Pressure Function and Limit Theorems for Almost Anosov Flows

Abstract

We obtain limit theorems (Stable Laws and Central Limit Theorems, both standard and non-standard) and thermodynamic properties for a class of non-uniformly hyperbolic flows: almost Anosov flows, constructed here. The link between the pressure function and limit theorems is studied in an abstract functional analytic framework, which may be applicable to other classes of non-uniformly hyperbolic flows.

1 Introduction and Summary of the Main Results

This paper is a contribution to the theory of limit theorems and thermodynamic formalism in the context of non-uniformly hyperbolic flows on manifolds. In particular we introduce a new class of ‘almost Anosov flows’, a natural analogue of almost Anosov diffeomorphisms introduced in [HY95], and prove limit theorems for natural observables $\psi$ with respect to the SRB measure, also giving the form of the associated pressure function. This example is presented in the context of a general framework, which may be applicable to a range of other non-uniformly hyperbolic flows. We recall that various statistical properties for several classes of Anosov flows are known [D98, D03, L04], but none of these results apply to the class of ‘almost Anosov flows’ considered here.

Given a flow $(\Phi _t)_t$ with a real-valued potential $\phi$, we suppose that there is an equilibrium state $\mu _\phi$. A standard way of studying the behaviour of averages of a real-valued observable $\psi$ is to consider a Poincaré section Y and study the induced first return map $F:Y\rightarrow Y$, the induced potentials ${\bar{\phi }}$ and ${\bar{\psi }}$, with the induced measure $\mu _{{\bar{\phi }}}$. In the discrete time case, in fact in the setting where all the dynamical systems are countable Markov shifts, [S06, Section 2] showed a one-to-one relation between limit laws for $\psi$ and the asymptotic form of the pressure function ${\mathcal {P}}(\phi +s\psi )$, as $s\rightarrow 0$. This function experiences a phase transition at $s=0$, and its precise form determines the type of limit laws (in particular, the index of the stable law), see [MTo04, Z07] and [S06, Theorem 7]. Furthermore, under certain conditions on $\psi$, [S06, Theorem 8] gave an asymptotically linear relation between the induced pressure ${\mathcal {P}}(\overline{\phi +s\psi })$ and ${\mathcal {P}}(\phi +s\psi )$. On the limit theorems side, [MTo04, Z07] and [S06, Theorem 7] show how one can go between results on the induced and on the original system: the first of these also applies in the flow setting. In these cases, the tail of the roof function/return time to the Poincaré section (both for the flow and the map) plays a major role in determining the form of the results. Here we will follow this paradigm in the setting of a new class of flows. The proofs of the main theorems are facilitated by corresponding theorems in an abstract functional analytic framework. Applying this to the considered example requires precise estimates on the tails of the roof function, which we prove for our main example.

Our central example is an almost Anosov flow, which is a flow having a continuous flow-invariant splitting of the tangent bundle $T\mathcal {M}= E^{\mathbf {u}} \oplus E^{\mathbf {c}} \oplus E^{\mathbf {s}}$ (where $E^{\mathbf {c}}_q$ is the one-dimension flow direction) such that we have exponential expansion/contraction in the directions $E^{\mathbf {u}}_q, E^{\mathbf {s}}_q$, except for a finite number of periodic orbits (in our case a single orbit). We can think of these as perturbed Anosov flows, where the perturbation is local around these periodic orbits, making them neutral. A precise description is given in Sect. 2. Almost Anosov diffeomorphisms have been introduced in [HY95] and sufficiently precise estimates on the tail of the return function to a ‘good’ set have been obtained in [BT17]. (These estimates are for both finite and infinite measure preserving almost Anosov diffeomorphisms.) We emphasise that the roof function is not constant on the stable direction, which is a main source of difficulty. When considering these examples we choose $\phi$ so that $\mu _\phi$ is the SRB measure. We build on the construction in [BT17] to obtain the asymptotic of the tail behaviour of the roof functions for almost Anosov flows (when viewed as suspension flows). The main technical results for these systems are Propositions 2.2 and 2.4. These are then used to prove the main result Theorem 2.5.

For a major part of the statements of the abstract theorems in this paper (in contexts not restricted to almost Anosov flows) we do not require any Markov structure for our system: it is only when we want to see our results in terms of the pressure function (making the connection between the leading eigenvalue of the twisted transfer operator of the base map and pressure as in [S99, Theorem 4]) that this is needed. Our setup requires good functional analytic properties of the Poincaré map in terms of abstract Banach spaces of distributions. Using the rather mild abstract functional assumptions described in Sects. 4, 5 we obtain stable laws, standard and non-standard CLT; this is the content of Proposition 5.1. In Sect. 6 we recall [S06, Theorem 7] and [MTo04, Theorem 1.3] to lift Proposition 5.1 to the flow, which allow us to prove Proposition 6.1.

In Sect. 7 we do exploit the assumption of the Markov structure to relate the definition of the pressure $\mathcal {P}({\bar{\phi }}+s{\bar{\psi }}), s\ge 0$, with that of the family of eigenvalues of the family of twisted transfer operators of the Poincaré map; the twist is in terms of the roof function of the suspension flow and the potential $\psi$. Using this type of identification, in Theorem 7.1 we relate the induced pressure $\mathcal {P}({\bar{\phi }}+s{\bar{\psi }})$ with the original pressure. Using the main result in Sects. 7, 8 we summarise the results for the abstract framework in the concluding Theorem 8.2, which gives the equivalence between the asymptotic behaviour of the pressure function $\mathcal {P}(\phi +s\psi )$ and limit theorems. It is this summarising result that can be viewed as a version of Theorems 2–4 and Theorem 7 of [S06] combined, for flows.

We note that limit theorems for the almost Anosov flows studied here could have been obtained via the very recent results of limit laws for invertible Young towers as in [MV19, Theorem 3.1] together with the arguments of lifting limit laws from the suspension to the flow in [MTo04, Z07] and [S06, Theorem 7]. Notably [MV19] applies to invertible Young towers and as such to several classes of billiard maps/flows. We conclude our introduction by noting some possible extensions: we do not approach the correspondence between the shape of the pressure and decay of correlations, but believe that such a correspondence can be established using the present results and arguments/estimates in [MTe17, BMT18, BMT]. Moreover, for a large class of discrete time, infinite measure preserving systems, a version of [S06, Theorem 8] along with connection with limit theorems has been obtained in [BTT18]: we expect this to carry over to the case where the almost Anosov flow has infinite measure.

Organisation of the paper. In Sect. 2 we give the setup of the almost Anosov flow as ODE, and then translate it to a suspension flow over a Poincaré section. Proposition 2.4 computes the tails of the associated roof function. Section 2.3 gives the main limit theorems for the almost Anosov flow. Section 3 gives the preliminaries of equilibrium states for flows. In Sect. 4 we give the abstract background, including abstract hypotheses, of the transfer operator approach. In this abstract setting, the main result Proposition 5.1 is stated and proved in Sect. 5. Section 6 states and proves the limit law for the flow, and Sect. 7 deals with the asymptotic shape of the pressure function. In Sect. 8 we formulate and prove the limit laws for equilibrium states of the flow. Finally, the appendix gives some technical proofs for Sect. 5, and then concludes by checking all the hypotheses of the abstract results. In “Appendix A” we verify the abstract hypotheses of Proposition 5.1 and Theorem 8.2 for the almost Anosov flows introduced in Sect. 2 which allows us to complete the proofs of Theorems 2.5.

Notation. We use “big O” and $\ll$ notation interchangeably, writing $a_n=O(b_n)$ or $a_n\ll b_n$ if there is a constant $C>0$ such that $a_n\le Cb_n$ for all $n\ge 1$. We write $a_n \sim b_n$ if $\lim _n a_n/b_n = 1$. Throughout we let $\rightarrow ^d$ stand for convergence in distribution. By F being piecewise Hölder, etc. we mean that F is Hölder on the elements $a \in {{\mathcal {A}}}$, with a uniform exponent, but the Hölder norm of $F|_a$ is allowed to depend on a.

2 The Setup for Almost Anosov Flows

2.1 Description via ODEs

Let $\mathcal {M}$ be an odd-dimensional compact connected manifold without boundary. An almost Anosov flow is a flow having continuous flow-invariant splitting of the tangent bundle $T\mathcal {M}= E^{\mathbf {u}} \oplus E^{\mathbf {c}} \oplus E^{\mathbf {s}}$ (where $E^{\mathbf {c}}_q$ is the one-dimension flow direction) such that we have exponential expansion/contraction in the direction of $E^{\mathbf {u}}_q, E^{\mathbf {s}}_q$, except for a finite number of periodic orbits (in our case a single orbit $\Gamma$). After restricting to an irreducible component if necessary, Anosov flows are automatically transitive, cf. [BS02, 5.10.3]. As noted in the introduction, we can think of almost Anosov flows as perturbed Anosov flows, where the perturbation is local around $\Gamma$, making $\Gamma$ neutral.

Let $\Phi _t:\mathcal {M}\times {{\mathbb {R}}}\rightarrow \mathcal {M}$ be an almost Anosov flow on a 3-dimensional manifold. We assume that the flow in local Cartesian coordinates near the neutral periodic orbit $\Gamma := (0,0) \times {{\mathbb {T}}}$ is determined by a vector field $X:\mathcal {M}\rightarrow T\mathcal {M}$ defined as:

$$\begin{aligned} \begin{pmatrix} \dot{x} \\ \dot{y}\\ \dot{z} \end{pmatrix} = X \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x(a_0 x^2 + a_2 y^2) \\ -y(b_0 x^2 + b_2 y^2) \\ 1 + w(x,y) \end{pmatrix} \end{aligned}$$

(2.1)

where¹$a_0, a_2, b_0, b_2 \ge 0$ with $\Delta = a_2b_0 - a_0b_2 \ne 0$, and w is a homogeneous function with exponent $\rho \ge 0$ as leading term, cf. Remark 2.6. For the Limit Theorem 2.5, we additionally require $a_2 > b_2$ (so that $\beta > 1$, the finite measure case). We will use a Poincaré section $\Sigma$ which is $\{ z = 0\}$ in local coordinates near $\Gamma$, but which is thought to be defined across the whole manifold so that the Poincaré maps is defined everywhere on $\Sigma$. The Poincaré map of the Anosov flow (i.e., before the perturbation rendering $\Gamma$ neutral) is Anosov itself, and hence has a Markov partition, for instance based on arcs in $W^s(p) \cup W^u(p)$ for $p = (0,0)$, but not $W_{loc}^s(p) \cup W_{loc}^u(p)$ because we want p in the interior of a partition element. As the perturbation is local around $\Gamma$ and leaves local stable and unstable manifolds of $\Gamma$ unchanged, the same Markov partition can be used for the almost Anosov flow.

Let us call the horizontal (i.e., (x, y)-component) of the flow $\Phi _t^{hor}$. This is the vector field of [BT17, Equation (4)], with $\kappa = 2$ and the vertical component is added as a skew product. Therefore we can take some crucial estimates from the estimates of $\Phi _t^{hor}$ in [BT17, Proposition 2.1].

The flow $\Phi _t$ has a periodic orbit $\Gamma = \{ p \} \times {{\mathbb {T}}}$ of period 1 (which is neutral because DX is zero on $\Gamma$), and it has local stable/unstable manifolds $W^{\mathbf {s}}_{loc}(\Gamma ) = \{ 0 \} \times (-\epsilon ,\epsilon ) \times {{\mathbb {T}}}^1$ and $W^{\mathbf {u}}_{loc}(\Gamma ) = (-\epsilon ,\epsilon ) \times \{ 0 \} \times {{\mathbb {T}}}^1$. It is an equilibrium point of neutral saddle type if we only consider $\Phi ^{hor}_t$. The time-1 map ${\hat{f}}$ of $\Phi _t^{hor}$ is an almost Anosov map, with Markov partition $\{ {\hat{P}}_i\}_{i \ge 0}$, where we assume that p is an interior point of ${\hat{P}}_0$.

Fig. 1

The first quadrant of the rectangle ${\hat{P}}_0$, with stable and unstable foliations of time-1 map ${\hat{f}} = \Phi _1^{hor}$ drawn vertically and horizontally, respectively. Also the integral curve of q is drawn

Given $q \in {\hat{f}}^{-1}({\hat{P}}_0)$, define

$$\begin{aligned} {\hat{\tau }}(q) := \min \{ t > 0 : \Phi _t^{hor}(q) \in {\hat{W}}^{\mathbf {s}}\}, \end{aligned}$$

(2.2)

where ${\hat{W}}^{\mathbf {s}}$ is the stable boundary leaf of ${\hat{P}}_0$, see Fig. 1. Let ${\hat{W}}^{\mathbf {u}}(y)$ denote the unstable leaf of ${\hat{f}}$ intersecting $(0,y) \in {\hat{W}}^{\mathbf {s}}(p)$. The function ${\hat{\tau }}$ is strictly monotone on ${\hat{W}}^{\mathbf {u}}(y)$. For $y > 0$ and $T \ge 1$, let $\xi (y,T)$ denote the distance between (0, y) and the (unique) point $q \in {\hat{W}}^{\mathbf {u}}(y)$ with ${\hat{\tau }}(q) = T$. The crucial information from [BT17, Proposition 2.1] is

$$\begin{aligned} |\xi (y,T) - \xi (y,T+1)| = \beta \xi _0(y)^{-\beta } T^{-(\beta +1)} (1+o(1)) \quad \text { as } T \rightarrow \infty , \end{aligned}$$

(2.3)

for $\beta = (a_2+b_2)/(2b_2)$ and specific values $\xi _0(y)$ given in [BT17, Proof Proposition 2.1]. The parameters $a_2,b_2 \ge 0$ can be chosen freely, so (2.3) holds for $\beta \in (\frac{1}{2},\infty )$, but in this paper we choose $a_2 > b_2$, and therefore $\beta \in (1,\infty )$. Also the “small tail” estimates of (2.3) are stronger than the “big tail” estimates needed in our main theorems, but are required for particular results in the infinite measure setting of [BT17].

Remark 2.1

In fact, for the most general vector fields treated in [BT17, Equation (3)] (i.e., vector fields with higher order terms), we cannot do better than “big tails” estimates:

$$\begin{aligned} \xi (y,T) = \xi _0(y) T^{-\beta }(1+ O(T^{-\beta _*}+T^{-\frac{1}{2}} \log T)), \end{aligned}$$

(2.4)

for $\beta _* = \min \{1, a_2/b_2, b_0/a_0\}$, see [BT17, Lemma 2.3]. This is due to the perturbation arguments in [BT17, Section 2.4] that become necessary when higher order terms are present and a precise first integral from [BT17, Lemma 2.2] is not available. Since the horizontal part of vector field of (2.1) here is the ‘ideal’ vector field from [BT17, Equation (4)], our small tail estimate (2.3) becomes possible.

The next proposition gives an estimate of integrals along such curves. This allows us to estimate the tail of the roof function (when viewing $\Phi _t$ as a suspension flow) in Proposition 2.4 and also, the tail of induced potentials (see Remark 2.6). In the first instance we use it to estimate the vertical component of the flow $\Phi _t$ (compared to t).

Proposition 2.2

Let $\theta :{{\mathbb {R}}}^2 \rightarrow {{\mathbb {R}}}$ be a homogeneous function with exponent $\rho \in {{\mathbb {R}}}$ such that $\theta (x,0) \not \equiv 0 \not \equiv \theta (0,y)$ for $x\ne 0 \ne y$. Then there is a constant $C_\rho > 0$ (given explicitly in the proof) such that for every q with ${\hat{\tau }}(q) = T$ as in (2.2),

$$\begin{aligned} \Theta (T) := \int _0^T \theta ( \Phi _t(q,0))\, dt = {\left\{ \begin{array}{ll} C_\rho T^{1-\frac{\rho }{2}} (1+o(1)) &{}\quad \text { if } \rho < 2, \\ C_\rho \log (T) (1+o(1)) &{}\quad \text { if } \rho = 2, \\ C_\rho (1+o(1)) &{}\quad \text { if } \rho > 2. \end{array}\right. } \end{aligned}$$

(2.5)

For the proof of Proposition 2.2 we recall some notation and the form of the first integral L(x, y) from [BT17, Lemma 2.2], replacing $\kappa$ from that paper by 2. Recall that $\Delta := a_2b_0 - a_0b_2 \ne 0$. Let $u,v \in {{\mathbb {R}}}$ be the solutions of the linear equation

$$\begin{aligned} {\left\{ \begin{array}{ll} (u+2) a_0 = v b_0 \\ (v+2)b_2 = u a_2 \end{array}\right. } \quad \text { that is:}\quad {\left\{ \begin{array}{ll} u = \frac{2 b_2}{\Delta }(a_0+b_0), \\ v = \frac{2 a_0}{\Delta } (a_2+b_2). \end{array}\right. } \end{aligned}$$

(2.6)

Note that u, v and $\Delta$ all have the same sign. Define:

$$\begin{aligned} \beta _0 := \frac{a_0+b_0}{2 a_0} = \frac{u+v+2 }{2 v}, \quad \beta := \beta _2 := \frac{a_2+b_2}{2 b_2} = \frac{u+v+2 }{2 u}, \quad \begin{array}{l} c_0 = a_0 + b_0 \\ c_2 = a_2+b_2 \end{array}.\nonumber \\ \end{aligned}$$

(2.7)

Note that $\frac{\beta _0}{\beta _2} = \frac{u}{v}$, $\frac{a_0 u}{b_2v} = \frac{c_0}{c_2}$ and $\beta _0, \beta _2 > \frac{1}{2}$ (or $=\frac{1}{2}$ if we allow $b_0=0$ or $a_2=0$ respectively). The content of [BT17, Lemma 2.2] is that

$$\begin{aligned} L(x,y) = {\left\{ \begin{array}{ll} x^u y^v ( \frac{a_0}{v}\ x^2 + \frac{b_2}{u}\ y^2 ) &{}\quad \text { if } \Delta > 0;\\ x^{-u} y^{-v} ( \frac{a_0}{v}\ x^2 + \frac{b_2}{u}\ y^2 )^{-1} &{}\quad \text { if } \Delta < 0, \end{array}\right. } \end{aligned}$$

(2.8)

is a first integral of (i.e., preserved by) $\Phi _t^{hor}$ (and therefore of $\Phi _t$).

For the proof of Proposition 2.2, we follow the proof of [BT17, Proposition 2.1]. In comparison, we have $\kappa$ from [BT17, Proposition 2.1] equal to 2, our current integrand is more complicated, but we only need first order error terms.

Proof of Proposition 2.2

Fix $\eta$ such that the local unstable leaf ${\hat{W}}^{\mathbf {u}}_{loc}(0,\eta )$ intersects $\overline{{\hat{f}}^{-1}({\hat{P}}_0) {\setminus } {\hat{P}}_0}$. Recall from the text below (2.2) that, given T large enough, there is a unique point $(\xi (\eta ,T), \eta ') \in {\hat{W}}^{\mathbf {u}}_{loc}(0,\eta )$ such that

$$\begin{aligned} (\zeta '_0, \omega (\eta ,T)) := \Phi _T^{hor}((\xi (\eta ,T), \eta ')) \in {\hat{W}}_{loc}^s(\zeta _0,0), \end{aligned}$$

where ${\hat{W}}_{loc}^s(\zeta _0,0)$ is the local stable leave forming the right boundary of $P_0$, see Fig. 2. Assuming ${\hat{W}}^{\mathbf {u}}_{loc}(0,\eta )$ and ${\hat{W}}_{loc}^s(\zeta _0,0)$ are straight horizontal and vertical lines respectively, induces a negligible error in these vertical coordinates, so in the rest of the proof, we write $\eta ' = \eta$ and $\zeta _0' = \zeta _0$.

Fig. 2

The first quadrant of the rectangle ${\hat{P}}_0$, with the integral curve I(T) connecting $(\xi (\eta ,T),\eta ) \in {\hat{W}}^{\mathbf {u}}_{loc}(0,\eta ))$ and $(\zeta _0',\omega (\eta ,T)) \in {\hat{W}}^{\mathbf {s}}(\zeta '_0,\omega (\eta ,T))$

For simplicity of notation, we will suppress the $\eta$ and T in $\xi (\eta ,T)$ and $\omega (\eta ,T)$. For $0 \le t \le T$, let $\Phi _t^{hor}(\xi ,\eta ) = (x(t), y(t))$, so $(x(0)), y(0)) = (\xi ,\eta )$ and $(x(T), y(T)) = (\zeta _0, \omega )$. We need to compute

$$\begin{aligned} \Theta (T) = \int _0^T\theta (x(t), y(t)) \, dt, \end{aligned}$$

where $\theta :{{\mathbb {R}}}^2 \rightarrow {{\mathbb {R}}}$ is homogeneous of exponent $\rho$.

New coordinates: We compute $\Theta (T)$ by introducing new coordinates (L, M) where L is the first integral from (2.8) (and hence independent of t), and $M(t) = y(t)/x(t)$, so $y(t) = M(t)x(t)$. For simplicity of notation, we suppress t in x, y and M. Differentiating to $\dot{y} = \dot{M} x + M \dot{x}(t)$, and inserting the values for $\dot{x}$ and $\dot{y}$ from (2.1), we get

$$\begin{aligned} \dot{M} = -M( c_0 + c_2 M^2) x^2. \end{aligned}$$

(2.9)

We carry out the proof for $\Delta > 0$ (the case $\Delta < 0$ goes likewise), so the first integral is $L(x,y) = x^u y^v ( \frac{a_0}{v}\, x^2 + \frac{b_2}{u}\, y^2)$ in (2.8). Since (x, y) is in the level set $L(x,y) = L(\xi ,\eta ) = \xi ^u \eta ^v(\frac{a_0}{v} \xi ^2 + \frac{b_2}{u}\eta ^2)$, we can solve for $x^2$ in the expression

$$\begin{aligned} \xi ^u\eta ^v\left( \frac{a_0}{v} \xi ^2 + \frac{b_2}{u}\eta ^2 \right) = x^u y^v \left( \frac{a_0}{v} x^2 + \frac{b_2}{u} y^2 \right) = x^{u+v+2 } M^v \left( \frac{a_0}{v} + \frac{b_2}{u} M^2\right) . \end{aligned}$$

Use (2.6) and (2.7) to obtain

$$\begin{aligned} \frac{a_0}{v} + \frac{b_2}{u} M^2 = \frac{\Delta }{2 c_0c_2} ( c_0+c_2M^2 ) \quad \text { and }\quad \frac{a_0 \xi ^2}{v} + \frac{b_2 \eta ^2 }{u} = \frac{\Delta }{2 c_0c_2} ( c_0 \xi ^2 +c_2 \eta ^2 ). \end{aligned}$$

Note also from (2.7) that $\frac{2v}{u+v+2} = \frac{1}{\beta _0}$, $\frac{2u}{u+v+2} = \frac{1}{\beta _2}$ and $\frac{2}{u+v+2} = 1-\frac{1}{2\beta _0} - \frac{1}{2\beta _2}$. This plus the previous two equations together gives

$$\begin{aligned} x^2 = G(T) M^{-\frac{1}{\beta _0} } (c_0+c_2M^2)^{\frac{1}{2\beta _0}+\frac{1}{2\beta _2}-1}, \end{aligned}$$

(2.10)

with

$$\begin{aligned} G(T) := G(\xi (\eta , T), \eta ) = \xi (\eta , T)^{\frac{1}{\beta _2}} \eta ^{\frac{1}{\beta _0}} (c_0 \xi (\eta ,T)^2 + c_2 \eta ^2)^{1-\frac{1}{2\beta _0} - \frac{1}{2\beta _2}}. \end{aligned}$$

(2.11)

We claim that $G(T) \sim G_0 T^{-1}$ as $T \rightarrow \infty$ for some $G_0 > 0$.

Proving the claim: First note that

$$\begin{aligned} T = \int _0^T dt = \int _{M(0)}^{M(T)} \, \frac{dM}{\dot{M}} = -\int _{M(T)}^{M(0)} \, \frac{dM}{\dot{M}}. \end{aligned}$$

(2.12)

Recall that $M(0) = \eta /\xi$ and $M(T) = \omega /\zeta _0$. Inserting this and (2.9) and (2.10) into (2.12) gives

$$\begin{aligned} \int _{\omega /\zeta _0}^{\eta /\xi } \frac{dM}{ M^{1-\frac{1}{\beta _0}} \left( c_0+c_2 M^2 \right) ^{ \frac{1}{2 \beta _0} + \frac{1}{2 \beta _2} } } = G(T) T. \end{aligned}$$

(2.13)

As T increases, the integral curve connecting $(\xi (\eta ,T), \eta )$ to $(\zeta _0,\omega (\eta ,T))$ tends to the union of the local stable and unstable manifolds of (0, 0), whilst $M(T) = \omega (\eta ,T)/\zeta _0 \rightarrow 0$ and $M(0) = \eta /\xi (\eta ,T) \rightarrow \infty$. From their definition, $\xi (\eta ,T)$ and $\omega (\eta ,T)$ are decreasing in T, so their T-derivatives $\xi '(\eta ,T), \omega '(\eta ,T) \le 0$.

