mathcal {B}}}({\widetilde{O}})J \end{aligned}$$

(2)

(if the local von Neumann algebras are factors), with J is the modular conjugation of the relative commutant von Neumann algebra ${{\mathcal {A}}}(O)'\cap {{\mathcal {A}}}({\widetilde{O}})$ associated with $\text{\O}mega$.

We then define the (canonical) entanglement entropy of ${{\mathcal {A}}}$ with respect to $O,{\widetilde{O}}$ as

$$\begin{aligned} S_{{\mathcal {A}}}(O,{\widetilde{O}}) = -{\hbox {tr}}(\rho _{{\mathcal {F}}}\log \rho _{{\mathcal {F}}}) \, , \end{aligned}$$

(3)

where ${{\mathcal {F}}}$ is the canonical intermediate type I factor (2). Here ${{\,\mathrm{Tr}\,}}$ is the trace of ${{\mathcal {F}}}$ (namely ${{\mathcal {F}}}= B({{\mathcal {H}}}_A)\otimes 1_{{{\mathcal {H}}}_B}$ and ${\hbox {tr}}$ corresponds to the usual trace on $B({{\mathcal {H}}}_A)$) and $\rho _{{\mathcal {F}}}$ is the vacuum density matrix relative to ${{\mathcal {F}}}$.

The above definition concerns a local net ${{\mathcal {A}}}$. If ${{\mathcal {A}}}$ if a Fermi net, graded locality rather than locality holds. In this case, the split property is still defined by (1) and the entanglement entropy by (3). However, the canonical intermediate type I factor is to be defined by a twisted version of formula (2), cf. equation 50 of [25].

A main result in [25] is that above defined canonical entanglement entropy is finite for the chiral conformal net ${{\mathcal {M}}}$ generated by a complex free fermion on $S^1$. Here, double cones are intervals $I \subset {\widetilde{I}}$ of $S^1$.

In fact in [25] an explicit formula for the von Neumann entropy is given, and its finiteness follows from observing the connection to the theory of Hankel operators . It is in fact the first known case where such canonical entropy is proved to be finite. It is therefore a natural question to estimate this finite entropy, in particular its asymptotic property as the cross ratio $(\sin (\eta /2))^2$ goes to zero or equivalently when the end points of the interval get close to each other (cf. Remark 3.9 in [25]) . Note that due to the monoticity of relative entropy the entropy is bounded below (cf. Lemma 2.4) by $\frac{1}{6} |\ln \sin (\eta /2)|$, so the real interest is about its upper bound. Our result (cf. Cor. 3.17 ) is that the upper bound is again a constant multiplied by $|\ln \eta |$ as $\eta \rightarrow 0$. The proof of this result is surprisingly delicate and rely on deep results in [21, 37] and [38]. The results of [21, 37] and [38] are motivated by questions of semi-classical analysis of entropy in QFT, and the context of these questions are very different from ours. In fact since our functions are not smooth on the circle, we have to modify the proof of some of the results in these papers for our analysis. These modifications include Lemma 3.11 which is based on a result in [38], but now applied in three different scales in the proof of Th. 3.15. By using properties of Hankel operators, it turns out that we can do our estimates by removing the poles of our functions inside the unit disk. We then use a change of part of the path to evaluate the fourier coefficients of our functions, first in the relatively easy case when our functions have no branch cuts. When our functions have branch cuts inside the unit disk, we reduce our analysis to the estimation of Besov quasinorm of these functions (cf. Section 3.3). We expect that our techniques will have applications in more general cases.

There are some similarities between our entropy and reflective entropy discussed in the physics literature (cf. [9] and references therein). In [9] there are also numerical computations of such reflective entropy and their numerical data agrees with our asymptotic analysis, but it is not clear at all that those numerical computations on finite lattices in [9] actually converge to our entropy. It is an interesting question to further understand this similarity. It is also an interesting question to improve our estimates in this paper, in particular to determine the constant in Cor. 3.17. See Remark (3.18) and Remark (3.8).

The rest of this paper is organized as follows. In Section 2 we recall the entropy defined in [25] in the context of chiral net of free fermion, and recall some basic facts related to Hankel operators in [30]. We begin our asymptotic analysis in Section 3. Our basic idea is explained at the beginning of Section 3.1. Roughly speaking we deform the path of integration, removing the poles of our functions inside the unit disk, and then estimate Besov quasi-norms of these functions (cf. the proof of Th. 3.7). We then use these results in Section 3.4 to give the Schatten norm of our functions. This is based on Lemma 3.11, and a key result in Th. 3.15. In Section 3.5 we prove Cor. 3.17 as a consequence of our results in the previous two sections and results of [21]. In the last Section we show that our entropy function is continuous in $\eta$ and goes to 0 as $\eta$ goes to $\pi$.

2 Preliminaries

2.1 Schatten-von Neumann Ideals

This paper relies on the results for general quasi-normed ideals of compact operators. Here we limit our attention to the case of Schatten-von Neumann operator ideals $S_q, q>0$. Detailed information on these ideals can be found e.g. in [30] and [36]. We shall point out only some basic facts. For a compact operator A on a separable Hilbert space H, denote by $s_n(A), n =1, 2, ...$ its n-th singular values, that is, the eigenvalues of the operator $|A| :=\sqrt{A^*A}$. Note that if $R_1, R_2$ are bounded operators, Then (cf. [36])

$$\begin{aligned} s_n(R_1 A R_2) \le ||R_1|| s_n(A) ||R_2|| \end{aligned}$$

(4)

where ||A|| to denote the norm of an operator A. We denote the identity operator on H by 1. The Schatten-von Neumann ideal $S_q, q>0$ consists of all compact operators A , for which $|A|_{S_q} := (\sum _{k=1}^\infty s_k(A)^q)^{\frac{1}{q}} < \infty$. Note that $|A|_{S_q}=|A^*|_{S_q}$.

If $q\ge 1$ , then the above functional defines a norm; if $0<q<1$ , then it is a so-called quasi-norm. There is nevertheless a convenient analogue of the triangle inequality, which is called the q-triangle inequality:

$$\begin{aligned} |A_1 + A_2|^q_{S_q} \le |A_1|^q_{S_q} + |A_2|^q_{S_q}, 0< q\le 1 \end{aligned}$$

(5)

We also have the Holder inequality:

$$\begin{aligned} |A_1 A_2|_{S_q} \le |A_1 |_{S_{q_1}} |A_2 |_{S_{q_2}}, 1/q = 1/q_1 + 1/q_2 , 0 < q_1, q_2 \le \infty \end{aligned}$$

(6)

See [19] and also [3]. In what follows we focus on the case $q\in (0, 1]$. We will use ||A|| to denote the norm of an operator, and $||A||_1$ the trace of |A|. By definition

$$\begin{aligned} |||A|^q||_1 =|A|^q_{S_q} \end{aligned}$$

Note that for a nonzero operator A, the singular values of A/||A|| is bounded above by 1, therefore if $0<p<q\le 1$ we have $|A/||A|||_{S_q}^q \le |A/||A|||_{S_p}^p$, from which we have

$$\begin{aligned} |A|_{S_q}^q \le |A|_{S_p}^p ||A||^{q-p} \end{aligned}$$

(7)

2.2 Basic Representation of $LU_1$ and Free Fermion net

Let H denote the Hilbert space $L^2(S^1; {\mathbb {C}})$ of square-summable ${\mathbb {C}}$-valued functions on the circle. The group $LU_1$ of smooth maps $S^1 \rightarrow U_1$, with $U_1$ the set of the unit circle in ${\mathbb {C}}$, acts on H as multiplication operators.

Let us decompose $H= H_+ \oplus H_-$, where

$$\begin{aligned} H_+ = \{\text {functions whose negative Fourier coeffients vanish}\} \, . \end{aligned}$$

We denote by P the Hardy projection from H onto $H_+$.

Denote by $U_{\mathrm{res}}(H)$ the group consisting of unitary operator A on H such that the commutator [P, A] is a Hilbert-Schmidt operator. Denote by $\mathrm{Diff}^+(S^1 )$ the group of orientation preserving diffeomorphism of the circle. It follows from Proposition 6.3.1 and Proposition 6.8.2 in [31] that $LU_1$ and $\mathrm{Diff}^+(S^1 )$ are subgroups of $U_{\mathrm{res}}(H)$. The basic representation of $LU_1$ is the representation on Fermionic Fock space $F_p = \Lambda (PH) \otimes \Lambda ((1-P)H)^*$ as defined in Section 10.6 of [31]. For more details, see [31] or [39]. Such a representation gives rise to a graded net as follows. Denote by ${{\mathcal {A}}}(I)$ the von Neumann algebra generated by $c(\xi )^\prime s$, with $\xi \in L^2 (I, {\mathbb {C}} )$. Here $c(\xi ) = a(\xi ) + a(\xi )^*$ and $a(\xi )$ is the creation operator defined as in Chapter 1 of [39]. Let $Z:F_p \rightarrow F_p$ be the Klein transformation given by multiplication by 1 on even forms and by i on odd forms. It follows from Section 15 of chapter 2 of [39] that ${{\mathcal {A}}}$ is a graded Möbius covariant net, and ${{\mathcal {A}}}$ will be called the net of free fermion. ${{\mathcal {A}}}$ is strongly additive, and the commutant of ${{\mathcal {A}}}(I)$ is $Z{{\mathcal {A}}}(I')Z^{-1}$ where $I'$ is the complement of I on the circle.

Let $I_1,I_2$ be two open intervals on the circle, and $I_1, I_2$ are disjoint, that is $\bar{I_1} \cap \bar{I_2} = \varnothing$, and $I=I_1\cup I_2$.

For bounded operators $A, B : F_p \rightarrow F_p$, we define $A^+ =\Gamma A \Gamma , \ A^- = A - A^+$, where $\Gamma$ is an operator on $F_p$ given by multiplication by 1 on even forms and $-1$ on odd forms. Note that $Z= \frac{1 - i\Gamma }{1 - i}$.

An operator A is called even (resp. odd) if $A = A^+$ (resp. $A = A^-$). Note that $\omega (a)=0$ if a is odd, where $\omega$ is the vacuum state corresponding to the vacuum vector $\text{\O}mega$.

We set

$$\begin{aligned} \omega _1\otimes _2 \omega _2 (AB)= \langle A\text{\O}mega , \text{\O}mega \rangle \langle B\text{\O}mega , \text{\O}mega \rangle ,\quad \forall A\in {{\mathcal {A}}}(I_1),\ B\in {{\mathcal {A}}}(I_2) \ . \end{aligned}$$

By (1) Lemma 3.1 in [24] this defines a normal state on the von Neumann algebra generated by ${{\mathcal {A}}}(I_1)$ and ${{\mathcal {A}}}(I_2)$.

The mutual information $S(\omega ,\omega _1\otimes _2 \omega _2)$ (cf. Definition 2.1 of [24]) is computed in Section 3 of [24].

2.3 von Neumann Entropy from Split Property

In this section we recall the von Neumann entropy defined in [25] from split property that we aim to compute in this paper.

2.3.1 General Symmetric Interval

We will focus on the one particle structure on $L^2(S^1; {\mathbb {C}})$ in this section. On $S^1$, we consider the following general four “symmetric intervals"

$$\begin{aligned} I_1 =\{ e^{i\theta } : 0<\theta<\phi \}\, , \quad I_2&=\{ e^{i\theta } : \phi -\pi<\theta < 0\}\, , \end{aligned}$$

(8)

$$\begin{aligned} -I_1 =\{ e^{i(\pi +\theta )} : 0<\theta< \phi \}\, , \quad -I_2&=\{ e^{i(\pi +\theta )} : \phi -\pi<\theta< 0\}\, , \quad 0<\phi <\pi \, . \end{aligned}$$

(9)

We will denote by $\eta = \pi -\phi$. See Figure 1.

Fig. 1

The symmetric intervals

Denote by $I_0:=\{ e^{i\theta } : 0<\theta < 2\phi \}$. For any interval I if we denote by $I^2$ the set of z such that $z=w^2$ for some $w\in I$, then it is clear that $I_0=I_1^2$. We shall consider the action of SU(1, 1) on $S^1$ which is given by $z\rightarrow \frac{az +b}{\bar{b} z+ \bar{a}}$ with $|a|^2-|b|^2= \pm 1$. The Möbius group $\mathrm{Mob}$ is the subgroup of SU(1, 1) of elements with determinant $|a|^2-|b|^2= 1.$ The action $z\rightarrow \frac{1}{z}$ is orientation reversing. This element has $a=d=0, b=c=-1$.

