Decreasing families of dynamically determined intervals in the power-law
  family
7 Decreasing families of dynamically determined
intervals in the power-law family
Waldemar Pa luba∗
Institute of Mathematics
Warsaw University
Banacha 2
02-097 Warsaw, Poland
e-mail: paluba@mimuw.edu.pl
Abstract
We study the rate of growth of ratios of intervals delimited by
the post-critical orbit of a map in the quasi-quadratic family x 7→
−|x|α+a. The critical order α is an arbitrary real number α > 1. The
range of the parameter a is confined to an interval (1, aα) of length
depending on the critical order. We prove that in every power-law
family there is a unique parameter pα corresponding to the kneading
sequence RLRRRLRC. Subsequently, we obtain monotonicity results
concerning ratios of all intervals labeled by infinite post-critical orbit
in the case of the kneading sequence RLRL... This extends the results
from [9], via refinement of the tools based on special properties of
power-law mappings in non-euclidean metric.
Mathematics Subject Classification (2000): Primary 37D05.
1 Introduction
In this paper we continue our work done in [9] on families of unimodal quasi-
quadratic maps of the form fa(x) = −|x|
α + a, with a real parameter a and
an arbitrary – in general non-integer – fixed exponent α > 1 .
∗Partially supported by a KBN grant no. 2 PO3A 010 22.
http://arxiv.org/abs/0704.1385v1
The problem of monotone behaviour of the dynamics in such a family has
been first successfully solved for the strictly quadratic case α = 2 . The tools
initially developed for the quadratic case (see e.g [2], [6], [10], also [11], and
an independent attempt, partly relying on numerical evidence in [1]) were
broadly generalized in the work of Kozlovski-Shen-van Strien, see [3]. There
are also very interesting recent results by G.Levin, concerning uniqueness
of appearance of periodic orbits of given multiplier in the quadratic family
z2 + c. Not only was he able to give a simple proof of Douady-Hubbard-
Sullivan theorem (cf.[4]), but he could continue somewhat beyond the hyper-
bolic domains in the Mandelbrot set also, see [5]. In this work, we focus on
questions closely related to these of Levin’s, though only orbits of periods 2
or 4 appear here. In return, working with real variable tools, we can do all
critical degrees, integer or non-integer, indiscriminately.
Despite of a great deal of progress achieved in the aforementioned papers,
and in other works as well, virtually all those developments are inherently
limited to the case of integer critical degrees. Non-integers clearly require
a fresh and different approach. For any real number α > 1 the power-law
map x 7→ |x|α has negative Schwarzian derivative, and hence it expands the
non-euclidean lengths. This observation has long become one of the key tools
in one-dimensional dynamics. However, the power-law is not just a negative
Schwarzian map. It is a homogeneous map, and in the Poincaré metric with
the element dt
on the positive half-line (0,∞), it is nothing but a linear
map acting as multiplication by the coefficient α, once we set the origin of
the Poincaré coordinates at 1. This simple fact is rather hard to make use
of in a direct way, but carries some strong consequences that can be applied
in a dynamical setting.
In our previous paper on this subject (see [9]), we introduced the tech-
nique of indirect use of linearity of the power-law map in the non-euclidean
metric and exemplified its usefulness in dynamics. There, we studied maps
in the one-parameter quasi-quadratic family fa with the kneading sequence
RRR . . . , that is for the value of the parameter a smaller than 1 . For the
infinite decreasing family of intervals with endpoints labeled by the succes-
sive points of the post-critical orbit we proved that the ratio of any two such
intervals is a function monotone in parameter a . This means, we studied the
situation which arises before the orbit of the critical point becomes a super-
stable orbit of period 2 . It is clear that this period 2 super-sink situation
arises only once in our family.
In the current work, we further develop our tools in order to examine
the case of some parameters greater than 1, where the length of the interval
(1, aα) to which those parameters are confined depends on the critical order
α > 1 . In particular we shall be able to deal with kneading sequences of the
form RLRL . . . , proving monotonicity of the ratios, in the respective decreas-
ing families, of intervals delimited by the post-critical orbit. As a step in the
build-up of the above techniques, we shall also establish uniqueness of the
period 8 super-sink, corresponding to the kneading sequence RLRRRLRC,
in every power-law family (even when it does not admit a holomorphic ex-
tension!); uniqueness of the period 4 super-sink RLRC is elementary and
follows along the way.
2 Notation and preliminaries
To begin with, we set some notation in conformance with that of [9]. The
names non-euclidean and Poincaré we be used interchangeably.
The Poincaré coordinate of a point x in an oriented, open interval (p, q)
will be denoted by
pp,q(x) = ln
q − x
and respectively pp,∞(x) = ln(x− p); also pp,−∞(x) = ln |x− p|.
To single out the non-euclidean metric on the half-line, which turns the
mapping h into a linear map, we will coin the term nonlinearity of an interval
for the length of this interval measured in the Poincaré metric on (0,∞).