Since $c_0, c_2 > 0$ (otherwise $\Delta = 0$), the integrand of (2.13) is $O(M^{\frac{1}{\beta _0}-1})$ as $M \rightarrow 0$ and $O(M^{-\frac{1}{\beta _2}-1})$ as $M \rightarrow \infty$. Hence the integral is increasing and bounded in T. But this means that G(T)T is increasing in T and bounded as well. Since by (2.11)

$$\begin{aligned} G(T) T = \left( \xi (\eta ,T) T^{\beta _2} \right) ^{\frac{1}{\beta _2}} \eta ^{\frac{1}{\beta _0}} (c_0 \xi (\eta ,T)^2 + c_2 \eta ^2 )^{1-\frac{1}{2 \beta _0} - \frac{1}{2 \beta _2}}, \end{aligned}$$

we find by combining with (2.13) that

$$\begin{aligned} \xi _0(\eta ) := \lim _{T\rightarrow \infty } \xi (\eta ,T) T^{\beta _2}= & {} \lim _{T \rightarrow \infty } (G(T)T)^{\beta _2} \eta ^{-\frac{\beta _2}{\beta _0}} (c_0 \xi (\eta ,T)^2 + c_2 \eta ^2 )^{-\beta _2(1-\frac{1}{2 \beta _0} - \frac{1}{2 \beta _2})} \\= & {} c_2^{-\frac{1}{u}} \eta ^{- \frac{a_2}{b_2} } \left( \int _0^\infty \frac{ dM }{ M^{1-\frac{1}{\beta _0}} \left( c_0 + c_2 M^2 \right) ^{ \frac{1}{2 \beta _0} + \frac{1}{2 \beta _2} } } \right) ^{\beta _2}, \end{aligned}$$

where we have used $-\beta _2(1-\frac{1}{2 \beta _0} - \frac{1}{2 \beta _2}) = -\frac{2 \beta _2}{u+v+2 } = -\frac{1}{u}$ for the exponent of $c_2$, and $-\frac{\beta _2}{\beta _0} - 2\beta _2(1-\frac{1}{2\beta _0}-\frac{1}{2\beta _2}) = 1-2\beta _2 = -\frac{a_2}{b_2}$ for the exponent of $\eta$. This shows that

$$\begin{aligned} G(T) \sim c_2^{1-\frac{1}{2\beta _0}-\frac{1}{2\beta _2}} \xi _0(\eta )^{\frac{1}{\beta _2}} \eta ^{2-\frac{1}{\beta _2}}\, T^{-1} =: G_0 T^{-1} \quad \text { as } T \rightarrow \infty . \end{aligned}$$

(2.14)

Estimating the integral $\Theta (T)$: Recalling $M = y/x$ and using homogeneity of $\theta$, we have $\theta (x,y) = x^\rho \theta (1,M)$, and $\theta _0 := \theta (1,0)$ and $\theta _\infty := \theta (0,1) = \lim _{M \rightarrow \infty } M^{-\rho } \theta (1,M)$ are non-zero by assumption. Inserting the above into the integral of (2.13), and using (2.10) to rewrite x, we obtain

$$\begin{aligned} \Theta (T) = G(T)^{\frac{\rho }{2}-1} \int _{\omega /\zeta _0}^{\eta /\xi } \frac{ M^{-\frac{\rho }{2\beta _0}} (c_0+c_2M^2)^{-\frac{\rho }{2}(1-\frac{1}{2\beta _0} -\frac{1}{2\beta _0})} \theta (1,M) }{M^{1-\frac{1}{\beta _0}} \left( c_0+c_2 M^2 \right) ^{ \frac{1}{2\beta _0} + \frac{1}{2\beta _2} } } \, dM. \end{aligned}$$

(2.15)

Set $\rho _0 := \frac{\rho }{2} - (1+\frac{\rho }{2}) (\frac{1}{2\beta _0} + \frac{1}{2\beta _2})$. The leading terms in the integrand of (2.15) are

$$\begin{aligned} {\left\{ \begin{array}{ll} \theta _0 c_0^{\rho _0} M^{-1+(1-\frac{\rho }{2})\frac{1}{\beta _0}} &{}\quad \text{ as } M \rightarrow 0;\\ \theta _\infty c_2^{\rho _0} M^{-1-(1-\frac{\rho }{2})\frac{1}{\beta _2}} &{}\quad \text{ as } M \rightarrow \infty . \end{array}\right. } \end{aligned}$$

(2.16)

The case $\rho < 2$: By (2.16), we have for $\rho < 2$ that the exponent of M is $> -1$ as $M \rightarrow 0$ and $< -1$ as $M \rightarrow \infty$. This means that the integral in (2.15) converges to some constant

$$\begin{aligned} C^* = \int _0^\infty \frac{ M^{-\frac{\rho }{2\beta _0}} (c_0+c_2M^2)^{-\frac{\rho }{2}(1-\frac{1}{2\beta _0}-\frac{1}{2\beta _0})} \theta (1,M) }{ M^{1-\frac{1}{\beta _0}} \left( c_0+c_2 M^2 \right) ^{ \frac{1}{2\beta _0} + \frac{1}{2\beta _2} } } \, dM \end{aligned}$$

as $T {\rightarrow } \infty$, and $\Theta \sim C^* G(T)^{\frac{\rho }{2}-1} \sim C_\rho T^{1{-}\frac{\rho }{2}}$ for $C_\rho {=}\left( c_2^{1{-}\frac{1}{2\beta _0}{-} \frac{1}{2\beta _2}} \xi _0(\eta )^{\frac{1}{\beta _0}} \eta ^{2-\frac{1}{\beta _2}}\! \right) ^{\frac{\rho }{2}-1} C^*$ by (2.11). This finishes the proof for $\rho < 2$.

The case $\rho > 2$: The value of $\Theta (T)$ based on the leading terms (2.16) of the integrand only is

$$\begin{aligned} \Theta (T) \sim \frac{G(T)^{\frac{\rho }{2}-1}}{\frac{\rho }{2}-1} \left( \beta _0 \theta _0 c_0^{\rho _0} \left( \frac{\omega }{\zeta _0} \right) ^{(1-\frac{\rho }{2})\frac{1}{\beta _0}} + \beta _2 \theta _\infty c_2^{\rho _0} \left( \frac{\eta }{\xi } \right) ^{-(1-\frac{\rho }{2})\frac{1}{\beta _2}} \right) . \end{aligned}$$

Insert $\xi = \xi _0(\eta ) T^{-\beta _2} (1+o(1))$ and $\omega = \omega _0(\eta ) T^{-\beta _0} (1+o(1))$ from [BT17, Proposition 2.1]:

$$\begin{aligned} \Theta (T) \sim \frac{ G(T)^{\frac{\rho }{2}-1}}{\frac{\rho }{2}-1} \left( \beta _0 \theta _0 c_0^{\rho _0} \left( \frac{\zeta _0}{\omega _0(\eta )} \right) ^{\frac{\rho /2-1}{\beta _0}} + \beta _2 \theta _\infty c_2^{\rho _0} \left( \frac{\eta }{\xi _0(\eta )} \right) ^{\frac{\rho /2-1}{\beta _2}} \right) T^{\frac{\rho }{2}-1}. \end{aligned}$$

Next, by inserting the asymptotics of G(T) from (2.14), the factor $T^{\frac{\rho }{2}-1}$ cancels:

$$\begin{aligned} \Theta (T) \sim C_\rho := \frac{ G_0^{\frac{\rho }{2}-1}}{\frac{\rho }{2}-1} \left( \beta _0 \theta _0 c_0^{\rho _0} \left( \frac{\zeta _0}{\omega _0(\eta )} \right) ^{\frac{\rho /2-1}{\beta _0}} + \beta _2 \theta _\infty c_2^{\rho _0} \left( \frac{\eta }{\xi _0(\eta )} \right) ^{\frac{\rho /2-1}{\beta _2}} \right) . \end{aligned}$$

The case $\rho = 2$: The factor $G(T)^{\frac{\rho }{2}-1}$ in (2.15) now disappears and the leading terms (2.16) are $\theta _0 c_0^{\rho _0} M^{-1}$ and $\theta _\infty c_2^{\rho _0} M^{-1}$ respectively. This gives

$$\begin{aligned} \Theta (T) \sim \int _{\omega /\zeta _0}^{\eta /\xi } \frac{\theta _0 c_0^{\rho _0}+\theta _\infty c_2^{\rho _0}}{M} \, dM = \left( \theta _0 c_0^{\rho _0}+\theta _\infty c_2^{\rho _0} \right) \left( \log \frac{\eta }{\xi } - \log \frac{\omega }{\zeta _0}\right) . \end{aligned}$$

Inserting again the values of $\xi$ and $\omega$ from [BT17, Proposition 2.1] gives

$$\begin{aligned} \Theta (T) \sim \left( \theta _0 c_0^{\rho _0}+\theta _\infty c_2^{\rho _0} \right) (\beta _0+\beta _2) \log T \quad \text { as } T \rightarrow \infty . \end{aligned}$$

This completes the proof. $\square$

2.2 Description via suspension flows, tail estimates of the roof function

The 3-dimensional time-1 map $\Phi _1$ preserves no 2-dimensional submanifold of $\mathcal {M}$. Yet in order to model $\Phi _t$ as a suspension flow over a 2-dimensional map, we need a genuine Poincaré map. For this we choose a section $\Sigma$ transversal to $\Gamma$ and containing a neighbourhood U of p. As an example, $\Sigma$ could be ${{\mathbb {T}}}^2 \times \{ 0 \}$, and the Poincaré map to ${{\mathbb {T}}}^2 \times \{ 0 \}$ could be (a local perturbation of) Arnol’d’s cat map; in this case (and most cases) $\mathcal {M}$ is not homeomorphic to ${{\mathbb {T}}}^3$ because the homology is more complicated, see [BF13, N76].

Let $h:\Sigma \rightarrow {{\mathbb {R}}}^+$, $h(q) = \min \{ t > 0 : \Phi _t(q) \in \Sigma \}$ be the first return time. Assuming that $\sup _\Sigma |w(x,y)| < 1$, the first return time h is bounded and bounded away from zero, i.e., $0< \inf _{\Sigma } h < \sup _{\Sigma } h$. There is no loss of generality in assuming that $\inf _{\Sigma }h \ge 1$.

The Poincaré map $f := \Phi _h: \Sigma \rightarrow \Sigma$ has a neutral saddle point p at the origin. Its local stable/unstable manifolds are $W^{\mathbf {s}}_{loc}(p) = \{ 0 \} \times (-\epsilon ,\epsilon )$ and $W^{\mathbf {u}}_{loc}(p) = (-\epsilon ,\epsilon ) \times \{ 0 \}$. Because the flow $\Phi _t$ is a perturbation of an Anosov flow, and f is a Poincaré map, it has a finite Markov partition $\{P_i\}_{i \ge 0}$ and we can assume that p is in the interior of $P_0$. In the sequel, let U be a neighbourhood of p that is small enough that (2.1) is valid on $U \times [0,1]$ but also that $f(U) \supset {\hat{P}}_0 \cup P_0$.

In order to regain the hyperbolicity lacking in f, let

$$\begin{aligned} r(q) := \min \{ n \ge 1 : f^n(q) \in Y \} \end{aligned}$$

(2.17)

be the first return time to $Y := \Sigma {\setminus } P_0$. Then the Poincaré map $F = f^{r} = \Phi _\tau$ of $\Phi _t$ to $Y \times \{ 0 \}$ is hyperbolic (see [H00, Section 7]), where

$$\begin{aligned} \tau (q) = \min \{ t > 0 : \Phi _t((q,0)) \in Y \times \{ 0 \} \} = \sum _{j=0}^{r-1} h \circ f^j \end{aligned}$$

is the corresponding first return time.

Consequently, the flow $\Phi _t:\mathcal {M}\times {{\mathbb {R}}}\rightarrow \mathcal {M}$ can be modelled as a suspension flow on $Y^\tau = \left( \bigcup _{q \in Y} \{ q \} \times [0,\tau (q)) \right) /(q,\tau (q)) \sim (F(q),0)$. Since the flow and section $Y \times \{ 0 \}$ are $C^1$ smooth, $\tau$ is $C^1$ on each piece $\{r= k\}$.

Lemma 2.3

In the notation of Proposition 2.2 with $\theta = w$ from the z-component in (2.1), we have $\tau (q) = {\hat{\tau }}(q) + O(1)$, $r= {\hat{\tau }}(q)+\Theta ({\hat{\tau }}(q))+O(1)$ and $\Theta ({\hat{\tau }}(q)) = o({\hat{\tau }}(q))$ as $q \rightarrow 0$.

Proof

By the definition of ${\hat{\tau }}$ we have $\Phi _{{\hat{\tau }}}^{hor}(q) \in {\hat{W}}^{\mathbf {s}}$. Therefore it takes a bounded amount of time (positive or negative) for $\Phi _{{\hat{\tau }}}(q,0)$ to hit $Y \times \{ 0 \}$, so $|\tau (q)-{\hat{\tau }}(q)| = O(1)$.

If in (2.5) we set $\theta = w$, then ${\hat{\tau }}(q) + \Theta ({\hat{\tau }}(q))$ indicates the vertical displacement under the flow $\Phi _t$. In particular, it gives the number of times the flow-line intersects $\Sigma$, and hence $r= {\hat{\tau }}(q) + \Theta ({\hat{\tau }}(q)) + O(1)$. $\square$

Fig. 3

The first quadrant of the rectangle $P_0$, with stable and unstable foliations of Poincaré map $f = \Phi _h$ drawn vertically and horizontally, respectively. Also one of the integral curves is drawn

Let $\phi :Y^\tau \rightarrow {{\mathbb {R}}}$, $\phi (q) = \lim _{t\rightarrow 0} -\frac{1}{t} \log | D\Phi _t|_{W^{\mathbf {u}}}(q)|$ be the potential for the SRB measure of the flow. Let $\mu _{{\bar{\phi }}}$ be the F-invariant equilibrium measure of the potential ${\bar{\phi }}:Y \rightarrow {{\mathbb {R}}}$, ${\bar{\phi }}(q) = \int _0^{\tau (q)} \phi \circ \Phi _t(q) \, dt= -\log | D F|_{W^{\mathbf {u}}}(q)|$; so ${\bar{\phi }}$ is the logarithm of the derivative in the unstable direction of the Poincaré map F. This is at the same time the SRB-measure for F and thus is absolutely continuous conditioned to unstable leaves.

Proposition 2.4

Recall that $\beta = \frac{a_2+b_2}{2b_2} \in (\frac{1}{2}, \infty )$. There exists $C^* > 0$ such that

$$\begin{aligned} \mu _{{\bar{\phi }}}(\{ \tau > t \}) = C^* t^{-\beta } (1+o(1)) \end{aligned}$$

(2.18)

for the F-invariant SRB measure $\mu _{{\bar{\phi }}}$.

Proof

The function $\tau$ is defined on $\Sigma {\setminus } P_0$ and $\tau \ge h_2 = h + h \circ f$ on $Y_{\{r\ge 2\}} := f^{-1}(P_0) {\setminus } P_0$. The set $Y_{\{r\ge 2\}}$ is a rectangle with boundaries consisting of two stable and two unstable leaves of the Poincaré map f. Let $W^{\mathbf {u}}(y)$ denote the unstable leaf of f inside $Y_{\{r\ge 2\}}$ with (0, y) as (left) boundary point. Let $y_1 < y_2$ be such that $W^{\mathbf {u}}(y_1)$ and $W^{\mathbf {u}}(y_2)$ are the unstable boundary leaves of $Y_{\{r\ge 2\}}$.

The unstable foliation of ${\hat{f}} = \Phi _1^{hor}$ does not entirely coincide with the unstable foliation of f. Let ${\hat{W}}^{\mathbf {u}}(y)$ denote the unstable leaf of ${\hat{f}}$ with (0, y) as (left) boundary point. Both ${\hat{W}}^{\mathbf {u}}(y)$ and $W^{\mathbf {u}}(y)$ are $C^1$ curves emanating from (0, y); let $\varvec{{\gamma }}(y)$ denote the angle between them. Then the lengths

$$\begin{aligned} \text{ Leb }(W^{\mathbf {u}}(y) \cap \{ \tau> t\})= & {} |\cos \varvec{{\gamma }}(y)|\ \text{ Leb }({\hat{W}}^{\mathbf {u}}(y) \cap \{ \tau > t\}) (1+o(1))\\= & {} |\cos \varvec{{\gamma }}(y)|\ \xi _0(y)\ t^{-\beta } (1+o(1)) \end{aligned}$$

as $t\rightarrow \infty$, where the last equality and the notation $\xi _0(y)$ and $\beta = (a_2+b_2)/(2b_2)$ come from (2.3).

We decompose the measure $\mu _{{\bar{\phi }}}$ on $Y_{\{r\ge 2\}}$ as

$$\begin{aligned} \int _{Y_{\{r\ge 2\}}} v\, d\mu _{{\bar{\phi }}} = \int _{y_1}^{y_2} \left( \int _{W^{\mathbf {u}}(y) } v\, d\mu ^{\mathbf {s}}_{W^{\mathbf {u}}(y)} \right) d\nu ^{\mathbf {u}}(y). \end{aligned}$$

The conditional measures $\mu _{W^{\mathbf {u}}(y)}$ on $W^{\mathbf {u}}(y)$ are equivalent to Lebesgue $m_{W^{\mathbf {u}}(y)}$ with density $h_0^{\mathbf {u}} = \frac{d\mu _{W^{\mathbf {u}}(y)}}{dm_{W^{\mathbf {u}}(y)}}$ tending to a constant $h_*(y)$ at the boundary point (0, y). Therefore, as $t \rightarrow \infty$,

$$\begin{aligned} \mu _{{\bar{\phi }}}(\tau> t)= & {} \int _{y_1}^{y_2} \mu _{W^{\mathbf {u}}(y)}(W^{\mathbf {u}}(y) \cap \{ \tau> t\} ) \, d\nu ^{\mathbf {u}}(y) \\= & {} \int _{y_1}^{y_2} h_0^{\mathbf {u}}\ m_{W^{\mathbf {u}}(y)}( W^{\mathbf {u}}(y) \cap \{ \tau> t\} ) \, d\nu ^{\mathbf {u}}(y) \\= & {} \int _{y_1}^{y_2} h_0^{\mathbf {u}} |\cos \varvec{{\gamma }}(y)|\ m_{{\hat{W}}^{\mathbf {u}}(y)}({\hat{W}}^{\mathbf {u}}(y) \cap \{ \tau > t\} ) (1+o(1)) \, d\nu ^{\mathbf {u}}(y) \\= & {} \int _{y_1}^{y_2} h_0^{\mathbf {u}} |\cos \varvec{{\gamma }}(y)|\ \xi _0(y)\ t^{-\beta } (1+o(1)) \, d\nu ^{\mathbf {u}}(y) = C^* t^{-\beta }(1+o(1)), \end{aligned}$$

for $C^* = \int _{y_1}^{y_2} h_*(y) |\cos \varvec{{\gamma }}(y)| \ \xi _0(y) \, d\nu ^{\mathbf {u}}(y)$, and using (2.3) in the third line. This proves the result. $\square$

2.3 Main results for the Poincaré map F and flow $\Phi _t$

Throughout this section we assume the setup and notation of Sect. 2.2. We emphasise that we are in the finite measure setting, so

$$\begin{aligned} \tau ^* := \int _Y \tau \, d\mu _{{\bar{\phi }}} < \infty . \end{aligned}$$

(2.19)

We recall that the natural potential associated to the SRB-measure for F is ${\bar{\phi }} = -\log | D F|_{W^{\mathbf {u}}}$, which is the induced version of $\phi = \lim _{t\rightarrow 0}\frac{-\log | D\Phi _t|_{W^{\mathbf {u}}}}{t}$. Our main result can be viewed as a version of the results in [S06] for the flow $\Phi _t$; it gives gives the link between limit theorems for $(\Phi _t,\psi )$ for (unbounded from below) potentials $\psi$ on $\mathcal {M}$ and pressure function ${\mathcal {P}}(\phi +s\psi ), s\ge 0$. We let $\overline{\phi +s \psi }, s\ge 0$ be the family of induced potentials and denote the associate pressure function by ${\mathcal {P}}(\overline{\phi +s\psi })$. Some background on equilibrium states and pressure function (along with their induced versions) is recalled in Sect. 3.

Theorem 2.5

Suppose ${\bar{\psi }}(y) := \int _0^\tau \psi \circ \Phi _t \, dt =C'-\psi _0$ where $C'>0$ and $\psi _0$ is positive piecewise $C^1(Y)$. Let $\beta > 1$ be as in Proposition 2.4, Furthermore, suppose that $\mu _{{\bar{\phi }}}(\psi _0 > t) \sim ct^{-\frac{\beta }{\kappa }}$ for some $c>0$ and $\kappa \in (1/\beta , \beta )$. Set $\psi ^* =\int _Y {\bar{\psi }}\,d \mu _{{\bar{\phi }}}$ and let $\psi _T= \int _0^T \psi \circ \Phi _t \, dt$. The following hold as $T\rightarrow \infty$, w.r.t.  $\mu _{{\bar{\phi }}}$.

• (a)

  • (i) When $\beta <2$ and $\kappa \in (\beta /2,\beta )$, set b(T) such that $\frac{T}{(c b(T))^{\frac{\beta }{\kappa }}}\rightarrow 1$.

    Then $\frac{1}{b(T)}(\psi _T-\psi ^*\cdot T)\rightarrow ^d G_{\beta /\kappa }$, where $G_{\beta /\kappa }$ is a stable law of index $\beta /\kappa$ and this is, further, equivalent to ${\mathcal {P}}(\overline{\phi +s\psi })=\psi ^*s+c s^{\beta /\kappa }(1+o(1))$, as $s\rightarrow 0$.

  • (ii) When $\beta \le 2$ but $\kappa =\beta /2$, set $b(T)\sim c^{-1/2} \sqrt{T\log T}$.

    Then $\frac{1}{b(T)}(\psi _T-\psi ^*\cdot T)\rightarrow ^d {{\mathcal {N}}}(0,1)$ and this is, further, equivalent to ${\mathcal {P}}(\overline{\phi +s\psi })=\psi ^* s+c s^2 \log (1/s)(1+o(1))$, as $s\rightarrow 0$.

• (b) Suppose that $\beta >2$. Then there exists $\sigma \ne 0$ such that $\frac{1}{\sqrt{T}}(\psi _T- \psi ^*\cdot T)\rightarrow ^d {{\mathcal {N}}}(0,\sigma ^2)$ and this is, further, equivalent to ${\mathcal {P}}(\overline{\phi +s\psi })=\psi ^* s+\frac{\sigma ^2}{2}s^2 (1+o(1))$ as $s\rightarrow 0$.

Moreover, ${\mathcal {P}}(\phi +s\psi )=\frac{1}{\tau ^*}{\mathcal {P}}(\overline{\phi +s\psi }) (1+o(1))$.

Remark 2.6

If $\psi :\mathcal {M}\rightarrow {{\mathbb {R}}}$ satisfies the following properties, then the condition on ${\bar{\psi }}$ in Theorem 2.5 holds. In local tubular coordinates near $\{ p \} \times {{\mathbb {T}}}$, $\psi = g_0-g(W(x,y))$ where $g_0$ is Hölder function vanishing on a neighbourhood of p, and W(x, y) is a linear combination of homogeneous functions depending only on (x, y), such that the term $W_\rho (x,y)$ with lowest exponent $\rho$ satisfies $W_\rho (0,1) \ne 0 \ne W_\rho (1,0)$, and $g:{{\mathbb {R}}}\rightarrow {{\mathbb {R}}}$ is analytic with $g(0) = 0$, $g'(0) \ne 0$. Then $\kappa = 1-\frac{\rho }{2}$ (see Proposition 2.2), so assuming that this lowest exponent $\rho \in (2(1-\beta ), 2(\beta -1)/\beta )$, then $\kappa \in (\frac{1}{\beta },\beta )$.

Proof of Theorems 2.5

This follow directly from applying Theorem 8.2$(\Phi _t,\phi )$, which uses a special case of Proposition 5.1. “Appendix A” verifies all the abstract hypotheses of Proposition 5.1 and Theorem 8.2 for the flow $\Phi _t$ with base map F and roof function $\tau$ introduced in Sect. 2 and potential $\psi$ defined in the statement of Theorem 2.5. The extra assumptions (including coboundary assumptions) that ensure $\sigma \ne 0$ in Proposition 5.1(ii) and Theorem 8.2 (b) are verified in “Appendix A.3”. $\square$

3 Background on Equilibrium States for Suspension Flows

Here we outline the standard general theory of thermodynamic formalism for suspension flows. We start with a discrete time dynamical system $F:Y\rightarrow Y$. Defining $\mathcal {M}_F$ to be the set of F-invariant probability measures, for a potential $\psi \!:\!Y\!\rightarrow \! [-\infty , \infty ]$ we define the pressure as

$$\begin{aligned} {\mathcal {P}}(\psi ):=\sup \left\{ h(\mu )+\int \psi ~d\mu :\ \mu \in \mathcal {M}_F \text { and } -\int \psi ~d\mu <\infty \right\} . \end{aligned}$$

Given a roof function $\tau :Y\rightarrow {{\mathbb {R}}}^+$, let $Y^\tau =\{(y,z)\in Y\times {{\mathbb {R}}}: 0\le z\le \tau (y)\}/\sim$ where $(y,\tau (y))\sim (Fy,0)$. Then the suspension flow $F_t:Y^\tau \rightarrow Y^\tau$ is given by $F_t(y,z)=(y,z+t)$ computed modulo identifications. We will suppose throughout that $\inf \tau >0$, but note that the case $\inf \tau =0$ can also be handled, see for example [Sav98, IJT15]. Barreira and Iommi [BI06] define pressure as

$$\begin{aligned} {\mathcal {P}}(\phi ):=\inf \{u\in {{\mathbb {R}}}:{\mathcal {P}}({\bar{\phi }}-u\tau )\le 0\} \end{aligned}$$

(3.1)

in the case that (Y, F) is a countable Markov shift and the potential $\phi :Y^\tau \rightarrow {{\mathbb {R}}}$ induces to ${\bar{\phi }}:Y\rightarrow {{\mathbb {R}}}$, where

$$\begin{aligned} {\bar{\phi }}(x):=\int _0^{\tau (x)}\phi (x, t)\, dt, \end{aligned}$$

(3.2)

has summable variations. This also makes sense for general suspension flows, so we will take (3.1) as our definition. In [AK42], there is a bijection between $F_t$-invariant probability measures $\mu$, and the corresponding F-invariant probability measures ${\bar{\mu }}$ given by the identification

$$\begin{aligned} \mu =\frac{{\bar{\mu }}\times m}{({\bar{\mu }}\times m)(Y^\tau )}, \end{aligned}$$

where m is Lebesgue measure. That is, whenever there is such a $\mu$ there is such a ${\bar{\mu }}$, and vice versa. Moreover, Abramov’s formula for flows [Ab59b] gives the following characterisation of entropy:

$$\begin{aligned} h(\mu )=\frac{h({\bar{\mu }})}{({\bar{\mu }}\times m)(Y^\tau )}, \end{aligned}$$

(3.3)

and clearly

$$\begin{aligned} \int \phi ~d\mu =\frac{\int {\bar{\phi }}~d{\bar{\mu }}}{({\bar{\mu }}\times m)(Y^\tau )} = \frac{\int {\bar{\phi }}~d{\bar{\mu }}}{\int \tau \, d{\bar{\mu }}}. \end{aligned}$$

(3.4)

One consequence of these formulas is that the pressure in (3.1) is independent of the choice of cross section Y. This follows essentially from the fact that if we choose a subset of Y and reinduce there, then Abramov’s formula gives the same value for pressure (here we can use the discrete version of Abramov’s formula [Ab59a]). Note that we say ‘Abramov’s formula’ in both the continuous and discrete time cases as the formulas are analogous [Ab59a, Ab59b], and similarly for integrals, where the formula holds by the ergodic and ratio ergodic theorems.

We say that $\mu _\phi$ is an equilibrium state for $\phi$ if $h(\mu _\phi )+\int \phi ~d\mu _\phi ={\mathcal {P}}(\phi )$. The same notion extends to the induced system $(Y, F, {\bar{\phi }})$. In that setting we may also have a conformal measure $m_{{\bar{\phi }}}$ for ${\bar{\phi }}$. This means that $m_{{\bar{\phi }}}(F(A))=\int _A e^{-{\bar{\phi }}}\, d m_{{\bar{\phi }}}$ for any measurable set A on which F is injective. Later on, in order to link our limit theorem results with pressure, we will assume that $F:Y\rightarrow Y$ is Markov. In that setting our assumptions on $\phi$ will be equivalent to assuming ${\mathcal {P}}(\phi )=0$, in which case we will have a equilibrium states and conformal measures $\mu _{{\bar{\phi }}}$ and $m_{{\bar{\phi }}}$.