If $m(z)= \frac{az +b}{{\bar{b}} z+ {\bar{a}}}$, the unitary action of m on $S^1$ is given by (See Section 4 of [39])

$$\begin{aligned} (U_m f) (z) = (a -\bar{b} z)^{-1} f(m^{-1}z) \, . \end{aligned}$$

(10)

Since $(a -\bar{b} z)^{-1}$ is holomorphic for $|z|<1$ and $|a|>|b|$, $U_m$ and its inverse preserves PH, and so $U_m$ commutes with the Hardy space projection P. The flip map $(F_1 f)(z)= \frac{1}{z}f(\frac{1}{z})$ clearly satisfies $PF_1P= 1-P$. By sending the orientation reversing element $z\rightarrow \frac{1}{z}$ in SU(1, 1) to $F_1$, we get an action of SU(1, 1) on ${{\mathcal {H}}}$ which is of the form

$$\begin{aligned} (U_m f) (z) = \alpha _m(z) f(m^{-1}z) \, , \end{aligned}$$

(11)

where $m (z)= \frac{az +b}{{\bar{b}} z+ {\bar{a}}}, \alpha _m(z)= (a -\bar{b} z)^{-1}$.

Let $m\in \mathrm{Mob}$ be such that $m I_0$ is the upper half circle. Let $m_1= m^{-1} F_1 m$. It is straightforward to see that

$$\begin{aligned} m_1(z)= \frac{z(1+e^{2i\phi })/2 - e^{2i\phi }}{z- (1+e^{2i\phi })/2 } \, . \end{aligned}$$

(12)

Define

$$\begin{aligned} (F_0 f) (z) = \alpha _{m_1}(z) f(m_1^{-1}z)\, . \end{aligned}$$

(13)

We have the following definition

$$\begin{aligned} (jf)(z) := \alpha _{m_1}(z^2) \left( \frac{1}{2} + \frac{z}{2u}\right) f(u) + \alpha _{m_1}(z^2) \left( \frac{1}{2} - \frac{z}{2u}\right) f(-u)\, , \end{aligned}$$

(14)

where

$$\begin{aligned} u^2= m_1(z^2) \end{aligned}$$

(15)

Note that $m_1(z^2)$ depends on $\eta$.

Note that j maps $L^2(I_1)$ to $L^2(I_2 \cup - I_2)$. We will denote by $M_I$ the multiplication operator by $\chi _I$, the characteristic function of interval I.

By definition, $M_{I_1}$ and $jM_{I_1}j$ are orthogonal projections whose ranges $L^2(I_1)$ and $jL^2(I_1)$ are also orthogonal. Hence $P_{12} := M_{I_1} + jM_{I_1}j$ is a projection. If we wish to emphasize the dependence on $\eta$ we will write $P_{12}$ as $P_{12}(\eta )$.

Note that $L^2(I_1)\oplus jL^2(I_1)$ is the canonical type I standard subspace which is intermediate between $L^2(I_1)$ and $L^2(I_1 \cup I_2 \cup - I_2)$. The following is theorem 3.3 from [25]:

Theorem 2.1

$$\begin{aligned} P_{12} = M_g + M_h R \quad \mathrm{on}\ L^2(S^1)\, , \end{aligned}$$

is a projection onto $L^2(I_1)\oplus jL^2(I_1)$ where

$$\begin{aligned} g(z)= & {} \frac{(u+z)^2}{4uz} \chi _{I_1}(u) - \frac{(u-z)^2}{4uz} \chi _{I_1}(-u) + \chi _{I_1}(z) \, , \end{aligned}$$

(16)

$$\begin{aligned} h(z)= & {} \frac{z^2-u^2}{4uz}( \chi _{I_1}(u)- \chi _{I_1}(-u)) \, , \end{aligned}$$

(17)

$$\begin{aligned} (R(f)(z)= & {} f(-z), u^2= m_1(z^2)\, , \end{aligned}$$

(18)

$$\begin{aligned} m_1(z)= & {} \frac{z(1+e^{2i\phi })/2 - e^{2i\phi }}{z- (1+e^{2i\phi })/2 }. \end{aligned}$$

(19)

We note that since g, h are invariant under $u\rightarrow -u$, these functions are independent of the choice of the square root u of $m_1(z^2)$ in Th. 2.1.

For simplicity we will write multiplication operator such as $M_g$ simply as g when no confusion arises. For an example we can write $P_{12}= g+hR$. Similarly we write $g^*$ the complex conjugate of a function such as g.

The von Neumann entropy $S(F,\eta )$ that comes from the canonical type I standard subspace $L^2(I_1)\oplus jL^2(I_1)$ is defined as follows. Let $f_0:=-x\ln x -(1-x)\ln (1-x), 0< x< 1$, and we define $f_0(0)=f_0(1)=0$. We make the following definition:

Definition 2.2

$$\begin{aligned} S(F,\eta ) = {\hbox {tr}}(f_0(PP_{12}(\eta )P)) \end{aligned}$$

where $P_{12}$ is as in Th. 2.1. For any operator self adjoint T and continuous f defined on the spectrum of T and PTP, if $f(PTP)-Pf(T)P$ is trace class, we use $\tau (T,f)$ to denote the trace of $f(PTP)-Pf(T)P$.

Since $P_{12}(\eta )$ is a projection and $f_0(P_{12}(\eta ))=0$, $S(F,\eta )=\tau (P_{12}(\eta ),f_0)$. We choose the notation $S(F,\eta )$ since S(F) is what we used in [25], and we put in extra $\eta$ to emphasize the dependence on $\eta$. Let us first explain why $S(F,\eta )=S(F)$ where S(F) is as in Th. 3.8 of [25]. Let us first recall how S(F) is defined. We shall denote by $F:=L^2(I_1)\oplus jL^2(I_1)$ the canonical type I standard subspace. Recall that on ${{\mathcal {H}}}= L^2(S^1 ; {\mathbb {C}})$, the complex structure on ${{\mathcal {H}}}$ is given by $i(2P -1)$, with P the projection onto the Hardy space. Let $F'$ be the orthogonal complement of $i(2P-1)F$. Let $P_F:=P_{12}(\eta ), P_{F'}$ be the projections onto $F,F'$ respectively. Then $P_{F'}= (2P-1)(1-P_F)(2P-1)$, and

$$\begin{aligned} P_FP_{F'}P_F = P_F (2P-1)(1-P_F)(2P-1) P_F = 4(P_FPP_F - (P_FPP_F)^2) \end{aligned}$$

By Lemma 2.4 of [25] $P_FP_{F'}P_F= 4 \frac{\Delta _F}{(\Delta _F + 1)^2} P_F$ where $\Delta _F$ is the modular operator associated with F, it follows that $P_FPP_F - (P_FPP_F)^2=\frac{\Delta _F}{(\Delta _F + 1)^2} P_F$. Let $0<\lambda _i<1, 1\le i<\infty$ be the list (counting multiplicities) of eigenvalues of $\Delta _F$ that is in the interval (0, 1). Define $\mu _i = \frac{1}{1+\lambda _i}$. By definition S(F) is the von Neumann entropy of state $\rho _F'= \frac{\widehat{\rho _F'}}{{\hbox {tr}}\widehat{\rho _F'}}$, where $\widehat{\rho _F'}$ is $\Lambda (\Delta _F|_{{{\mathcal {H}}}_F(0,1)})$ by Cor. 2.10 of [25]. By using Prop. 2.11 of [25] and a straightforward computation we have

$$\begin{aligned} S(F)= -\sum _{1\le i <\infty } (\mu _i \ln \mu _i +(1-\mu _i )\ln (1-\mu _i)) \end{aligned}$$

By examing the spectrum of $\frac{\Delta _F}{(\Delta _F + 1)^2} P_F$ as in Lemma 2.7 in [25] we see that the list (counting multiplicities) of eigenvalues of $\frac{\Delta _F}{(\Delta _F + 1)^2} P_F$ is $\frac{\lambda _i}{(\lambda _i + 1)^2}=\mu _i(1-\mu _i), 1\le i<\infty$. Since $(P_FPP_F - (P_FPP_F)^2)=\frac{\Delta _F}{(\Delta _F + 1)^2} P_F$, the list (counting multiplicities) of eigenvalues of $P_FPP_F$ is $\mu _i'$, where $\mu _i'=\mu _i$ or $\mu _i'= 1-\mu _i, 1\le i<\infty$. Therefore

$$\begin{aligned} S(F)= -\sum _{1\le i <\infty } (\mu _i' \ln \mu _i' +(1-\mu _i' )\ln (1-\mu _i')) \end{aligned}$$

Since (cf. Chapter 1 of [36]) the set of nonzero elements in the spectrum of $T^*T$ is the same as the set of nonzero elements in the spectrum of $TT^*$ for a bounded operator T, it follows that the list of nonzero eigenvalues (counting multiplicities) of $PP_FP = PP_{12}(\eta )P =PP_F P_FP$ is the same as that of $P_FPPP_F= P_FPP_F$. So we have $S(F,\eta )=S(F)$.

In [25] we proved that $S(F,\eta )$ is finite by observing its connection with Hankel operators. This relies on the growth of fourier coefficients of g, h. We recall the following result which is proved in [25] and follows essentially from an observation of [18], see Th. 3.7 in [25].

Lemma 2.3

Suppose $f =\sum _n f_n z^n$.

If $|f_n|\le C n^{-\alpha }$ with $\alpha > \frac{3}{2}, n\ge 0$, then $|P f (1-P)|_{S_q}$ is bounded by a constant which only depends on C and $1> q > \frac{1}{\alpha -\frac{1}{2}}$ ; If $|f_n| \le C |n|^{-\alpha }$ with $\alpha > \frac{3}{2}, n< 0$, then $|(1-P) f P |_{S_q}$ is bounded by a constant which only depends on C and $1> q > \frac{1}{\alpha -\frac{1}{2}}$.

Note that g, h are invariant under $u\rightarrow -u$. Note that $u(z)\in I_1\cup -I_1$ if an only if $u(z)^2 \in I_1^2=I_0$. Since $u(z)^2= m_1(z^2)$, it follows that $u(z)^2 \in I_1^2=I_0$ if and only if $m_1(z^2) \in I_0$. By definition of $m_1(z)$, we have $m_1^{-1}(I_0)$ is the complement of $I_0$, which is $I_2^2$. It follows that $u(z)\in I_1\cup -I_1$ if an only if $z\in I_2\cup -I_2$. So $h= \frac{z^2-u^2}{4uz} \chi _{I_2\cup -I_2}(z).$ Note that this matches with equation (34) of [25]. For the reader who may be confused with the equation (34) of [25], we note that in the definition of h in 2.1 it is important that we have $\chi _{I_1}(u)$, not $\chi _{I_1}(z)$.

If $u(z)\in I_1$, then $g(z)=\frac{(u+z)^2}{4uz} = \frac{1}{2}+ \frac{1}{4}(\frac{u}{z} +\frac{z}{u})$. Note that $|u|=|z|=1$, it follows that $g\ge 0.$ Similarly $g\ge 0$ if $u(z)\in -I_1$. When $z\in I_1$, $g=1$, and $g=0$ when $z\in -I_1$. So $g\ge 0$. Similarly we can check that ih is real. $g-1/2$ and h are both odd functions of z, and $g^2-g=h^2.$ To do computations for $\eta$ close to 0, it is convenient to choose an analytic continuation of u inside the unit disk. Recall that $u^2= m_1(z^2)= \frac{z^2\cos \eta -e^{-i\eta }}{e^{i\eta } z^2-\cos \eta }$. Note that the roots of $z^2\cos \eta -e^{-i\eta }$ are outside the unit disk. For the square root of $z^2\cos \eta -e^{-i\eta }$, we can choose any branch cut outside the closed unit disk, for an example, two half lines coming out of the two roots of the equation $z^2\cos \eta -e^{-i\eta }=0$ which do not intersect the closed unit disk. The two roots of $e^{i\eta } z^2-\cos \eta$ are inside the disk, and for the square root of $e^{i\eta } z^2-\cos \eta$, we can choose the branch cut to be the closed line segment connecting $\pm e^{-i\eta /2} \sqrt{\cos \eta }$ which are the poles of u (cf. Page 128 of [1] for such a choice of branch cut for square root of a quadratic function). u is then the quotient of these two functions. u is an analytic function in the unit disk minus the branch cut. We will see that this branch cut is important for our analysis when $\eta \rightarrow 0$.