Under this convention, the integral of nonlinearity of h over an interval (p, q)
equals, up to a multiplicative constant, to the nonlinearity of the domain of
integration.
Given an orientation preserving homeomorphism ϕ : (p, q) → (r, s) we
shall observe the ‘bar’ notation for its counterpart in the non-euclidean co-
ordinates, i.e. the mapping ϕp,q : R → R defined by the formula
ϕp,q(t) = pr,s
p−1p,q(t)
The non-euclidean push of ϕ at a point x ∈ (p, q) is, by definition, the
quantity
pr,s (ϕ(x)) − pp,q(x) .
By the strength of a push we mean its absolute value.
By ϕ+p,q and ϕ
p,q we denote the finite or infinite limits
ϕ+p,q = lim
(ϕ(t) − t)
ϕ−p,q = lim
(ϕ(t) − t) ,
provided they exist.
When ϕ is the restriction of the homogeneous map h(x) = xα to an inter-
val (p, q) ⊆ (0,∞) we shall always put h or h in place of ϕ or ϕ respectively.
For a fixed exponent α > 1, let fa = −|x|
α + a , and the successive points
of the orbit of the critical point will simply be denoted by na = f
a (0) .
Moreover, homogeneity of the power-law map allows for the linear change
of coordinates, na 7→ na/1a , so that we can set 0a = 0, 1a = 1 and the
dependence on the parameter a turns into the dependence on the value of
2a in these new coordinates. For a > 1 this rescaled value of 2a is in the
interval (−1, 0) – so long as the post-critical orbit does not escape to infinity
– and the quantity p0,1(|2a|), which for obvious reason we will denote by ā , is
increasing simultaneously with a. Throughout this work, this very quantity
will be chosen as our new parameter, and it is always tacitly assumed that
the rescaling na 7→ na/1a has been done.
We now record several observations concerning one-dimensional non-euclidean
coordinates. Below, they are stated as propositions, verifiable by elementary
calculations derived directly from the definition of the Poincaré metric.
Proposition 2.1 For any x ∈ (−∞, 0) the following two Poincaré coordi-
nates coincide
p0,−∞(x) = px,1(0).
Proof. We have p0,−∞(x) = ln (−x) = ln
= px,1(0). �
Proposition 2.2 For any x ∈ (−1, 0) the following two Poincaré coordi-
nates coincide
p1,−1(x) = px,−x
p−1x,1 (p0,−1(x))
Proof. The identity in question is tantamount to ln x−1
= ln c−x
1 + x
x + c
, (2.1)
where c = p−1x,1(p0,−1(x)), i.e. px,1(c) = p0,−1(x). But this last equality means
, and further c = x2, so that (2.1) follows. �
Given a point x ∈ (−∞, 0), we then pick a point y ∈ (x, 0). We shall let
the point x vary, by which we mean a choice of another point x̃ ∈ (−∞, 0).
The discrepancy in the non-euclidean coordinate will be denoted by
∆ϑ = p0,−∞(x̃) − p0,−∞(x).
A broader version of Proposition 2.1 is the following
Proposition 2.3 In the above notation we have
px̃,0
p−1x̃,1 (px,1(y) + ∆ϑ)
− px,0(y) = p1,−∞(x̃) − p1,−∞(x).
Proof. We have ∆ϑ = ln x̃
and p1,−∞(x̃) − p1,−∞(x) = ln
. Denote
c = p−1x̃,1(px,1(y) + ∆ϑ), a point characterized by
c− x̃
1 − c
y − x
1 − y
. (2.2)
We will be done once we show c−x̃
= 1−x̃
c− x̃
y − x
1 − x
1 − x̃
. (2.3)
From (2.2) we get x̃
= 1−y
and (2.3) can now be checked im-
mediately. �
Proposition 2.1 is what we get of Proposition 2.3, in place of subtracting
two infinite terms, when we set y = 0. We generalize Proposition 2.2 in a
similar way. Suppose we are given a point x ∈ (−1, 0), and a point y ∈
(x,−x). Again, we let the point x vary by choosing a new point x̃ ∈ (−1, 0).
The discrepancies in the appropriate Poincaré coordinates of the two points
will be denoted by
∆t = p1,−1(x̃) − p1,−1(x) ,
and by
∆θ = p0,−1(x̃) − p0,−1(x)
respectively. A statement parallel to Proposition 2.3 is the following
Proposition 2.4 In the above notation we have
px̃,−x̃
p−1x̃,1 (px,1(y) + ∆θ)
− px,−x(y) = ∆t.