4 Abstract Setup

We start with a flow $f_t:\mathcal {M}\rightarrow \mathcal {M}$, where $\mathcal {M}$ is a manifold. Let Y be a co-dimension 1 section of $\mathcal {M}$ and define $\tau :Y\rightarrow {{\mathbb {R}}}_{+}$ to be a roof function. We think of $\tau$ as a first return of $f_t$ to Y and define $F:Y\rightarrow Y$ by $F=f_\tau$. The flow $f_t$ is isomorphic with the suspension flow $F_t:Y^\tau \rightarrow Y^\tau$, $Y^\tau =\{(y,z)\in Y\times {{\mathbb {R}}}: 0\le z\le \tau (z)\}/\sim$, as described in Sect. 3. Throughout, we assume that $\tau$ is bounded from below.

Given the potential $\phi :\mathcal {M}\rightarrow {{\mathbb {R}}}$ and its induced version ${\bar{\phi }}:Y\rightarrow {{\mathbb {R}}}$ defined in (3.2), we assume that there is a conformal measure $m_{{\bar{\phi }}}$ for $(F,{\bar{\phi }})$. In the rest of this section we recall the abstract framework and hypotheses in [LT16] as relevant for studying limit theorems.

4.1 Banach spaces and equilibrium measures for $(F,{\bar{\phi }})$ and $(f_t,\phi )$.

Let $R_0\!:\!L^1(m_{{\bar{\phi }}})\!\rightarrow \! L^1(m_{{\bar{\phi }}})$ be the transfer operator for the first return map $F:Y\rightarrow Y$ w.r.t.  conformal measure $m_{{\bar{\phi }}}$, defined by duality on $L^1(m_{{\bar{\phi }}})$ via the formula $\int _Y R_0v\,w\,d m_{{\bar{\phi }}}= \int _Y v\,w\circ F\,d m_{{\bar{\phi }}}$ for every bounded measurable w.

We assume that there exist two Banach spaces of distributions ${{\mathcal {B}}}$, ${{\mathcal {B}}}_w$ supported on Y such that for some $\alpha ,\alpha _1>0$

(H1):

(i):

$C^\alpha \subset {{\mathcal {B}}}\subset {{\mathcal {B}}}_w\subset (C^{\alpha _1})'$, where $(C^{\alpha _1})'$ is the dual of $C^{\alpha _1} = C^{\alpha _1} (Y,{\mathbb {C}})$.²

(ii):

$hg\in {{\mathcal {B}}}$ for all $h\in {{\mathcal {B}}}$, $g\in C^\alpha$.

(iii):

The transfer operator $R_0$ associated with F maps continuously from $C^\alpha$ to ${{\mathcal {B}}}$ and $R_0$ admits a continuous extension to an operator from ${{\mathcal {B}}}$ to ${{\mathcal {B}}}$, which we still call $R_0$.

(iv):

The operator $R_0:{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}$ has a simple eigenvalue at 1 and the rest of the spectrum is contained in a disc of radius less than 1.

A few comments on (H1) are in order and here we just recall the similar ones in [LT16]. We note that (H1)(i) should be understood in terms of the following usual convention (see, for instance, [GL06, DL08]): there exist continuous injective linear maps $\pi _i$ such that $\pi _1(C^\alpha )\subset {{\mathcal {B}}}$, $\pi _2({{\mathcal {B}}})\subset {{\mathcal {B}}}_w$ and $\pi _3({{\mathcal {B}}}_w)\subset (C^{\alpha _1})'$. Throughout, we leave such maps implicit, but recall their meaning here. In particular, we assume that $\pi =\pi _3\circ \pi _2\circ \pi _1$ is the usual embedding, i.e., for all $h\in C^\alpha$ and $g\in C^{\alpha _1}$

$$\begin{aligned} \langle \pi (h),g\rangle _0=\int _Y h g\, d m_{{\bar{\phi }}}. \end{aligned}$$

Via the above identification, the conformal measure $m_{{\bar{\phi }}}$ can be identified with the constant function 1 both in $(C^{\alpha _1})'$ and in ${{\mathcal {B}}}$ (i.e., $\pi (1)=m_{{\bar{\phi }}}$). Also, by (H1)(i), ${{\mathcal {B}}}'\subset (C^\alpha )'$, hence the dual space can be naturally viewed as a space of distributions. Next, note that ${{\mathcal {B}}}'\supset (C^{\alpha _1})''\supset C^{\alpha _1}\ni 1$, thus we have $m_{{\bar{\phi }}}\in {{\mathcal {B}}}'$ as well. Moreover, since $m_{{\bar{\phi }}}\in {{\mathcal {B}}}$ and $\langle 1,g\rangle _0=\langle g,1\rangle _0=\int g\, d\mu ^{\mathbf {s}}_{{\bar{\phi }}}$, $m_{{\bar{\phi }}}$ can be viewed as the element 1 of both spaces ${{\mathcal {B}}}$ and $(C^{\alpha _1})'$.

By (H1), the spectral projection $P_0$ associated with the eigenvalue 1 is defined by $P_0=\lim _{n\rightarrow \infty }R_0^n$. Note that by (H1)(ii), for each $g\in C^\alpha$, $P_0 g\in {{\mathcal {B}}}$ and

$$\begin{aligned} \langle P_0g,1\rangle _0=m_{{\bar{\phi }}}(P_0 g)=\lim _{n\rightarrow \infty }m_{{\bar{\phi }}}(1\cdot R_0^n g)=m_{{\bar{\phi }}}(g)=\langle g,1\rangle _0. \end{aligned}$$

By (H1)(iv), there exists a unique $\mu _{{\bar{\phi }}}\in {{\mathcal {B}}}$ such that $R_0\mu _{{\bar{\phi }}}=\mu _{{\bar{\phi }}}$ and $\langle \mu _{{\bar{\phi }}},1\rangle =1$. Thus, $P_0g=\mu _{{\bar{\phi }}} \langle g,1\rangle _0$. Also $R_0'm_{{\bar{\phi }}}=m_{{\bar{\phi }}}$ where $R_0'$ is dual operator acting on ${{\mathcal {B}}}'$. Note that for any $g\in C^{\alpha _1}$,

$$\begin{aligned} |\langle \mu _{{\bar{\phi }}}, g\rangle _0|=|\langle P_01, g\rangle _0|=\left| \lim _{n\rightarrow \infty }R_0^n m_{{\bar{\phi }}}(g)\right| = \lim _{n\rightarrow \infty }\left| m_{{\bar{\phi }}}(g\circ F^n)\right| \le |g|_\infty . \end{aligned}$$

Hence $\mu _{{\bar{\phi }}}$ is a measure. For each $g\ge 0$,

$$\begin{aligned} \langle P_01, g\rangle _0=\lim _{n\rightarrow \infty } \langle R_0^n1, g\rangle =\lim _{n\rightarrow \infty } \langle 1, g\circ F^n\rangle _0\ge 0. \end{aligned}$$

It follows that $\mu _{{\bar{\phi }}}$ is a probability measure.

Summarising the above, the eigenfunction associated with the eigenvalue $\lambda =1$ is an invariant probability measure for F given by $d\mu _{{\bar{\phi }}}=h_0 d m_{{\bar{\phi }}}$ where $P_01=h_0$. Using that $\pi (1)=m_{{\bar{\phi }}}$ we also have $P_01=\mu _{{\bar{\phi }}}$. Under (H1), the standard theory summarised in Sect. 3 applies ${\mathcal {P}}({\bar{\phi }})=\log \lambda =0$. So, under (H1), $\mu _{{\bar{\phi }}}$ is an equilibrium (probability) measure for $(F,{\bar{\phi }})$. Further, via (3.4),

$$\begin{aligned} \mu _\phi =\frac{\mu _{{\bar{\phi }}}\times m}{(\mu _{{\bar{\phi }}}\times m)(Y^\tau )} \end{aligned}$$

gives an equilibrium (probability) measure for $(f_t, \phi )$.

4.2 Transfer operator R defined w.r.t. the equilibrium measure $\mu _{{\bar{\phi }}}$

Given the equilibrium measure $\mu _{{\bar{\phi }}}\in {{\mathcal {B}}}$, we let $R:L^1(\mu _{{\bar{\phi }}})\rightarrow L^1(\mu _{{\bar{\phi }}})$ be the transfer operator for the first return map $F:Y\rightarrow Y$ w.r.t.  the invariant measure $\mu _{{\bar{\phi }}}$ given by $\int _Y R v\,w\,d \mu _{{\bar{\phi }}}= \int _Y v\,w\circ F\,d \mu _{{\bar{\phi }}}$ for every bounded measurable w.

Under (H1)(i)–(iv), we further assume

(H1)(v):

The transfer operator R maps continuously from $C^\alpha$ to ${{\mathcal {B}}}$ and R admits a continuous extension to an operator from ${{\mathcal {B}}}$ to ${{\mathcal {B}}}$, which we still call R.

By (H1)(iv) and (H1)v, the operator $R:{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}$ has a simple eigenvalue at 1 and the rest of the spectrum is contained in a disc of radius less than 1. While the spectra of $R_0$ and R are the same, the spectral projection $P=\lim _{n\rightarrow \infty }R^n$ acts differently on ${{\mathcal {B}}},{{\mathcal {B}}}_w$. In particular, $P1=1$, while $P_01=\mu _{{\bar{\phi }}}$, $P_01=h_0$. Throughout the rest of this paper, for any $g\in C^\alpha$, we let

$$\begin{aligned} \langle g, 1\rangle :=\langle g, P_0 1\rangle _0=\langle P_0 1, g\rangle _0=\int _Y g\, d \mu _{{\bar{\phi }}} \end{aligned}$$

and note that $P_0g=h\langle g, 1\rangle _0$ and $Pg=\langle g, 1\rangle$.

4.3 Further assumptions on the transfer operator R

Given R as defined in Sect. 4.2, for $u\ge 0$ and $\tau :Y\rightarrow {{\mathbb {R}}}_{+}$ we define the perturbed transfer operator

$$\begin{aligned} {\hat{R}}(u) v := R(e^{-u\tau }v). \end{aligned}$$

By [BMT, Proposition 4.1], a general proposition on twisted transfer operators that holds independently of the specific properties of F, we can write for sufficiently small positive u,

$$\begin{aligned} {\hat{R}}(u) = r_0(u)\int _0^\infty M(t)\, e^{-ut}\, dt \quad \text {with} \quad M(t) := R(\omega (t-\tau )), \end{aligned}$$

(4.1)

where $\omega :{{\mathbb {R}}}\rightarrow [0, 1]$ is an integrable function with ${\text {supp}}\omega \subset [-1,1]$ and $r_0$ is analytic such that $r_0(0)=1$.

Remark 4.1

In fact, $r_0(u)=1/{\hat{\omega }}(u)$ for ${\hat{\omega }}(u)=\int _{-1}^1 e^{-ut} \omega (t)\, dt$. Therefore $\frac{d}{du}r_0(u)$ and $\frac{d}{du}r_0(u)$ are continuous functions, and for later use, we set $\gamma _1 := \frac{d}{du} r_0(0)$, $\gamma _2 := \frac{d^2}{du^2} r_0(0)$.

Since it is short, for the reader’s convenience we include the proof of (4.1) (as in the proof of [BMT, Proposition 4.1]).

Proof

Let $\omega$ be an integrable function supported on $[-1, 1]$ such that $\int _{-1}^1 \omega (t)\, dt = 1$, and set ${\hat{\omega }}(u)=\int _{-1}^1 e^{-ut} \omega (t)\, dt$. Note that ${\hat{\omega }}(u)$ is analytic and ${\hat{\omega }}(0)=1$. Since $\tau \ge h \ge 1$ (see the assumption on h in Sect. 2.2) and ${\text {supp}}\omega \subset [-1,1]$,

$$\begin{aligned} \int _0^\infty \omega (t-\tau ) e^{-ut}\, dt=e^{-u\tau }\int _{-\tau }^\infty \omega (t)\, e^{-ut}\, dt=e^{-u\tau }{\hat{\omega }}(u). \end{aligned}$$

Hence,

$$\begin{aligned} \int _0^\infty R(\omega (t-\tau )v) e^{-ut}\, dt=R\left( \int _0^\infty \omega (t-\tau ) v e^{-ut}\, dt\right) ={\hat{\omega }}(u){\hat{R}}(u) v. \end{aligned}$$

Formula (4.1) follows with $r_0(u)=1/{\hat{\omega }}(u)$, so $r_0(0) = 1$. $\square$

For most results we require that

(H2):

There exists a function $\omega$ satisfying (4.1) and $C_\omega > 0$ such that

(i):

for any $t\in {{\mathbb {R}}}_{+}$ and for all $v\in C^\alpha$, $h\in {{\mathcal {B}}}$, we have $\omega (t-\tau )h \in {{\mathcal {B}}}_w$ and

$$\begin{aligned} \langle \omega (t-\tau )vh, 1\rangle \le C_\omega \Vert v\Vert _{C^\alpha }\Vert h\Vert _{{{\mathcal {B}}}_w}\mu _{{\bar{\phi }}}(\omega (t-\tau )); \end{aligned}$$

(ii):

for all $T>0$,

$$\begin{aligned} \int _T^\infty \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt\le C_\omega \mu _{{\bar{\phi }}}(\tau >T). \end{aligned}$$

Further, we assume the usual Doeblin-Fortet inequality:

(H3):

There exist $\sigma _0>1$ and $C_0, C_1>0$ such that for all $h\in {{\mathcal {B}}}$, for all $n\in {{\mathbb {N}}}$ and for all $u>0$,

$$\begin{aligned} \Vert {\hat{R}}(u)^n h\Vert _{{{\mathcal {B}}}_w}\le C_1 e^{-un} \Vert h\Vert _{{{\mathcal {B}}}_w},\quad \Vert {\hat{R}}(u)^n h\Vert _{{{\mathcal {B}}}} \le C_0 e^{-un} \sigma _0^{-n}\Vert h\Vert _{{{\mathcal {B}}}}+C_1\Vert h\Vert _{{{\mathcal {B}}}_w}. \end{aligned}$$

4.4 Refined assumptions on $\tau$

For the purpose of obtaining limit laws for F (and in the end $f_t$) we assume that

(H4):

One of the following holds as $t\rightarrow \infty$,

(i):

$\mu _{{\bar{\phi }}}(\tau >t)=\ell (t) t^{-\beta }$, for some slowly varying function $\ell$ and $\beta \in (1,2]$. When $\beta =2$, we assume $\tau \notin L^2(\mu _{{\bar{\phi }}})$ and $\ell$ is such that the function ${\tilde{\ell }}(t)=\int _1^t\frac{\ell (x)}{x}\, dx$ is unbounded and slowly varying.

(ii):

$\tau \in L^2(\mu _{{\bar{\phi }}})$. We do not assume³$\tau \in {{\mathcal {B}}}$, but require that for any $h\in {{\mathcal {B}}}$ and $\kappa =1,2$, $R(\tau ^\kappa h)\in {{\mathcal {B}}}$.

5 Limit Laws for $(\tau ,F)$

In this section we obtain limit theorems for the Birkhoff sum $\tau _n=\sum _{j=0}^{n-1}\tau \circ F^j$. Under the abstract assumptions formulated in Sect. 4, we obtain the asymptotics of the leading eigenvalue of ${\hat{R}}(u)$ (as in Sects. 5.1 and 5.2 below). In turn, as clarified in Sect. 5.3, the asymptotics of this (family of) eigenvalues give the asymptotics of the Laplace transform $\mathbb {E}_{\mu _{{\bar{\phi }}}} (e^{-ub_n^{-1}\tau _n})$, for suitable normalising sequences $b_n$, proving the claimed limit theorems. Our result reads as follows.

Proposition 5.1

Assume (H1) and (H2). The following hold as $n\rightarrow \infty$, w.r.t. any probability measure $\nu$ such that $\nu \ll \mu _{{\bar{\phi }}}$.

• (i) Assume (H4)(i). When $\beta <2$, set $b_n$ such that $\frac{n\ell (b_n)}{b_n^\beta }\rightarrow 1$. Then $\frac{1}{b_n}(\tau _n- \frac{\tau ^* \cdot n}{\nu (Y)})\rightarrow ^d G_\beta$, where $G_\beta$ is a stable law of index $\beta$.

  When $\beta =2$, set $b_n$ such that $\frac{n{\tilde{\ell }}(b_n)}{b_n^2}\rightarrow c>0$. Then $\frac{1}{b_ n}(\tau _n- \frac{\tau ^* \cdot n}{\nu (Y)})\rightarrow ^d {{\mathcal {N}}}(0,c)$.

• (ii) Assume (H4)(ii) and $\tau -\tau ^* \ne h - h \circ F$ for any $h \in {{\mathcal {B}}}$. Then there exists $\sigma \ne 0$ such that $\frac{1}{\sqrt{n}}(\tau _n-\frac{\tau ^* \cdot n}{\nu (Y)})\rightarrow ^d {{\mathcal {N}}}(0,\sigma ^2)$.

Note here that the slowly varying function $\ell$ from the general theory later on reduces to $\ell (n) = C^*$, because the tail estimates (2.3) have this form.

Remark 5.2

The above result holds for all piecewise $C^1(Y)$ observables v in the space ${{\mathcal {B}}}_0(Y)$ defined in “Appendix A.3”. For item (i), we need to assume that v is in the domain of a stable law with index $\beta \in (1,2]$ with $v\notin L^2(\mu _{{\bar{\phi }}})$ and adjusted $C^* = C^*(v)$, while for item (ii) we assume $v\in L^2(\mu _{{\bar{\phi }}})$ and a non-coboundary condition precisely formulated in Proposition 6.1 (ii).

The rest of this section is devoted to the proof of the above result.

5.1 Family of eigenvalues for ${\hat{R}}(u)$

By Eq. (4.1), Remark 4.1 and (H2)v(ii), ${\hat{R}}(u)$ is an analytic family of bounded linear operators on ${{\mathcal {B}}}_w$ for $u>0$ and this family can be continuously extended as $u\rightarrow 0$. As in Remark 4.1, $\lim _{u\rightarrow 0}\frac{d}{du}r_0(u), \frac{d^2}{du^2}r_0(u)$ exist and we let $\gamma _1 := \frac{d}{du} r_0(0)$, $\gamma _2 := \frac{d^2}{du^2} r_0(0)$.

Lemma 5.3

Assume (H2). Then

$$\begin{aligned} \Vert {\hat{R}}(u)-{\hat{R}}(0)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\ll u+\mu _{{\bar{\phi }}} (\tau >1/u)=:q(u). \end{aligned}$$

Proof

Using (4.1), we write

$$\begin{aligned} {\hat{R}}(u)&=(r_0(u)-1)\int _0^\infty M(t) e^{-ut}\, dt+r_0(0)\int _0^\infty M(t) (e^{-ut}-1)\, dt+r_0(0)\int _0^\infty M(t)\, dt\\&=(r_0(u)-1)\int _0^\infty M(t) e^{-ut}\, dt+\int _0^\infty M(t) (e^{-ut}-1)\, dt+{\hat{R}}(0). \end{aligned}$$

Since $|(r_0(u)-1|\le u|\gamma _1|\ll u$, we have that as $u\rightarrow 0$,

$$\begin{aligned} \Vert {\hat{R}}(u)-&{\hat{R}}(0)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\ll u\int _0^\infty \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt+\int _0^{1/u}(e^{-ut}-1)\Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt\\&+\int _{1/u}^\infty \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt\ll u + \mu _{{\bar{\phi }}} (\tau >1/u)+u\int _0^{1/u}t\Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt. \end{aligned}$$

We continue using (H2)(ii). Let $S(t)=\int _{t}^\infty \Vert M(x)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dx$ and note that

$$\begin{aligned} \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\ll \int _t^{t+1}\Vert M(x)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dx \ll S(t+1)-S(t). \end{aligned}$$

(5.1)

Rearranging,

$$\begin{aligned} u&\int _0^{1/u}t\Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt \ll u\int _0^{1/u}t(S(t+1)-S(t))\, dt\\&=u\Big (\int _1^{1/u+1}(t-1)S(t))\, dt-\int _0^{1/u}tS(t)\, dt\Big )\\&=u\int _0^1 tS(t)\, dt-u\int _1^{1/u+1}S(t)\, dt-u\int _{1/u}^{1/u+1}t S(t)\, dt\ll u+S(1/u). \end{aligned}$$

By (H2)(ii), $S(1/u)\ll \mu _{{\bar{\phi }}} (\tau >1/u)$ and the conclusion follows. $\square$

By (H1)(iv), 1 is an eigenvalue for ${\hat{R}}(0)$. By (H3), there exists a family of eigenvalues $\lambda (u)$ well-defined for $u\in [0,\delta _0)$, for some $\delta _0>0$. Also, for $u\in [0,\delta _0)$, ${\hat{R}}(u)$ has a spectral decomposition

$$\begin{aligned} {\hat{R}}(u)=\lambda (u) P(u)+ Q(u),\quad \Vert Q(u)^n\Vert _{{{\mathcal {B}}}}\le C\sigma _0^n, \end{aligned}$$

(5.2)

for all $n\in {{\mathbb {N}}}$, some fixed $C>0$ and some $\sigma _0<1$. Here, P(u) is a family of rank one projectors and we write $P(u)=\mu _{{\bar{\phi }}}(u)\otimes v(u)$, normalising such that $\mu _{{\bar{\phi }}}(u)(v(u))=1$, where $\mu _{{\bar{\phi }}}(0)$ is the invariant measure and $v(0)=1\in {{\mathcal {B}}}$. Equivalently, we normalise such that $\langle v(u),1\rangle =1$ and write

$$\begin{aligned} \lambda (u)=\langle {\hat{R}}(u)v(u), 1\rangle . \end{aligned}$$

(5.3)

The result below gives the continuity of the families P(u) and v(u) (in ${{\mathcal {B}}}_w$).

Lemma 5.4

Assume (H1), (H2) and (H3). Let q(u) be as defined in Lemma 5.3. Then there exist $0 < \delta _1\le \delta _0$ and $C>0$ such that

$$\begin{aligned} \Vert P(u)-P(0)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\le C q(u)|\log (q(u))| \end{aligned}$$

for all $0 \le u < \delta _1$. The same continuity estimate holds for v(u) (viewed as an element of ${{\mathcal {B}}}_w$).

Proof

Given the continuity estimate in Lemma 5.3, we can apply [KL99, Remark 5] (an improved version of [KL99, Corollary 1] under less general assumptions). Assumption (H3) here corresponds to [KL99, Hypotheses 2, 3] and the estimate in Lemma 5.3 corresponds to [KL99, Hypothesis 5]. The extra assumptions of [KL99, Remark 5] are satisfied in the present setup, since by (H3), ${\hat{R}} (u)$ is a family of $\Vert \cdot \Vert _{{{\mathcal {B}}}}$ contractions. Going from P(u) to v(u) is a standard step (see, for instance, [LT16, Proof of Lemma 3.5]).   $\square$

Since $\lambda (0)=1$ is a simple eigenvalue (by (H1)), Lemma 5.4 ensures that $\lambda (u)$ is a family of simple eigenvalues (see [KL99, Remark 4]). The next result gives precise information on the asymptotic and continuity properties of $\lambda (u)$, $u\rightarrow 0$ (a version of the results in [AD01a]) in the present abstract framework.

Lemma 5.5

Assume (H1) and (H2). Let q(u) be as defined in Lemma 5.3. Set $\Pi (u)=\mu _{{\bar{\phi }}} (1-e^{-u\tau })$. Then as $u\rightarrow 0$,

$$\begin{aligned} 1-\lambda (u) =\Pi (u)\bigg (1+O\big (q(u)|\log (q(u))|\big )\bigg ). \end{aligned}$$

Proof

By (5.3), $\lambda (u) v(u)={\hat{R}}(u)v(u)$, $u\in [0,\delta _0)$ with $\lambda (0)=1$ and $v(0)=\mu _{{\bar{\phi }}}$. Then

$$\begin{aligned} \lambda (u)=\langle {\hat{R}}(u)v(u),1\rangle =\mu _{{\bar{\phi }}}(e^{-u\tau })+\langle ({\hat{R}}(u)-{\hat{R}}(0))(v(u)-v(0)),1\rangle . \end{aligned}$$

because $\langle v(u),1\rangle =1$. Since (H2) holds (so, (4.1) holds),

$$\begin{aligned} \lambda (u) = \mu _{{\bar{\phi }}}(e^{-u\tau })+ V(u)&:= \mu _{{\bar{\phi }}}(e^{-u\tau })+ \int _0^\infty (e^{-ut}-1) \langle \omega (t-\tau ) [v(u)-v(0)],1\rangle \, dt\\&\quad +(r_0(u)-1)\int _0^\infty e^{-ut}\langle \omega (t-\tau ) [v(u)-v(0)],1\rangle \, dt. \end{aligned}$$

Thus, $1-\lambda (u)=\Pi (u)-V(u)$. By (H2)(i),

$$\begin{aligned} \langle \omega (t-\tau )[v(u)-v(0)], 1\rangle \le C \Vert v(u)-v(0)\Vert _{{{\mathcal {B}}}_w}\mu _{{\bar{\phi }}}(\omega (t-\tau )). \end{aligned}$$

Recalling that ${\text {supp}}\omega \subset [-1,1]$, for all $u\in [0,\delta _0)$ and some $C_1>0$,

$$\begin{aligned}&\left| \int _0^\infty e^{-ut}\langle \omega (t-\tau ) [v(u)-v(0)],1\rangle \, dt \right| \\&\quad \le C \Vert v(u)-v(0)\Vert _{{{\mathcal {B}}}_w} \left| \int _0^\infty e^{-ut}\int _Y \omega (t-\tau )\, d\mu _{{\bar{\phi }}}\, dt\right| \\&\quad \le C \Vert v(u)-v(0)\Vert _{{{\mathcal {B}}}_w}\left| \int _Y \int _{\tau -1}^{\tau +1} e^{-ut}\omega (t-\tau )\, dt\, d\mu _{{\bar{\phi }}}\right| \\&\quad \le C \Vert v(u)-v(0)\Vert _{{{\mathcal {B}}}_w} \int _Ye^{-u\tau }\left| \int _{-1}^{1}\omega (t)e^{ut}\, dt \right| \, d\mu _{{\bar{\phi }}}\\&\quad \le C_1 \Vert v(u)-v(0)\Vert _{{{\mathcal {B}}}_w} \int _Ye^{-u\tau }\, d\mu _{{\bar{\phi }}}=C_1 \Vert v(u)-v(0)\Vert _{{{\mathcal {B}}}_w} (\Pi (u)+1). \end{aligned}$$

In the previous to last inequality, we have used that $\Big |\int _{-1}^{1}\omega (t)e^{ut}\, dt \Big | \ll 1$ for $u \in [0,\delta _0)$. By a similar argument,

$$\begin{aligned} \Big |\int _0^\infty (e^{-ut}-1)\langle \omega (t-\tau )&[v(u)-v(0)],1\rangle \, dt \Big |\le C_1 \Vert v(u)-v(0)\Vert _{{{\mathcal {B}}}_w}\Pi (u). \end{aligned}$$

The previous two displayed equations imply that for $u\in [0,\delta _0)$,

$$\begin{aligned} |V(u)|&\le C_1 \Vert v(u)-v(0)\Vert _{{{\mathcal {B}}}_w} |r_0(u)-1|(\Pi (u)+1) +C_1 \Vert v(u)-v(0)\Vert _{{{\mathcal {B}}}_w}\Pi (u)\\&\le C_1\Vert v(u)-v(0)\Vert _{{{\mathcal {B}}}_w}\Pi (u). \end{aligned}$$

By Lemma 5.4, $\Vert v(u)-v(0)\Vert _{{{\mathcal {B}}}_w}=O(q(u)|\log (q(u))|)$. Hence $V(u)=O(\Pi (u)q(u)|\log (q(u)|)$. $\square$

A first consequence of the above result is

Corollary 5.6

Assume (H1) and (H2). The following hold as $u\rightarrow 0$.