On $I_2$

$$\begin{aligned} g'(z)= & {} \frac{(u^2-z^2)(zu'-u)}{4z^2u^2}, h'(z)= \frac{(u^2+z^2)(zu'-u)}{4z^2u^2} \end{aligned}$$

(20)

$$\begin{aligned} m_1(z^2)= & {} \frac{z^2\cos \eta -e^{-i\eta }}{e^{i\eta } z^2-\cos \eta }, m_1(z)'= \frac{-\sin ^2(\eta )e^{i\eta }}{({e^{i\eta } z-\cos \eta })^2}, m_1(z)''= \frac{-\sin ^2(\eta ) e^{2i\eta }}{({e^{i\eta } z-\cos \eta })^3} \end{aligned}$$

(21)

Let $L (z) := |z^2 -e^{-i\eta } \cos \eta |$. Let us explain how to estimate the derivatives of g, h when $\eta$ is sufficiently small. We will use g as an example since h is similar. If $u(z)\in I_1$, then $g(z)=\frac{(u+z)^2}{4uz} = \frac{1}{2}+ \frac{1}{4}(\frac{u}{z} +\frac{z}{u})$. Hence if $u(z)\in I_1$ the derivatives of g are linear combinations of derivatives of $\frac{u}{z}$ and $\frac{z}{u}$. To compute the derivatives of u(z), by definition $u(z)^2= m_1(z^2)$. So by Chain Rule $u(z)' u(z)= m_1'(z^2) z$. Keep in mind $|u|=|z|=1$. It follows that $g'$ is up to addition by a bounded function $m_1'(z^2)$ multiplied by a bounded function. $g''$ is the sum of a bounded function, $((m_1'(z^2)))^2$ multiplied by a bounded function and $m_1''(z^2)$ multiplied by a bounded function. The same idea applies to all other cases. Note that when $\eta$ is sufficiently small $L(z)\ge 1/2 \eta ^2$, and therefore from (21) we have

$$\begin{aligned} |((m_1'(z^2)))^2|= \frac{(\sin \eta )^4}{L(z)^4}\le 2\frac{(\sin \eta )^2}{L(z)^3} \end{aligned}$$

From this we have

$$\begin{aligned} |h'|\le & {} O(1)+ C \frac{\sin ^2(\eta )}{L(z)^2}, |h''|\le O(1)+ C \frac{\sin ^2(\eta )}{L(z)^{3}}, |g'|\le O(1)\nonumber \\&+\, C \frac{\sin ^2(\eta )}{L(z)^2}, |g''|\le O(1)+ C \frac{\sin ^2(\eta )}{L(z)^{3}} \end{aligned}$$

(22)

where O(1) and C are constants independent of $\eta$.

Note that the minimal value of $L(z^2)$ is when $z= e^{-i\eta /2}$, i.e. when z is at the middle point of $I_2$, and $L(e^{-i\eta /2}) = |1-\cos (\eta )| \thicksim \frac{1}{2} \eta ^2$ when $\eta$ is close to 0, it follows that on $I_2\cup -I_2$

$$\begin{aligned} |h'|\le C \frac{1}{L(z)}, |h''|\le C \frac{1}{L(z)^{2}}, |g'|\le C \frac{1}{L(z)}, |g''|\le C \frac{1}{L(z)^{2}} \end{aligned}$$

(23)

where C is independent of $\eta$ and $\eta$ is close to 0. Note that since $g=1$ on $I_1$, the above formula for g also holds on $I_1$.

Note that $m_1(z)$ is conjugate to the flip, and fix the end points of $I_0$. When z is at the end points of $I_2$ or $-I_2$, $z^2$ takes values at the end points of $I_0$. It follows that at the end points of $I_2$ or $-I_2$, $u(z)^2= m_1(z^2)=z^2$. From formula (20) we can see that $g'$ is continuous, and $g'= 0$ on the boundary of $I_2, -I_2.$ $g''$ exists at all points on the circle except the four boundary points of $I_2, -I_2$ and is bounded. Since $g'= 0$ on the boundary of $I_2, -I_2,$ it follows that the second derivative of g in the distribution sense agrees with $g''$, and in particular it is essentially bounded. Hence $g\in W^{2, \infty }$. Similarly from formula (20) we see that $h'$ is not continuous, but $h'(z)$ on $I_2 \cup -I_2$ is bounded when z is close to the boundary of $I_2, -I_2$.

The image of $I_2$ (resp. $-I_2$) under V is interval $(1, \tan \psi )$ (resp. $(-1, -\cot \psi )$) with $\psi =(\pi /4 + \eta /2)$. The cross ratio of the interval $-I_1, I_1$ in the clockwise order is $\frac{1}{\sin ^2(\eta /2)}.$

First we recall the lower bound of $S(F,\eta )$:

Lemma 2.4

$S(F,\eta )\ge \frac{-1}{6} \ln (\sin (\eta /2))$.

Proof

Since ${{\mathcal {A}}}(I_1)\subset F\subset {{\mathcal {A}}}(I_1\cup I_2 \cup -I_2)$, and the commutant of ${{\mathcal {A}}}(I_1\cup I_2 \cup -I_2)$ is $Z{{\mathcal {A}}}(-I_1)Z^{-1}$, where $Z= \frac{1 - i\Gamma }{1 - i}$ is the Klein operator (cf. Section 2.2), we have ${{\mathcal {A}}}(-I_1)\subset Z F'Z^{-1}$.

Denote by $\omega _F, \omega _{F'}$ the restriction of the vacuum state to F and its commutant $F'$. Let $\hat{\omega }$ be the tensor state $\omega _F\otimes \omega _{F'}$, namely $\hat{\omega }(m_1m_2)= \omega (m_1)\omega (m_2), \forall m_1\in F, m_2\in F'$. Let us show that when restricting $\hat{\omega }$ to ${{\mathcal {A}}}(I_1) \vee {{\mathcal {A}}}(-I_1)$, this is the same as $\omega _1\otimes _2 \omega _2$ as in Sect. 2.2. Since $F,F'$ are type I factors and ${\mathrm{Ad}}\Gamma$ are automorphisms of F (resp. $F'$) of order two, it follows that $\Gamma = u_1u_2$ where $u_1$ (resp. $u_2$) is unitary element in F (resp. $F'$). Multiplying by a phase factor if necessary we can choose $u_1^2=1, u_2^2=1$. Since $\Gamma u_1\Gamma =u_1^3= u_1,$ it follows that $u_1$ is an even element, and similarly $u_2$ is an even element.

By definition $\hat{\omega }$ is the same as $\omega _1\otimes _2 \omega _2$ on elements of the form $ab^{+}$, where $a\in {{\mathcal {A}}}(I_1)$, and $b^+$ is an even element of ${{\mathcal {A}}}(-I_1)$, i.e., $\Gamma b^+ \Gamma = b^+$. It is sufficient to check $\hat{\omega }(a b^{-}) =0$ if $\Gamma b^- \Gamma =- b^-, b^-\in {{\mathcal {A}}}(-I_1)$, i.e., $b^-$ is an odd element.

Note that $Zb^-Z^{-1}= -i\Gamma b^- \in F'$, and so $b^- = i \Gamma Zb^-Z^{-1}= i u_1 u_2 Zb^-Z^{-1}$. By definition of $\hat{\omega }$ we have

$$\begin{aligned} \hat{\omega }(a b^{-}) = \hat{\omega }(i au_1 u_2 Zb^-Z^{-1})= \omega (i au_1)\omega (u_2 Zb^-Z^{-1}) \end{aligned}$$

Note that $u_2 Zb^-Z^{-1}$ is an odd element in $F'$ and so $\omega (u_2 Zb^-Z^{-1})=0$.

By monotonicity of relative entropy (cf. Chapter 5 of [26])

$$\begin{aligned} S(\omega ,\omega _1\otimes _2 \omega _2)\le S(\omega , \omega _F\otimes \omega _{F'} ) =2 S(F, \eta ) \end{aligned}$$

By Th. 3.16 in [24] we have proved the Lemma. $\square$

2.4 An Inequality from Besov Quasinorm

We proceed now to the Besov classes $B^{\frac{1}{p}}_p$ for $0< p < 1$. Let F be an infinitely differentiable function on the real line such that $F \ge 0$, with support in [1/2, 2], and

$\sum _{n\ge 0} F (\frac{x}{2^n}) = 1, \forall x\ge 1$. It is very easy to construct such a function. We can take a nonnegative smooth function F on the interval [1/2, 1] such that $F(1/2) = 0, F(1) = 1, F^{(k)}(1/2) = F^{(k)}(1) = 0, \forall k \ge 1$. Then we can put $F(x)=1-F(x/2), x\in [1, 2]$ and $F(x) = 0$, when x is outside [1/2, 2]. Given an analytic function G in the unit disk with continuous extension to the boundary, assume that $G(z)-G(0)= \sum _{n> 0} G_n z^{n},$ define $F_n*G(z) = \sum _{j\ge 1} F(j/n) G_j z^j$ where $n\ge 1$ is an integer. Note that $F_n*G(z)$ is a trignometric polynomial of degree less than 2n. By definition we have $G(z)-G(0)= \sum _{m\ge 0, n=2^m} F_n *G$. Let $z=e^{2\pi it}$.

It follows from Page 250 of [30] that if $0<q<1$

$$\begin{aligned} |PG (1-P)|_{S_q}^q \le \sum _{m=0,n=2^m}^\infty 2^{1-q}2^m\int _{-1/2\le t\le 1/2} |F_n*G(z)|^q dt \end{aligned}$$

(24)

This inequality will play a crucial role in our paper. We shall refer to $\sum _{m=0,n=2^m}^\infty 2^m \int _{-1/2\le t\le 1/2} |F_n*G(z)|^q dt$ as the Besov quasinorm of G. The definition depends on the choice of F, and all choices give equivalent seminorms, but we shall not make use of this fact. The other direction that

$$\begin{aligned} \sum _{m=0, n=2^m}^\infty 2^m\int _{-1/2\le t\le 1/2} |F_n*G(z)|^q dt < \infty \end{aligned}$$

implies $|PG(1-P)|_{S_q}^q < \infty$ is also true, see Th. 3.1 of [30].

2.5 Poisson Summation

Let F be an infinitely differentiable function on the real line with support in [1/2, 2], and let m be a positive integer. Consider $F_m(z) = \sum _{k\in {{\mathbb {Z}}}} F(k/m) z^k$ where $z=e^{2\pi it}$ with t a point on the upper half plane (including the boundary real line).

Recall that ${{\mathcal {F}}}F$ is the Fourier transform of F, ${{\mathcal {F}}}F(s) =\int F(x) e^{-2\pi ixs } dx$. The proof of the following Lemma is essentially the same as the proof of Lemma 3.3 in [30]:

Lemma 2.5

Let $G_m(t) = F_m(e^{2\pi it}) - m({{\mathcal {F}}}F)(-mt)$ where t is on the upper half plane with real part in $[-1/2, 1/2]$. Denote by $G_m$ the maximum of $|G_m(t)|$ on $[-1/2, 1/2]$. Then $G_m m^N \rightarrow 0$ for all $N>0$.

Proof

Let $\psi (x)= F(x/m)e^{2\pi xit}$. Note that $\psi (x)$ is a Schwartz function since t is on the upper half plane, and the support of F is in [1/2, 2]. The rest of the proof using Poisson summation formula is exactly the same as the proof of Lemma 3.3 in [30]. $\square$

It is interesting to note that if t is real, $F_m(e^{2it})$ is a periodic function in t and therefore can be thought as a function on the unit circle, but $m({{\mathcal {F}}}F)(-mt)$ is not. Nevertheless $m({{\mathcal {F}}}F)(-mt)$ captures the dominating part of $F_m(z)$ when $m\rightarrow \infty$ as the above Lemma shows. By repeatedly using integration by parts we have

$$\begin{aligned} ({{\mathcal {F}}}F)(-mt) \le C_N \frac{e^{-ms}}{1+ m^N |t|^N}, \forall N>0 \end{aligned}$$

(25)

where s is the imaginary part of t, and $C_N$ is a constant which only depends on F.

3 Asymptotic Analysis

We will determine the upper bound of $S(F,\eta )$ in the next few sections. Let us first describe the basic ideas.

We note that

$$\begin{aligned} PP_{12}P= PgP+ PhPR \end{aligned}$$

, and both PgP and PhP (cf. Th. 2.1 for definitions) are Toeplitz operators. When $\eta$ is small, the support of h shrinks to zero size, so we expect the main contribution to $S(F,\eta )$ should come from PgP. To do this we first need to have a good control on the Schattern-von Neumann norm of $Ph(1-P)$, this is done in Sect. 3.3. There is also a further complication concerning PgP. It turns out that $f_0(PgP)$ is not trace class, but $f_0(PgP)- Pf_0(g)P$ is. This problem is addressed in Sect. 3.4.

Suppose a function $F(z,\eta ), z\in S^1$ is defined on the circle which depends also on a parameter $0<\eta <\pi$. We always assume that F is bounded, i.e., $|F(z,\eta )|\le M$ for some constant M which is independent of $z,\eta$. We are interested in the property of $F(z,\eta )$ when $\eta \rightarrow 0$.

Definition 3.1

A bounded function $F(z,\eta )$ is said to be very good if both $|PF(1-P)|_{S_q}$ $|(1-P)FP|_{S_q}$ are O(1) when $\eta \rightarrow 0$ for a $0<q<1$. A function $F(z,\eta )$ is said to be good if both $|PF(1-P)|_{S_q}^q$ and $|(1-P)FP|_{S_q}^q$ are $o(-\ln \eta )$ when $\eta \rightarrow 0$. We write $F(z,\eta )\thicksim G(z,\eta )$ if there exist two positive constants $C_1, C_2$ such that

$$\begin{aligned} C_1 |G(z,\eta )| \le |F(z,\eta )| \le C_2 |G(z,\eta )| \end{aligned}$$

Proposition 3.2

• (1) If F is good (resp. very good) , then $|FP-PF|_{S_q}^q=o(-\ln \eta )$ (resp. O(1)) ;

• (2) If F and G are good (resp. very good), then both $F+G$ and FG are good (resp. very good).