Proof. The point c = p−1x̃,1(px,1(y) + ∆θ) satisfies
= y−x
· 1+x
, which
can be transformed into
c− x̃
1 − y
y − x
1 + x̃
1 − x̃
1 − x
1 − x̃
. (2.4)
We will be done if we show that c−x̃
· x+y
= x̃−1
· 1+x
, which is the same as
x̃ + c
c− x̃
y + x
y − x
1 + x̃
x̃− 1
1 + x
. (2.5)
Since x̃+c
= 1 + 2x̃
, equation (2.5) follows immediately from (2.4). �
Proposition 2.5 Suppose x, x′ ∈ (0, 1) and y ∈ (x, 1). Let y′ be such a point
in (x′, 1) that px′,1(y
′) = px,1(y). Then
p1,0(x
′) − p1,0(x) = py′,0(x
′) − py,0(x). (2.6)
Proof. The point y′ is chosen in such a way that y
= y−x
, or 1−x
= 1−x
Identity (2.6) is now immediate. �
3 The period 4 super-sink
In this short section we describe the behavior of the point 4a when we let
the parameter ā vary in such a range, that 3a ∈ (0, 1) and the point 4a stays
within the interval (2a,−2a).
Let a positive number t be the Poincaré coordinate of 2a in the oriented
interval (1,−1), and we set
g(t) = p2a,−2a(4a).
The following theorem holds true.
Theorem 3.1 The inverse function g−1 : R → R+ is strictly increasing, and
g′(t) > 1. In particular, the value g(t) = 0, corresponding to the super-stable
orbit with the kneading sequence RLRC is assumed only once.
Proof. Consider a pair of admissible parameter values ā and ā′, i.e. such
that the orbits 2a, 3a, 4a (and respectively 2a′ , 3a′, 4a′) satisfy the restrains
on the dynamics we set above. Then
∆t = p1,−1(2a′) − p1,−1(2a), while ∆g = p2
(4a′) − p2a,−2a(4a). (3.1)
Applying Proposition 2.4 to this case we get
(p−12
,1(p2a,1(4a) + (p0,−1(2a′) − p0,−1(2a)))) − p2a,−2a(4a) = ∆t, (3.2)
so, because of monotonicity of the coordinate functions, we only need to
establish that
,1(4a′) − p2a,1(4a) > p0,−1(2a′) − p0,−1(2a). (3.3)
This inequality becomes clear once we split the procedure leading from point
2a (respectively 2a′) to 4a (respectively to 4a′) into three steps. In the
first step, we act on the interval (0,−1) by the restriction of the power-law
map. Thus, due to negative Schwarzian derivative, the initial discrepancy
(p0,−1(2a′) − p0,−1(2a)) in the Poicaré coordinates gets increased. So we see
(3a′) − p1,2a(3a) ≥ p0,−1(2a′) − p0,−1(2a).
In the second step, we turn the interval (0,−1) over, onto the interval
(1, 2a), or onto (1, 2a′) respectively, and then we truncate the image at the
critical point 0. This cut-off only increases the Poincaré coordinate of ev-
ery point, which after the turnover landed in (1, 0), because we now read
the Poincaré coordinate in the interval (1, 0) rather than in a larger domain
(1, 2a), or (1, 2a′) respectively. Moreover, the increase in the Poincaré coordi-
nate inflicted by cutting the domain interval short, is in the case of point 3a
smaller then in the case of 3a′ . This is so, because the endpoint 2a is closer to
the critical point, while the endpoint 2a′ is further away to the left, so of two
corresponding points with identical Poincaré coordinate within the respec-
tive domain intervals (with the other endpoint at 1), the gain in the latter
situation is larger than in the former. But instead of equal coordinates, we
have even better inequality p1,2
(3a′) > p1,2a(3a), which further enlarges the
gain. Thus, in this second step, made of the turnover followed by truncation,
the initial discrepancy grows even further and so
p1,0(3a′) − p1,0(3a) > p1,2
(3a′) − p1,2a(3a).
In the last step, we again act by a negative Schwarzian map stretching
the discrepancy between the Poincaré coordinates yet further, and finally we
make the turnover onto (1, 2a), and respectively onto (1, 2a′), to arrive at
(3.3). Therefore ∆g > ∆t and the proof is complete. �
4 The period 8 super-sink
In the previous section we have established that, when we vary the param-
eter ā , the position of the point 4a within the interval (2a,−2a) changes
monotonically, with the derivative greater than 1. It clearly follows from the
proof, that this derivative actually stays bounded away from 1, in a way that
depends on the critical order α . In section 5 we will study in detail the case
of p2a,−2a(4a) < 0, and describe the behavior of the intervals delimited by the
post-critical orbit with the kneading sequence RLRL . . . .