• (i) If (H4)(i) holds with $\beta <2$, then $1-\lambda (u)=\tau ^* u+u^\beta \ell (1/u)(1+o(1))$.

• (ii) If (H4)(i) holds with $\beta =2$, then $1-\lambda (u)=\tau ^* u+u^2 L(1/u)(1+o(1))$, where $L(t)=\frac{1}{2}{\tilde{\ell }}(t)$.

Proof

Under (H4)(i) with $\beta <2$, $\Pi (u)=\tau ^* u+u^\beta \ell (1/u)(1+o(1))$; we refer to, for instance, [F66, Ch. XIII] and [AD01b]. This together with Lemma 5.5 (with $q(u)\sim u$) ensures that $1-\lambda (u)=\tau ^* u+u^\beta \ell (1/u)+O(u^2\log (1/u))+O(u^{\beta +1}\ell (1/u))$. Item (i) follows.

Under (H4)(i) with $\beta =2$, it follows from [AD01b, Theorem 3.1 and Corollary 3.7] that $\Pi (u)=\tau ^* u+u^2 L(1/u)(1+o(1))$, where $L(t)\sim \ell (t)\log (t)$ with ${\tilde{\ell }}(t) = \int _1^t \ell (x)/x \, dx$ as in (H4)(i) for $\beta =2$. The estimate for $\Pi (u)$ together with Lemma 5.5 (with $q(u)\sim u$) implies that $1-\lambda (u)=\tau ^* u+u^2L(1/u)+ O(u\log (1/u)\Pi (u))$.

In the case of (H4)(i) with $\beta =2$ and $\ell (1/u)\rightarrow \infty$ as $u\rightarrow 0$, we have $\log (1/u)=o(L(1/u))$. As a consequence, $u\log (1/u)\Pi (u)=O(u^{2}\log (1/u))=o(u^2L(1/u))$ and the conclusion follows. In the general case (which allows $\ell$ to be asymptotically constant), for fixed small $\delta >0$ and $\epsilon >0$, we write

$$\begin{aligned} u\log (1/u)\Pi (u)&=u\log (1/u)\Big (\Pi (u)-\int _{\tau \le \delta /u} (e^{-u^{2+\epsilon }\tau ^{2+\epsilon }}-1)\,d\mu _{{\bar{\phi }}}\Big )\\&\quad +u\log (1/u)\int _{\tau \le \delta /u} (e^{-u^{2+\epsilon }\tau ^{2+\epsilon }}-1)\,d\mu _{{\bar{\phi }}}= u\log (1/u)(I_1(u)+I_2(u)). \end{aligned}$$

First,

$$\begin{aligned} u\log (1/u)|I_2(u)|&\ll u\log (1/u)u^{2+\epsilon }\int _{\tau \le \delta /u}\tau ^{2+\epsilon }\, d\mu _{{\bar{\phi }}}\\&\le \log (1/u)u^{3+\epsilon }u^{-2\epsilon }\delta ^{2\epsilon }\int _Y\tau ^{2-\epsilon }\, d\mu _{{\bar{\phi }}}=o(u^2L(1/u)). \end{aligned}$$

Next, compute that

$$\begin{aligned} I_1(u)&=\int _{\tau \le \delta /u} (e^{-u\tau }-e^{-u^{2+\epsilon }\tau ^{2+\epsilon }})\,d\mu _{{\bar{\phi }}}+\int _{\tau> \delta /u} (e^{-u\tau }-1)\,d\mu _{{\bar{\phi }}}\\&=\int _{\tau \le \delta /u} (e^{-u\tau }-e^{-u^{2+\epsilon }\tau ^{2+\epsilon }})\,d \mu _{{\bar{\phi }}}+O(\mu _{{\bar{\phi }}}(\tau >\delta /u))\\&=\int _{\tau \le \delta /u} e^{-u\tau }(1-e^{-u^{1+\epsilon }\tau ^{1+\epsilon }})\,d \mu _{{\bar{\phi }}}+O(u^{2}\ell (\delta /u)). \end{aligned}$$

Let $G(x)=\mu _{{\bar{\phi }}}(\tau <x)$ and compute that

$$\begin{aligned} \int _{\tau \le \delta /u}&e^{-u\tau }(1-e^{-u^{1+\epsilon }\tau ^{1+\epsilon }}) \,d\mu _{{\bar{\phi }}}\ll u^{1+\epsilon }\int _{\tau \le \delta /u} \tau ^{1+\epsilon } e^{-u\tau }\,d\mu _{{\bar{\phi }}}\\&\ll u^{1+\epsilon }\int _0^\infty x^{1+\epsilon }e^{-ux}\, dG(x)= -u^{1+\epsilon }\int _0^\infty x^{1+\epsilon }e^{-ux}\, d(1-G(x))\\&=u^{2+\epsilon }\int _0^\infty x^{1-\epsilon }(1-G(x))\, e^{-ux}\, dx+u^{1+\epsilon }\int _0^\infty x^{-\epsilon }(1-G(x))\, e^{-ux}\, dx\ll u^{1+\epsilon }. \end{aligned}$$

Altogether, $I_1(u)\ll u^{1+\epsilon }$. Hence, $u\log (1/u)|I_1(u)|=o(u^2L(1/u))$. This together with the estimate for $u\log (1/u)|I_2(u)$ implies that $u\log (1/u)\Pi (u)=o(u^2L(1/u))$ and item (ii) follows. $\square$

5.2 Estimates required for the CLT under (H4)(ii)

For the CLT case we need the following

Proposition 5.7

Assume (H1), (H2) and (H4)(ii). Suppose that $\tau \ne h\circ F-h$, for any $h\in {{\mathcal {B}}}$. Then there exists $\sigma \ne 0$ such that $1-\lambda (u)=\tau ^* u+\frac{\sigma ^2}{2}u^2(1+o(1))$.

We need to ensure that under the assumptions of Proposition 5.7, which do not require that $\tau \in {{\mathcal {B}}}$, there exists the required $\sigma \ne 0$. The argument goes by and large as [G04, Proof of Theorem 3.7] (which works with a different Banach space) with the exception of estimating the second derivative of the eigenvalue ${\tilde{\lambda }}$ defined below. The argument in [G04, Proof of Theorem 3.7] for estimating such a derivative does not directly apply to our setup due to: (i) the two Banach spaces ${{\mathcal {B}}}, {{\mathcal {B}}}_w$ at our disposal are not embedded in $L^p$, $p>1$; (ii) the presence of $\log q(u)$ in Lemma 5.4. Our estimates below rely heavily on (H2) and (H4)(ii).

As usual, we can can reduce the proof to the case of mean zero observables. Let ${\tilde{\tau }}=\tau -\tau ^*$. We recall that under (H4)(ii), $R(\tau h)\in {{\mathcal {B}}}$, for every in $h\in {{\mathcal {B}}}$ and the same holds for ${\tilde{\tau }}$. By H1(v), $(I-R)$ is invertible on the space of functions inside ${{\mathcal {B}}}$ of zero integral. Thus, as in [G04, Proof of Theorem 3.7], we can set

$$\begin{aligned} \chi := (I-R)^{-1}R{\tilde{\tau }} \in {{\mathcal {B}}}. \end{aligned}$$

(5.4)

Define ${\tilde{R}}(u)=R(e^{-u{\tilde{\tau }}})$. Clearly, ${\tilde{R}}(u)$ has the same continuity properties as ${\hat{R}}(u)$. Let ${\tilde{\lambda }}(u)$ be the associated family of eigenvalues. Recall that $\lambda (u)$ is the family of eigenvalues associated with ${\hat{R}}(u)$. Hence, $\lambda (u)=e^{u\tau ^*}{\tilde{\lambda }}(u)$. As in the previous sections, let v(u) be the family of associated eigenvectors normalised such that $\langle v(u),1\rangle =1$. The next three results are technical tools required in the proof of Proposition 5.7; the third, which has a longer proof, is postponed to Sect. 5.2.1.

Lemma 5.8

Suppose that (H1)–(H3) and (H4)(ii) hold, and recall ${\tilde{\tau }}=\tau -\tau ^*$. Then

$$\begin{aligned} \frac{d^2}{du^2}{\tilde{\lambda }} (u)=-\int _Y{\tilde{\tau }}^2\, d\mu _{{\bar{\phi }}} - 2\left\langle \frac{d}{du}{\tilde{R}}(u)\frac{d}{du}v(u) , 1\right\rangle +T(u), \end{aligned}$$

where $T(u)\rightarrow 0$, as $u\rightarrow 0$.

Proof

Set ${\tilde{\Pi }} (u)=\mu _{{\bar{\phi }}}(1-e^{-u{\tilde{\tau }}})$. By the calculation used in the proof of Lemma 5.5, $1-{\tilde{\lambda }}(u)={\tilde{\Pi }}(u)-\langle ({\tilde{R}}(u)-{\tilde{R}}(0))({\tilde{v}}(u)-v(0)),1\rangle$. Differentiating twice,

$$\begin{aligned} \frac{d^2}{du^2}{\tilde{\lambda }} (u)-\frac{d^2}{du^2}{\tilde{\Pi }} (u)&= \left\langle \frac{d^2}{du^2}{\tilde{R}}(u)(v(u)-v(0),1\right\rangle +2\left\langle \frac{d}{du}{\tilde{R}}(u)\frac{d}{du} v(u),1\right\rangle \\&\quad +\left\langle ({\tilde{R}}(u)-{\tilde{R}}(0))\frac{d^2}{du^2} v(u),1\right\rangle . \end{aligned}$$

Under (H2)(i) and (H4)(ii), arguments similar to the ones used in Lemma 5.5 imply that as $u\rightarrow 0$,

$$\begin{aligned} \left| \left\langle ({\tilde{R}}(u)-{\tilde{R}}(0))\frac{d^2}{du^2} v(u),1\right\rangle \right|&\ll \left\| \frac{d^2}{du^2}v(u)\right\| _{{{\mathcal {B}}}_w} \int _0^\infty (e^{-ut}-1) \mu _{{\bar{\phi }}}(\omega (t-{\tilde{\tau }}))\, dt\\&+|r_0(u)-1|\left\| \frac{d^2}{du^2}v(u)\right\| _{{{\mathcal {B}}}_w} \int _0^\infty e^{-ut} \mu _{{\bar{\phi }}}(\omega (t-{\tilde{\tau }}))\, dt\ll u\log (1/u)=o(1). \end{aligned}$$

(H4)(ii) and (5.4) imply that $| \langle \frac{d^2}{du^2}{\tilde{R}}(u)(v(u)-v(0),1\rangle |\ll u\log (1/u)=o(1)$. Finally, $\frac{d^2}{du^2}{\tilde{\Pi }} (u)=-\int _Y{\tilde{\tau }}^2\, d\mu _{{\bar{\phi }}}(1+o(1))$. The conclusion follows by putting the above estimates together. $\square$

Lemma 5.9

Assume the setup of Proposition 5.7. Then

$$\begin{aligned} \left\langle \frac{d}{du}{\tilde{R}}(u)\frac{d}{du}v(u), 1\right\rangle =-\int {\tilde{\tau }} \chi \, d\mu _{{\bar{\phi }}}+D(u), \end{aligned}$$

where $D(u)=o(1)$ as $u\rightarrow 0$, and $\chi$ is defined as in (5.4).

Proof of Proposition 5.7

By Lemmas 5.8 and 5.9,

$$\begin{aligned} \frac{d^2}{du^2}{\tilde{\lambda }}(u)=-\int _Y{\tilde{\tau }}^2\, d\mu _{{\bar{\phi }}}-2\int {\tilde{\tau }} \chi \, d\mu _{{\bar{\phi }}}+{\tilde{T}}(u), \end{aligned}$$

(5.5)

where ${\tilde{T}}(u)\rightarrow 0$, as $u\rightarrow 0$. Hence, $-\frac{d^2}{du^2}{\tilde{\lambda }}(u)|_{u=0}=\int _Y{\tilde{\tau }}^2\, d\mu _{{\bar{\phi }}}+2\int {\tilde{\tau }} \chi \, d\mu _{{\bar{\phi }}}$ and we can set $\sigma ^2=\int _Y{\tilde{\tau }}^2\, d\mu _{{\bar{\phi }}}+2\int {\tilde{\tau }} \chi \, d\mu _{{\bar{\phi }}}$. From here on the proof goes word for word as [G04, Proof of Theorem 3.7], which shows that given the previous formula for $\sigma$, the only possibility for $\sigma =0$ is when ${\tilde{\tau }}=h-h\circ F$, for some density $h\in {{\mathcal {B}}}$. This is ruled out by assumption.

To conclude recall that $\lambda (u)=e^{u\tau ^*}{\tilde{\lambda }}(u)$. By (5.5), $1-{\tilde{\lambda }}(u)=\frac{\sigma ^2}{2}u^2(1+o(1))$, as required. $\square$

5.2.1 Proof of Lemma 5.9

Proof of Lemma 5.9

Let $W(u):=\frac{d}{du}{\tilde{R}}(u)$. Since ${\tilde{R}}(u) = R(e^{-u {\tilde{\tau }}})$, we have that for any $h\in {{\mathcal {B}}}$,

$$\begin{aligned} W(0)h:=\left. \frac{d}{du}{\tilde{R}}(u)\right| _{u=0} h=-R({\tilde{\tau }} h). \end{aligned}$$

(5.6)

By (H4)(ii), $W(0)h\in {{\mathcal {B}}}$. Recall that P(u) is the eigenprojection for ${\hat{R}}(u)$, so for ${\tilde{R}}(u)$ as well. For $\delta$ small enough (independent of u), we can write

$$\begin{aligned} P(u)=\int _{|\xi -1|=\delta }(\xi I-{\tilde{R}}(u))^{-1}\, d\xi . \end{aligned}$$

Recall that $v(u)=\frac{P(u)1}{\langle P(u)1, 1\rangle }=\frac{P(u)1}{m_{{\bar{\phi }}}(u)1}$; in particular, $\mu _{{\bar{\phi }}}(u)1=\langle P(u)1, 1\rangle$. Compute

$$\begin{aligned} \frac{d}{du}v(u)=\frac{\frac{d}{du}P(u)1}{\mu _{{\bar{\phi }}}(u)1}-\frac{P(u)1}{(\mu _{{\bar{\phi }}} (u)1)^2}\frac{d}{du}\mu _{{\bar{\phi }}}(u)1. \end{aligned}$$

(5.7)

Also, compute that by (5.6)

$$\begin{aligned} \frac{d}{du} P(u)&=\int _{|\xi -1|=\delta }(\xi I-{\tilde{R}}(u))^{-1}W(u)(\xi I-{\tilde{R}}(u))^{-1}\, d\xi \nonumber \\&=\int _{|\xi -1|=\delta }(\xi I-R)^{-1}W(0)(\xi I-R)^{-1}\, d\xi +Z_0(u)\nonumber \\&=-\int _{|\xi -1|=\delta }(\xi I-R)^{-1}R{\tilde{\tau }}(\xi I-R)^{-1}\, d\xi +Z_0(u), \end{aligned}$$

(5.8)

where the first term is well defined in ${{\mathcal {B}}}$ by (H4)(ii) and

$$\begin{aligned} \Vert Z_0(u)\Vert _{{{\mathcal {B}}}}&\ll \left\| \int _{|\xi -1|=\delta } (\xi I-R)^{-1} \Big (W(u)-W(0)\Big )(\xi I-R)^{-1}\, d\xi \right\| _{{{\mathcal {B}}}}\\&\quad + \left\| \int _{|\xi -1|=\delta }\Big ((\xi I-R(u))^{-1}-(\xi I-R)^{-1}\Big )W(u)(\xi I-R)^{-1}\, d\xi \right\| _{{{\mathcal {B}}}}\\&\quad + \left\| \int _{|\xi -1|=\delta }(\xi I-R)^{-1}W(u)\Big ((\xi I-R(u))^{-1}-(\xi I-R)^{-1}\Big )\, d\xi \right\| _{{{\mathcal {B}}}}. \end{aligned}$$

Again using (H4)(ii), $\frac{d^2}{du^2}{\tilde{R}}(u)$ is bounded in $\Vert \cdot \Vert _{{{\mathcal {B}}}}$; so, $\Vert W(u)-W(0)\Vert _{{{\mathcal {B}}}}\ll u$. Hence, each one of three terms of the previous displayed equation contains an O(u) factor. By the argument recalled in the proof of Lemma 5.4, for any $\xi$ such that $|\xi -1|=\delta$, the integrand of each one of these three terms is o(1). Altogether, $\Vert Z_0(u)\Vert _{{{\mathcal {B}}}}=o(1)$. Also, by (5.8) and Lemma 5.10 below,

$$\begin{aligned} \frac{d}{du} P(u)1 = -\chi -\xi _0 \int _Y \chi \, d\mu _{{\bar{\phi }}}+Z_0(u), \end{aligned}$$

(5.9)

for some $\xi _0 \in {\mathbb {C}}$ and $\AA = (I-R)^{-1} R {\tilde{\tau }}$ as in (5.4). Plugging (5.9) into (5.7),

$$\begin{aligned} \frac{d}{du}v(u)&=\frac{1}{\mu _{{\bar{\phi }}}(0)1}\left( -\chi -\xi _0 \int _Y \chi \, d\mu _{{\bar{\phi }}}\right) \\&-\frac{P(0)1}{(\mu _{{\bar{\phi }}}(0)1)^2}\left\langle -\chi -\xi _0 \int _Y \chi \, d\mu _{{\bar{\phi }}}, 1\right\rangle +Z_1(u)+Z_0(u)\\&=-\chi +{\tilde{\xi }}\int _Y \chi \, d\mu _{{\bar{\phi }}} +Z_1(u)+Z_0(u), \end{aligned}$$

where ${\tilde{\xi }}$ is linear combination of $\xi _0$ and

$$\begin{aligned} \Vert Z_1(u)\Vert _{{{\mathcal {B}}}_w} \ll&\, \frac{|\mu _{{\bar{\phi }}}(u)1-\mu _{{\bar{\phi }}}(0)1|}{m_{{\bar{\phi }}}(u)1} +\frac{\Vert P(u)1- P(0)1\Vert _{{{\mathcal {B}}}_w}}{(m_{{\bar{\phi }}}(u)1)^2} +\left| \frac{1}{\mu _{{\bar{\phi }}}(u)1}-\frac{1}{\mu _{{\bar{\phi }}}(0)1} \right| \\&+\left| \frac{1}{(\mu _{{\bar{\phi }}}(u)1)^2} -\frac{1}{(\mu _{{\bar{\phi }}}(0)1)^2}\right| . \end{aligned}$$

By Lemma 5.4, $\Vert P(u)- P(0)\Vert _{{{\mathcal {B}}}_w}\ll u\log (1/u)$ and as a consequence, $|\mu _{{\bar{\phi }}}(u)1-\mu _{{\bar{\phi }}}(0)1|\ll u\log (1/u)$. Thus, $\Vert Z_1(u)\Vert _{{{\mathcal {B}}}_w}=o(1)$. Recalling that $\Vert Z_0(u)\Vert _{{{\mathcal {B}}}_w}=o(1)$, we have

$$\begin{aligned} \frac{d}{du}v(u)=-\chi +{\tilde{\xi }}\int _Y \chi \, d\mu _{{\bar{\phi }}} +Z(u), \end{aligned}$$

where $\Vert Z(u)\Vert _{{{\mathcal {B}}}_w}=o(1)$. Using this expression for $\frac{d}{du}v(u)$, we obtain

$$\begin{aligned}&\left\langle \frac{d}{du}{\tilde{R}}(u)\frac{d}{du}v(u),1\right\rangle =-\left\langle R{\tilde{\tau }} \frac{d}{du}v(u),1\right\rangle +\left\langle (W(u)-W(0))\, \frac{d}{du}v(u),1\right\rangle \\&\quad =-\int _Y R{\tilde{\tau }} \chi \, d\mu _{{\bar{\phi }}} +{\tilde{\xi }}\int {\tilde{\tau }} \, d\mu _{{\bar{\phi }}}\int _Y \chi \, d\mu _{{\bar{\phi }}}+\left\langle R Z(u),1\rangle +\langle (W(u)-W(0))\, \frac{d}{du}v(u),1\right\rangle \\&\quad =-\int _Y {\tilde{\tau }} \chi \, d\mu _{{\bar{\phi }}} +\langle R Z(u),1\rangle +\left\langle (W(u)-W(0))\, \frac{d}{du}v(u),1\right\rangle , \end{aligned}$$

where we have used $\int {\tilde{\tau }} \, d\mu _{{\bar{\phi }}}=0$. Finally, using (H2), we compute that

$$\begin{aligned} |\langle R Z(u),1\rangle |\ll \Vert Z(u)\Vert _{{{\mathcal {B}}}_w}\int _0^\infty \int _Y \omega (t-{\tilde{\tau }})\, d\mu _{{\bar{\phi }}}\, dt\ll \Vert Z(u)\Vert _{{{\mathcal {B}}}_w}=o(1). \end{aligned}$$

Similarly, $|\langle (W(u)-W(0)\, w(u),1\rangle |=o(1)$, since we already know that $\Vert W(u)-W(0)\Vert _{{{\mathcal {B}}}_w}=o(1)$. Thus,

$$\begin{aligned} \left\langle \frac{d}{du}{\tilde{R}}(u)\frac{d}{du}v(u),1\right\rangle =-\int _Y {\tilde{\tau }} \chi \, d\mu _{{\bar{\phi }}} +o(1), \end{aligned}$$

ending the proof. $\square$

Lemma 5.10

Assume the setup and notation of Lemma 5.9. Then there exists $\xi _0 \in {\mathbb {C}}$ with such that

$$\begin{aligned} \int _{|\xi -1|=\delta } (\xi I-R)^{-1}R{\tilde{\tau }}\,(\xi I-R)^{-1} 1\, d\xi =\chi +\xi _0 \int _Y \chi \, d\mu _{{\bar{\phi }}}. \end{aligned}$$

Proof

We note that for each $\xi$ such that $\xi -1|=\delta$, $R{\tilde{\tau }}\,(\xi I-R)^{-1} 1$ is well defined in ${{\mathcal {B}}}$ by (H4)(ii) and that $\int _{|\xi -1|=\delta }R{\tilde{\tau }}\,(\xi I-R)^{-1} 1\, d\xi =0$. Hence, $\int _{|\xi -1|=\delta }(I-R)^{-1}R{\tilde{\tau }}\,(\xi I-R)^{-1} 1\, d\xi \in {{\mathcal {B}}}$.

By the first resolvent identity and the Mean Value Theorem,

$$\begin{aligned}& \int _{|\xi -1|=\delta } (\xi I-R)^{-1}R{\tilde{\tau }}\,(\xi I-R)^{-1} 1\, d\xi \\&=\int _{|\xi -1|=\delta }(I-R)^{-1}R{\tilde{\tau }}\,(\xi I-R)^{-1} 1\, d\xi \\&\qquad + \int _{|\xi -1|=\delta }\Big ((\xi I-R)^{-1}-(I-R)^{-1}\Big )R{\tilde{\tau }}\,(\xi I-R)^{-1} 1\, d\xi \\&= \chi P(0)1 + \int _{|\xi -1|=\delta } (1-\xi ) (\xi I-R)^{-1}\chi \,(\xi I-R)^{-1} 1\, d\xi \\&= \chi +\xi _1 \int _{|\xi -1|=\delta }(\xi I-R)^{-1}\chi \,(\xi I-R)^{-1} 1\, d\xi , \end{aligned}$$

where $\xi _1$ is a complex constant with $0<|\xi _1|\le \delta$.

Write $(\xi I-R)^{-1}-I=-(\xi I-R)^{-1}( (\xi -1) I -R)$. The Mean Value Theorem gives

$$\begin{aligned} \int _{|\xi -1|=\delta }&(\xi I-R)^{-1}\chi \,(\xi I-R)^{-1} 1\, d\xi = \int _{|\xi -1|=\delta }(\xi I-R)^{-1}a\,\Big (I+(\xi I-R)^{-1}-I\Big ) 1\, d\xi \\&= \int _{|\xi -1|=\delta }(\xi I-R)^{-1}\chi \, d\xi - \int _{|\xi -1|=\delta }(\xi I-R)^{-1}\chi \,(\xi I-R)^{-1}( (\xi -1) I - R)1\, d\xi \\&=P(0) \chi -\int _{|\xi -1|=\delta }(\xi I-R)^{-1}\chi \, (\xi I-R)^{-1} (\xi -2) \, d\xi \\&=\int _Y \chi \, d\mu _{{\bar{\phi }}}-(\xi _2-2)\int _{|\xi -1|=\delta }(\xi I-R)^{-1}\chi \,(\xi I-R)^{-1} 1\, d\xi , \end{aligned}$$

for some $\xi _2 \in {\mathbb {C}}$ with $|\xi _2-1| \le \delta$. Thus,

$$\begin{aligned} (\xi _2-1)\int _{|\xi -1|=\delta }(\xi I-R)^{-1}\chi \,(\xi I-R)^{-1} 1\, d\xi =\int _Y \chi \, d\mu _{{\bar{\phi }}}. \end{aligned}$$

Since $\int _Y \AA \, d\mu _{{\bar{\phi }}} > 0$, we have $\xi _2 \ne 1$. The conclusion follows with $\xi _0 =\xi _1(\xi _2-1)^{-1}$.

$\square$

5.3 Proof of Proposition 5.1

We start with part (i), i.e., the stable and non standard Gaussian laws. Let $\tau _n$ and $b_n$ be as in Proposition 5.1. Using (5.2) and Corollary 5.6 (based on [AD01b]) we obtain that the Laplace transform $\mathbb {E}_{\mu _{{\bar{\phi }}}} (e^{-ub_n^{-1}\tau _n})$ converges (as $n\rightarrow \infty$) to the Laplace transform of either the stable law or of the ${{\mathcal {N}}}(0,1)$ law. The conclusion w.r.t.  $\mu _{{\bar{\phi }}}$ follows from the theory of Laplace transforms (as in [F66, Ch. XIII]). The conclusion w.r.t.  $\nu$ follows from [E04, Theorem 4].

For part (ii), i.e., the CLT, let $\tau _n$ and $b_n$ be as in Proposition 5.1 under (H4)(ii). Proposition 5.7 shows that the Laplace transform $\mathbb {E}_{\mu _{{\bar{\phi }}}} (e^{-ub_n^{-1}\tau _n})$ converges (as $n\rightarrow \infty$) to the Laplace transform of ${{\mathcal {N}}}(0,\sigma ^2)$. The conclusion w.r.t.  $\mu _{{\bar{\phi }}}$ follows from the theory of Laplace transforms (as in [F66, Ch. XIII]). The conclusion w.r.t.  $\nu$ follows from [E04, Theorem 4].