Proof

By equation (7) we may assume that F, G are good for the same q.

Ad (1): $FP-PF= -PF(1-P)+ (1-P)FP$

So by q-triangle inequality

$$\begin{aligned} |FP-PF|_{S_q}^q \le |(1-P)FP|_{S_q}^q + |PF(1-P)|_{S_q}^q \end{aligned}$$

and (1) follows.

Ad (2): The statement for $F+G$ follows from the q-triangle inequality as in (5).

Note that $PFG(1-P)= (PF-FP) G(1-P) + FPG(1-P)$. So we have

$$\begin{aligned} |PFG(1-P)|_{S_q}^q\le & {} |(PF -FP) G(1-P)|_{S_q}^q +|F P G(1-P)|_{S_q}^q \\\le & {} |(P F -F P)|_{S_q}^q ||G||^q +|P G(1-P)|_{S_q}^q ||F||^q \end{aligned}$$

and (2) is proved. $\square$

3.1 Deformation of path

We’d like to use Lemma 2.3 to do estimation. For this purpose it is important to estimate the growth of the Fourier coefficients of our functions such as h, g. Unfortunately $h''$ grows like (cf. equation (22)) $\frac{\eta ^2}{L(z)^3}$ on $I_2$, and $L(z)\thicksim (\theta + \eta ^2)$ where $\theta$ is the distance between z and the middle point of $I_2$. This makes it difficult or even impossible to obtain $O(n^{-2})$ type estimate. One simple idea is to see if we can use Cauchy’s theorem to deform the path $I_2$ to a path where $h''$ is better controlled. A natural such path is the path N which join the ends of $I_2$ inside the unit disk with property $|L(z)|= \sin \eta$. On this path N, $h''\thicksim \frac{1}{\eta }$, and when integrated over N which has length $\thicksim \pi \eta$ will give us O(1). But we have to pay close attention to possible poles and branch cuts enclosed by $I_2$ and N. We will see that ultimately it is the branch cut that is responsible for the asymptotic growth of our entropy.

3.2 Deformation of path: The case with no Branch cut

Lemma 3.3

Assume that $F(z,\eta )$ is analytic in the interior bounded by $I_2$ and N in the unit disk, and has continuous first derivative on $I_2$. In addition assume on the circle F is 0 at the boundary of $I_2$, and $F'$ is O(1) on the boundary of $I_2$. If $\int _N |F''| = O(1)$ where N is the path which join the ends of $I_2$ inside the unit disk with property $|L(z)|= \sin \eta$. Then $a_n (F): = \int _{I_2} F z^n dz = O(n^{-2}), \forall n\ge 0$.

Proof

By our assumptions on F and integration by parts

$$\begin{aligned} a_n = \int _{I_2} F z^n dz = \frac{1}{(n+1)(n+2)}\int _{I_2} F'' z^n dz + O(n^{-2}) \end{aligned}$$

It is sufficient to check that

$$\begin{aligned} \int _{I_2} F'' z^n dz = O(1) \end{aligned}$$

Since F is analytic in the unit disk, deforming the path $I_2$ to N, and keep in mind $|z^n|\le 1, n\ge 0$ when $|z|\le 1$ we have

$$\begin{aligned} |\int _{I_2} F'' z^n dz| = |\int _{N} F'' z^n dz| \le \int _N |F''| = O(1) \end{aligned}$$

$\square$

Proposition 3.4

Both $u^2$ and $u^{-2}$ are very good.

Proof

By definition

$$\begin{aligned} u^2= & {} m_1(z^2)= \frac{z^2\cos \eta -e^{-i\eta }}{e^{i\eta } z^2-\cos \eta } \\ m_1(z^2)= & {} \frac{z^2\cos \eta -e^{-i\eta }}{e^{i\eta } z^2-\cos \eta }= e^{-i\eta } \cos \eta - \frac{e^{-i2\eta } \sin ^2(\eta )}{z^2- e^{-i\eta } \cos \eta } \end{aligned}$$

It follows that

$$\begin{aligned} P u^2 (1-P )= 0 \end{aligned}$$

since $u^2$ is analytic outside the unit disk, and by using Laurent series for $u^2$

$$\begin{aligned} (1-P) u^2 P = (1-P) T_1 P -(1-P) T_2 P \end{aligned}$$

where

$$\begin{aligned} T_1= & {} \frac{1/2 e^{-3/2 i\eta } (\sin (\eta ))^2 (\cos \eta )^{-1/2}}{z- e^{-1/2 i\eta }\sqrt{\cos \eta }} \\ T_2= & {} \frac{1/2 e^{-3/2 i\eta } (\sin (\eta ))^2 (\cos \eta )^{-1/2}}{z+ e^{-1/2 i\eta }\sqrt{\cos \eta }} \end{aligned}$$

Note that $(1-P )T_1 P$ is a rank one operator by using Laurent series for $T_1$, and the norm of $P T_1 (1-P)$ is given by the maximum of $|T_1|$ on the circle. These are very special cases of finite rank Hankel operators, cf. 1.3 of [30] for more details. The maximum of $|T_1|$ on the circle is

$$\begin{aligned} (\cos \eta )^{-1/2}\frac{(\sin (\eta ))^2}{1-\sqrt{\cos {\eta }}} = O(1) \end{aligned}$$

as $\eta \rightarrow 0$. It follows that

$$\begin{aligned} |(1-P )T_1 P)|_{S_q}^q = O(1) \end{aligned}$$

for any $q>0$. Similarly

$$\begin{aligned} |(1-P) T_2 P|_{S_q}^q = O(1) \end{aligned}$$

for any $q>0$ and we have proved $u^2$ is very good. Note that $[Pu^{-2}(1-P)]^* =(1-P) u^2 P$ It follows that

$$\begin{aligned} | Pu^{-2}(1-P) |_{S_q}^q = | (1-P)u^{2} P |_{S_q}^q \end{aligned}$$

and the Proposition is proved. $\square$

Proposition 3.5

$(z^{-2}- u^{-2})(\chi _{I_2} + \chi _{-I_2})$ and its complex conjugate are very good.

Proof

It is sufficient to prove that $(z^{-2}- u^{-2})\chi _{I_2}$ and its complex conjugate $(z^{2}- u^{2})\chi _{I_2}$ are very good. The proof for $(z^{-2}- u^{-2})\chi _{-I_2}$ and its complex conjugate is similar. To simplify writing we will denote the function $(z^{-2}- u^{-2})\chi _{I_2}$ by $h_1$ only in the proof of this proposition. We first show that $| (1-P) h_1 P |_{S_q} = O(1)$ for some $0\le q <1$. By Lemma 2.3 it is enough to show that $\int _{I_2} h_1 z^{n} dz = O(n^{-2}), \forall n\ge 0$.

On N we shall need an estimate of derivatives of $h_1$ similar to that of formula (22) for h on the circle. Note that when $\eta$ is small enough, |z| is close to 1 on N. Since $|z^2-e^{-i\eta }\cos \eta |=\sin \eta$ on N, $\cos \eta \thicksim 1-\frac{1}{2}\eta ^2$, $\cos \eta -(\cos \eta )^{-1}\thicksim \eta ^2$, we have $|z^2e^{i\eta }- (\cos \eta )^{-1}|\thicksim \eta$ when $\eta$ is sufficiently small, and hence

$|m_1(z^2)|= |\frac{z^2e^{i\eta }- (\cos \eta )^{-1}}{\sin \eta }|$ is close to 1 when $\eta$ is sufficiently small. Now the comments before formula (22) applies verbatim, We have both $h_1$ and $h_1'$ are O(1) and $|h_1''| \thicksim 1/\eta ,$ and it follows that

$$\begin{aligned} \int _N |h_1''| ds \le C \pi \eta /\eta = O(1) \end{aligned}$$

By Lemma 3.3 it follows that $| (1-P) h_1 P |_{S_q} = O(1), 2/3<q<1$. Note that $h_1^*$ has poles inside the unit disk, so our deformation of path argument above does not work for $h^*$. But $h_1^* = -z^2u^2 h_1$, hence if we multiply $h_1^*$ by $\frac{1}{-z^2u^2}$ then we get $h_1$, and we have removed the poles of $h_1^*$. The function $\frac{1}{-z^2u^2}$ has modulus 1 and both $\frac{1}{-z^2u^2}$ and its complex conjugate are very good by Prop. 3.4, it will follow that $h_1^*$ is very good. This will be called as “the trick of removing poles ". In more explicit terms,

$(Ph_1(1-P))^* = (1-P) (-z^2 u^2) h_1 P$,

$$\begin{aligned} | P h_1 (1-P) |_{S_q} = |[(Ph_1(1-P))]^*|_{S_q} = |(1-P) (-z^2 u^2) h_1P|_{S_q} = O(1) \end{aligned}$$

by Prop. 3.2, Prop. 3.4 and $| (1-P) h_1 P |_{S_q} = O(1), 2/3<q<1$ that has already been proved.

Note that $| P h_1^* (1-P) |_{S_q} =| (1-P) h_1 P |_{S_q}$ and similarly with P replaced by $1-P$, and the proposition is proved. $\square$

3.3 Deformation of path: The case with Branch cut

Our goal in this section is to show that h is good. Note that h is independent of the choice of branch cut of u inside unit disk , and in this section we choose the branch cut to be the closed line segment with end points $e^{-i \eta /2}\sqrt{\cos \eta }$ and $-e^{-i \eta /2} \sqrt{\cos \eta }$. Here $\eta$ is very close to 0 so that $\cos \eta \thicksim 1$.

Note that h has poles inside the unit disk, but since by Prop. 3.4$u^{-2}$ and $u^2$ are very good, by Prop. 3.2 it is sufficient to show that $h_0:= u^{-2} h$ is good, and $u^{-2} h$ has no poles in the unit disk, but has branch cut. This is another example of the trick of removing poles. The branch cut is important, because without the branch cut we c could conclude as in the previous section that by using the trick of removing poles, both g, h are very good, but this would contradict the lower bound in Lemma 2.4. First we have :

$$\begin{aligned} z^2- u^2= -\frac{(z^2-e^{-2i\eta })(z^2-1) e^{i\eta }}{e^{i\eta } z^2- \cos \eta } \end{aligned}$$

Let $z_1= e^{i\theta _1}$ with $\theta _1= \eta + 2 \theta ,$ then

$$\begin{aligned} h_0= Cz^{-3}(\cos \eta )^{-1/2} \frac{\frac{1}{2}(z_1+ z_1^{-1})-\cos \eta }{1-z_1\cos \eta } \sqrt{\frac{z_1-\cos \eta }{z_1-(\cos \eta )^{-1}}} \end{aligned}$$

where C is a constant with $|C|=1$. Hence it is enough to check that

$$\begin{aligned} h_1= \frac{\frac{1}{2}(z_1+ z_1^{-1})- \cos \eta }{1-z_1\cos \eta } \sqrt{\frac{z_1-\cos \eta }{z_1- (\cos \eta )^{-1}}} \end{aligned}$$

is good. Note that $z_1=e^{i\eta } z^2$ and we think of $h_1$ as a function of z. Note that $h_0^* = u^2 h = -u^4 h_0$ since $h^*=-h$, it follows that $h_1^*= C_1 z^6 u^4 h_1$ where $C_1$ is a nonzero constant. Hence if we can show that $|(1-P) h_1 P|_{S_q}^q =o(-\ln \eta )$, then it follows

$$\begin{aligned} |P h_1 (1-P)|_{S_q}^q =|(1-P) h_1^* P|_{S_q}^q =|(1-P) z^6 u^4 h_1 P|_{S_q}^q =o(-\ln \eta ) \end{aligned}$$

since $z^6 u^4$ is very good by Prop. 3.4 and Prop. 3.2.

In the following we will only consider $h_1$ restricted to $I_2$ and show $|(1-P) h_1 P|_{S_q}^q =o(-\ln \eta )$. Exactly the same argument also shows that $h_1$ restricted to $-I_2$ verifies similar inequality.

Denote by $a_n =\int _{I_2} h_1 z^n dz, n\ge 0$. Consider the function $h_2(z)= \sum _{n\ge 0} a_n z^{-n-1}$ on the circle. We will write $h_2$ as a sum of three functions. Note that $h_1 z^n$ is analytic in the unit disk except along the branch cut. We will write the $\int _{I_2} h_1 z^n dz$ as the integral of $h_1 z^n$ on three paths on the z plane. To describe these paths, note that we will be doing integrals in a small neighborhood of 1 when $\eta$ is close to 0. In this small neighborhood the map $z\rightarrow z_1=e^{i\eta } z^2$ is certainly one to one. Hence it is enough to describe these paths under the map $z\rightarrow z_1=e^{i\eta } z^2$.

The imagine of these three paths are easier to describe in terms of $z_1=e^{i\eta } z^2$ on the $z_1$ plane: first the path on the upper half of $z_1$ plane with $|z_1-\cos \eta |= \sin \eta$ from $e^{i\eta }$ to $\cos \eta -\sin \eta$; We denote this quarter of the circle by $\hat{J_1}$.