In here, we will focus on these admissible parameters ā , for which
p2a,−2a(4a) > 0 and p4a,−4a(8a) ≤ 0 , i.e. we are past the (unique) parameter
corresponding to RLRC, but we do not cover the critical point yet another
time. From now on, we are making our choice of the parameter subject to
this restriction. We shall see that, as long as the above condition on the
dynamics is satisfied, the movement of the point 8a is also monotone in pa-
rameter, and in the non-euclidean metric in (4a,−4a) this point moves with
the derivative strictly positive. It will follow that the RLRRRLRC super-
stable orbit appears uniquely in every power-law family. It is a subject of an
ongoing work, that goes beyond the scope of this paper, to examine whether
a claim analogous to that of Theorem 3.1 can be fully extended to larger set
of parameters.
In our current case, the scheme of the argument we used to prove Theorem
3.1, alone will not suffice, and a more delicate technique must be employed.
Yet, some understanding of the way Poincaré coordinates vary remains an
important component. Since, due to the more intricate dynamics, the re-
quired property of the non-euclidean coordinates becomes less self-evident,
we state it as a separate lemma. The points x, y, z below will correspond
to the points 2a, 4a, 8a of the post-critical orbit. The origin of the sum-
mands, which do not have equivalent in the statement of Proposition 2.4 will
be explained later, in the course of the proof of Theorem 4.1 below. Here,
we only indicate that the last term has to do with the limit strength of a
non-euclidean push.
Lemma 4.1 Suppose we are given two triples of points, (x, y, z) and (x̃, ỹ, z̃),
satisfying the following conditions:
(i) x, x̃ ∈ (0,−1) and p0,−1(x̃) > p0,−1(x) ,
(ii) y ∈ (0,−x), ỹ ∈ (0,−x̃) and px̃,1(ỹ) ≥ px,1(y) + (p0,−1(x̃) − p0,−1(x)) ,
(iii) z ∈ (y, 0], z̃ ∈ (ỹ,−ỹ) and pỹ,x̃(z̃) ≥ py,x(z) + (p0,−x̃(ỹ) − p0,−x(y)) +
ln ỹ−x̃
− ((px̃,1(ỹ) − px,1(y)) − (p0,−1(x̃) − p0,−1(x))) .
Then pỹ,−ỹ(z̃) > py,−y(z) .
Proof. It is immediate to check that for arbitrary y, ỹ ∈ (0, 1) one has
pỹ,−1(0) = py,−1(0) + (p0,1(ỹ) − p0,1(y)) + ln
1 + ỹ
1 + y
− (p−1,1(ỹ) − p−1,1(y)) .
(4.1)
We now assume ỹ > y, and allowing z 6= 0 we verify, that for any z ∈ [0, y)
the following generalization of (4.1) holds
pỹ,−ỹ(p
ỹ,−1(py,−1(z) + (p0,1(ỹ) − p0,1(y)) + ln
1 + ỹ
1 + y
−(p−1,1(ỹ) − p−1,1(y)))) ≥ py,−y(z). (4.2)
In order to see this, notice that
(p0,1(ỹ) − p0,1(y)) + ln
1 + ỹ
1 + y
− (p−1,1(ỹ) − p−1,1(y)) = ln
and denote c = p−1ỹ,−1(py,−1(z) + ln
), which means c−ỹ
−ỹ−c
= z−y
c− ỹ
1 + ỹ
1 + z
z − y
1 + ỹ
. (4.3)
We will be done if we show that c−ỹ
−ỹ−c
≥ z−y
, being equivalent to 2ỹ
+1 ≥ y+z
or ỹ
. The last inequality follows from (4.3), once we recall ỹ ≥ y.
In the next step we extend formula (4.2), allowing x̃ 6= −1. Assuming
1 > −x̃ > ỹ > y > z ≥ 0, we will now show that
pỹ,−ỹ(p
ỹ,x̃(py,x̃(z) + (p0,−x̃(ỹ) − p0,−x̃(y))
ỹ − x̃
y − x̃
− (px̃,1(ỹ) − px̃,1(y)))) > py,−y(z). (4.4)
We emphasize that the inequality in formula (4.4) is always sharp, even for
z = 0.
This time, we set
c = py,x̃(z) + (p0,−x̃(ỹ) − p0,−x̃(y)) + ln
ỹ − x̃
y − x̃
− (px̃,1(ỹ) − px̃,1(y)), (4.5)
which means
c− ỹ
x̃− c
z − y
x̃− z
−x̃− ỹ
−x̃− y
1 − ỹ
1 − y
We transform this identity into
x̃− ỹ
c− ỹ
− 1 =
x̃− z
z − y
x̃ + ỹ
x̃ + y
1 − y
1 − ỹ
and further into
c− ỹ
x̃− ỹ
x̃− z
z − y
x̃ + ỹ
x̃ + y
1 − y
1 − ỹ
We will be done if we show pỹ,−ỹ(c) > py,−y(z), i.e.