6 Limit Laws for the Flow $f_t$

The result below generalises Proposition 5.1 to the flow $f_t$. Given an $\alpha$-Hölder observable $g:\mathcal {M}\rightarrow {{\mathbb {R}}}$, define ${\bar{g}}:Y\rightarrow {{\mathbb {R}}}$ as in (3.2) and let $g_T=\int _0^T g\circ f_t\, dt$.

Proposition 6.1

Assume (H1). Let $g:\mathcal {M}\rightarrow {{\mathbb {R}}}$ and let $g^*=\int _Y {\bar{g}}\,d \mu _{{\bar{\phi }}}$. Suppose that the twisted operator ${\hat{R}}_g(u)v=R(e^{-u{\bar{g}}}v)$ satisfies (H2) (with $\tau$ replaced by ${\bar{g}}$). Then the following hold as $T\rightarrow \infty$, w.r.t.  $\mu _{{\bar{\phi }}}$ (or any probability measure $\nu \ll \mu _{{\bar{\phi }}}$).

• (i) Assume (H4)(i) and $\mu _{{\bar{\phi }}}({\bar{g}}>T)\sim \mu _{{\bar{\phi }}}(\tau >T)$.

  When $\beta <2$, set b(T) such that $\frac{T\ell (b(T))}{b(T)^\beta }\rightarrow 1$. Then $\frac{1}{b(T)}(g_T-g^*\cdot T)\rightarrow ^d G_\beta$, where $G_\beta$ is a stable law of index $\beta$.

  When $\beta =2$, set b(T) such that $\frac{T{\tilde{\ell }}(b(T))}{b(T)^2}\rightarrow c>0$. then $\frac{1}{b(T)}(g_T-g^*\cdot T)\rightarrow ^d {{\mathcal {N}}}(0,c)$.

• (ii) Suppose that (H4)(ii) holds. If ${\bar{g}}\notin {{\mathcal {B}}}$, we assume that $R{\bar{g}}\in {{\mathcal {B}}}$. We further assume that ${\bar{g}}\ne h-h\circ F$ with $h\in {{\mathcal {B}}}$. Then there exists $\sigma >0$ such that $\frac{1}{\sqrt{T}}(g_T- g^* \cdot T)\rightarrow ^d {{\mathcal {N}}}(0,\sigma ^2)$.

Remark 6.2

The assumption $\mu _{{\bar{\phi }}}({\bar{g}}>T)\sim \mu _{{\bar{\phi }}}(\tau >T)$ is satisfied for all $g$ bounded above with $\int _0^{h(x)} g\circ f_t(p) \, dt = c > 0$ (where $p \in \Sigma$ is the fixed point of the Poincaré section) and such that $\int _0^{h(x)} g\circ f_t(q) \, dt$ is $C^\alpha$ in q near p. The proof of this is similar to [G04, Proof of Theorem 1.3] where the source of non-hyperbolicity was a neutral fixed point $x_0$ instead of a neutral periodic orbit $\Gamma = \{ p \} \times {{\mathbb {T}}}^1$. Therefore, we just need to replace $g(x_0)$ in [G04, Proof of Theorem 1.3] by $\int _0^{h(p)} g\circ f_t(p) \, dt$

In Theorem 8.2 below, we consider potentials of the form ${\bar{g}}=C'-C\tau ^\kappa (1+o(1))$, $\kappa \in (0,\beta )$ and obtain specific form of (i) and/or (ii) for different values of $\kappa$.

Proof

We use that $f_t:\mathcal {M}\rightarrow \mathcal {M}$ can be represented as a suspension flow $F_t:Y^\tau \rightarrow Y^\tau$. Under the present assumptions, ${\bar{g}}$ satisfies all the assumptions of Proposition 5.1 (with $\tau$ replaced by ${\bar{g}}$).

Let ${\bar{g}}_n=\sum _{j=0}^{n-1}{\bar{g}}\circ F^j$ and recall $\tau _n=\sum _{j=0}^{n-1}\tau \circ F^j$. Proposition 5.1 (i) applies to ${\bar{g}}_n$; the argument goes word for word as in the proof of Proposition 5.1 (i) for $\tau _n$. Under the present assumptions on ${\bar{g}}$, item (ii) of Proposition 5.1 applies to ${\bar{g}}_n$ with the argument used in the proof of Proposition 5.7 applying word for word with ${\bar{g}}$ instead of $\tau$.

Item (i) follows from this together with Lemma 6.3 below (correspondence between stable laws/non standard Gaussian for the base map F and suspension flow. Item (ii) follows in the same way using [MTo04, Theorem 1.3] instead of Lemma 6.3. $\square$

The next result is a version of [S06, Theorem 7] (generalising [MTo04, Theorem 1.3]) for suspension flows which holds in a very general setup; in particular, it is totally independent of method used to prove limit theorems for the base map.

Lemma 6.3

Assume $\int _Y \tau d\mu _{{\bar{\phi }}}<\infty$ and let $g\in L^q(\mu _\phi )$, for $q>1$. Suppose that there exists a sequence $b_n=n^{-\rho }\ell (n)$ for $\rho \in (1,2]$ and $\ell$ a slowly varying function such that $b_n^{-1}\left( \tau _n-n\int _Y \tau d\mu _{{\bar{\phi }}}\right)$ is tight on $(Y, \mu _{{\bar{\phi }}})$. Then the following are equivalent:

• (a) $b_n^{-1}{\bar{g}}_n$ converges in distribution on $(Y,\mu _{{\bar{\phi }}})$.

• (b) $b(T)^{-1} g_T$ converges in distribution on $(Y^\tau , \mu _\phi )$, where $b(T)=T^{-\rho }\ell (T)$.

Proof

The fact the (b) implies (a) is obvious. The implication from (a) to (b) is contained in the proofs of [MTo04, Theorem 1.3] for the standard CLT case and in [S06, Theorem 7] for the stable and nonstandard CLT. We sketch the argument for completeness.

Following [MTo04], let n[x, T] denote the largest integer such that $\tau _n(x)\le T$; that is,

$$\begin{aligned} \tau _{n[x,T]}(x)\le T\le \tau _{n[x,T]+1}(x). \end{aligned}$$

By the ergodic theorem, $\lim _{T\rightarrow \infty }\frac{1}{T}n[x,T]=\int _Y \tau \,d\mu _\phi =: {\bar{\tau }}$, a.e. Hence, $n[x,T]=[{\bar{\tau }} T](1+o(1))$, a.e. as $T\rightarrow \infty$.

Recall that $g:Y^\tau \rightarrow {{\mathbb {R}}}$ and define ${\hat{g}}_T(y,u)=g_T(y)$. Recall ${\bar{g}}(y)=\int _0^{\tau (y)} g(y,u)\, du$. By assumption, $b_n^{-1}{\bar{g}}_n$ converges in distribution on $(Y,\mu _\phi )$. Hence, (as in [MTo04, Lemma 3.1]), $b_n^{-1} {\hat{g}}_n$ converges in distribution on $(Y^\tau , \mu _\phi ^\tau )$; this is a consequence of [E04, Theorem 4].

In what follows we adopt the convention $b_T=b(T)$. Write

$$\begin{aligned} \frac{g_T}{b_T}&=\frac{b_{[{\bar{\tau }} T]}}{b_T}\Big (\frac{{\bar{g}}_{[{\bar{\tau }} T]}}{b_{[{\bar{\tau }} T]}}+ \frac{1}{b_{[{\bar{\tau }} T]}}\Big ({\bar{g}}_{n[x,T]}-{\bar{v}}_{[{\bar{\tau }} T]}\Big )\Big ) +\Big (\frac{{\hat{g}}_T}{b_T}-\frac{g_{n[x,T]}}{b_{n[x,T]}}\Big )\\&\quad +\Big (\frac{{\hat{g}}_{n[x,T]}}{b_{n[x,T]}}-\tau _{n[x,T]}\circ F^{\tau _{n[x,T]}}\Big ). \end{aligned}$$

Since $\ell$ is slowly varying, $\frac{b_{[{\bar{\tau }} T]}}{b_T}\rightarrow {\bar{\tau }}^\rho$. Since $g\in L^1(\mu _\phi ^\tau )$, [S06, Step 2 of the the proof of Theorem 7] (a generalization of [MTo04, Lemma 3.4]) applies and thus, $\frac{1}{b_{[{\bar{\tau }} T]}}\Big ({\bar{g}}_{n[x,T]}-{\bar{g}}_{[{\bar{\tau }} T]}\Big )\rightarrow _d 0$ on $(Y, \mu _\phi )$.

Next, since $b_n^{-1}(\tau _n-n\mu _\phi (Y))$ is tight on $(Y, \mu _\phi )$, we have that $\tau \in L^{1/\rho }(\mu _\phi )$. By assumption, $g\in L^q(\mu _\phi ^\tau )$, for $q>1$. Hence, the assumptions of [MTo04, Lemma 2.1] are satisfied with $a=p=1/\rho$ and any $b:=q>1$ (with a, b, p as there). By [MTo04, Lemma 2.1 (b)], $\frac{g_T}{b_T}-\frac{{\hat{g}}_{n[x,T]}}{b_{n[x,T]}}\rightarrow _d 0$ on $(Y^\tau , \mu _\phi ^\tau )$.

To conclude, note that ${\hat{g}}_{n[x,T]}(y,u)=g_{n[x,T]}(y)$. By [S06, Step 3 of the the proof of Theorem 7], $\frac{{\hat{g}}_{n[x,T]}}{b_{n[x,T]}}-\tau _{n[x,T]}\circ F^{\tau _{n[x,T]}}\rightarrow _d 0$ on $(Y, \mu _\phi )$, as required. $\quad \square$

7 Asymptotics of ${\mathcal {P}}(\phi +s\psi )$ for the Flow $f_t$

In this section and the next we shall assume that $F:Y\rightarrow Y$ is Markov, which allows us to express our results in terms of pressure. We will also assume that $P(\phi ) = 0$ for all the potential functions $\phi$ involved. [S06, Theorem 8] gives a link between the shape of the pressure of a given discrete time finite measure dynamical system and an induced version (this was extended in [BTT18] to some infinite measure settings). Here we give a version of this result in the abstract setup of Sect. 4 along with suitable assumptions on a second potential $\psi$. Note that our assumptions are not directly comparable with those in [S06, Theorem 8].

Throughout this section, we let $\phi :\mathcal {M}\rightarrow {{\mathbb {R}}}$ and $\mu _\phi$ be as Sect. 4. In particular, we recall that $\tau ^*= \int _Y \tau \, d\mu _{{\bar{\phi }}} < \infty$ and assume that the conformal measure $m_{{\bar{\phi }}}$ satisfies (H1). This ensures that $\mu _{{\bar{\phi }}}$ is an equilibrium measure for $(F,{\bar{\phi }})$ and $\mu _{\phi }$ is an equilibrium measure for $(f_t,\phi )$. We consider the potential $\psi :\mathcal {M}\rightarrow {{\mathbb {R}}}$ and its induced version ${\bar{\psi }}:Y\rightarrow {{\mathbb {R}}}$ defined in (3.2) and require:

(H5):

There exists $C, C'>0$ such that for $y\in Y$, ${\bar{\psi }}(y)=C'-\psi _0(y)$, where $\psi _0(y) = C \tau ^\kappa (y) (1+o(1))$ for some $\kappa >0$. For $\kappa >1$ we further require that $\int _Y\tau ^\kappa \, d\mu _{{\bar{\phi }}}<\infty$.

Given $\psi :\mathcal {M}\rightarrow {{\mathbb {R}}}$ and its induced version ${\bar{\psi }}$ on Y, we define the ‘doubly perturbed’ operator

$$\begin{aligned} {\hat{R}}(u,s)v := R(e^{-u\tau }e^{s{\bar{\psi }}}v). \end{aligned}$$

Under (H5), we require the following extended version of (H3).

(H6):

There exist $\sigma _1>1$, constants $C_0, C_1, \delta >0$ such that for all $h\in {{\mathcal {B}}}$, for all $n\in {{\mathbb {N}}}$ and $u\ge 0$, $s\in [0,\delta )$,

$$\begin{aligned} \Vert {\hat{R}}(u,s)^n h\Vert _{{{\mathcal {B}}}_w}\le C_1\Vert h\Vert _{{{\mathcal {B}}}_w},\quad \Vert {\hat{R}}(u,s)^n h\Vert _{{{\mathcal {B}}}} \le C_0\sigma _1^{-n}\Vert h\Vert _{{{\mathcal {B}}}}+C_1\Vert h\Vert _{{{\mathcal {B}}}_w}. \end{aligned}$$

To ensure that we can understand the pressure ${\mathcal {P}}(\overline{\phi +s\psi })$ in terms of the eigenvalues of ${\hat{R}}(u,s)$, under (H1) (in particular, for $\alpha$ as in (H1)(i)) and (H5) we require that

(H7):

There exist $\gamma \in (0,1]$ (with $\gamma <1$ when $\kappa >1$) and $C_2, C_3, \delta >0$ such that for all $u\ge 0$, $s\in (0, \delta )$, $h \in {{\mathcal {B}}}$ and every element a of the (Markov) partition ${{\mathcal {A}}}$ we have:

$$\begin{aligned} \Vert (e^{-u\tau +s{\bar{\psi }}}-1)1_a h\Vert _{{{\mathcal {B}}}_w} \le \left( C_2 s^\gamma \sup _a\psi _0^\gamma + C_3 u^\gamma \sup _a \tau ^\gamma \right) \Vert h \Vert _{{{\mathcal {B}}}_w}. \end{aligned}$$

(7.1)

In (7.1), multiplication by $1_a$ means that we restrict $\tau ,{\bar{\psi }}$ to a. The main result of this section reads as follows.

Theorem 7.1

Assume (H2) and suppose that $\psi :\mathcal {M}\rightarrow {{\mathbb {R}}}$ satisfies (H5) with $C' > C \int \tau ^\kappa \, d\mu _{{\bar{\phi }}}$. Assume (H6) and (H7). Then

$$\begin{aligned} {\mathcal {P}}(\phi +s\psi )= \frac{1}{\tau ^*}{\mathcal {P}}(\overline{\phi +s\psi })(1+o(1))\text { as } s\rightarrow 0^+. \end{aligned}$$

7.1 Technical tools, family of eigenvalues of ${\hat{R}}(u,s)$

Note that ${\hat{R}}(u,0) v={\hat{R}}(u) v$ and recall from Sect. 5.1 that under (H1)–(H3), the family of eigenvalues $\lambda (u)$ is well-defined on $[0,\delta _0)$ with $\lambda (0)=1$. Recall that (H6) and (H7) hold for some $\delta >0$. Throughout the rest of this work we let $\delta _1=\min \{\delta , \delta _0\}$.

Lemma 7.2

Assume (H1), (H2), (H5) and (H7). Then for all $s\in (0,\delta _1)$, there exists $c>0$ such that

$$\begin{aligned} \Vert {\hat{R}}(u,s)-{\hat{R}}(u,0)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\le c s^\gamma . \end{aligned}$$

Proof

Using (4.1), write

$$\begin{aligned} {\hat{R}}(u,s)-{\hat{R}}(u,0)= r_0(u)\int _0^\infty R(\omega (t-\tau ) (e^{s{\bar{\psi }}}-1)) e^{-ut}\, dt. \end{aligned}$$

(7.2)

By (H7), $(e^{s{\bar{\psi }}}-1)1_a\in {{\mathcal {B}}}$ for any element a in the (Markov) partition ${{\mathcal {A}}}$. Without loss of generality we assume that ${\text {supp}}\omega (t-\tau )$ is a subset of a finite union $\cup _{a \in A_t} a$ of elements in ${{\mathcal {A}}}$. Thus, $\tau$ and t are of the same order of magnitude for $a \in A_t$ and $\Vert (e^{s{\bar{\psi }}}-1) 1_{{\text {supp}}\, \omega (t-\tau )}\Vert _{{{\mathcal {B}}}_w} \ll \max _{a \in A_t}\Vert (e^{s{\bar{\psi }}}-1) 1_a\Vert _{{{\mathcal {B}}}_w}$.

Note that

$$\begin{aligned} \Vert R(\omega (t-\tau ) (e^{s{\bar{\psi }}}-1))v\Vert _{{{\mathcal {B}}}_w}\le C\Vert R(\omega (t-\tau ))\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w} \max _{a \in A_t}\Vert (e^{s{\bar{\psi }}}-1) 1_a\Vert _{{{\mathcal {B}}}_w}. \end{aligned}$$

By (H5) and (H7) (with $u=0$),

$$\begin{aligned} \Vert (e^{s{\bar{\psi }}}-1) 1_{{\text {supp}}\, \omega (t-\tau )}\Vert _{{{\mathcal {B}}}_w} \ll \Vert (e^{s{\bar{\psi }}}-1) 1_a\Vert _{{{\mathcal {B}}}_w} \ll s^\gamma \sup _{a \in A_t}\psi _0^\gamma \ll s^\gamma t^{\kappa \gamma }. \end{aligned}$$

Recall that $\gamma <1$ if $\kappa >1$. Putting the above together and recalling $R(\omega (t-\tau ))=M(t)$, we obtain that for any $\epsilon \in (0,1-\gamma )$,

$$\begin{aligned} \Vert {\hat{R}}(u,s)-{\hat{R}}(u, 0)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w} \ll s^\gamma \int _0^\infty t^{\kappa \gamma } \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt \ll s^\gamma \int _0^\infty t^{\kappa -\epsilon } \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt. \end{aligned}$$

Proceeding as in the proof of Lemma 5.3, in particular using (5.1), we have $\int _0^\infty t^{\kappa -\epsilon } \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt\ll \int _0^\infty t^{\kappa -\epsilon } (S(t+1)-S(t))\, dt$ for $S(t){=}\int _t^\infty \Vert M(x)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dx$. This gives

$$\begin{aligned}&\int _0^\infty t^{\kappa -\epsilon } \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt \ll \int _0^\infty t^{\kappa -\epsilon } S(t)\, dt - \int _0^\infty t^{\kappa -\epsilon } S(t+1)\, dt \\&= \int _0^\infty t^{\kappa -\epsilon } S(t)\, dt - \int _1^\infty t^{\kappa -\epsilon } (1-\frac{1}{t})^{\kappa -\epsilon } S(t)\, dt \\&\ll \int _0^1 t^{\kappa -\epsilon } S(t)\, dt + (\kappa -\epsilon )\int _1^\infty t^{\kappa -1-\epsilon } S(t)\, dt \ll \int _1^\infty t^{\kappa -1-\epsilon } S(t)\, dt. \end{aligned}$$

By assumption, $\tau \in L^{\kappa }(\mu _{{\bar{\phi }}})$, so $t^\kappa \mu _{{\bar{\phi }}}(\tau >t)\le \int _{\tau \ge t}\tau ^\kappa \, d\mu _{{\bar{\phi }}} \le \int _Y\tau ^\kappa \, d\mu _{{\bar{\phi }}}<\infty$. By (H2)(ii), $S(t)\ll \mu _{{\bar{\phi }}}(\tau >t)$. Thus,

$$\begin{aligned} \int _0^\infty t^{\kappa -\epsilon } \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt \ll \int _1^\infty t^{\kappa -1-\epsilon }\mu _{{\bar{\phi }}}(\tau >t) \, dt \le \int _Y \tau ^\kappa \, d\mu _{{\bar{\phi }}} \int _1^\infty t^{-(1+\epsilon )} \, dt<\infty , \end{aligned}$$

(7.3)

which ends the proof. $\quad \square$

Lemmas 5.3 and 7.2 ensure that $(u,s)\rightarrow {\hat{R}}(u,s)$ is analytic in u and continuous in s, for all $s\in [0,\delta _0)$, when viewed as an operator from ${{\mathcal {B}}}$ to ${{\mathcal {B}}}_w$ with

$$\begin{aligned} \Vert {\hat{R}}(u,s)-{\hat{R}}(0, 0)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\ll s^\gamma +q(u), \end{aligned}$$

where q(u) is as defined in Lemma 5.3.

Recall that $\lambda (u)$ is well-defined for all $u\in [0,\delta _1)$ with $\delta _1=\min \{\delta , \delta _0\}$. By (H6), there exists a family of eigenvalues $\lambda (u,s)$ well-defined for $u,s\in [0,\delta _1)$ with $\lambda (0,0)=1$. Throughout, let v(u, s) be a family of eigenfunctions associated with $\lambda (u,s)$. The next result gives the continuity properties of v(u, s).

Lemma 7.3

Assume (H1), (H2), (H5) and (H6). Then there exists $\delta _2\le \delta _1$ such that for all $u,s\in [0,\delta _2)$, there exists $c>0$ such that

$$\begin{aligned} \Vert v(u,s)-v(u,0)\Vert _{{{\mathcal {B}}}_w}\le c s^\gamma \log (1/s). \end{aligned}$$

Proof

This follows from Lemma 7.2 and (H6) (which holds for $u\ge 0$ and $s\in [0,\delta _1)$) together with the argument recalled in the proof of Lemma 5.4. $\quad \square$

We recall $\Pi (u)=\mu _{{\bar{\phi }}} (1-e^{-u\tau })$and set $\Pi _0(s):=\mu _{{\bar{\phi }}} (e^{s{\bar{\psi }}}-1)$. The result below gives the asymptotic behaviour and continuity properties of $\lambda (u,s)$.

Lemma 7.4

Assume (H1), (H2), (H5), (H6) (for some range of $\kappa$) and (H7). Then as $u, s\rightarrow 0$,

$$\begin{aligned} 1-\lambda (u, s) =\Pi (u)+\Pi _0(s)+D(u,s), \end{aligned}$$

where $|\frac{d}{du}D(u,s)|\ll (\log (1/u))^{\kappa \gamma +2}(s^\gamma +q(u))+ \mu _{{\bar{\phi }}}(\tau >\log (1/u))$ with q(u) as defined in Lemma 5.3.

Proof

Proceeding as in the proof of Lemma 5.5, write $\lambda (u, s) v(u, s)={\hat{R}}(u, s)v(u, s)$ for all $u, s\in [0,\delta _1)$ with $\lambda (0, 0)=1$ and $v(0, 0)=\mu _{{\bar{\phi }}}$. Normalising such that $\langle v(u, s),1\rangle =1$,

$$\begin{aligned} 1 -\lambda (u,s)&= \Pi (u)+\Pi _0(s)-W(u,s)-V(u,s) \quad \text { for } \\ W&= \int _Y (e^{-u\tau }-1)(e^{s{\bar{\psi }}}-1)\, d\mu _{{\bar{\phi }}} \ \text { and }\ V(u,s) = \langle ({\hat{R}}(u, s)-{\hat{R}}(0, 0))(v(u, s)-v(0, 0)),1\rangle . \end{aligned}$$

We first deal with V(u, s). Let $Q(u,s):=\langle ({\hat{R}}(u, 0)-{\hat{R}}(0, 0))(v(u, s)-v(0, 0)),1\rangle$. Using (7.2),

$$\begin{aligned} V(u,s)= r_0(u) \int _0^\infty e^{-ut} \langle (e^{s{\bar{\psi }}}-1)\omega (t-\tau ) [v(u)-v(0)],1\rangle \, dt+Q(u,s). \end{aligned}$$

By the argument used in the proof of Lemma 5.5,

$$\begin{aligned} |Q(u,s)|\ll \Vert v(u,s)-v(0, 0)\Vert _{{{\mathcal {B}}}_w} \Pi (u). \end{aligned}$$

By (H2)(i),

$$\begin{aligned} \langle (e^{s{\bar{\psi }}}-1) \omega (t-\tau )[v(u)-v(0)], 1\rangle&\ll \Vert (e^{s{\bar{\psi }}}-1)(v(u,s )-v(0, 0)\Vert _{{{\mathcal {B}}}_w}\mu _{{\bar{\phi }}}(\omega (t-\tau )), \end{aligned}$$

which together with (H7) (with $u=0$) gives

$$\begin{aligned} \langle (e^{s{\bar{\psi }}}-1) \omega (t-\tau )[v(u)-v(0)], 1\rangle \ll s^\gamma t^{\kappa \gamma } \Vert v(u,s )-v(0, 0)\Vert _{{{\mathcal {B}}}_w}\mu _{{\bar{\phi }}}(\omega (t-\tau )). \end{aligned}$$

Proceeding as in the proof of Lemma 5.5 and using that $\int _Y \tau ^\kappa \, d\mu _{{\bar{\phi }}}<\infty$,

$$\begin{aligned} \Big |\int _0^\infty e^{-ut}&\langle (e^{s{\bar{\psi }}}-1)\omega (t-\tau ) [v(u,s)-v(0, 0)],1\rangle \, dt \Big |\\&\ll s^\gamma \Vert v(u,s)-v(0,0)\Vert _{{{\mathcal {B}}}_w} \int _Y \tau ^{\kappa \gamma } e^{-u\tau }\, d\mu _{{\bar{\phi }}}\ll s^\gamma \Vert v(u,s)-v(0, 0)\Vert _{{{\mathcal {B}}}_w}. \end{aligned}$$

By a similar argument,

$$\begin{aligned} \left| \int _0^\infty (e^{-ut}-1)\langle (e^{s{\bar{\psi }}}-1) \omega (t-\tau ) [v(u, s)-v(0, 0)],1\rangle \, dt \right| \ll s^\gamma \Vert v(u,s)-v(0, 0)\Vert _{{{\mathcal {B}}}_w}. \end{aligned}$$

By Lemmas 5.4 and 7.3, $\Vert v(u,s)-v(0, 0)\Vert _{{{\mathcal {B}}}_w}\ll s^\gamma \log (1/s)+ q(u)|\log q(u)|$. This together with the previous two displayed equations gives $|V(u,s)|\ll s^\gamma (s^\gamma \log (1/s)+ q(u)|\log q(u)|)$.