The second path is along part of the branch cut $[\cos \eta -\sin \eta , \cos \eta ]$, and then turning in the opposite direction along the same closed interval. We denote this interval by $\hat{J_2}$. The last part is in the lower half of $z_1$ plane from $\cos \eta -\sin \eta$ to $e^{-i\eta }$, and we denote this quarter of the circle by $\hat{J_3}$. See Figure 2 for the image of the three paths on the $z_1$ plane. In Figure 2 points 2, 3 correspond to $\cos \eta -\sin \eta ,\cos \eta$ respectively on the $z_1$ plane. The small arc part of the unit circle from $e^{-i\eta }$ to $e^{i\eta }$ is the image of $I_2$ on the $z_1$ plane. We will denote by $J_1, J_2, J_3$ the pre-images of $\hat{J_1}, \hat{J_2}, \hat{J_3}$ on the z plane.

Fig. 2

Image of a contour

3.3.1 The part from Integral Along a Quarter of a Circle

Let us first show that $b_n = \int _{J_1} h_1 z^n dz= O(n^{-2}).$ Note that $h_1$ is equal to 0 at $z_1=e^{i\eta }$.

$$\begin{aligned} \cos \eta -\sin \eta + \frac{1}{\cos \eta -\sin \eta } = 2+\eta ^2 +o(\eta ^2) \end{aligned}$$

Recall that

$$\begin{aligned} h_1= \frac{\frac{1}{2}(z_1+ z_1^{-1})- \cos \eta }{1-z_1\cos \eta } \sqrt{\frac{z_1-\cos \eta }{z_1- (\cos \eta )^{-1}}} \end{aligned}$$

When $z_1=\cos \eta -\sin \eta$ and $\eta$ is sufficiently close to 0, we have $z_1\thicksim 1-\eta -\frac{1}{2}\eta ^2, \cos \eta \thicksim 1-\frac{1}{2}\eta ^2$. It follows that $\frac{1}{2}(z_1+ z_1^{-1})- \cos \eta \thicksim \eta ^2$, $1-z_1\cos \eta \thicksim \eta$, $z_1-\cos \eta \thicksim \eta$, $z_1- (\cos \eta )^{-1}\thicksim \eta$, $|\cos \eta -\sin \eta |\le 1-\frac{1}{2}\eta$,

we conclude that the value of $h_1 z^n$ at $z_1= z^2 e^{i\eta }= \cos \eta -\sin \eta$ is bounded by an absolute constant multiplied by

$$\begin{aligned} \eta (1-\frac{\eta }{2})^{\frac{n}{2}} \end{aligned}$$

We need the following:

Lemma 3.6

$(1-\eta )^n \eta = O(n^{-1})$ uniformly in $\forall 0< \eta < 1$.

Proof

Let $f(\eta )=(1-\eta )^n \eta .$ Note that $f(0)=f(1)=0$, and so the maximum of f is attained at the critical point of f. Let $f'= (1-\eta )^n -n\eta (1-\eta )^{n-1}=0$ we get $\eta = \frac{1}{n+1}$, and so the maximum of f is

$$\begin{aligned} (1-\frac{1}{n+1})^n \frac{1}{n+1} \end{aligned}$$

Note that

$$\begin{aligned} \lim _{n\rightarrow \infty } (1-\frac{1}{n+1})^n =e^{-1} \end{aligned}$$

and the Lemma is proved. $\square$

By using integration by parts, $h_1$ is equal to 0 at $z_1=e^{i\eta }$, and Lemma 3.6 we have

$$\begin{aligned} \int _{J_1} h_1 z^n dz= O(n^{-2}) - \frac{1}{n+1}\int _{J_1} h_1' z^{n+1} dz \end{aligned}$$

On $J_1$ we shall need an estimate of derivatives of $h_1$ similar to that of formula (22) for h on the circle. Note that when $\eta$ is small enough, |z| is close to 1 on $J_1$. Since $|z_1-\cos \eta |=\sin \eta$ on $J_1$, $|z_1-(\cos \eta )^{-1}|\thicksim \eta$ when $\eta$ is sufficiently small, and hence

$|m_1(z^2)|= |\frac{z_1\cos \eta -1}{z_1-\cos \eta }|$ is close to 1 when $\eta$ is sufficiently small. Now the comments before formula (22) applies verbatim, and we find that $h_1' z^{n}$ is O(1) on the boundary of $J_2$, and

$$\begin{aligned} \int _{J_1} h_1 z^n dz= O(n^{-2}) - \frac{1}{(n+1)(n+2)}\int _{J_1} h_1'' z^{n+2} dz \end{aligned}$$

Note that

$$\begin{aligned} \int _{J_1} |h_1''| \thicksim \int _0^\eta \frac{\eta ^2}{\eta ^3} d\theta = O(1) \end{aligned}$$

we have shown that

$$\begin{aligned} \int _{J_1} h_1 z^n dz = O(n^{-2}) \end{aligned}$$

Similarly we have

$$\begin{aligned} \int _{J_3} h_1 z^n = O(n^{-2}) \end{aligned}$$

3.3.2 The part from Integrals Along the Branch cut

Set $c_n := \int _{J_2} h_1 z^n dz$. Since u changes signs from the upper part of the branch cut to the lower part, we should actually consider $2c_n$. But since our estimate is up to multiplication by a positive constant, we can ignore this constant 2 in the following. We need to show $h_3(z):= \sum _{n\ge 0} c_n z^{-n-1}$ is good.

Since $|(1-P)h_3 P|_{S_q}^q = |Ph_3^* (1-P)|_{S_q}^q$, We will use inequality (24) for $h_3^*$.

Let ${\hat{f}}_n(t)= F_n* h_3^*$, and $f_n(t)= {\hat{f_n}}^*$. Then

$$\begin{aligned} f_n(t)= \int _{J_2} \sum _{j\ge 1} h_1 z^{j-1} F(j/n) e^{-2\pi i j t} dz \end{aligned}$$

Note that $z_1= z^2 e^{i\eta },$ $z= e^{-i\eta /2} \sqrt{z_1}, dz= \frac{e^{-i\eta /2}}{2 \sqrt{z_1}}dz_1$, and

$$\begin{aligned}&f_n(t)= \int _{\cos \eta -\sin \eta }^{\cos \eta }\frac{\frac{1}{2}(z_1+ z_1^{-1})- \cos \eta }{1-z_1\cos \eta } \sqrt{\frac{z_1-\cos \eta }{z_1- (\cos \eta )^{-1}}} \\&\qquad \qquad \sum _{j\ge 1} [\sqrt{z_1}e^{-i\eta /2}]^{j} F(j/n) e^{-2\pi i j t} \frac{1}{2z_1} dz_1 \end{aligned}$$

Set $t_1=-t-\frac{\eta }{4\pi }$. Note we have

$$\begin{aligned} \int _{-1/2\le t\le 1/2} |{\hat{f}}_n (t)|^p dt=\int _{-1/2\le t\le 1/2} |f_n(t)|^p dt =\int _{-1/2\le t_1\le 1/2} |f_n(t)|^p dt \end{aligned}$$

where the second equality follows since $f_n(t)$ is a function of t with period 1.

We need to estimate $\int _{-1/2\le t_1\le 1/2} |f_n(t)|^p dt$. On $J_2$,

$$\begin{aligned}\sqrt{\frac{z_1-\cos \eta }{z_1- (\cos \eta )^{-1}}} = O(1), \frac{1}{2}(z_1+ z_1^{-1})- \cos \eta = O(\eta ^2) \end{aligned}$$

and

$$\begin{aligned} \int _{\cos \eta -\sin \eta }^{\cos \eta } \frac{1}{1-z_1\cos \eta } dz_1= O(-\ln \eta ) \end{aligned}$$

When $\eta$ is sufficiently small we have $\frac{1}{2z_1}\le 1$.

Note that $\sum _{j\ge 1} [\sqrt{z_1}e^{-i\eta /2}]^{j} F(j/n) e^{-2\pi i j t} = \sum _{j\ge 1} F(j/n) e^{2\pi i j{\hat{t}}}$ with ${\hat{t}}=t_1+ \frac{s}{2\pi }, s= -i \frac{1}{2}\ln z_1$, and on $J_2$, $\frac{1}{2}\ln z_1 <0$. When $-1/2 \le t_1\le 1/2$ , apply Lemma 2.5 to $\sum _{j\ge 1} F(j/n) e^{2\pi i j{\hat{t}}}$, we have that up to $O(n^{-N})$ term for any $N>0$, we can replace $f_n(t)$ by

$$\begin{aligned} g_n(t_1) =\int _{\cos \eta -\sin \eta }^{\cos \eta } \frac{\frac{1}{2}(z_1+ z_1^{-1})-\cos \eta }{1-z_1\cos \eta } \sqrt{\frac{z_1-\cos \eta }{z_1- (\cos \eta )^{-1}}} n {{\mathcal {F}}}F (-n (t_1+ \frac{s}{2\pi })) \frac{1}{2z_1}dz_1 \end{aligned}$$

where $s= -i \frac{1}{2}\ln z_1$. We will choose N large enough such that

$$\begin{aligned} \sum _{m=0, n=2^m}^\infty n\int _{ |t_1 | \le \frac{1}{2}}|f_n(t)|^p dt = O(1)+ \sum _{m=0, n=2^m}^\infty n\int _{ |t_1 | \le \frac{1}{2}}|g_n(t_1)|^p dt_1 \end{aligned}$$

Now it is sufficient to evaluate $\sum _{m=0, n=2^m}^\infty n\int _{ |t_1 | \le \frac{1}{2}}|g_n(t_1)|^p dt_1$.

Note that by inequality (25)

$$\begin{aligned} |{{\mathcal {F}}}F (-n (t_1+ \frac{s}{2\pi }))|\le C_N \frac{e^{\frac{n}{2} \ln (z_1)}}{1+ n^N(|t_1+\frac{s}{2\pi }|))^N}, \forall N\ge 0 \end{aligned}$$

where the constant $C_N$ depends on N and F.

It follows that when $\eta$ is sufficiently small

$$\begin{aligned}&|g_n(t_1)|\le n \int _{\cos \eta -\sin \eta }^{\cos \eta }\frac{|\frac{1}{2}(z_1+ z_1^{-1})- \cos \eta |}{1-z_1\cos \eta } \sqrt{\frac{z_1-\cos \eta }{z_1- (\cos \eta )^{-1}}} \\&\quad C_N \frac{e^{\frac{n}{2} \ln (\cos \eta )}}{(1+ n^N(|t_1|+\frac{-\ln (\cos \eta )}{2\pi }))^N} dz_1 \end{aligned}$$

We note that the exponential decay factor $e^{\frac{n}{2} \ln (\cos \eta )}$ is due to the fact that the branch cut is inside the unit disk.

Recall that on $J_2$,

$$\begin{aligned} \sqrt{\frac{z_1-\cos \eta }{z_1- (\cos \eta )^{-1}}} = O(1), \frac{1}{2}(z_1+ z_1^{-1})- \cos \eta = O(\eta ^2) \end{aligned}$$

and

$$\begin{aligned} \int _{\cos \eta -\sin \eta }^{\cos \eta } \frac{1}{1-z_1\cos \eta } = O(-\ln \eta ) \end{aligned}$$

It follows that

$$\begin{aligned} |g_n(t_1)|\le C_N n (-\ln \eta ) \eta ^2 \frac{e^{\frac{n}{2} \ln (\cos \eta )}}{1+ n^N(|t_1|+\frac{-\ln (\cos \eta )}{2\pi })^N} \end{aligned}$$

(26)

To evaluate $\int _{-1/2\le t_1 \le 1/2} |g_n(t_1)|^p dt_1, t_1=t-\frac{\eta }{4\pi }$, we break this integral into two parts. Set $\delta :=-\ln (\cos \eta )$. First we evaluate

$$\begin{aligned} \int _{|t_1 |\le \delta } |g_n(t_1)|^p dt_1 \end{aligned}$$

Choose $N=1$ in (26). Since $p<1$ we have $\int _{|t_1 |\le \delta } |g_n(t_1)|^p dt_1\le C_1 \eta ^{2p} (-\ln \eta )^p \delta ^{1-p} e^{-\delta \frac{n}{2} p}$

Hence

$$\begin{aligned} \sum _{m=0, n=2^m}^\infty n\int _{|t_1 |\le \delta } |g_n(t_1)|^p dt_1 \le C_1 \sum _{m=0, n=2^m}^\infty n \eta ^{2p} (-\ln \eta )^p \delta ^{1-p} e^{-\delta \frac{n}{2} p} \end{aligned}$$

Note that $\eta ^2 =O(\delta )$ and by Lemma 3.10 we have proved that

$$\begin{aligned} \sum _{m=0, n=2^m}^\infty n \int _{|t_1 |\le \delta } |g_n(t_1)|^p dt_1 = O( (-\ln \eta )^p) \end{aligned}$$