−ỹ−c
> z−y
, which is
equivalent to ỹ
, and so it is enough to verify that
x̃− ỹ
x̃− z
z − y
x̃ + ỹ
x̃ + y
1 − y
1 − ỹ
z − y
This inequality can be rewritten as
x̃− z
z − y
x̃ + ỹ
x̃− ỹ
x̃ + y
1 − y
1 − ỹ
z − y
x̃− ỹ
or (recall that y < z, x̃ < 0, ỹ > 0)
(x̃− z)(x̃ + ỹ)y(1 − y) < (x̃ + y)(1 − ỹ)(yx̃− zỹ),
and further
x̃y(1 + x̃)(ỹ − y) < z(ỹ − y)(x̃ỹ + x̃y − x̃ + yỹ).
To conclude, we cancel out (ỹ − y), and observe that
x̃y + x̃ỹ − x̃ + yỹ > x̃(1 + x̃). (4.6)
This is so because (4.6) boils down to the inequality x̃2+x̃(2−y−ỹ)−yỹ < 0,
which is elementarily true for all x̃ ∈ (−1, 0) and y, ỹ ∈ (0, 1). For completion
of the proof we now consider an arbitrary point x ∈ (0, x̃), such that y < −x.
We consider the movement of x-variable from position x to x̃ and apply
Proposition 2.4 twice, first to the induced movement of y-variable, then to
the consequent movement of z-variable. By virtue of that Proposition, we
see that points ŷ and ẑ, determined by the identities
px̃,1(ŷ) = px,1(y) + (p1,−1(x̃) − p1,−1(x))
pŷ,x̃(ẑ) = py,x(z) + (p1,−1(x̃) − p1,−1(x))
satisfy ŷ < ỹ and pŷ,−ŷ(ẑ) = py,−y(z) + (p1,−1(x̃)− p1,−1(x)) > py,−y(z). Thus
obviously ln ỹ−x̃
> ln ỹ−x̃
ŷ−x̃
If ẑ ≥ 0, i.e. pŷ,−ŷ(ẑ) ≤ 0, then keeping x̃ fixed, we then apply formula
(4.4) with ŷ, ẑ in place of y, z, to the effect of yet further increase of the
Poincaré coordinate of z̃ compared to that of ẑ (and so of z itself), measured
within respective symmetric y-domains. In case of pŷ,−ŷ(ẑ) > 0 the image of
point z has already past the midpoint of the (varying) symmetric y-domain
interval while y-variable has been changing from y to ŷ. Again, we then keep
x̃ fixed, to move the y-variable further, from ŷ to ỹ. This time, application
of formula (4.4) can induce some decrease in the Poincaré coordinate of the
outcome – the resulting point p−1ỹx̃ (c), with c as in (4.5), can divide the y-
domain interval (ỹ,−ỹ) in smaller proportion than ẑ did in (ŷ,−ŷ). Anyway,
due to sharp inequality in (4.4) for all z such that py,−y(z) < 0, the midpoint
could only be attained from the other side. In other words, inequality (4.4)
guarantees that the derivative of the induced z-movement, measured in the
respective Poincaré coordinates, is positive (and actually bounded away from
0) as long as the values assumed by the z-variable are non-positive. Thus, in
particular the value 0 can be attained only once, and so if we put a point ẑ
with py,−y(ẑ) > 0 into the formula at the left-hand side of inequality (4.4), we
necessarily end up with a point on the same side of 0. Because the starting
point z was on the other side, the lemma holds in this case too. �
With lemma 4.1 in place, we are in the position to state and prove the
main result of this section.
Theorem 4.1 In the power-law family fa : x 7→ −|x|
α +a, with α > 1, there
exists unique parameter a = a(α) corresponding to the kneading sequence
RLRRRLRC.
Proof. In the course of the proof we make use of the tools developed in
section 2 of [9], where we pointed to some consequences of homogeneity of
the power-law mappings. In particular, we had Lemma 2.1 there, asserting
that for any two points q, q̃ ∈ (0, 1) one has
q,1 − h
q̃,1 = (p0,1(h(q)) − p0,1(h(q̃))) − (p0,1(q) − p0,1(q̃)). (4.7)
Speaking colloquially, identity (4.7) tells, that when we move the endpoint
of an interval (1, q) in (1, 0) towards the critical point, then an extra gain in
the Poincaré coordinate, coming from the successive action of the power-law
map, is just enough to make up for the loss (measured in non-euclidean met-
ric in (1, q) and (1, q̃) respectively) suffered because of the simultaneously
increased strength of the limit non-euclidean push towards that moving end-
point. Other propositions and lemmas of section 2 of [9] served to establish,
that this limit situation, corresponding to Poincaré coordinate close to −∞,
is essentially the worst possible, and when we consider an interior point of
a definite Poincaré coordinate rather than the limit case, then the balance
of gains vs. losses is in our favor (”we are never in the red”). We will be
sending upon those properties when necessary, without reproducing them in
this paper.