We continue with $|\frac{d}{du}V(u,s)|$. Compute that

$$\begin{aligned} \frac{d}{du}V(u,s)=\left\langle \frac{d}{du}{\hat{R}}(u, s)(v(u, s)-v(0, 0)),1\right\rangle +\left\langle \left( {\hat{R}}(u, s)-{\hat{R}}(0, 0)\right) \frac{d}{du}v(u, s),1\right\rangle . \end{aligned}$$

By (H5), (H7) and the argument used above in estimating V(u, s), we get

$$\begin{aligned} \left| \left\langle \frac{d}{du}{\hat{R}}(u, s)(v(u, s)-v(0, 0)),1\right\rangle \right|&\ll \left| \left\langle \frac{d}{du}{\hat{R}}(u, 0)(v(u, s)-v(0, 0)),1\right\rangle \right| \\&\ll \Vert v(u,s)-v(0, 0)\Vert _{{{\mathcal {B}}}_w}\Big |\int _0^\infty t\omega (t-\tau ) e^{-ut}\,dt \Big |\\&\ll \Vert v(u,s)-v(0, 0)\Vert _{{{\mathcal {B}}}_w}\ll \log (1/u)(s+q(u)). \end{aligned}$$

By (H5), (H7) and Lemma 5.4, $\Vert \frac{d}{du}v(u, s)\Vert _{{{\mathcal {B}}}_w}\ll \Vert \frac{d}{du}v(u, 0)\Vert _{{{\mathcal {B}}}_w}\ll \log (1/u)$. Thus,

$$\begin{aligned} \left| \left\langle ({\hat{R}}(u, s)-{\hat{R}}(0, 0))\frac{d}{du}v(u, s),1\right\rangle \right| \ll (s^\gamma +q(u))\log (1/u) \end{aligned}$$

and $|\frac{d}{du}V(u,s)|\ll (s^\gamma +q(u))\log (1/u)$. We briefly estimate W(u, s). Compute that

$$\begin{aligned} \left| \frac{d}{du}W(u,s)\right|&\ll \int _{\{\tau<\log (1/u)\}}\tau e^{-\tau }|1-e^{s{\bar{\psi }}}|\, d\mu _{{\bar{\phi }}}+\int _{\{\tau \ge \log (1/u)\}}\tau \, d\mu _{{\bar{\phi }}}\\&\ll s^\gamma \int _{\{\tau <\log (1/u)\}}\tau ^{\kappa \gamma +1}\, d\mu _{{\bar{\phi }}}+\mu _{{\bar{\phi }}}(\tau \ge \log (1/u))\\&\ll s^\gamma (\log (1/u))^{\kappa \gamma +2}+\mu _{{\bar{\phi }}}(\tau \ge \log (1/u)). \end{aligned}$$

The conclusion follows with $D(u,s)=-(W(u,s)+V(u,s))$. $\quad \square$

For a further technical result exploiting (H7) we introduce the following notation. For $u, s\ge 0$, set $g=s{\bar{\psi }}-u\tau$, so ${\hat{R}}(u,s)=R(e^g)$. For $N>1$ and $x \in [a]$, define the ‘lower’ and ‘upper’ flattened versions of g as

$$\begin{aligned} g_N^-(x)={\left\{ \begin{array}{ll} g(x) &{}\quad \text { if } \tau (y)\le N \text { for all } y\in a, \\ \inf _{y\in [a]}g(y) &{}\quad \text { otherwise}, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} g_N^+(x)={\left\{ \begin{array}{ll} g(x) &{}\quad \text { if } \tau (y)\le N \text { for all } y\in a, \\ \sup _{y\in [a]}g(y) &{}\quad \text { otherwise}. \end{array}\right. } \end{aligned}$$

Since $\psi$ is bounded above by (H5), monotonicity of the pressure function gives

$$\begin{aligned} {\mathcal {P}}({\bar{\phi }}+g_N^-)\le {\mathcal {P}}({\bar{\phi }}+g) \le {\mathcal {P}}({\bar{\phi }}+g_N^+) \le {\mathcal {P}}({\bar{\phi }}) + s \, \sup {\bar{\psi }} < \infty \end{aligned}$$

(7.4)

for all $u,s \ge 0$. For fixed $u,s\in [0,\delta _0)$, set ${\hat{R}}_N^\pm (u,s)=R(e^{g_N^\pm })$. By (H6), the associated eigenvalues $\lambda _N^\pm (u,s)$ exist for all N and $u,s\in [0,\delta _0)$.

Lemma 7.5

Assume (H2), (H5), (H6) and (H7). Then for fixed $u,s\in [0,\delta _0)$,

$$\begin{aligned} \lim _{N\rightarrow \infty }|\lambda (u,s)-\lambda _N^\pm (u,s)|\rightarrow 0. \end{aligned}$$

The reason to introduce $g^\pm _N$ is that, unlike g, the potentials $g_N^\pm$ have summable variation, and therefore, $P({\bar{\phi }}+g_N^\pm )=\log \lambda _N^\pm (u, s)$ by [S99, Lemma 6]. Here [S99, Theorem 3] is used to equate Gurevich pressure in [S99, Lemma 6] and variational pressure in this paper, and then [S99, Theorem 4] together with the existence of the eigenfunctions $v^\pm _N(u,s)$ and eigenmeasures $m^\pm _N(u,s)$ of $R^\pm _N(u,s)$ and its dual, to conclude that ${\bar{\phi }}+g^\pm _N$ are positively recurrent. From this, together with (7.4) and Lemma 7.5, we conclude

$$\begin{aligned} {\mathcal {P}}(\overline{\phi +s\psi -u}) = \log \lambda (u, s). \end{aligned}$$

(7.5)

We note that this connection between the eigenvalues of a perturbed transfer operator and the pressure can be seen in similar contexts in [S06].

Proof of Lemma 7.5

Assume $u\ne 0$ and/or $s\ne 0$. Proceeding similarly to the argument of Lemma 7.2, we write

$$\begin{aligned} {\hat{R}}(u,s)-{\hat{R}}_N^\pm (u,s)= \int _0^\infty R(\omega (t-\tau ) (e^g -e^{{\tilde{g}}_N}))\, dt. \end{aligned}$$

(7.6)

Without loss of generality we assume that ${\text {supp}}\omega (t-\tau )$ is a subset of a finite union $\cup _{b\in B_t} b$ of elements $b \in {{\mathcal {A}}}$. Note that for any $v\in C^\alpha (Y)$,

$$\begin{aligned} \Vert R(\omega (t-\tau ) (e^g -e^{\pm g_N}))v\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\le C\Vert R(\omega (t-\tau ))\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\max _{b\in B_t}\Vert (e^g -e^{{\tilde{g}}_N}) 1_b v\Vert _{{{\mathcal {B}}}_w}. \end{aligned}$$

We only have to consider those $b\in B_t$ with $b \cap \{\tau >N\}\ne \emptyset$, otherwise $(e^g -e^{g^\pm _N}) 1_b=0$. This together with (H7) gives

$$\begin{aligned} \max _{b\in B_t}\Vert (e^g -&e^{\pm g_N}) 1_{b}v\Vert _{{{\mathcal {B}}}_w} \ll \max _{b\in B_t}\sup _b|g -\pm g_N|^\gamma \ll t^{\gamma \kappa } 1_{\{t>N\}}. \end{aligned}$$

Recall $\gamma <1$ for $\kappa >1$. By (H2) and the previous displayed equation, for any $\epsilon \in (0,1-\gamma )$,

$$\begin{aligned}&\Vert {\hat{R}}(u,s)-{\hat{R}}_N^\pm (u,s)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\nonumber \\&\ll \int _N^\infty t^{\gamma \kappa }\Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}~dt =\int _N^\infty t^{-(1-\gamma -\epsilon )\kappa } t^{\kappa -\epsilon } \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}~dt\nonumber \\&\le N^{-(1-\gamma -\epsilon )\kappa }\int _N^\infty t^{\kappa -\epsilon } \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}~dt. \end{aligned}$$

(7.7)

By equation (7.3), $\int _0^\infty t^{\kappa -\epsilon } \Vert M(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt<\infty$. Using (7.7) and (7.3),

$$\begin{aligned} \Vert {\hat{R}}(u,s)-{\hat{R}}_N^\pm (u,s)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w} \ll N^{-(1-\gamma -\epsilon )\kappa }. \end{aligned}$$

This together with the argument recalled in the proof of Lemma 5.4 gives

$$\begin{aligned} \Vert v(u,s)-v_N^{\pm }(u,s) \Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\ll N^{-(1-\gamma -\epsilon )\kappa }\log N. \end{aligned}$$

where for fixed $u, s\in [0,\delta _0)$, v(u, s) and $v_N^\pm (u,s)$ are the eigenfunctions associated with $\lambda (u,s)$ and $\lambda _N^\pm (u,s)$, respectively. Thus,

$$\begin{aligned} \Vert {\hat{R}}(u,s)-{\hat{R}}_N^\pm (u,s) \Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\rightarrow 0, \quad \Vert v(u,s)-v_N^{\pm }(u,s) \Vert _{{{\mathcal {B}}}_w} \rightarrow 0 \text{ as } N\rightarrow \infty . \end{aligned}$$

The conclusion follows from these estimates together with the argument used in Lemma 5.5 and the first part of the proof of Lemma 7.4. $\square$

7.2 Proof of Theorem 7.1

Lemma 7.6

Assume that ${\bar{\psi }} \le C'$ (as in (H5)) and that ${\mathcal {P}}(\phi +s\psi )>0$ for $s>0$. Then ${\mathcal {P}}(\overline{\phi + s\psi -{\mathcal {P}}(\phi + s\psi )})=0$.

Proof

Set $u_0={\mathcal {P}}(\phi + s\psi )$ which is positive for $s > 0$ by assumption. We first show that the pressures we are dealing with are finite. As in [S99, Lemma 2], finiteness of the transfer operator acting on constant functions is sufficient to give finiteness of the pressure, i.e., since we are dealing with positive $u_0$, we just need to show $\sup _{x\in Y}| ({\hat{R}}(u, s)1)(x)|<\infty$ for $u \ge 0$. We estimate

$$\begin{aligned} ({\hat{R}}(u, s)1)(x)&= \sum _{F(y)=x} e^{(\overline{\phi + s\psi -u})(y)}\lesssim \sum _{F(y)=x} e^{\overline{\phi }(y)} = (R_01)(x)<\infty , \end{aligned}$$

because ${\bar{\psi }} \le C'$ by (H5).

To show that ${\mathcal {P}}(\overline{\phi + s\psi -u_0})\le 0$, suppose by contradiction that ${\mathcal {P}}(\overline{\phi + s\psi -u_0})> 0$. We use the ideas of [S99, e.g. Theorem 2] to truncate the whole system, i.e., restrict to the system to the first n elements of the partition ${{\mathcal {A}}}$ and the corresponding flow. We write the pressure of this system as ${\mathcal {P}}_n(\cdot )$, so [S99, Theorem 2] says that ${\mathcal {P}}_n(\overline{\phi +s\psi -u_0})> 0$ for all large n. By the definition of pressure, there exists a measure ${\bar{\mu }}$ so that

$$\begin{aligned} h({\bar{\mu }})+\int \overline{\phi +s\psi -u_0}\, d{\bar{\mu }}>0. \end{aligned}$$

Note that $\int \tau ~d{\bar{\mu }}<\infty$ since $\tau$ is bounded on this subsystem. Abramov’s formula (3.3) implies that the projected measure $\mu$ [which relates to ${\bar{\mu }}$ via (3.4)] has

$$\begin{aligned} h(\mu )+\int \phi + s\psi -u_0 ~d\mu >0, \end{aligned}$$

contradicting the choice $u_0 = {\mathcal {P}}(\phi + s\psi )$.

To show that ${\mathcal {P}}(\overline{\phi + s\psi -u_0})\ge 0$, we will use the continuity of the pressure function $u\mapsto {\mathcal {P}}(\overline{\phi + s\psi -u})$ wherever this is finite (recall from above that we have finiteness for any $u\in [0, u_0)$). By the definition of pressure, for any $\epsilon \in (0, u_0)$, there exists $\mu '$ with

$$\begin{aligned} h(\mu ')+\int \phi +s\psi - (u_0-\epsilon )~d\mu '>0. \end{aligned}$$

By (3.4), any invariant probability measure for the flow corresponds to an invariant probability measure ${\bar{\mu }}'$ for the map F. By Abramov’s formula,

$$\begin{aligned} h({\bar{\mu }}')+\int \overline{\phi +s\psi - (u_0-\epsilon )}~d{\bar{\mu }}' = \left( h(\mu ')+\int \phi +s\psi - (u_0-\epsilon )~d\mu '\right) \int \tau ~d{\bar{\mu }}' >0 \end{aligned}$$

regardless of whether $\int \tau ~d{\bar{\mu }}'$ is finite or not. By continuity in u, ${\mathcal {P}}(\overline{\phi + s\psi -u_0}) \ge 0$ as required. $\quad \square$

The following result is a consequence of (7.5) and of Lemma 7.4.

Lemma 7.7

Assume the setting of Lemma 7.4. Suppose that there exists $a>0$ such that $s=O(u^a)$, as $u\rightarrow 0$. Then as $u\rightarrow 0^+$,

$$\begin{aligned} \frac{d}{du}{\mathcal {P}}(\overline{\phi +s\psi -u})=- \tau ^* (1+o(1)). \end{aligned}$$

Proof

By (7.5) and Lemma 7.4,

$$\begin{aligned} \frac{d}{du}{\mathcal {P}}(\overline{\phi +s\psi -u})=\frac{d}{du} \log \lambda (u,s)=-\frac{d}{du}\left( \Pi (u)+D(u,s)\right) , \end{aligned}$$

where $\frac{d}{du} D(u,s)=O\Big ((\log (1/u))^{\kappa \gamma +2} (s^\gamma +q(u))+\mu _{{\bar{\phi }}}(\tau >\log (1/u))\Big )$. Also, compute that

$$\begin{aligned} \frac{d}{du}\Pi (u)=\int _Y\tau e^{-u\tau }\, d\mu _{{\bar{\phi }}}=\int _Y\tau \, d\mu _{{\bar{\phi }}}+\int _Y\tau (e^{-u\tau }-1)\, d\mu _{{\bar{\phi }}}. \end{aligned}$$

Since $\tau |e^{-u\tau }-1|$ is bounded by the integrable function $\tau$ and it converges pointwise to 0, it follows from the Dominated Convergence Theorem that $\int _Y\tau (e^{-u\tau }-1)\, d\mu _{{\bar{\phi }}}\rightarrow 0$, as $u\rightarrow 0^+$. Hence, $\frac{d}{du}\Pi (u)=\tau ^*(1+o(1))$.

Since $\tau \in L^1(\mu _{{\bar{\phi }}})$ by assumption, $\mu _{{\bar{\phi }}}(\tau >\log (1/u))\rightarrow 0$, as $u\rightarrow 0$. To conclude, note that since for some $a>0$, $s=O(u^a)$, as $u\rightarrow 0$, we have $D(u,s)=o(1)$. $\square$

We can now complete

Proof of Theorem 7.1

Set $r(u,s) = \frac{d}{du} {\mathcal {P}}(\overline{\phi + s \psi - u})$. By Lemma 7.7, as $u\rightarrow 0$ and $s=O(u^a)$ for some $a>0$,

$$\begin{aligned} r(u,s)=-\tau ^* (1+o(1)). \end{aligned}$$

For any small $u_0 > 0$, integration gives

$$\begin{aligned} {\mathcal {P}}(\overline{\phi + s\psi -u_0}) - {\mathcal {P}}(\overline{\phi +s\psi }) = \int _0^{u_0} r(u,s) \, du = -\tau ^* u_0(1+o(1)). \end{aligned}$$

(7.8)

The convexity of $s \mapsto {\mathcal {P}}(\phi +s\psi )$ implies that $\frac{d}{ds} {\mathcal {P}}(\phi +s\psi ) = \int \psi \, d\mu _\phi = \frac{1}{\tau ^*} \int {\bar{\psi }} \, d\mu _{{\bar{\phi }}}$. Thus ${\mathcal {P}}(\phi +s\psi ) \ge \frac{s}{\tau ^*}\int {\bar{\psi }} \, d\mu _{{\bar{\phi }}}$ since $\int {\bar{\psi }} \, d\mu _{{\bar{\phi }}} > 0$; this is guaranteed by our assumption that $C' > C \int \tau ^\kappa \, d\mu _{{\bar{\phi }}}$. Hence, $u_0 = u_0(s) ={\mathcal {P}}(\phi +s\psi )\gg s$ and such $u_0$ satisfies the assumptions of Lemma 7.7 with $a=1$. Thus, (7.8) holds for this $u_0$.

By Lemma 7.6 applied to $u_0 = u_0(s) ={\mathcal {P}}(\phi +s\psi )$, we obtain ${\mathcal {P}}(\overline{\phi +s\psi -u_0}) = 0$. Thus, the left hand side of (7.8) is $-{\mathcal {P}}(\overline{\phi + s\psi )}$. By assumption, $u_0(s)>0$, for $s>0$. The continuity property of the pressure function gives $u_0(s)\rightarrow 0$ as $s\rightarrow 0$. Hence, (7.8) applies to $u_0(s)={\mathcal {P}}(\phi +s\psi )$. Thus, ${\mathcal {P}}(\phi +s\psi )=\frac{1}{\tau ^*}{\mathcal {P}}(\overline{\phi +s\psi }) (1+o(1))$, which ends the proof. $\quad \square$

8 Pressure Function and Limit Theorems

For $\psi : \mathcal {M}\rightarrow {{\mathbb {R}}}$, define ${\bar{\psi }}:Y\rightarrow {{\mathbb {R}}}$ as in (3.2) and let $\psi _T=\int _0^T \psi \circ f_t\, dt$. Throughout this section we assume that ${\bar{\psi }}$ satisfies (H5); in particular, we require that ${\bar{\psi }}=C'-\psi _0$, with $\psi _0 \sim C\tau ^\kappa (1+o(1))$, $\int \tau ^\kappa \, d\mu _{{\bar{\phi }}}<\infty$ for some $C,C'>0$. For s real, $s\ge 0$, define

$$\begin{aligned} {\hat{R}}_\psi (s) v := R(e^{s{\bar{\psi }}}v)=e^{sC'}R(e^{-s\psi _0}v) =: e^{sC'}{\hat{R}}_1(s). \end{aligned}$$

As in Sect. 4.3 (when making assumptions on ${\hat{R}}(u)$), we write

$$\begin{aligned} {\hat{R}}_1(s) =r_0(s)\int _0^\infty R(\omega (t-\psi _0)) e^{-st}\, dt = r_0(s)\int _0^\infty M_0(t)\, e^{-st}\, dt, \end{aligned}$$

(8.1)

where $M_0(t) := R(\omega (t-\psi _0))$ and $\omega :{{\mathbb {R}}}\rightarrow [0, 1]$ is an integrable function with ${\text {supp}}\omega \subset [-1, 1]$ and $r_0(0)=1$. Similarly to (H2), we require that

(H8):

There exists a function $\omega$ satisfying  (8.1) and $C_\omega >0$ such that

• (i) for any $t\in {{\mathbb {R}}}_{+}$ and for all $h\in {{\mathcal {B}}}$, we have $\omega (t-\psi _0)h \in {{\mathcal {B}}}_w$ and for all $v\in C^\alpha (Y)$,

  $$\begin{aligned} \langle \omega (t-\psi _0)vh, 1\rangle \le C_\omega \Vert v\Vert _{C^\alpha (Y)}\Vert h\Vert _{{{\mathcal {B}}}_w}\mu _{{\bar{\phi }}}(\omega (t-\psi _0)). \end{aligned}$$

• (ii) there exists $C>0$ such that for all $T>0$,

  $$\begin{aligned} \int _T^\infty \Vert M_0(t)\Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w}\, dt\le C \mu _{{\bar{\phi }}}(\psi _0>T). \end{aligned}$$

Remark 8.1

In practice, checking (H8) requires no extra difficulty compared to checking (H2). But in the generality of the present abstract framework, there is no obvious way of deriving (H8) from (H2).

Summarising the arguments used in the proof of Proposition 5.1 and Theorem 7.1 together with Proposition 6.1, in this section we obtain

Theorem 8.2

Assume (H1) and (H8) and $C' > C \int \tau \, d\mu$ as in Theorem 7.1. Let $\psi :\mathcal {M}\rightarrow {{\mathbb {R}}}$ and assume that ${\bar{\psi }}$ satisfies (H5), (H6) and (H7) (with $u=0$). Set $\psi ^* =\int _Y {\bar{\psi }}\, d\mu _{{\bar{\phi }}}$. The following hold as $T\rightarrow \infty$.

• (a) Suppose that (H4)(i) holds and that $\mu (\psi _0>t)\sim \mu (\tau ^\kappa >t)$ with $\kappa$ in (some subset of) $(0,\beta )$.

  • (i) When $\beta <2$ and $\kappa \in (\beta /2,\beta )$, set b(T) such that $T\ell (b(T))/b(T)^{\frac{\beta }{\kappa }}\rightarrow 1$. Then $\frac{1}{b(T)}(\psi _T-\psi ^*\cdot T)\rightarrow ^d G_{\beta /\kappa }$, where $G_{\beta /\kappa }$ is a stable law of index $\beta /\kappa$. This is further equivalent to ${\mathcal {P}}(\overline{\phi +s\psi })=\psi ^*s+s^{\beta /\kappa } \ell (1/s)(1+o(1))$, as $s\rightarrow 0$.

  • (ii) If $\beta \le 2$ but $\kappa =\beta /2$, set b(T) such that $T{\tilde{\ell }}(b(T))/b(T)^{\frac{2}{\kappa }}\rightarrow c>0$. Then $\frac{1}{b(T)}(\psi _T-\psi ^*\cdot T)\rightarrow ^d {{\mathcal {N}}}(0,c)$ and this is further equivalent to ${\mathcal {P}}(\overline{\phi +s\psi })=\psi ^*s+s^2 L(1/s)(1+o(1))$, as $s\rightarrow 0$ with $L(x)=\frac{1}{2}{\tilde{\ell }}(x)$.

• (b) Suppose that (H4)(ii) holds and further assume that $\mu (\psi _0>t)\ll \mu (\tau ^\kappa >t)$ with $\kappa \in (0,\beta /2)$, and $R{\bar{\psi }}\in {{\mathcal {B}}}$. Then there exists $\sigma \ne 0$ such that $\frac{1}{\sqrt{T}}(\psi _T- \psi ^*\cdot T)\rightarrow ^d {{\mathcal {N}}}(0,\sigma ^2)$ and this is further equivalent to ${\mathcal {P}}(\overline{\phi +s\psi })=\psi ^*s+\frac{\sigma ^2}{2}s^2 (1+o(1))$ as $s\rightarrow 0$.

Moreover, if (H6) and (H7) hold for all $u,s \in [0,\delta )$, then in both cases (a) and (b) we have ${\mathcal {P}}(\phi +s\psi )=\frac{1}{\tau ^*}{\mathcal {P}}(\overline{\phi +s\psi }) (1+o(1))$, for $\tau ^*=\int _Y \tau \,d \mu _{{\bar{\phi }}}$.

Proof

We start by noting that the proofs of the results used in the proofs of Theorems 7.1 and Propositions 5.1 and 6.1 essentially go through when replacing $\tau$ by $\tau ^\kappa$. We briefly summarize the required arguments.

For $s>0$, let $\lambda (s)$, $\lambda _1(s)$ be the families of eigenvalues associated with ${\hat{R}}_\psi (s)$ and ${\hat{R}}_1(s)$, respectively. Note that $\lambda (s)=e^{sC'}\lambda _1(s)$. Hence,

$$\begin{aligned} \lambda (s)-1=sC'+\lambda _1(s)-1+s^2C'+O(s^3). \end{aligned}$$

(8.2)

Further, by (7.5), ${\mathcal {P}}(\overline{\phi +s\psi })=\log \lambda (s)$. Similar to the proofs of Proposition 5.1 and Proposition 6.1, to conclude that (a) and/or (b) hold, we just need to estimate $1-\lambda (s)$.

For the proof of (a) and (b), we note that under (H5)–(H8) (where (H8) and (H6) with $u=0$ are the analogues of (H2) and (H3) with $\tau$ there replaced by $\psi _0$), Lemma 5.5 gives $1-\lambda _1(s)=\Pi _1(s)(1+O(q_1(s)|\log (q_1(s)|))$, where $q_1(s)=s+\mu (\psi _0>1/s)=s+s^{\beta /\kappa }\ll s^{\beta /\kappa }$ and $\Pi _1(s)=\mu _{{\bar{\phi }}} (e^{s\psi _0}-1)$ . Hence,

$$\begin{aligned} 1-\lambda _1(s)=\Pi _1(s)(1+O(s^{\beta /\kappa }\log (1/s))). \end{aligned}$$

In case (a) (i), the argument recalled in the proof of Corollary 5.6 (i) gives $\Pi _1(s)=s\int _Y\psi _0\,d\mu _{{\bar{\phi }}}+s^{\beta /\kappa } \ell (1/s)(1+o(1))$. This together with (8.2) gives

$$\begin{aligned} \lambda (s)-1=\left( C'-\int _Y\psi _0\,d\mu _{{\bar{\phi }}}\right) s+s^{\beta /\kappa } \ell (1/s)(1+o(1))=\psi ^*s+s^{\beta /\kappa } \ell (1/s)(1+o(1)). \end{aligned}$$

As a consequence, ${\mathcal {P}}(\overline{\phi +s\psi })=\psi ^*s+s^{\beta /\kappa } \ell (1/s)(1+o(1))$. Similarly, by the argument used in the proof of Proposition 5.1 (i), this expansion of the eigenvalue/pressure is equivalent to $\frac{1}{b(n)}({\bar{\psi }}_n-\psi ^*\cdot n)\rightarrow ^d G_{\beta /\kappa }$, where ${\bar{\psi }}_n=\sum _{j=0}^{n-1} {\bar{\psi }}\circ F^j$. Further, by the argument used in Proposition 6.1 (i) with $\beta <2$ this is further equivalent to $\frac{1}{b(T)}(\psi _T-\psi ^*\cdot T)\rightarrow ^d G_{\beta /\kappa }$, which completes the proof of (a) (i).

The proof of a(ii) goes similar with the versions of the proofs of Proposition 5.1 (i), Proposition 6.1 (i) with $\beta <2$ replaced by the argument used in the proofs of Proposition 5.1 (i), Proposition 6.1 (i) with $\beta =2$.

The proof of (b) goes again similarly with the proofs of Proposition 5.1 (i), Proposition 6.1 (i) replaced by the argument used in the proofs of Proposition 5.1 (ii), Proposition 6.1 (ii).

Finally, if (H6) for all $u,s\in [0,\delta _0)$ and (H7) also hold, then Theorem 7.1 applies and ensures that ${\mathcal {P}}(\phi +s\psi )=\frac{1}{\tau ^*}{\mathcal {P}}(\overline{\phi +s\psi }) (1+o(1))$, ending the proof. $\quad \square$

Checking (H1)–(H8) for the Almost Anosov Flow

We start with a technical result that will be essential in verifying (H2) and (H6).

Lemma A.1

Assume that w(x, y) is a homogeneous function of degree $\rho$ as in Proposition 2.2. Then

$$\begin{aligned} \sup _k \left\{ \sup _{\{r= k\}} \tau - \inf _{\{r= k\}}\tau \right\} < \infty . \end{aligned}$$

Proof

Recall that ${\hat{\tau }}(x,y) = \min \{ t > 0 : \Phi _t^{hor}(x,y) \in {\hat{W}}^{\mathbf {s}}\}$ and that $|\tau - {\hat{\tau }}| = O(1)$. From Lemma 2.3 we know that $r= {\hat{\tau }} + \Theta ({\hat{\tau }}) + O(1)$. We will first show that if ${\hat{\tau }}$ varies between, say, n and $n+1$, then the corresponding $r$ varies by an amount of O(1) as well.

Indeed, take (x, y) and $(x', y')$ arbitrary such that $n \le {\hat{\tau }}(x,y) , {\hat{\tau }}(x',y') \le n+1$. Define $T = {\hat{\tau }}(x,y)$ and abbreviate $q(t) = (x(t), y(t)) = \Phi _t^{hor}(x,y)$ for $0 \le t \le T$. There is $t_0 \in {{\mathbb {R}}}$ with $|t_0| \le 1$ such that the y-component $\Phi _{-t_0}^{hor}({\hat{x}},{\hat{y}})$ is y. Also write $q'(t) = (x'(t),y'(t)) = \Phi _{t-t_0}^{hor}(x',y')$, and the Euclidean distance $\epsilon (t) = |q'(t)-q(t)|$.