Next we evaluate $\int _{\frac{1}{2}\ge |t_1 |\ge \delta } |g_n(t_1)|^p dt_1$. This time we choose N in (26) such that $1+p> Np>1$. Note that when $\eta$ is small enough we have

$$\begin{aligned} \int _{\frac{1}{2}\ge |t_1 |\ge \delta } |t_1|^{-Np} dt_1 \le 2\delta ^{1-Np} \end{aligned}$$

We get

$$\begin{aligned} \sum _{m=0, n=2^m}^\infty n\int _{\frac{1}{2}\ge |t_1 |\ge \delta } |g_n(t_1)|^p dt_1 \le 2 C_N \sum _{m=1, n=2^m}^\infty n^{1+p-Np} \eta ^{2p} (-\ln \eta )^p \delta ^{1-Np} e^{-\delta \frac{n}{2} p} \end{aligned}$$

Note that $\eta ^2 =O(\delta )$ and by Lemma 3.10 we prove that

$$\begin{aligned} \sum _{m=0, n=2^m}^\infty n\int _{\frac{1}{2}\ge |t_1 |\ge \delta } |g_n(t_1)|^p dt_1 = O( (-\ln \eta )^p) \end{aligned}$$

By inequality (24) we have proved

$$\begin{aligned} |(1-P)h_3 P|_{S_p}^p = O( (-\ln \eta )^p), \forall 0<p<1. \end{aligned}$$

Putting together these three parts from Sects. 3.3.1 and 3.3.2, and use Lemma 2.3,

we prove the following Theorem:

Theorem 3.7

If $2/3<p <1$, then

$$\begin{aligned} |(1-P)h P|_{S_p}^p = O( (-\ln \eta )^p), |Ph (1-P)|_{S_p}^p = O( (-\ln \eta )^p) \end{aligned}$$

Remark 3.8

Though the above theorem is sufficient for our purpose, we can actually show that $||(1-P)h P||_1= O(1)$. It is an interesting question to see if one can improve the above theorem to show that h is very good.

Corollary 3.9

If $2/3<p <1$, then

$$\begin{aligned} |(1-P)gh P|_{S_p}^p = O( (-\ln \eta )^p), |Phg (1-P)|_{S_p}^p = O( (-\ln \eta )^p) \end{aligned}$$

Proof

By definition

$$\begin{aligned} gh= \frac{z^2+u^2}{16u^2z^2}(z^2-u^2)\chi _{I_2\cup -I_2} +\frac{1}{2}h \end{aligned}$$

The corollary follows from Th. 3.7 and Proposition 3.4, Prop. 3.5 and Prop. 3.2. $\square$

Lemma 3.10

Assume that $x>0, p>0,\delta >0$. Then

$\sum _{m=0}^\infty 2^{mx} e^{-2^m \delta p} \delta ^x = O(1)$ when $\delta \rightarrow 0$.

Proof

Let $F(y):= y^x e^{-yp}.$ Then $\lim _{y\rightarrow 0} F(y)= 0 = \lim _{y\rightarrow \infty } F(y)$. The only critical point of F(y) is at $y=x/p$ and the maximum of F(y) is $F(x/p)< \infty .$ F is increasing when $y< x/p$ and decreasing when $y>x/p.$

It follows that

$$\begin{aligned} \sum _{m=0}^\infty 2^{mx} e^{-2^m \delta p} \delta ^x \le 2 F(x/p)\delta ^x + \int _0^\infty 2^{wx} e^{-2^w\delta p} \delta ^x dw \end{aligned}$$

Set $2^w \delta = w_1$, and since $x>0$ we have

$$\begin{aligned} \int _0^\infty 2^{wx} e^{-2^w\delta p} \delta ^x dw =\int _{\delta }^\infty w_1^{x-1} e^{-w_1p} \frac{1}{\ln 2} dw_1 = O(1) \end{aligned}$$

and the Lemma follows.

3.4 Estimation of Entropy

Since $f_0(PgP), P_0f(g)P$ are not trace class operators, but $f_0(PgP)- Pf_0(g)P$ is, this makes it very delicate to show that $S(F,\eta ) -{\hbox {tr}}(f_0(PgP)- Pf_0(g)P)$ is small. This is proved in several steps in this section.

We begin this section with a generalization of Lemma 2.2 in [38].

Lemma 3.11

Suppose that $R_1(z), R_2(z), z>0, V$ are bounded operators, and $W= R_1(z) V R_2 (z).$ Assume that as $z\rightarrow 0$, $||R_1(z)|| \thicksim \frac{1}{z^{t_1}}$, $||R_2(z)||\thicksim \frac{1}{z^{t_2}}$, $||W ||\thicksim \frac{1}{z^{t_0}}$, where $t_0, t_1,t_2$ are positive. Let $0< \sigma <1$. Then

$$\begin{aligned} ||W||_1 \le ||W||^{1-\sigma } ||R_1||^\sigma ||R_2||^\sigma |V|_{S_\sigma }^\sigma \thicksim \frac{1}{z^{(t_0(1-\sigma )+ (t_1+t_2)\sigma )}} |V|_{S_\sigma }^\sigma \end{aligned}$$

Proof

First by (6)

$||W||_1 =|||W|^{1-\sigma } |W|^\sigma ||_1 \le ||W||^{1-\sigma } |||W|^\sigma ||_1$

Note that by (4) $s_n(R_1 V R_2)\le ||R_1|| s_n(V) ||R_2||$, and so $s_n(R_1 V R_2)^\sigma \le ||R_1||^\sigma s_n(V)^\sigma ||R_2||^\sigma$ and the Lemma follows. $\square$

In our applications in this section $0\le t_0 \le 2, 0\le t_1 \le 1,0\le t_2 \le 1$. So the maximum of $t_0(1-\sigma )+ (t_1+t_2)\sigma )$ is 2. There are two different cases that are important in the following: The first case is when we need $t_0(1-\sigma )+ (t_1+t_2)\sigma <2$ to make sure our integral is convergent: in this case if we can manage to find one of the $t_i, i=0,1,2$ which do not take their maximal value then we will achieve our goal. The second case is when $0\le t_0 \le 1, 0\le t_1+t_2 \le 1$. In this case the maximum of $t_0(1-\sigma )+ (t_1+t_2)\sigma \le 1$, and we need to get $t_0(1-\sigma )+ (t_1+t_2)\sigma < 1$. Again this can be done if we can manage to get $t_0$ or $t_1+t_2$ to take values less than their allowed maximum , then we can make sure that our integral is convergent. We will see three different such “savings" of the exponents in the following.

First we will use an integral representation for $f_0(T)$ (cf. [7]).

Lemma 3.12

Suppose $0\le T\le 1$ is an operator. Then

$$\begin{aligned} f_0(T)=\int _{\frac{1}{2}}^\infty \frac{2\beta }{\beta +\frac{1}{2}} \frac{\widetilde{T}}{z+ \widetilde{T}} d\beta \end{aligned}$$

where $\widetilde{T}= T(1-T),z=\beta ^2-\frac{1}{4}$.

Proof

Using Fundamental Theorem of Calculus one checks that if $0\le x\le 1$,

$$\begin{aligned} f_0(x)=\int _{\frac{1}{2}}^\infty \frac{2\beta }{\beta +\frac{1}{2}} \frac{x(1-x)}{\beta ^2-\frac{1}{4}+ x(1-x)} d\beta \end{aligned}$$

and the Lemma follows from functional calculus for self-adjoint operators. $\square$

Let $A=PP_{12}P, {\widetilde{A}}= PP_{12}P(1-PP_{12}P)$. See Th. 2.1 for definition of $P_{12}$.

$\widetilde{B_1}= Pg^2P- (PgP)^2, {\widetilde{B}}= PgP- (PgP)^2$.

Note that since $P_{12}$ is a projection, ${\widetilde{A}}= PP_{12}P(1-PP_{12}P)= PP_{12}(1-P)P_{12}P$. By Prop. 3.5 and Th. 3.7 of [25] ${\widetilde{A}}$ is of trace class. Similarly $\widetilde{B_1}$ if of trace class but ${\widetilde{B}}$ is not.

By definition 2.2 we have

$$\begin{aligned} S(F,\eta )= \tau (P_{12}(\eta ), f_0) = {\hbox {tr}}(f_0(PAP)) \end{aligned}$$

Lemma 3.13

$$\begin{aligned} |\widetilde{B_1} - {\widetilde{A}}|_{S_q}^q = O((-\ln \eta )^q), 2/3< q<1 \end{aligned}$$

Proof

Note that since $P_{12}$ is a projection, ${\widetilde{A}}= PP_{12}^2P- (PP_{12}P)^2= PP_{12}(1-P)P_{12}P$. Also $\widetilde{B_1}= Pg(1-P)gP$.

So we have

$$\begin{aligned} \widetilde{B_1} - {\widetilde{A}}= -PhR(1-P)P_{12} P- Pg (1-P)hRP \end{aligned}$$

Note $[R,P]=0$, and $Pg (1-P)hRP= Pgh(1-P)RP + Pg[1-P, h]RP$. Note that $[1-P, h]= -[P,h]= (1-P)h P - Ph(1-P)$ so Th. 3.7 applies.

By Th. 3.7, Cor. 3.9 and Prop. 3.2 the lemma is proved. $\square$

Note that

$$\begin{aligned} \frac{\widetilde{A}}{z+ {\widetilde{A}}} - \frac{\widetilde{B_1}}{z+ {\widetilde{B_1}}} = z\times \frac{1}{z+{\widetilde{A}}} (\widetilde{B_1} - {\widetilde{A}})\frac{1}{z+{\widetilde{B_1}}} \end{aligned}$$

Apply Lemma 3.11 for $W=\frac{1}{z+ {\widetilde{A}}} - \frac{1}{z+ {\widetilde{B_1}}}= \frac{1}{z+{\widetilde{A}}} (\widetilde{B_1} - {\widetilde{A}})\frac{1}{z+{\widetilde{B_1}}}$, with $t_0=1, t_1=t_2=1$, we have

$$\begin{aligned} ||\frac{\widetilde{A}}{z+ {\widetilde{A}}} - \frac{\widetilde{B_1}}{z+ {\widetilde{B_1}}}||_1 \le C|\widetilde{B_1} - {\widetilde{A}}|_{S_\sigma }^\sigma \frac{1}{z^\sigma } \end{aligned}$$

It follows that by Lemma 3.13

$$\begin{aligned} \int _{1/2}^\infty |{\hbox {tr}}(\frac{2\beta }{\beta +1/2} (\frac{\widetilde{A}}{z+ {\widetilde{A}}} - \frac{\widetilde{B_1}}{z+ {\widetilde{B_1}}})) | d\beta \le C \int _{1/2}^\infty \frac{1}{z^\sigma } \times O((-\ln \eta )^\sigma ) \end{aligned}$$

Recall that $z= \beta ^2-1/4,$ and so $\int _{1/2}^\infty \frac{1}{z^\sigma } = O(1)$ if $1> \sigma > 1/2$. It follows that

$$\begin{aligned} S(F,\eta ) - \int _{1/2}^\infty {\hbox {tr}}(\frac{2\beta }{\beta +1/2} \frac{\widetilde{B_1}}{z+ {\widetilde{B_1}}} ) d\beta = O((-\ln \eta )^\sigma ), 2/3< \sigma <1 \end{aligned}$$

(27)

where $S(F,\eta )$ is the first term in above integral by Lemma 3.12.

Next we estimate

$$\begin{aligned} \int _{1/2}^\infty {\hbox {tr}}(\frac{2\beta }{\beta +1/2} \frac{\widetilde{B_1}}{z+ \widetilde{B_1}}) d\beta - \tau (g,f_0) \end{aligned}$$

First we introduce some notations that will simplify writing. These notations will only be used in this section.

Let $X= PgP - (PgP)^2, Y:= \widetilde{B_1}= Pg(1-P)gP.$ $X-Y= P(g-g^2)P = Ph_1^2P, h_1=ih$. $h_1^*=h_1$. Here h is as in Th. 2.1. Note that $X\ge 0, Y\ge 0$.

$$\begin{aligned} h_1 Y = h_1 Pg(1-P)gP = [h_1, P] g (1-P)gP + Ph_1g (1-P)gP \end{aligned}$$

. It follows from Th. 3.7, Cor. 3.9 and Prop. 3.2 that

$$\begin{aligned} |h_1Y|_{S_q}^q = O((-\ln \eta )^q), 2/3< q<1 \end{aligned}$$

Recall by definition

$$\begin{aligned} \tau (g,f_0) = \int _{1/2}^\infty \frac{2\beta }{\beta +1/2} {\hbox {tr}}\left( \frac{X}{X+ z} - P \frac{W}{W+ z}P\right) \end{aligned}$$

The estimate of

$$\begin{aligned} \int _{1/2}^\infty \frac{2\beta }{\beta +1/2} {\hbox {tr}}(\frac{\widetilde{B_1}}{z+ \widetilde{B_1}}) d\beta - \tau (g,f_0) \end{aligned}$$

reduces to estimate

$$\begin{aligned} \frac{Y}{Y+ z}- (\frac{X}{X+ z} - P \frac{W}{W+ z}P) \end{aligned}$$

where $W=h_1^2$.