Proceeding similarly to what we did in the proof of Theorem 3.1, we split
the procedure leading from 4a to 8a , and respectively from 4a′ to 8a′ , into
several steps. First, we increase ā to ā′. Theorem 3.1 yields, in particular,
that p0,−2
(4a′) − p0,−2a(4a) > 0. Next, we act upon 4a′ , and 4a, by the map
h, and under the action of h the above discrepancy gets enlarged. This is
so, because due to homogeneity, we may for the purpose of performing this
step, tentatively set each of the endpoints, −2a′ and respectively −2a, at 1.
Then each of the Poincaré coordinates p0,−2
(4a′), p0,−2a(4a), is transformed
by same, fixed negative Schwarzian map h0,1. In the following step, we
turn each of the intervals (0, h(2a′)), (0, h(2a)) over, and stretch them onto
(1, 3a′) and respectively (1, 3a). The image of 4a′ is 5a′ , and by the so far
described steps, it is clear that p1,3
(5a′)−p1,3−a(5a) > p0,−2
(4a′)−p0,−2a(4a).
By the truncation argument from the proof of Theorem 3.1, we know that
p1,0(3a′) − p1,0(3a) > ā − a . In particular, the nonlinearity of the interval
(1, 3a′) is larger than that of (1, 3a). Now, we act by the homogeneous map
h again. Notice, that unlike in the case of h0,2
, this time the mapping h1,3
does not coincide with h1,3a . Anyway, we can still claim that in this step the
discrepancy of the respective Poincaré coordinates grows again, i.e
p1,h(3
)(h(5a′)) − p1,h(3a)(h(5a)) > p1,3a′ (5a′) − p1,3a(5a) . (4.8)
To this end, we invoke Propositions 2.5 and 2.4 of [9]. From the former, it
follows that the strength of the non-euclidean push generated by h restricted
to some domain, is a monotone function of the nonlinearity of that domain,
when measured for a fixed Poincaré coordinate within the varying domain.
From the latter, we derive that when the domain stays fixed, the strength
of the non-euclidean push of h is monotone in the Poincaré coordinate of
the argument. We have noticed already that the nonlinearity of (1, 3a) is in-
creasing in parameter ā, and also that p1,3
(5a′) > p1,3a(5a) , so the principle
of monotone behaviour of the strength of non-euclidean push can be applied
to the triples of points we consider. This immediately implies the desired
increase in the discrepancy of appropriate Poincaré coordinates, as stated in
(4.8).
Making the next step, we turn the obtained triples (1, h(5a′), h(3a′)) and
(1, h(5a), h(3a)) over, onto (2a′, 6a′ , 4a′), and respectively onto (2a, 6a, 4a), and
then truncate them at the critical point 0. In the proof of Theorem 3.1, as
well as in a step above, we were satisfied to ascertain that this truncation
increases the Poincaré coordinates discrepancy, which in current step would
yield p2
,0(6a′) − p2a,0(6a) > p2a′ ,4a′ (6a′) − p2a,4a(6a), because by Theorem
3.1 we know that p2
(4a′) > p2a,−2a(4a). To proceed further, one more
observation is needed. It is fairly clear that we have following lower bound
on the increase of the Poincaré coordinates discrepancy, generated by the
cut-off at 0:
,0(6a′) − p2a,0(6a) > p2a′ ,4a′ (6a′) − p2a,4a(6a) + ln
4a′ − 2a′
4a − 2a
. (4.9)
The equality in (4.9) is the limit case, attained for infinitesimally short in-
tervals placed at the left-hand endpoints, i.e. when p2a,4a(6a) → −∞ and
simultaneously p2
(6a′) → −∞. For non-infinitesimal intervals satisfying
(6a′) > p2a,4a(6a), the same argument as in the proof of Theorem 3.1
obviously yields sharp inequality in (4.9), and so the growth of the discrep-
ancy gained in the cut-off step is strictly larger than the logarithmic term.
In the following step we once more act by homogeneous map h, and
because h0,2
coincides with h0,2a , the same argument as before gives
,0)(h(6a′)) − ph(2a),0(h(6a)) > p2a′ ,0(6a′) − p2a,0(6a) . (4.10)
This adds yet an extra amount to the discrepancy we consider. We again turn
the intervals (h(2a′), 0) and (h(2a), 0) over and stretch them onto (1, 3a′) and
(1, 3a), with 6a′ going onto 7a′ and 6a going onto 7a respectively. It remains to
examine what happens in the last step, when we act by the respective (non-
coinciding!) restrictions of h to the obtained intervals, before we eventually
return onto (2a′, 4a′) and onto (2a, 4a) by linear rescaling. This is what we
need Lemma 4.1 for. In what follows we verify its assumptions are fulfilled
in our setting.