Clearly $| w(q'(t)) - w(q(t))| \ll | \nabla w(q(t))| \epsilon (t)$, so $\epsilon := \epsilon (0) \ll T^{-(1+\beta )}$ by (2.3). However, since ${\hat{q}}(0)-q(0)$ is roughly in the unstable direction, $\epsilon (t)$ will increase to size O(1) as t increases to ${\hat{\tau }}(q')$. This is too large, but we can set $T_1 := \min \{ t > 0 : x(t) = y(t)\}$, and then estimate $\epsilon (T_1)$ as follows. Take $T'_1 = \min \{ t > 0 : x'(t) = y'(t)\}$, then $|T'_1 - T_1| = O(1)$ and we can write $q(T_1) = (\delta ,\delta )$, $q'(T'_1) = (\delta ',\delta ')$ with $\epsilon ' := \delta '-\delta$. Furthermore, since L from (2.8) is constant on integral curves, we have

$$\begin{aligned} L(x,y) = x^uy^v \left( \frac{a_0}{v} x^2+\frac{b_2}{u} y^2\right) = \delta ^{u+v+2} \left( \frac{a_0}{v}+\frac{b_2}{u}\right) , \end{aligned}$$

(A.1)

and

$$\begin{aligned} L(q'(0))\! =\! L(x+\epsilon , y) \!=\! (x+\epsilon )^uy^v \left( \frac{a_0}{v} (x+\epsilon )^2+ \frac{b_2}{u} y^2\right) \!=\! (\delta +\epsilon ')^{u+v+2} \left( \frac{a_0}{v}+\frac{b_2}{u}\right) . \end{aligned}$$

Taking the difference of both expressions, we obtain

$$\begin{aligned} \frac{\epsilon }{x}\, u\, L(x,y) \left( 1 + \frac{2a_0}{u a_0 x^2 \!+\! v b_2 y^2} + O\left( \frac{\epsilon }{x}\right) \right) \!=\! \frac{\epsilon '}{\delta } \, (u+v+2)\, L(x,y)\left( 1 + O\left( \frac{\epsilon '}{\delta }\right) \right) . \end{aligned}$$

Divide by $(u+v+2)\, L(x,y)/\delta$ and ignore the quadratic error terms. Then we have

$$\begin{aligned} \epsilon ' \sim \frac{u}{u+v+2} \left( 1+\frac{2a_0}{v b_2y^2} \right) \frac{ \delta \ \epsilon }{x}. \end{aligned}$$

But $\delta$ can be estimated in terms of x using (A.1), and since $\frac{u}{u+v+2} = \frac{1}{2\beta }$ and $\frac{a_0 u}{b_2v} = \frac{c_0}{c_2}$, we obtain

$$\begin{aligned} \epsilon ' \sim \frac{\epsilon }{2\beta } x^{\frac{1}{2\beta }-1} y^{1-\frac{1}{2\beta }} \left( \left( 1+\frac{c_0}{c_2}\right) \left( 1+\frac{c_0}{c_2 y^2}\right) \right) ^{-\frac{1}{u+v+2}} \quad \text { as } n \rightarrow \infty . \end{aligned}$$

Rewriting x and $\epsilon$ in terms of T using (2.3), we obtain $\epsilon ' \ll T^{-\frac{3}{2}}$. Next, because the vector-valued function $\nabla w$ is homogeneous with exponent $\rho -1$, Proposition 2.2 gives for $\rho < 2$:

$$\begin{aligned} \int _0^{T_1} | w(q'(t))-w(q(t)) | \, dt\ll & {} \sup _{t \in [0,T_1]} \epsilon (t) \int _0^{T_1} | \nabla w(q(t)) | \, dt + O(1) \\\ll & {} \epsilon '\ T^{1-\frac{\rho -1}{2}} + O(1) = T^{-\frac{\rho }{2}} + O(1). \end{aligned}$$

The integral $\int _{T_1}^T \Vert w(q'(t))-w(q(t)) \Vert \, dt$ is dealt with in the same way, but now integrating backwards from $\Phi _{{\hat{\tau }}(x,y)}^{hor}(x,y)$ and $\Phi _{{\hat{\tau }}(x',y')}^{hor}(x',y')$. Therefore $\int _0^T w(q'(t))-w(q(t)) \, dt$ varies at most O(1) on the region $\{ (x,y) : n \le {\hat{\tau }}(x,y) \le n+1\}$.

Since also $\int _0^T w(q(t)) \, dt = o(T)$, we can find $N \in {{\mathbb {N}}}$ independent of n such that $\tau = \int _0^T 1+w(q(t)) \, dt + O(1)$ varies by at least 1 (but at most by O(N)) over $\{ n \le {\hat{\tau }} \le n+N\}$, and therefore $r$ varies by at least 1 on this region. It follows that for each k, $\{ r= k\} \subset \{ n \le {\hat{\tau }} \le n+N\}$ for some $n=n(k)$ and $\sup _{\{r= k\}} \tau - \inf _{\{r= k \}} \tau$ is bounded, uniformly in k.

The proof for $\rho \ge 2$ goes likewise. $\square$

1.1 Verifying (H1): recalling previously used Banach spaces

Charts and distances: Throughout, ${{\mathcal {W}}}^{\mathbf {s}}$ denotes the set of admissible leaves, which consists of maximal stable leaves in elements of the partition ${{\mathcal {Y}}}= \{ Y_j \}_j = \{ P_i\}_i \vee \{ \{ r= k\} \}_{k \ge 2}$. It is convenient to arrange the enumeration of ${{\mathcal {Y}}}$ such that $Y_j = \{ r= j\}$ for $j \ge 2$, and use indices $j \le 1$ for the remaining (parts of) $P_i$. To define distances between leaves, as in [BT17, Section 3.1], we use charts $\chi _j:[0,L_{\mathbf {u}}(Y_j)] \times [0,1] \rightarrow Y_j$, where $L_{\mathbf {u}}(Y_j)$ is the length of the (largest) unstable leaf in $Y_j$. For any leaf $W \subset Y_j$, we have the parametrisation

$$\begin{aligned} \chi _j^{-1}(W)=\{(g_{Y_j,W}(\rho )), \rho \in [0,1]\}, \end{aligned}$$

(A.2)

so $g_{Y_j}$ is a parametrisation of W in the chart. Suppressing $Y_j$ and using g and ${\tilde{g}}$ for close-by stable leaves $W \subset Y_k$ and ${\tilde{W}}\in {{\mathcal {W}}}^{\mathbf {s}} \in Y_\ell$ we define their by

$$\begin{aligned} d(W,{\tilde{W}})= {\left\{ \begin{array}{ll} \sup _{x \in [0,1]}|g(x)-{\tilde{g}}(x)| &{}\quad \text { if } k = \ell ;\\ \sup _{x \in [0,1]}|g(x)-L_{\mathbf {u}}(Y_k)-{\tilde{g}}(x)| + \sum _{j=k+1}^{l-1} L_{\mathbf {u}}(Y_j) &{}\quad \text { if } 2 \le k < \ell ;\\ \infty &{}\quad \text { otherwise,} \end{array}\right. } \end{aligned}$$

where g and ${\tilde{g}}$ are parametrisations of W and ${\tilde{W}}$ in the appropriate chart. Here an empty sum $\sum _{j=k+1}^k$ is 0 by convention.

From here on, in the notation $\int _W \cdot ~d\mu ^{\mathbf {s}}$, the measure $\mu ^{\mathbf {s}}$ refers to the SRB measure $\mu _{{\bar{\phi }}}$ conditioned on the stable leaf W: similarly for $m^{\mathbf {s}}$ and Lebesgue measure. Hence the norms defined below are w.r.t. the invariant measure, not the conformal measure as in [BT17]. It is possible to do this due to Lemma A.2, specifically (A.7).

Definition of the norms Given $h\in C^1(Y,{\mathbb {C}})$, define the weak norm by

$$\begin{aligned} \Vert h\Vert _{{{\mathcal {B}}}_w}:=\sup _{W\in {{\mathcal {W}}}^{\mathbf {s}}}\;\sup _{|\varphi |_{ C^1(W)}\le 1 }\int _W h\varphi \, d\mu ^{\mathbf {s}}. \end{aligned}$$

(A.3)

Given $q\in [0,1)$ we define the strong stable norm by

$$\begin{aligned} \Vert h\Vert _{\mathbf {s}}:=\sup _{W\in {{\mathcal {W}}}^{\mathbf {s}}}\; \sup _{|\varphi |_{ C^q(W)}\le 1 }\int _W h\varphi \, d\mu ^{\mathbf {s}}. \end{aligned}$$

(A.4)

Finally, we define the strong unstable norm by

$$\begin{aligned} \Vert h\Vert _{\mathbf {u}}:=\sup _\ell \sup _{W,{\tilde{W}}\in {{\mathcal {W}}}^{\mathbf {s}} \cap Y_\ell } \ \sup _{{\mathop {d(\varphi ,{\tilde{\varphi }}) \le d(W,{\tilde{W}})}\limits ^{|\varphi |_{C^1(W)}, |{\tilde{\varphi }}|_{C^1({\tilde{W}})} \le 1}}} \frac{1}{d(W,{\tilde{W}})}\left| \int _{W} h\varphi \, d\mu ^{\mathbf {s}} -\int _{{\tilde{W}}} h{\tilde{\varphi }}\, d\mu ^{\mathbf {s}}\right| ,\nonumber \\ \end{aligned}$$

(A.5)

where $d(\varphi ,{\tilde{\varphi }}) = | \varphi \circ \chi _\ell ( g(\eta ), \eta ) - {\tilde{\varphi }} \circ \chi _\ell ({\tilde{g}}(\eta ), \eta )|_{C^1([0,1])}$.

The strong norm is defined by $\Vert h\Vert _{{{\mathcal {B}}}}=\Vert h\Vert _{\mathbf {s}}+\Vert h\Vert _{\mathbf {u}}$.

Definition of the Banach spaces We define ${{\mathcal {B}}}$ to be the completion of $C^1$ in the strong norm and ${{\mathcal {B}}}_w$ to be the completion in the weak norm. As clarified in [BT17, Lemma 3.2, Lemma 3.3, Proposition 3.2], (H1) holds with $\alpha = \alpha _1 = 1$.

The spaces ${{\mathcal {B}}}$ and ${{\mathcal {B}}}_w$ defined above are simplified versions of functional spaces defined in [DL08], adapted to the setting of (F, Y). The main difference in the present setting is the simpler definition of admissible leaves and the absence of a control on short leaves. This is possible due to the Markov structure of the diffeomorphism.

As this is the same Banach space as in [BT17], most parts of (H1) can be taken from there. For H1(v), which is concerned with the transfer operator w.r.t. $\mu _{{\bar{\phi }}}$, we first need a lemma.

Lemma A.2

Let $\varvec{{\gamma }}= \varvec{{\gamma }}(q)$ be the angle between the stable and unstable leaf at q and $h_0^{\mathbf {u}} = \frac{d\mu ^{\mathbf {u}}}{dm^{\mathbf {u}}}$ be the density of $\mu _{{\bar{\phi }}}$ conditioned on unstable leaves. Then

$$\begin{aligned} \int _{W^{\mathbf {s}}} R v \cdot \varphi \, d\mu ^{\mathbf {s}} = \sum _j \int _{W_j} v \cdot \varphi \circ F \cdot K^{\mathbf {u}}_{W_j} \, d\mu ^{\mathbf {s}}, \end{aligned}$$

(A.6)

where the sum is over all preimage leaves $W_j = F^{-1}(W^{\mathbf {s}}) \cap \{ r= j\}$ and

$$\begin{aligned} K^{\mathbf {u}}_{W_j} = \frac{J_{\mu _{{\bar{\phi }}}}^{\mathbf {s}} F}{J_{\mu _{{\bar{\phi }}}}F} = \frac{1}{J_{m_{{\bar{\phi }}}}^{\mathbf {u}} F} \frac{(h_0^{\mathbf {u}} \sin \varvec{{\gamma }}) \circ F}{h_0^{\mathbf {u}} \sin \varvec{{\gamma }}} \end{aligned}$$

(A.7)

is piecewise $C^1$ on unstable leaves.

Proof

We have the pointwise formulas for $R_0:L^1(m_{{\bar{\phi }}})\rightarrow L^1(m_{{\bar{\phi }}})$ and $R:L^1(\mu _{{\bar{\phi }}})\rightarrow L^1(\mu _{{\bar{\phi }}})$:

$$\begin{aligned} R_0v = \mathbf{1}_Y \frac{v}{|DF|}\circ F^{-1},\quad Rv = \mathbf{1}_Y \frac{h_0 \circ F^{-1}}{h_0} \, \frac{v}{|DF|}\circ F^{-1}, \end{aligned}$$

where $|DF|=J_{m_{{\bar{\phi }}}} = |\det (DF)|$. This also shows that $J_{\mu _{{\bar{\phi }}}}F = J_{m_{{\bar{\phi }}}} \frac{h_0}{h_0 \circ F^{-1}}$ and analogous formulas hold for $J_{\mu ^{\mathbf {s}}}F$ and $J_{\mu ^{\mathbf {u}}}F$. Integration over a stable leaf $W \in {{\mathcal {W}}}^{\mathbf {s}}$ with preimage leaves $W_j$ gives

$$\begin{aligned} \int _W R_{\mu _{{\bar{\phi }}}}v \, \varphi \, d\mu ^{\mathbf {s}}= & {} \int _W \frac{h_0 \circ F^{-1}}{h_0} \, \frac{v}{|DF|}\circ F^{-1} \, \varphi \, h_0^{\mathbf {s}} \, dm^{\mathbf {s}} \\= & {} \sum _j \int _{W_j} \frac{J_{m^{\mathbf {s}}}F}{J_{m_{{\bar{\phi }}}}F} \, \frac{h_0}{h_0 \circ F}\, v\, \varphi \circ F \, \frac{h_0^{\mathbf {s}} \circ F}{h_0^{\mathbf {s}}} \, h_0^{\mathbf {s}} dm^{\mathbf {s}} \\= & {} \sum _j \int _{W_j} \frac{J_{\mu ^{\mathbf {s}}}F}{J_{\mu _{{\bar{\phi }}}}F} \, v\, \varphi \circ F \, d\mu ^{\mathbf {s}} = \sum _j \int _{W_j} v \, \varphi \circ F \, K^{\mathbf {u}}_{W_j}\, d\mu ^{\mathbf {s}}. \end{aligned}$$

Since $d\mu ^{\mathbf {u}} = h_0^{\mathbf {u}} \, dm^{\mathbf {u}}$ for a $C^1$ density $h_0^{\mathbf {u}}$, and $dm_{{\bar{\phi }}} = dm^{\mathbf {s}} dm^{\mathbf {u}} \sin \varvec{{\gamma }}$, whence $J_{m_{{\bar{\phi }}}}F = J_{m^{\mathbf {s}}}F \cdot J_{m^\mathbf {u}}F \cdot \frac{\sin \varvec{{\gamma }}\circ F}{\sin \varvec{{\gamma }}}$, the other formula $K^{\mathbf {u}}_W = \frac{1}{J_{m_{{\bar{\phi }}}}^{\mathbf {u}} F} \frac{(h_0^{\mathbf {u}} \sin \varvec{{\gamma }}) \circ F}{h_0^{\mathbf {u}} \sin \varvec{{\gamma }}}$ follows as well. Then [BT17, Equation (18)], leading to the parametrisation of the unstable foliation of the induced map over ${\hat{f}} = \Phi _1^{hor}$, shows that this foliation is $C^1$. The analogous statement holds for the stable foliation. To obtain the unstable/stable foliations of F, we need to flow a bounded and piecewise $C^1$ amount of time (see Lemma A.1). Therefore $q \mapsto \varvec{{\gamma }}(q)$ is piecewise $C^1$. Hence the latter expression of $K^{\mathbf {u}}_W$ shows that it is piecewise $C^1$ on unstable leaves. $\quad \square$

Now (H1)(v) follows as in [BT17, Lemma 3.3], with $J_{\mu _{{\bar{\phi }}}}$ and $K^{\mathbf {u}}_{W_j}$ instead of $|DF|^{-1}$ and $|DF|^{-1} J_{W_j}F$; this can be done because $K^{\mathbf {u}}_{W_j}$ is piecewise $C^1$ on unstable leaves by Lemma A.2.

1.2 Verifying (H2) and (H8)

Let $\omega :{{\mathbb {R}}}\rightarrow [0,1]$ be a function with ${\text {supp}}\omega \subset [-1,1]$ and $\int \omega (x) \, dx = 1$. To fix a choice that suffices for the present purpose we take $\omega = 1_{[0,1]}$; then $\gamma _1 = \frac{1}{2}$ and $\gamma _2 = \frac{1}{6}$ (see Remark 4.1) and in particular, $\omega (t-\tau )= 1_{\{t\le \tau \le t+1\}}$. The first result below verifies assumption (H2)(i).

Proposition A.3

There is $C_\omega$ such that $\int _Y \omega (t-\tau ) v\, d\mu _{{\bar{\phi }}} \le C_\omega \Vert v \Vert _{{{\mathcal {B}}}_w} \int _Y \omega (t-\tau ) d\mu _{{\bar{\phi }}}$ for all $t \ge 0$ and $v \in {{\mathcal {B}}}_w$.

Proof

By Lemma A.1, there is N (independent of t) and k(t) such that

$$\begin{aligned} E := \{ q \in f^{-1}(P_0){\setminus } P_0 : k(t) \le r(q) \le k(t)+N \} \end{aligned}$$

contains ${\text {supp}}(\omega (t-\tau ))$. We split $v = v^+-v^-$ into its positive and negative parts and treat them separately. Also we assume without loss of generality that $0 \le \omega \le 1$ (otherwise we split $\omega$ in a positive and negative part as well). Then $\int \omega (t-\tau ) \, v^+ \, d\mu _{{\bar{\phi }}} \le \int _E v^+ \, d\mu _{{\bar{\phi }}}$.

Let ${{\mathcal {W}}}^{\mathbf {s}}$ denote the stable foliation of F and decompose the measure $\mu _{{\bar{\phi }}}$ as $\int v^+ \, d\mu _{{\bar{\phi }}} = \int _{{{\mathcal {W}}}^{\mathbf {s}}} \int _{W^{\mathbf {s}}} v^+ d\mu _{W^{\mathbf {s}}} \, d\nu ^{\mathbf {u}}$. Note that E is the union of leaves in $\tilde{{\mathcal {W}}}^{\mathbf {s}}$, so $\mathbf{1}_E$ is constant on each $W^{\mathbf {s}} \in {{\mathcal {W}}}^{\mathbf {s}}$. Take $\varphi :E \rightarrow {{\mathbb {R}}}^+$ arbitrary such that $\varphi |_{W^{\mathbf {s}}} \in C^1(W^{\mathbf {s}})$ with $\Vert \varphi \Vert _{C^1(W^{\mathbf {s}})} \le 1$ for each $W^{\mathbf {s}} \in {{\mathcal {W}}}^{\mathbf {s}}\cap E$. Similar to [BT17, Proposition 3.1], we get

$$\begin{aligned} \int _E v^+ \varphi \, d\mu _{{\bar{\phi }}}= & {} \int _{{{\mathcal {W}}}^{\mathbf {s}} \cap E} \int _{W^{\mathbf {s}}} v^+ \varphi \, d\mu _{W^{\mathbf {s}}} \, d\nu ^{\mathbf {u}} \le \int _{{{\mathcal {W}}}^{\mathbf {s}} \cap E} \Vert v^+ \Vert _{{{\mathcal {B}}}_w} \Vert \varphi \Vert _{C^1(W^{\mathbf {u}})} \, d\nu ^{\mathbf {u}}\\\le & {} \Vert v^+\Vert _{{{\mathcal {B}}}_w} \mu _{{\bar{\phi }}}(E) \ll \Vert v^+\Vert _{{{\mathcal {B}}}_w} \int _{k(t)}^{k(t)+N} \int _Y \omega (s-\tau )\, d\mu _{{\bar{\phi }}} \, ds \\\ll & {} \Vert v^+\Vert _{{{\mathcal {B}}}_w} N \int _Y \omega (t-\tau )\, d\mu _{{\bar{\phi }}}. \end{aligned}$$

The same holds for $v^-$, and this ends the proof. $\quad \square$

The next result verifies assumption (H2)(ii).

Proposition A.4

Let $M(t) = R(\omega (t-\tau ))$. Then $\int _T^\infty \Vert M(t) \Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w} \, dt \ll T^{-\beta }$.

Proof

We need to estimate the ${{\mathcal {B}}}_w$-norm of $M(t)v = R(\omega (t-\tau ) v)$. This means that we need to take some leaf $W^{\mathbf {s}} \in {{\mathcal {W}}}^{\mathbf {s}}$, the collection of stable leaves in Y stretching across an element of the Markov partition of F, and compute (as in Lemma A.2) that

$$\begin{aligned} \int _{W^{\mathbf {s}}} M(t) v \cdot \varphi \, d\mu ^{\mathbf {s}} = \sum _j \int _{W^{\mathbf {s}}_j} \omega (t-\tau ) v \cdot \varphi \circ F \cdot K^{\mathbf {u}}_{W_j}\, d\mu ^{\mathbf {s}}. \end{aligned}$$

As in the proof of Proposition A.3, we can split $v = v^+-v^-$ and use that $\{ t-1 \le \tau \le t+1\} \subset E := \{k(t) \le r\le k(t)+N\}$. Thus, following [BT17, Proposition 3.2 (weak norm)] and [BT17, Equation (42)], the integral for $v^+$ is bounded by

$$\begin{aligned} \sum _{j=k(t)}^{k(t)+N} \int _{W_j} v^+ J_{W^j} F |DF|^{-1} \cdot \varphi \circ F \, d\mu ^{\mathbf {s}}\le & {} \Vert v^+\Vert _{{{\mathcal {B}}}_w} |\varphi |_{C^1(W^{\mathbf {s}})} \sum _{j=k(t)}^{k(t)+N} \int _{W_j} K^{\mathbf {u}}_{W_j} \, d\mu ^{\mathbf {s}} \\\ll & {} \Vert v^+\Vert _{{{\mathcal {B}}}} \sum _{j=k(t)}^{k(t)+N} \mu _{{\bar{\phi }}}(\{r= j\})\\\ll & {} \Vert v^+\Vert _{{{\mathcal {B}}}} \, N\, \mu _{{\bar{\phi }}}(\{r= k(t)\}). \end{aligned}$$

For $v^+$ and $v^-$ together, this gives $\int _T^\infty \Vert M(t) \Vert _{{{\mathcal {B}}}\rightarrow {{\mathcal {B}}}_w} dt \ll \int _T^\infty \mu _{{\bar{\phi }}}(\{r= k(t)\})\, dt \ll \mu _{{\bar{\phi }}}(\{\tau \ge T \})$ and the proposition follows. $\quad \square$

Assumption (H8) is the same as (H2), with $\tau$ replaced by $\psi _0$ and $e^{-u\tau }$ by $e^{-s\psi _0}$. Its verification is entirely analogous to the above.

1.3 Verifying that $R\zeta \in {{\mathcal {B}}}$ for a large class of $\zeta$ (including ${\bar{\psi }}$ in Theorem 2.5 (b)) and completing the verification of (H4)

Assumption (H4)(i) and the tail estimate part of (H4)(ii) (so that $\tau \in L^2(\mu _{{\bar{\phi }}})$) is verified in Proposition 2.4. To check the remaining part of (H4)(ii) and assumption on ${\bar{\psi }}$ in Theorem 2.5 (b) we give a more general result in Lemma A.5 below. This result ensures that given ${\bar{\psi }}$ as in Theorem 2.5 (b) we have $R{\bar{\psi }}\in {{\mathcal {B}}}$. This is needed to verify the abstract assumptions of Theorem 8.2(b) for ${\bar{\psi }}$ as in Theorem 2.5(b).

Lemma A.5

For $0< \kappa < \beta$ and $\zeta : Y \rightarrow {{\mathbb {R}}}$ piecewise $C^1$, define

$$\begin{aligned} \Vert \zeta \Vert _0 := \sup _{j \ge 1} \frac{1}{j^{\kappa }}\left( \Vert \zeta \, 1_{Y_j} \Vert _{{{\mathcal {B}}}_w} + \frac{1}{j^{1+\beta }} \Vert \zeta \, 1_{Y_j} \Vert _{\mathbf {u}} \right) \end{aligned}$$

and the Banach space ${{\mathcal {B}}}_0 = \{ \zeta :Y \rightarrow {{\mathbb {R}}}: \zeta \text { is piecewise } C^1 \text { and } \Vert \zeta \Vert _0 < \infty \}$. Then $R({{\mathcal {B}}}_0) \subset {{\mathcal {B}}}$.

Proof

Recall that, by Proposition 2.2 and Lemma 2.3, $r= O(\tau )$ and vice versa. Abbreviate $Y_j = \{ r= j\}$; for large j these are strips close to the stable manifold $W^{\mathbf {s}}_p$ of the neutral fixed point, and bounded by stable and unstable curves and both $F^{-1}$ and $|DF|^{-1}$ are $C^1$ on each $\overline{F(Y_j)}$.

For the stable norm $\Vert \ \Vert _{\mathbf {s}}$, choose an arbitrary stable leaf W and q-Hölder function $\varphi \in C^q(W)$ with $|\varphi |_{C^q(W)} \le 1$. Let $W_j = F^{-1}(W) \cap Y_j$. By Lemma A.2, and the fact that $K^{\mathbf {u}} \ll j^{-(1+\beta )}$ and $|\varphi \circ F|_{C^1} \ll |\varphi |_{C^1}$ on each $W_j$, we have

$$\begin{aligned} \int _W R\zeta \, \varphi \, d\mu ^{\mathbf {s}}= & {} \sum _j \int _{W_j} \zeta \, \varphi \circ F \,K^\mathbf {u}\, d\mu ^{\mathbf {s}} \ll \sum _j j^{-(1+\beta )} \int _{W_j} \zeta \, \varphi \circ F \, d\mu ^{\mathbf {s}} \nonumber \\\ll & {} \sum _j j^{-(1+\beta )} \Vert \zeta \, 1_{Y_j} \Vert _{{{\mathcal {B}}}_w} \ll \sum _j j^{-1+\kappa -\beta } < \infty . \end{aligned}$$

(A.8)

The stable part $\Vert \, \Vert _{\mathbf {s}}$ of $\Vert \, \Vert _{{{\mathcal {B}}}}$ is treated in the same way.