As a first step we estimate

$$\begin{aligned} \frac{X}{X+ z} - P \frac{W}{W+ z}P= zP\frac{1}{X+ z} (X-PW) \frac{1}{W+ z}P \end{aligned}$$

We have $X-PW= Y+ PWP-PW$.

First we need a simple Lemma:

Lemma 3.14

If S is a positive operator, and $TT^* \le S, z>0$, then

$$\begin{aligned} ||\frac{1}{z+S} T || =||T^*\frac{1}{z+S} || \le \frac{1}{\sqrt{2} z^{\frac{1}{2}}} \end{aligned}$$

Proof

Since $||Q||=||Q^*||$, it is sufficient to prove

$$\begin{aligned} ||\frac{1}{z+S} T || \le \frac{1}{\sqrt{2} z^{\frac{1}{2}}} \end{aligned}$$

We have

$$\begin{aligned} \frac{1}{z+S} TT^* \frac{1}{z+S} \le \frac{1}{z+S} S \frac{1}{z+S} \le \frac{1}{2z} \end{aligned}$$

Hence

$$\begin{aligned} ||\frac{1}{z+S} T ||^2 = ||\frac{1}{z+S} TT^* \frac{1}{z+S}|| \le \frac{1}{2z} \end{aligned}$$

and the Lemma is proved. $\square$

Let us show that

$$\begin{aligned} \int _{1/2}^\infty z \frac{2\beta }{\beta +1/2} {\hbox {tr}}( P\frac{1}{X+ z} (PWP-PW) \frac{1}{W+ z}P) dz = O((-\ln \eta )^\sigma ) \end{aligned}$$

We will apply Lemma 3.11 with $R_1 =\frac{1}{X+ z} , V= PWP-PW, R_2=\frac{1}{W+ z}P$. It is clear that $t_1=t_2=1$, and we need choose $t_0$ small enough. The key observation is that

$$\begin{aligned} R_1 V R_2=\frac{1}{X+ z} (Ph_1 h_1P) \frac{1}{W+ z}P -\frac{1}{X+ z} PW \frac{1}{W+ z}P \end{aligned}$$

and since $X\ge Ph_1^2P$, from Lemma 3.14 we have

$$\begin{aligned} ||\frac{1}{X+ z} (Ph_1 h_1P) \frac{1}{W+ z}P||\le \frac{1}{\sqrt{2}z^{1/2+1}} \end{aligned}$$

It is also clear that

$$\begin{aligned} ||\frac{1}{X+ z} PW \frac{1}{W+ z}P||\le \frac{1}{z} \end{aligned}$$

and we achieve our goal with

$$\begin{aligned} ||R_1 V R_2||\le C \frac{1}{z^{3/2}} \end{aligned}$$

Now we can apply Lemma 3.11 with $t_0= 3/2, t_1=t_2=1$ to obtain

$$\begin{aligned} |z{\hbox {tr}}(R_1 V R_2)| \le \frac{1}{z^{\frac{1+\sigma }{2}}} |||V|^\sigma ||_1 \end{aligned}$$

Since $h_1^2= -h^2$, By Th. 3.7 we get

$$\begin{aligned}&\int _{1/2}^\infty \frac{2\beta }{\beta +1/2} {\hbox {tr}}(\frac{1}{X+ z} (Ph_1 h_1P) \frac{1}{W+ z}P -\frac{1}{X+ z} PW \frac{1}{W+ z}P)d\beta \nonumber \\&\quad = O((-\ln \eta )^\sigma )) , 2/3< \sigma <1 \end{aligned}$$

(28)

Let us consider

$$\begin{aligned} P\frac{1}{X+ z} Y (1- \frac{W}{z+W})P \end{aligned}$$

We write

$$\begin{aligned} P\frac{1}{X+ z} Y \frac{W}{z+W}P = P\frac{1}{X+ z} Pg(1-P) (1-P) g P h_1 \frac{h_1}{z+h_1^2}P \end{aligned}$$

and apply Lemma 3.11 with $R_1 =P\frac{1}{X+ z} Pg(1-P), V=(1-P) g P h_1, R_2= \frac{h_1}{z+h_1^2}P$

Note that by Lemma 3.14 we have

$$\begin{aligned} ||R_1||\thicksim z^{-1/2}, ||R_2||\thicksim z^{-1/2}, ||R_1V R_2||\thicksim z^{-1/2} \end{aligned}$$

We have $t_0=1/2, t_1=t_2=1/2$, again with savings on exponents. We have

$$\begin{aligned} ||P\frac{1}{X+ z} Y \frac{W}{z+W}P||_1 \le C |(1-P) g P h_1|_{S_\sigma }^\sigma \frac{1}{z^{\frac{1}{2}+ \frac{\sigma }{2}}} \end{aligned}$$

Note that $|(1-P) g P h_1|_{S_\sigma }^\sigma \le |(1-P) gh_1 P |_{S_\sigma }^\sigma + ||(1-P) g|| |[P,h_1]|_{S_\sigma }^\sigma$

By Th. 3.7 and Cor. 3.9, the same argument as above shows that

$$\begin{aligned} \int _{1/2}^\infty \frac{2\beta }{\beta +1/2} {\hbox {tr}}\left( P\frac{1}{X+ z} Y \frac{W}{z+W}P\right) d\beta = O((-\ln \eta )^\sigma ) , 2/3< \sigma <1 \end{aligned}$$

(29)

Finally we are left with

$$\begin{aligned} \left( \frac{1}{z+Y}-\frac{1}{z+X}\right) Y = \frac{1}{z+X} (z+X- z-Y)\frac{1}{z+Y} Y = \frac{1}{z+X} PWP \frac{Y}{z+Y} \end{aligned}$$

By choosing $R_1 =\frac{1}{z+X} Ph_1, V= h_1P g(1-P), R_2= (1-P)gP \frac{1}{z+Y}$ and use Lemma 3.14 we find that $t_0=1/2, t_1=t_2= 1/2$, again with savings as the preceding case to complete the proof that

$$\begin{aligned} \int _{1/2}^\infty \frac{2\beta }{\beta +1/2} {\hbox {tr}}\left( \left( \frac{1}{Y+ z} - \frac{1}{z+X}\right) Y\right) d\beta = O((-\ln \eta )^\sigma ), 2/3< \sigma <1 \end{aligned}$$

(30)

To summarize, we first prove that

$$\begin{aligned} S(F,\eta ) - \int _{1/2}^\infty {\hbox {tr}}(\frac{2\beta }{\beta +1/2} \frac{\widetilde{B_1}}{z+ \widetilde{B_1}} ) d\beta = O((-\ln \eta )^\sigma ), 2/3< \sigma <1 \end{aligned}$$

which is equation (27).

Next we estimate

$$\begin{aligned} \int _{1/2}^\infty \frac{2\beta }{\beta +1/2} {\hbox {tr}}(\frac{\widetilde{B_1}}{z+ \widetilde{B_1}}) d\beta - \tau (g,f_0) \end{aligned}$$

In addition to multiplication by $\frac{2\beta }{\beta +1/2}$, the integrand in the above integral is

$$\begin{aligned} \frac{Y}{Y+ z}- \left( \frac{X}{X+ z} - P \frac{W}{W+ z}P\right) \end{aligned}$$

where $W=h_1^2$. We show that

$$\begin{aligned} \frac{X}{X+ z} - P \frac{W}{W+ z}P= zP\frac{1}{X+ z} (Y+PWP-PW) \frac{1}{W+ z}P \end{aligned}$$

and the integral corresponds to the integrand

$$\begin{aligned} zP\frac{1}{X+ z} (PWP-PW) \frac{1}{W+ z}P \end{aligned}$$

is $O((-\ln \eta )^\sigma ), 2/3< \sigma <1$ as in equation (28).

Then we show the integral corresponds to the integrand

$$\begin{aligned} P\frac{1}{X+ z} YW \frac{1}{W+ z}P \end{aligned}$$

is $O((-\ln \eta )^\sigma ), 2/3< \sigma <1$ as in equation (29). This shows that up to $O((-\ln \eta )^\sigma ), 2/3< \sigma <1$, the integrand $(\frac{X}{X+ z} - P \frac{W}{W+ z}P)$ can be replaced by $-\frac{1}{z+X} Y$. And finally we show that the integral corresponds to the integrand ($\frac{1}{z+Y}-\frac{1}{z+X}) Y$ is also $O((-\ln \eta )^\sigma ), 2/3< \sigma <1$ as in equation (30).

So we have proved the following theorem:

Theorem 3.15

$$\begin{aligned} S(F,\eta )- \tau (g, f_0) =O((-\ln \eta )^\sigma ), 2/3< \sigma <1 \end{aligned}$$

where $S(F,\eta )$ and $\tau (g, f_0)$ are defined as in definition 2.2.

3.5 Upper Bound for Entropy

Th. 3.15 reduce the estimation of $S(F,\eta )$ to $\tau (g;f_0)$. $f_0(PgP)- Pf_0(g)P$ is called truncated Wiener-Hopf operators in [21]. There is a remarkable formula for $\tau (g;f_0)$ going back to H. Widom (cf. [21] and references therein). The more general version that we will use can be found in [37] and [21]. To describe this formula, we recall some basic definitions from [37].

For any complex valued function $f:{{\mathbb {C}}}\rightarrow {\mathbb C}$ and $s_1,s_2$ define

$$\begin{aligned} U(s_1,s_2;f)= \int _0^1 \frac{f((1-t)s_1+ts_2)- ((1-t)f(s_1) + tf(s_2))}{t(1-t)}dt \end{aligned}$$

and introduce

$$\begin{aligned} B(a;f)= \frac{1}{8\pi ^2} \int \int \frac{U(a(\psi _1),\alpha (\psi _2);f)}{|\psi _1-\psi _2|^2} d\psi _1d\psi _2 \end{aligned}$$

where a is another function $a:{{\mathbb {C}}}\rightarrow {{\mathbb {C}}}$.

The quantity B(a; f) is an object that appears very often in the theory of Wiener-Hopf operators.

We denote by

$$\begin{aligned} B_\epsilon (a;f): =\frac{1}{8\pi ^2} \int \int _{|\psi _1-\psi _2|< \epsilon } \frac{U(a(\psi _1,\psi _2;f)}{|\psi _1-\psi _2|^2} d\psi _1d\psi _2 \end{aligned}$$

and

$$\begin{aligned}&B_{\epsilon _1, \epsilon _2}(a;f): =\frac{1}{8\pi ^2} \int \int _{\epsilon _1< |\psi _1-\psi _2|< \epsilon _2} \frac{U(a(\psi _1,\psi _2;f)}{|\psi _1-\psi _2|^2} d\psi _1d\psi _2 \\&B_{\ge \epsilon }(a;f): =\frac{1}{8\pi ^2} \int \int _{\epsilon \le |\psi _1-\psi _2|} \frac{U(a(\psi _1,\psi _2;f)}{|\psi _1-\psi _2|^2} d\psi _1d\psi _2 \end{aligned}$$

Now we will use Cayley transformation to identify the unit circle with the extended real line, and to think our function g as a function ${\hat{g}}$ on the real line, that is ${\hat{g}}(x)= g(C(x))$. Recall Cayley transform $V(x) = i (x+i) / (x - i)$, which carries the (one point compactification of the) real line onto the circle and the upper half plane onto the unit disk. It induces a unitary map

$$\begin{aligned} {\displaystyle Uf(x)=\pi ^{-{\frac{1}{2}}}(x-i)^{-1}f(V(x))} \end{aligned}$$

of $L^2(S^1, {\mathbb {C}})$ onto $L^2({\mathbb {R}},{\mathbb {C}} )$. The operator U carries the Hardy space on the circle onto the Hardy space on the real line (cf. Chapter one of [30]). We will use the Cayley transform to identify intervals on the circle with one point removed to intervals on the real line. Note that ${\hat{g}}\in W^{2, \infty }$, and has compact support.

The length function is $L(z)= |z^2-e^{-i\eta }\cos \eta |$. Notice that $|L(z_1)-L(z_2)|\le |z_1^2-z_2^2| \le 2 |z_1-z_2|$.

Define scale function $\tau$ and amplitude function v ( cf. Sect. 3 of [21]) as follows:

$$\begin{aligned} \tau (x) = \frac{1}{5}L(C(x)), v(x)= (1+|x|)^{-2} \end{aligned}$$

Note that

$$\begin{aligned} C(x)= & {} i \frac{x+i}{x-i} \\ |C'(x)|= & {} |\frac{2}{(x-i)^2}|\le 2, |C''(x)|= |\frac{4}{(x-i)^3}|\le 4 \\ |\tau (x_1)-\tau (x_1)|\le & {} \frac{2}{5}|C(x_1)-C(x_2)|\le \frac{4}{5}|x_1-x_2| \end{aligned}$$

One can check directly from (23) that

$$\begin{aligned} |{\hat{g}}(x)|\le C v(x), |{\hat{g}}^{(k)}(x)|\le C_k \frac{v(x)}{\tau (x)^k}, k=1,2 \end{aligned}$$

where $C, C_k$ are constants. Note that the minimum of $\tau (x)$ is $\tau _{\mathrm{{min}}}= 1-\cos \eta \thicksim \eta ^2$.