In this last step we perform, the strength of the non-euclidean push
induced by h| (1,3
) , measured at 7a′ , can be greater than the respective
strength of h| (1,3a) at 7a. This means that the discrepancy accumulated in
all the so far steps can now diminish. However, the identity (4.7) provides a
bound from the above on the amount of possible loss. To see this, we recall
ā′ − ā = p0,−1(2a′) − p0,−1(2a) < p1,0(3a′) − p1,0(3a) , (4.11)
and according to 4.7 we have
,1(4a′) − p2a,1(4a)) − (p1,0(3a′) − p1,0(3a)) = (h
) (4.12)
We know that p3
,1(7a′) > p3a,1(7a) and the interval (1, 3a′) has larger non-
linearity than (1, 3a), so we are in a position to invoke Propositions 2.5 and
2.4 of [9] once more. By them we have
,1(7a′) − p3a,1(7a)) − (p4a′ ,2a′ (8a′) − p4a,2a(8a)) < (h
). (4.13)
The inequalities (4.9), (4.11) and (4.13) put together, provide for fulfillment
of condition (iii) of Lemma 4.1, with the points x, y and z assuming values
2a, 4a and 8a, as indicated before the statement of the lemma. Now the claim
of Theorem 4.1 follows directly from Lemma 4.1, and so we are done. �
We complete this section explicitly recording one extra property, which
we actually proved along the way. Denote the variable τ = p4a,−4a(8a) and
let γ = p2a,−2a(4a). From the proofs of Theorem 4.1 and Lemma 4.1 there
immediately follows
Corrolary 4.1 The function γ = γ(τ) : R− → R+ is strictly increasing in
τ , with the derivative γ′(τ) bounded away from 0 and +∞.
5 RLRLRLRL . . .
In this section we let the parameter ā vary in a range such that the kneading
sequence is RLRL . . . . From Theorem 3.1 it follows immediately that the
range of admissible ā’s is always a half-line (−∞, ā1), with the specific value
of ā1 depending on the critical order α. Upholding the normalization 0a = 0 ,
1a = 1 we have set before, this means the post-critical orbit begins with
2a ∈ (0,−1) , 3a ∈ (1, 0) and 4a ∈ (2a, 0). Then, we get two sequences of
nested intervals, the odds: (1, 3a), (3a, 5a), (5a, 7a) . . . , and the evens: (0, 2a),
(2a, 4a), (4a, 6a) . . . . In terms of multipliers, we either have a period 2 orbit
with negative multiplier, or this periodic orbit had turned into a repeller
and, by bifurcation, there was born a period 4 attracting periodic orbit with
positive multiplier. In what follows, we shall see that the ratios of consecutive
intervals within each of the two decreasing families are functions strictly
monotone in parameter ā. Moreover, the initial increase of the parameter,
i.e. ā′ − ā, does not eventually vanish, but a definite part of it is preserved
through all the steps. This will further provide, with some extra work, for
monotonicity of the multipliers, also in the case of repelling period 2 orbit.
This is a work in preparation. The remaining part of this paper is devoted
to the proof of the following claim.
Theorem 5.1 For ā ∈ (−∞, ā1) and for all non-negative integers n, the
ratio functions
rne =
|(2n + 4)a − (2n + 2)a|
|(2n + 2)a − (2n)a|
and rno =
|(2n + 5)a − (2n + 3)a|
|(2n + 3)a − (2n + 1)a|
(5.1)
are strictly increasing in ā.
Moreover, when the parameter increases from ā to ā′, then for every n ∈
Z+ the induced discrepancy of the Poincaré coordinates satisfies
p(n+2)
((n + 4)a′) − p(n+2)a,na((n + 4)a) > (p1,−1(2
a) − p1,−1(2a)) . (5.2)
Proof. As before, we divide the procedure into steps. Once we cover the most
delicate step, which turns out to be the passage from (5a′ , 7a′) to (6a′ , 8a′),
we will be in a position to continue inductively. We begin by moving the
initial point 2a to a new position 2a′ , with ā
′ > ā. Then, by the truncation
argument from the proof of Theorem 3.1, we have
(p1,0(3a′) − p1,0(3a)) > ā
′ − ā = ∆ā > ∆t, (5.3)
where we denoted ∆t = (p1,−1(2
a) − p1,−1(2a)). Since we then act by the
homogeneous map h, by (4.7) we get
,1(4a′) − p2a,1(4a) = (p1,0(3a′) − p1,0(3a)) + (h
,1) . (5.4)
Passing from 3a′ to 4a′ , we cannot directly apply the truncation argument
again, because in this step the Poincaré coordinate p1,2
(0) of the cut-off
point decreases (cf. Proposition 2.1). That can be fixed by decomposing
the step in two, and simultaneous use of Proposition 2.4, identity (4.7) and
truncation. According to (5.3) and Proposition 2.4,
,1(p1,0(3a′))
− p2a,−2a
p−12a,1(p1,0(3a))
> ∆t. (5.5)
Truncation at 0 obviously gives
,1(p1,0(3a′))
− p2a,0
p−12a,1(p1,0(3a))
> ∆t. (5.6)
Then, to the Poincaré coordinate of the point (p−12
,1(p1,0(3a′)), read in the
domain (2a′ , 1), we add the extra gain of (h
,1). The non-euclidean
length of this same extra interval, read in the domain (2a′ , 0) rather than in
(2a′ , 1), is of course larger, because of truncation. Thus
,0(4a′) − p2a,0(4a) > ∆t + (h
,1) . (5.7)
Doing the homogeneous mapping again, by (4.7) and (5.7) we get
,1(5a′) − p3a,1(5a) = h2a′ ,0(p2a′ ,0(4a′)) − h2a,0(p2a,0(4a)) >
> ∆t + (h
,1) + (h
4a,2a
) . (5.8)
Now, similarly to the final step in the proof of Theorem 4.1, we can argue
that the so far acquired gain in the Poincaré coordinate is enough to make
up for possible losses in the next two steps. This is fairly clear. The interval
(3a′ , 1) has larger nonlinearity than (3a, 1), and p3
,1(5a′) > p3a,1(5a), so
Propositions 2.5 and 2.4 of [9] do apply when we act by h | (1,3
) and h | (1,3a).