Now for the unstable part $\Vert \, \Vert _{\mathbf {u}}$, let $\varphi$ such that $|\varphi |_{C^1(W)}\le 1$. Using [BT17, Equation (43)], which expresses $\int _{W_j} h \phi \, dm$ in terms of the parametrisation of $W_j$, for any nearby leaves $W,{\tilde{W}}\in Y$, we define a diffeomorphism $v:{\tilde{W}} \rightarrow W$ as in [BT17, Proof of Proposition 3.2 (unstable norm part)]. Let $v_j = F^{-1} \circ v \circ F:{\tilde{W}}_j \rightarrow W_j$ be the corresponding bijection between the preimage leaves $W_j, {\tilde{W}}_j \subset Y_j$. Then we compute,

$$\begin{aligned} \left| \int _{{\tilde{W}}} R\zeta \varphi \, d\mu ^{\mathbf {s}} - \int _W R\zeta \varphi \, d\mu ^{\mathbf {s}} \right|&\le \sum _j \left| \int _{{\tilde{W}}_j} \zeta \varphi \circ F \, K^{\mathbf {u}}_{{\tilde{W}}_j} \, d\mu ^{\mathbf {s}} - \int _{W_j} \zeta \varphi \circ F \, K^{\mathbf {u}}_{W_j} \, d\mu ^{\mathbf {s}} \right| \\&\le \sum _j \int _{{\tilde{W}}_j} \zeta |\varphi \circ v \circ F - \varphi \circ F| \, K^{\mathbf {u}}_{{\tilde{W}}_j} \, d\mu ^{\mathbf {s}} \\&\quad + \sum _j \int _{{\tilde{W}}_j} \zeta \, |\varphi \circ F| \, \Big | K^{\mathbf {u}}_{W_j} \circ v_j - K^{\mathbf {u}}_{{\tilde{W}}_j} \Big | \, d\mu ^{\mathbf {s}} \\&\quad + \sum _j \int _{W_j} |\zeta \circ v_j - \zeta | |\varphi \circ F| \, K^{\mathbf {u}}_{W_j} \, d\mu ^{\mathbf {s}} \\&= S_1 + S_2 + S_3. \end{aligned}$$

Next

$$\begin{aligned} S_1 \le | \varphi |_{C^1}\, d(W, {\tilde{W}}) \sum _j |\zeta |_{{\tilde{W}}_j}|_\infty \, |K^{\mathbf {u}}_{{\tilde{W}}_j} |_\infty \ll \Vert \varphi \Vert _{C^1}\, d(W, {\tilde{W}}), \end{aligned}$$

because as in the first part of this proof, the sum in the above expression is bounded.

For the sum $S_2$, using (A.7) we split

$$\begin{aligned} \Big |K^{\mathbf {u}}_{{\tilde{W}}_j} - K^{\mathbf {u}}_{W_j} \circ v_j \Big |&= \frac{1}{J^{\mathbf {u}}_{m_{{\bar{\phi }}}}} \frac{| (h^{\mathbf {u}} \sin \varvec{{\gamma }}) \circ F \circ v_j -(h_0^{\mathbf {u}} \sin \varvec{{\gamma }}) \circ F |}{h_0^{\mathbf {u}} \sin \varvec{{\gamma }}}\\&\quad + \frac{1}{ J^{\mathbf {u}}_{m_{{\bar{\phi }}}} } \left| \frac{J^{\mathbf {u}}_{m_{{\bar{\phi }}}} }{J^{\mathbf {u}}_{m_{{\bar{\phi }}}} \circ v_j} - 1 \right| \frac{(h_0^{\mathbf {u}} \sin \varvec{{\gamma }}) \circ F \circ v_j}{h_0^{\mathbf {u}} \sin \varvec{{\gamma }}}\\&\quad + \frac{1}{J^{\mathbf {u}}_{m_{{\bar{\phi }}}}} \frac{(h^{\mathbf {u}} \sin \varvec{{\gamma }}) \circ F \circ v_j}{(h_0^{\mathbf {u}} \sin \varvec{{\gamma }})\circ v_j} \left| \frac{(h^{\mathbf {u}} \sin \varvec{{\gamma }}) \circ v_j}{h_0^{\mathbf {u}} \sin \varvec{{\gamma }}} - 1 \right| . \end{aligned}$$

By distortion estimate [BT17, Equation (40)], this is bounded by $C d(W,{\tilde{W}})\frac{1}{J^{\mathbf {u}}_{m_{{\bar{\phi }}}}}$ for some uniform distortion constant $C > 0$. Therefore

$$\begin{aligned} S_2 \le C \, d(W,{\tilde{W}}) \sum _j \Vert \zeta \, 1_{Y_j} \Vert _{{{\mathcal {B}}}_w} \, \Big | \frac{1}{J^{\mathbf {u}}_{m_{{\bar{\phi }}}}} \Big |_\infty \ll d(W,{\tilde{W}}) \sum _j j^\kappa j^{-(1+\beta )} < \infty \end{aligned}$$

as before.

Now for $S_3$, the weighted $\Vert \, \Vert _{\mathbf {u}}$ part of the norm $\Vert \ \Vert _0$ gives $|\zeta \circ v_j - \zeta | \ll j^{\kappa +1+\beta } d(W_j, {\tilde{W}}_j) \ll j^{\kappa +1+\beta } d(W, {\tilde{W}}) L_{\mathbf {u}}(Y_j)$. As in the first half of the proof, $|K^{\mathbf {u}}_{W_j}|_\infty \ll j^{-(1+\beta )}$. We have $L_{\mathbf {u}}(Y_j) \ll \tau ^{1+\beta }$ due to the small tail estimates (2.3), so

$$\begin{aligned} S_3 \ll |\varphi |_\infty d(W, {\tilde{W}}) \sum _j j^{\kappa } L_{\mathbf {u}}(Y_j) \ll |\varphi |_\infty d(W, {\tilde{W}}). \end{aligned}$$

Therefore $\Vert R\zeta \Vert _{{\mathcal {B}}}< \infty$ and the proof is complete. $\square$

Corollary A.6

$R(\tau ^\kappa h) \in {{\mathcal {B}}}$ for each $0< \kappa < \beta$ and $h \in {{\mathcal {B}}}$ (in particular, (H4)(ii) holds).

Since $C' - {\bar{\psi }} \sim C \tau ^\kappa$, this corollary can also be used to show that $R({\bar{\psi }} h) \in {{\mathcal {B}}}$.

Proof

Since $\tau$ is the return time of a $C^1$ flow to a $C^1$ Poincaré section, it is piecewise $C^1$, and we know $\tau |_{Y_j} \ll j$. Taking $h \in {{\mathcal {B}}}$ and $\varphi \in C^1(W)$ for an arbitrary stable leaf $W \subset Y_j$, we have $\int _W R(\tau ^ h) \, \varphi \, d\mu ^{\mathbf {s}} \le j^\kappa \int _W h \varphi \, d\mu ^{\mathbf {s}} = j^\kappa | h \Vert _{{{\mathcal {B}}}_w}$, so $\frac{1}{j^\kappa } \Vert \tau ^\kappa h \Vert _{{{\mathcal {B}}}_w} \ll \Vert h \Vert _{{{\mathcal {B}}}_w} < \infty$.

Using (2.3), we have $\epsilon \le \xi (y,T+1) - \xi (y,T) \ll T^{-(1+\beta )}$ for $T = \tau (x,y) + O(1)$ and $\xi (y,T)$ plays the role of x. By (2.4), $x = \xi _0(y) \tau (x,y)^{-\beta }(1+o(1))$ as $x \rightarrow 0$, so $\epsilon \ll x^{\frac{1+\beta }{\beta }}$. Using (2.4) again, we can estimate that as $\tau = \tau (x) \rightarrow \infty$ (so as $x \rightarrow 0$ and $\epsilon = o(x)$),

$$\begin{aligned} |\tau ^\kappa (x+\epsilon ,y) - \tau ^\kappa (x,y)|= & {} \xi _0(y)^\kappa |(x+\epsilon )^{-\frac{\kappa }{\beta }} - x^{-\frac{\kappa }{\beta }}|(1+o(1)) \nonumber \\= & {} \xi _0(y)^\kappa x^{-\frac{\kappa }{\beta }} \frac{\kappa }{\beta } \frac{\epsilon }{x}\left( 1+o(1)+O\left( \frac{\epsilon }{x}\right) \right) \nonumber \\= & {} \frac{\kappa }{\beta } \xi _0(y)^{-\frac{\kappa }{\beta }} \tau (x,y)^{\kappa +\beta } \epsilon \left( 1+o(1)+O\left( \frac{\epsilon }{x}\right) \right) .\quad \end{aligned}$$

(A.9)

Let $W, {\tilde{W}}$ nearby stable leaves in $Y_j$ with $\epsilon = d(W,{\tilde{W}})$ and $v:{\tilde{W}} \rightarrow W$ a diffeomorphism. Then, for $\varphi \in C^1(W)$, ${\tilde{\varphi }} \in C^1({\tilde{W}})$ with $d(\varphi , {\tilde{\varphi }}) \le d(W,{\tilde{W}})$,

$$\begin{aligned}&\int _{{\tilde{W}}} \tau ^k \, h \, {\tilde{\varphi }} \, d\mu ^{\mathbf {s}} - \int _{W} \tau ^k \, h \, \varphi \, d\mu ^{\mathbf {s}}\\&\quad \ll \int _{{\tilde{W}}} |\tau ^\kappa - \tau ^\kappa \circ v_j| \, h \, \varphi \, d\mu ^{\mathbf {s}} + j^{\kappa } \left( \int _{{\tilde{W}}} h \, {\tilde{\varphi }} \, d\mu ^{\mathbf {s}} - \int _{W} h \, \varphi \, d\mu ^{\mathbf {s}} \right) \\&\quad \ll j^{\kappa +\beta } \Vert h \Vert _{{{\mathcal {B}}}_w} + j^{\kappa } \Vert h \Vert _{\mathbf {u}}. \end{aligned}$$

Therefore $\frac{1}{j^{\kappa +\beta }} \Vert \tau ^\kappa h \Vert _{\mathbf {u}} \ll \Vert h \Vert _{{{\mathcal {B}}}_w} < \infty$, showing that $\tau ^\kappa h \in {{\mathcal {B}}}_0$. Combined with Lemma A.5, the result follows. $\quad \square$

As in the statement of Proposition 5.1 (ii) and Proposition 5.7 we need

Corollary A.7

${\tilde{\tau }} := \tau -\tau ^*$ cannot be written as $h\circ F-h$ for any $h\in {{\mathcal {B}}}$.

Proof

By Corollary A.6, $R{\tilde{\tau }}\in {{\mathcal {B}}}$. Because $\tau (z) \rightarrow \infty$ as $z \rightarrow W^s(p)$, $z \in F^{-1}(P_0) {\setminus } P_0$, we have $\sup _{W \in {{\mathcal {W}}}^{\mathbf {s}}} \int _W {\tilde{\tau }} \, d\mu ^{\mathbf {s}} = \infty$. Therefore ${\tilde{\tau }} \notin {{\mathcal {B}}}_w$ and hence ${\tilde{\tau }}$ is not a coboundary.

$\square$

1.4 Verifying (H7)

Extending the inequality $|1-e^{-x}| \le x^\gamma$ for all $\gamma \in (0,1]$ and $x \ge 0$, we can find $C_\gamma$ depending only on $s\sup _a{\bar{\psi }}$ such that

$$\begin{aligned} |e^{-u\tau + s{\bar{\psi }}}-1| 1_a \le C_\gamma (u\tau + s\psi _0)^\gamma \le C_\gamma \left( u^\gamma \sup _a \tau ^\gamma + s^\gamma \sup _a \psi _0^\gamma \right) . \end{aligned}$$

Therefore, for each $h \in {{\mathcal {B}}}$, and $\varphi \in C^1(W)$, $W \in {{\mathcal {W}}}^{\mathbf {s}}$,

$$\begin{aligned} \int _W \left| (e^{-u\tau + s{\bar{\psi }}}-1) 1_a h \, \varphi \right| \, d\mu ^{\mathbf {s}} \le C_\gamma \left( u^\gamma \sup _a \tau ^\gamma + s^\gamma \sup _a \psi _0^\gamma \right) \int _W h \, \varphi \, d\mu ^{\mathbf {s}}, \end{aligned}$$

so (H7) follows for $C_2 = C_3 = C_\gamma$.

1.5 Verifying (H6) (and thus, (H3))

In this section we verify (H6) for $\kappa >1/\beta$.

Proposition A.8

Assume that ${\bar{\psi }}$ satisfies (H5) and let ${\hat{R}}(u,s)v = R(e^{-u\tau }e^{s{\bar{\psi }}} v)$. Then there exists $\sigma _1 \in (0,1)$ and $\delta , C_0, C_1 > 0$ such that for all $h\in {{\mathcal {B}}}$, $n\in {{\mathbb {N}}}$, $0 \le s < \delta$, and $u \ge 0$,

$$\begin{aligned} \Vert {\hat{R}}(u,s)^n h\Vert _{{{\mathcal {B}}}_w} \le C_1 e^{-un} \Vert h\Vert _{{{\mathcal {B}}}_w},\quad \Vert {\hat{R}}(u,s)^n h\Vert _{{{\mathcal {B}}}} \le e^{-un}(C_0 \sigma _1^{-n}\Vert h\Vert _{{{\mathcal {B}}}}+C_1 \Vert h\Vert _{{{\mathcal {B}}}_w}). \end{aligned}$$

Before turning to the proof, we need another lemma (which we will apply with $g \equiv 1$, but the general g is needed for the induction in the proof).

Lemma A.9

Assume that ${\bar{\psi }}$ satisfies (H5) (in particular, ${\bar{\psi }}< C'-C^{-1}\tau ^\kappa < \infty$ for some $C',C>0$ and $\kappa > 1/\beta$). There exists $N \in {{\mathbb {N}}}$ depending only of $F:Y \rightarrow Y$ such that for all positive integrable functions g that are bounded and bounded away from zero, the following holds. There exist $\delta > 0$ such that for all $s \in [0,\delta ]$, all admissible stable leaves $W \subset Y$ and all $n \in N$,

$$\begin{aligned} \sum _{{\mathop {\tiny \text { of } F^{-n}(W)}\limits ^{W' \tiny \, \text { component}}}} \int _{W'} e^{s{\bar{\psi }}_n} \, g \circ F^n \, K^\mathbf {u}_{W'}F^n \, d\mu ^{\mathbf {s}} \le e^{sN \sup {\bar{\psi }}} \int _W g\, d\mu ^{\mathbf {s}}, \end{aligned}$$

where $K^\mathbf {u}_{W'}F^n$ equals the analogue of $K^\mathbf {u}_{W'}$ from (A.7) for the iterate $F^n$ on the preimage leaf $W'$.

Proof

Let $P_1$ be the partition element containing the images $F(\{ r= k\})$ of the strips $\{ r=k\}_{k \ge 2}$, see Fig. 3. Let $N \in {{\mathbb {N}}}$ be such that $F^{N-1}(P_1)$ intersects each $P_i$, $i \ge 1$. Let W be an arbitrary stable leaf in $P_i$ for some $i \ge 1$, and let ${{\mathcal {W}}}$ be the collection of preimage leaves under $F^{-N}$. By a change of coordinates,

$$\begin{aligned} \sum _{W' \in {{\mathcal {W}}}} \int _{W'} g \circ F^N \, K^\mathbf {u}_{W'}F^n \, d\mu ^{\mathbf {s}} = \int _W g \, d\mu ^{\mathbf {s}} \end{aligned}$$

for any integrable function g.

Given $a \ge 1$ to be chosen later, set ${{\mathcal {W}}}^+ = \{ W' \in {{\mathcal {W}}}: \inf _{W'} \tau ^\kappa \ge 2NaC \sup {\bar{\psi }} \}$ and ${{\mathcal {W}}}^- = {{\mathcal {W}}}{\setminus } {{\mathcal {W}}}^+$. Then there is $\epsilon > 0$ (depending only on the geometry of the Markov map F and $\frac{\sup g}{\inf g}$) such that $\sum _{W' \in {{\mathcal {W}}}^+} \int _{W'} g \circ F^N\, K^\mathbf {u}_{W'}F^N \, d\mu ^{\mathbf {s}} \ge \epsilon \int _W g\, d\mu ^{\mathbf {s}}$. Since $\int _{W'}\, d\mu ^{\mathbf {s}}$ is proportional to the measure of the element in $\bigvee _{i=0}^{N-1} F^{-i}({\mathcal {P}})$ that $W'$ belongs to, Proposition 2.4 gives $\epsilon \gg \mu _{{\bar{\phi }}}(\{ \tau > (2NaC \sup {\bar{\psi }})^{\frac{1}{\kappa }} \}) \gg Ba^{-\frac{1}{\beta '}}$ for some $B > 0$ and $\beta ' := \kappa \beta > 1$. We have ${\bar{\psi }}_N \le (N-1) \sup {\bar{\psi }} - C^{-1}\tau ^\kappa \le (N-1-2Na) \sup {\bar{\psi }} \le -Na\sup {\bar{\psi }}$ on ${{\mathcal {W}}}^+$, and ${\bar{\psi }}_N \le N \sup {\bar{\psi }}$ on ${{\mathcal {W}}}^-$. Therefore

$$\begin{aligned}&\sum _{W' \in {{\mathcal {W}}}} \int _{W'} e^{s{\bar{\psi }}_N} \, g \circ F^N \, K^{\mathbf {u}}_{W'}F^N\, d\mu ^{\mathbf {s}} \le \sum _{W' \in {{\mathcal {W}}}^+} \int _{W'} e^{- s N a \sup {\bar{\psi }}} \,g \circ F^N \, K^{\mathbf {u}}_{W'}F^N\, d\mu ^{\mathbf {s}} \\&\quad + \sum _{W' \in {{\mathcal {W}}}^-} \int _{W'} e^{s N \sup {\bar{\psi }}} \, g \circ F^N \, K^{\mathbf {u}}_{W'}F^N\, d\mu ^{\mathbf {s}} \\&\quad \le \epsilon e^{-s N a \sup {\bar{\psi }}} \int _W g \, d\mu ^{\mathbf {s}} + (1-\epsilon ) e^{s N \sup {\bar{\psi }}} \int _W g \, d\mu ^{\mathbf {s}} \\&\quad = \left( \epsilon x^{-a} + (1-\epsilon )x \right) \int _W g \, d\mu ^{\mathbf {s}} =: b(x) \int _W g \, d\mu ^{\mathbf {s}} \end{aligned}$$

for $x = e^{s N \sup {\bar{\psi }}}$. Clearly $b(1) = 1$ and $b'(x) = -a\epsilon x^{-(a+1)} + (1-\epsilon ) < 0$ whenever $0 < x \le (a\epsilon )^{\frac{1}{1+a}}$. Since $\epsilon \ge B a^{-\frac{1}{\beta '}}$, we find $b'(x) < 0$ for all $x \le a_0 := B^{-\frac{1}{1+a}} a^{\frac{\beta '-1}{\beta '(1+a)}}$. Choose $a > \max \{1, B^{\frac{\beta '}{\beta '-1}} \}$, so that $a_0 > 1$. This means that $b(x) \le 1$ for all $1 \le x \le a_0$, i.e., for all $0 \le s \le \frac{\log a_0}{N \sup {\bar{\psi }}}$.

Now for general $n = pN + q$ with $0 \le q < N$, we use induction on p. Let ${{\mathcal {W}}}_p(W)$ be the collection of preimage leaves of W under $F^{-pN}$. Assume by induction that

$$\begin{aligned} \sum _{W' \in {{\mathcal {W}}}_{p-1}(W_1)} \int _{W'} e^{s{\bar{\psi }}_{(p-1)N}} \, g \circ F^{(p-1)N} \, K^{\mathbf {u}}_{W'}F^{(p-1)N} \, d\mu ^{\mathbf {s}} \le \int _W g \, d\mu ^{\mathbf {s}} \end{aligned}$$

for all stable leaves $W_1$ and g as above. Then

$$\begin{aligned}&\sum _{W' \in {{\mathcal {W}}}_p(W)} \int _{W'} e^{s{\bar{\psi }}_{pN}} \, g \circ F^{pN} \, K^{\mathbf {u}}_{W'}F^{pN} \, d\mu ^{\mathbf {s}} \\&\quad \le \sum _{W_1 \in {{\mathcal {W}}}_1(W)} \sum _{W' \in {{\mathcal {W}}}_{p-1}(W_1)} \int _{W'} e^{s{\bar{\psi }}_{(p-1)N}} K^{\mathbf {u}}_{W'}F^{(p-1)N} \left( e^{s{\bar{\psi }}_N} \, g \circ F^N K^{\mathbf {u}}_{W'}F^N \right) \circ F^{(p-1)N} \, d\mu ^{\mathbf {s}} \\&\quad \le \sum _{W_1 \in {{\mathcal {W}}}_1(W)} \int _{W_1} e^{s{\bar{\psi }}_N} \, g \circ F^N \, K^{\mathbf {u}}_{W_1}F^N\, d\mu ^{\mathbf {s}} \le \int _W g\, d\mu ^{\mathbf {s}}. \end{aligned}$$

This way we proved the lemma for all multiples of N. For $0< q < N$, we get an extra factor $e^{sq \sup {\bar{\psi }}}$. This proves the lemma. $\square$

Proof of Proposition A.8

This proof goes as in [BT17, Proposition 3.2], but with some changes. The factor $z^n$ is to be replaced with $e^{-u\tau _n}$ where $\tau _n = \sum _{i=0}^{n-1} \tau \circ F^i$ and the factor $e^{s{\bar{\psi }}_n}$ for $\psi _n = \sum _{i=0}^{n-1} \psi \circ F^i$ is dealt with using Lemma A.9. Let W and ${\tilde{W}}$ be two stable leaves in the same partition element of $\{ P_i \}_{i \ge 1}$. Since $\tau$ and $\psi$ are not constant on partition elements, we get a third and fourth term in the strong unstable norm part of [BT17, Proposition 3.2] expressing the difference of $\tau _n$ on nearby preimage leaves ${\tilde{W}}_j$ and $W_j = v_j({\tilde{W}}_j)$ of W and ${\tilde{W}}$:

$$\begin{aligned} S_3 = \sum _j \left| \int _{{\tilde{W}}_j} h {\tilde{\varphi }} \circ F^n e^{s {\bar{\psi }}_n}\, (e^{-u \tau _n}-e^{-u(\tau _n \circ v_j)} ) \, K^{\mathbf {u}}_{W_j}F^n\, d\mu ^{\mathbf {s}} \right| \end{aligned}$$

(A.10)

and

$$\begin{aligned} S_4 = \sum _j \left| \int _{{\tilde{W}}_j} h {\tilde{\varphi }} \circ F^n e^{-u\tau _n \circ v_j}\, (e^{s{\bar{\psi }}_n}-e^{s {\bar{\psi }}_n \circ v_j})\, K^{\mathbf {u}}_{W_j}F^n\, d\mu ^{\mathbf {s}} \right| , \end{aligned}$$

(A.11)

for some ${\tilde{\varphi }}$ with $|{\tilde{\varphi }}|_{C^1({\tilde{W}})} \le 1$.

For $S_3$, without loss of generality, we can assume that $\tau _n \circ v_j \ge \tau _n$, so $|e^{-u \tau _n}-e^{-u(\tau _n \circ v_j)}| \le ue^{-u\tau _n} (\tau _n \circ v_j - \tau _n)$. By (A.9), we have $|\tau (x+\epsilon ,y) - \tau (x,y)| = \beta ^{-1} \xi _0(y)^{-\frac{1}{\beta }} \tau (x,y)^{1+\beta } \epsilon (1+o(1)+O(\frac{\epsilon }{x}))$. Applied to $\epsilon _i := d(F^i(W_j), F^i({\tilde{W}}_j)) \ll \tau ^{-(1+\beta )} \circ F^i \epsilon _{i+1}$ this gives

$$\begin{aligned} \sum _{i=0}^{n-1} | \tau \circ F^i \circ v_j - \tau \circ F^i|&\ll \sum _{i=0}^{n-1} \tau ^{1+\beta } \circ F^i\ d(F^i(W_j), F^i({\tilde{W}}_j)) \nonumber \\&\ll \sum _{i=0}^{n-1} \epsilon _{i+1} \ll \epsilon _n = d(W,{\tilde{W}}), \end{aligned}$$

(A.12)

and therefore $|e^{-u\tau _n}-e^{-u \tau _n \circ v_j} | \ll u e^{-u\tau _n} d(W,{\tilde{W}})$. Combining the above with (A.10), we obtain

$$\begin{aligned} S_3&\le u\ d(W,{\tilde{W}}) \sum _j \left| \int _{{\tilde{W}}_j} h {\tilde{\varphi }} \circ F^n e^{-u\tau _n + s \sup {\bar{\psi }}_n} \, K^{\mathbf {u}}_{W_j}F^n\, d\mu ^{\mathbf {s}} \right| \\&\ll \Vert h \Vert _{{{\mathcal {B}}}_w} \, u e^{-un}\, d(W,{\tilde{W}}) \sum _j \left| \int _{{\tilde{W}}_j} e^{s{\bar{\psi }}_n} \, K^{\mathbf {u}}_{W_j}F^n\, d\mu ^{\mathbf {s}} \right| \\&\le \Vert h \Vert _{{{\mathcal {B}}}_w} \, u e^{-un}\, d(W,{\tilde{W}}) e^{sN \sup {\bar{\psi }}}, \end{aligned}$$

where we used that $\tau _n \ge n$, and the last step follows from Lemma A.9 with the N (independent of n, u, s) taken from that lemma too.

For $S_4$, using (H5) and (A.9), we obtain:

$$\begin{aligned} |e^{s{\bar{\psi }}(x+\epsilon )}-e^{s{\bar{\psi }}(x))}|\ll & {} e^{s \sup {\bar{\psi }}} e^{-s\tau ^\kappa /C} s|\tau ^\kappa (x+\epsilon ,y) - \tau ^\kappa (x,y)| \\= & {} e^{s \sup {\bar{\psi }}} s e^{-s\tau ^\kappa /C}\, \frac{\kappa }{\beta } \, \xi _0(y) \tau ^{\kappa +\beta } \epsilon (1+o(1)+O(\epsilon )) \\\ll & {} e^{s \sup {\bar{\psi }}} \underbrace{s e^{-s\tau ^\kappa /C} \tau ^{\kappa -1}}_K \, \tau ^{1+\beta } \epsilon , \end{aligned}$$

where we compute the supremum over $\tau$ to conclude that $K \le e^{-\kappa } (C\kappa )^{\frac{\kappa -1}{\kappa }} s^{\frac{1}{\kappa }}$. Apply the above with $\epsilon = \epsilon _i := d(F^i(W_j), F^i({\tilde{W}}_j))$ as in (A.12). Then we can estimate $S_4$ in the same way as $S_3$:

$$\begin{aligned} S_4\le & {} s^{\frac{1}{\kappa }}\ d(W,{\tilde{W}}) \sum _j \left| \int _{{\tilde{W}}_j} h {\tilde{\varphi }} \circ F^n e^{-u\tau _n + s{\bar{\psi }}_n} \, K^{\mathbf {u}}_{W_j}F^n\, d\mu ^{\mathbf {s}} \right| \\\ll & {} \Vert h \Vert _{{{\mathcal {B}}}_w} e^{-un} e^{s N \sup {\bar{\psi }}}\, s^{\frac{1}{\kappa }} \, d(W,{\tilde{W}}). \end{aligned}$$

$\square$