Our ${\hat{g}}\in W^{2,\infty }$, has support in $[-2,2]$ and verifies conditions 4.1 in [21]. Hence Th. 3.2 in [37] applies to ${\hat{g}}$.

Lemma 3.16

$$\begin{aligned} \tau (g;f_0)= B({\hat{g}};f_0) \end{aligned}$$

Proof

By Prop. 4.1 of [21] $\tau (g;f)= B({\hat{g}};f)$ if $f\in C_0^4({{\mathbb {R}}})$. Now choose a sequence of smooth functions $f_n$ to approximate $f_0$ in the norm defined in Th. 3.2 of [37], and our lemma follows from Prop. 2.2 of [21] and Th. 3.2 of [37]. $\square$

By Th. 6.1 of [37] we have

$$\begin{aligned} |B({\hat{g}};f_0) | \le C \int \frac{v(x)}{\tau (x)} dx \end{aligned}$$

for some constant $C>0$.

Let us evaluate

$$\begin{aligned} \int \frac{v(x)}{\tau (x)} dx \end{aligned}$$

Change coordinate to $z=C(x)=e^{i\theta }$, we have

$$\begin{aligned} \int \frac{v(x)}{\tau (x)} dx\le C \int _0^{2\pi } \frac{1}{|z^2-e^{-i\eta }\cos \eta |}d\theta =2C \int _0^{2\pi } \frac{1}{|z-e^{-i\eta }\cos \eta |}d\theta \end{aligned}$$

By Prop. 1.4.10 of [35] we have

$$\begin{aligned} \int \frac{1}{|z-e^{-i\eta }\cos \eta |}d\theta \thicksim -\ln (1-\cos ^2(\eta ))\thicksim -2 \ln \eta \end{aligned}$$

Hence

$$\begin{aligned} |\tau (C_1) | \le C (-\ln \eta ) \end{aligned}$$

for some constant $C>0$.

By Th. 3.15 and Lemma 3.16 we have therefore proved

Corollary 3.17

$S(F,\eta )\le C (-\ln \eta )$ for some constant $C>0$ when $\eta \rightarrow 0$, where $S(F,\eta )$ is as in definition 2.2.

Remark 3.18

In fact we can write

$$\begin{aligned} B({\hat{g}};f_0)= B_{\eta ^2}({\hat{g}},f_0)+B_{\eta ^2,\eta }({\hat{g}},f_0)+ B_{\ge \eta }({\hat{g}},f_0) \end{aligned}$$

By Section 9 of [21] we have $B_{\eta ^2}({\hat{g}};f_0)=O(1)$, and $B_{\ge \eta }({\hat{g}};f_0) = \frac{-1}{6} \ln \eta + O(1)$. Unfortunately it is not clear if one can show $B_{\eta ^2, \eta }({\hat{g}};f_0)= o(-\ln \eta )$ since our function ${\hat{g}}$ is not smooth as the functions considered in Section 9 of [21]. If $B_{ \eta ^2,\eta }({\hat{g}},f_0)=o(-\ln \eta )$, then it follows that

$$\begin{aligned} S(F,\eta ) =\frac{1}{6} (-\ln \eta ) + o(-\ln \eta ) \end{aligned}$$

as $\eta \rightarrow 0$.

3.6 Continuity

Note that in general von Neumann entropy does not behave as well as relative entropy. It is therefore interesting to examine the properties of $S(F,\eta )$ as functions of $\eta$ using our explicit formula. In this section we prove that $S(F,\eta )$ is continuous and $\lim _{\eta \rightarrow \pi } S(F,\eta ) = 0$. First we have the following Lemma:

Lemma 3.19

Suppose $f(z,v) =\sum _n f_n(v) z^n$, $|f_n(v)| \le C |n|^{-\alpha }$ with $\alpha > \frac{3}{2}$ where C only depends on the neighborhood V of $v_0$. In addition assume that for each $n\ne 0$, $\lim _{ v\rightarrow v_0} f_n =0$. Then $\lim _{v\rightarrow v_0}|P f (1-P)|_{S_p}^p=0, \lim _{v\rightarrow v_0}|(1-P) f P|_{S_p}^p=0, 1> p> \frac{1}{\alpha -\frac{1}{2}}$.

Proof

Note that $P (f-C) (1-P) =P f (1-P)$ for any constant C. That is why $n\ne 0$ in the Lemma. We prove $\lim _{v\rightarrow v_0}|P f (1-P)|_{S_p}^p=0$. The proof of $\lim _{v\rightarrow v_0}|(1-P) f P|_{S_p}^p=0$ is similar. Given any $\epsilon >0$. We first write $f= f_N + f_{\ge N}$ where $f_N= \sum _{|n|\le N} f_n(v) z^n$.

As in [18],

$$\begin{aligned} P {f} (1-P)(z^n) = \sum _{k \ge 0} f_{k-n}(v) z^k = \xi _{-n},\quad n<0\ . \end{aligned}$$

It follows that

$$\begin{aligned} P f (1-P)= & {} \sum _{n< 0} (\cdot \, ,z^n )\xi _{-n} \\ ||P f (1-P)||^p_{S_p}\le & {} \sum _{n< 0} ||\xi _{-n}||^p\ \end{aligned}$$

where $(\cdot \, ,z^n )$ is the inner product with $z^n$.

Note that $||\xi _{-n}|| = \big (\sum _{k\ge 0}|f_{k-n}(v)|^2\big )^\frac{1}{2} = O(|n|^{-\alpha + \frac{1}{2}})$, by choosing N sufficiently large we have $|P f_{\ge N} (1-P)|_{S_p}^p < \epsilon /2$. Since

$$\begin{aligned} |P f_N (1-P)|_{S_p}^p \le 2N (\sum _{1\le |n|\le N} |f_n(v) |)^p \end{aligned}$$

By assumption we can choose v close enough to $v_0$ such that

$$\begin{aligned} |P f_N (1-P)|_{S_p}^p \le \epsilon /2 \end{aligned}$$

and the Lemma is proved. $\square$

Proposition 3.20

Suppose $T_1, T_2$ are projections, and let $\gamma := |P (T_1-T_2) (1-P)|_p^p +|(1-P) (T_1-T_2)P|_p^p< \infty$, for some $0<p<1$. Then $f_0(PT_1P)- f_0(PT_2P)$ is trace class and

$$\begin{aligned} |{\hbox {tr}}(f_0(PT_1P)- f_0(PT_2P))|\le C_p \gamma \end{aligned}$$

where $C_p$ is a constant which only depends on p.

Proof

By inequality (7) we can assume that $p>1/2$.

The idea of the proof is already present after Lemma 3.13.

By using Lemma 3.12

$$\begin{aligned} f_0(PT_1P)- f_0(PT_2P)= \int _{1/2}^\infty (\frac{2\beta }{\beta +1/2} (\frac{\widetilde{PT_1P}}{z+ \widetilde{PT_1P}} - \frac{\widetilde{PT_2P}}{z+ \widetilde{PT_2P}})) d\beta \end{aligned}$$

Where $\widetilde{T}:= T(1-T)$. Apply Lemma 3.11 for $W=\frac{1}{z+ \widetilde{PT_1P}} - \frac{1}{z+ \widetilde{PT_2P}}= \frac{1}{z+\widetilde{PT_1P}} (\widetilde{PT_2P} - \widetilde{PT_1P})\frac{1}{z+\widetilde{PT_2P}}$, with $t_0=1, t_1=t_2=1$ exactly as after Lemma 3.13, we have

$$\begin{aligned} \left\| \frac{\widetilde{PT_1P}}{z+ \widetilde{PT_1P}} - \frac{\widetilde{PT_2P}}{z+ \widetilde{PT_2P}}\right\| _1 \le C |\widetilde{PT_1P} - \widetilde{PT_2P}|_{S_p}^p \frac{1}{z^p} \end{aligned}$$

Recall that $z= \beta ^2-1/4,$ and so $\int _{1/2}^\infty \frac{1}{z^p} = O(1)$ if $p > 1/2$. Finally notice that since $T_1$ is a projection,

$$\begin{aligned} PT_1P(1-PT_1P)= PT_1^2P -(PT_1P)^2= PT_1(1-P)T_1P \end{aligned}$$

Denote by $T:=T_1-T_2$. Then we have $T_1= T+T_2$, and $PT_1(1-P)T_1P- PT_2(1-P)T_2P = PT_2 (1-P)TP + PT(1-P)T_1P$.

It follows that

$$\begin{aligned} |\widetilde{PT_1P} - \widetilde{PT_2P}|_{S_p}^p\le \gamma \end{aligned}$$

and the Proposition is proved. $\square$

Theorem 3.21

$S(F,\eta )$ is a continuous function of $\eta \in (0,\pi )$ and $\lim _{\eta \rightarrow \pi ^-}S(F,\eta )=0$.

Proof

Let $\eta = \pi - \phi$, and assume that $\phi \rightarrow 0$.

First from the formula (22) we see that both $g'', h''$ are bounded, up to addition of constants, by constants multiplied by $\frac{\phi ^2}{ L^3}$ where L is the distance between $z^2$ and $e^{-i\eta } \cos \eta$. Note that as $\phi \rightarrow 0$, the smallest L is reached at end points of $I_2$ and this value is $\thicksim \phi$. If we use angle $\theta$ between points in $I_2$ and the end points of $I_2$ where L attains its minimum as an integration parameter, then we have as $\phi \rightarrow 0$

$$\begin{aligned} \int _{I_2} |h''| d\theta \thicksim \int _{0}^\pi \frac{\phi ^2}{(\theta + \phi )^3} d\theta = O(1) \end{aligned}$$

Similarly

$$\begin{aligned} \int _{I_2} |g''| d\theta \thicksim \int _{0}^\pi \frac{\phi ^2}{(\theta + \phi )^3}d\theta = O(1) \end{aligned}$$

The same is also true for the integrals of $h'', g''$ over $-I_2$.

It follows by integration by parts that the Fourier coefficients of h, g are of $O(n^{-2})$ as $n\rightarrow \infty$. Moreover for $n\ge 0$ and remember h is an odd function we have

$$\begin{aligned} 2 \int _{I_2} h(z) z^{2n} dz=\int _{I_2\cup -I_2} h(z) z^{2n} dz= -\int _{J\cup -J} h(z) z^{2n} dz \end{aligned}$$

where J is a path connecting end points of $I_2$ and $-I_2$ with $|e^{i\eta }z^2-\cos \eta |= \sin (\phi )$, since $h(z) z^{2n}$ is analytic in the region bounded by $I_2\cup -I_2\cup J\cup -J$. Here we have used the fact that h is independent of the choice of analytical continuation of u and we can choose branch cut of u which is outside the region bounded by $I_2\cup -I_2\cup J\cup -J$.

It follows that for

$$\begin{aligned} \left| \int _{I_2} h(z) z^{2n} dz\right| = \left| \int _J h(z) z^{2n} dz\right| \le C \phi \end{aligned}$$

as $\phi \rightarrow 0$, where C is a constant. It is also clear that

$$\begin{aligned} \int _{I_2} h(z) z^{2n} dz \end{aligned}$$

is continuous in $\eta$. Since ih is real it follows all fourier coefficients of h goes to 0 when $\phi \rightarrow 0$. Similarly since $g-\frac{1}{2}$ is odd we have

$$\begin{aligned} \left| \int _{I_2} (g-\frac{1}{2}) z^{2n} dz\right| = \left| \int _J (g-\frac{1}{2}) z^{2n} dz\right| \rightarrow 0 \end{aligned}$$

as $\phi \rightarrow 0$. Since on $I_1$, $g=1$, $|\int _{I_1} z^{2n} dz| = O(\phi )$. Moreover since g is real, it follows that all fourier coefficients of $(g-\frac{1}{2})$ goes to 0 when $\phi \rightarrow 0$. By applying Lemma 3.19 (note that $n\ne 0$ in Lemma 3.19 ) and Prop. 3.20 with $T_1= P_{12}(\phi ), T_2=0$ we conclude that $\lim _{\eta \rightarrow \pi ^-} S(F,\eta ) = 0$. To prove continuity, we observe if we fix a small neighborhood V of $\eta _0$ in $(0,\pi )$, then on V we have $|h_n |\le C n^{-2}, |g_n |\le C n^{-2}$ where C only depends on the neighborhood V. Since $h_n, g_n$ are obviously continuous in $\eta$, the continuity of $S(F)(\eta )$ follows again by applying Lemma 3.19 and Prop. 3.20, with $T_1= P_{12}(\phi ), T_2=P_{12}(\eta _0)$. $\square$