Therefore, in this step the discrepancy (p3
,1(5a′) − p3a,1(5a)) can only be
diminished by an amount smaller than (h
,1), yielding
(6a′) − p4a,2a(6a) > ∆t + (h
4a,2a
), (5.9)
By (5.7), the nonlinearity of (4a′, 2a′) is larger than that of (4a, 2a), and also
(6a′) > p4a,2a(6a). Thus, when we act by h | (4
), and respectively by
h | (4a,2a), we certainly do not lose more than (h
4a,2a
) in the outgoing
discrepancy. Hence, by ( 5.9)
(7a′) − p5a,3a(7a) > ∆t. (5.10)
We can now make a shortcut towards completion of the current cycle. The
nonlinearity of (3a′ , 1) is larger than that of (3a, 1) and p3
,1(5a′) > p3a,1(5a),
which in turn gives that the nonlinearity of (5a′ , 3a′) is larger than that of
(5a, 3a). Also p5
(7a′) > p5a,3a(7a), so we can apply the argument about
monotonicity of the strength of the non-euclidean push, which we recalled in
the proof of Theorem 4.1, immediately arriving at
(8a′) − p6a,4a(8a) > p5a′ ,3a′ (7a′) − p5a,3a(7a) > ∆t. (5.11)
However, the above argument alone turns out to be insufficient, when we want
to do further iterates. To obtain an inequality which we could use inductively
at all steps, we need more subtle understanding at this particular stage of
our procedure. Here we go.
From (5.3) and Proposition 2.5 it follows that
(p3a,1(5a)),0
(3a′) − p5a,0(3a) > ∆ā , (5.12)
so by p3
,1(5a′) > p3a,1(5a) we have
,0(3a′) − p5a,0(3a) > ∆ā. (5.13)
By the same argument applied to (5a′ , 3a′) rather than (1, 3a′), we get
(p5a,3a (7a)),0
(3a′) − p7a,0(3a) > ∆ā. (5.14)
Now we do the homogeneous mapping h, and rescale the image onto (1, 2a′).
The image of 3a′ is 4a′ , and by a version of the truncation argument alike
that used before in the step leading from (1, 3a′) to (2a′ , 4a′), we use
,1(p1,0(3a′))) − p2a,0(p
(p1,0(3a))) > ∆t (5.15)
and (5.14) to get
(h1,0(p1,0(p
(p5a,3a (7a))))),0
(4a′) − p8a,0(4a) > ∆t. (5.16)
This is so, because (5.14) implies
ph(p−1
(p5a,3a (7a))),0
(h(3a′)) − ph(7a),0(h(3a)) > ∆ā, (5.17)
and when we consider the interval (c, d), where d = p−12
1(p2a,1(4a)+∆ā), and
the point c is defined so that
pc,1(d) − p8a,1(4a) = ∆ā (5.18)
then, according to Proposition 2.5 applied to the domain (1, 2a′) in place of
(0, 1), and with the points d and c singled out, we see that point c divides the
interval (d, 2a′) at the same proportion as 8a divided (4a, 2a). Re-applying
Proposition 2.5 to the domain (0, 2a′) with the same singled out pair of points,
we further see that
pc,0(d) > p8a,0(4a) + ∆t, (5.19)
because by Proposition 2.4 p2
,0(d) − p2a,0(4a) > ∆t. Recalling (5.17) and
taking into account that p2
,1(4a′) > p2
,1(d), which in turn gives p1,4
(0) >
p1,d(0), we can now do the standard truncation argument, cutting-off at 0 to
arrive at (5.16).
This formula could do for the iterative procedure if we cared only for
some, indefinite growth. To obtain definite growth, claimed in the statement
of Theorem 5.1, we need to work harder.
In the next step of the proof, we will see that the extra amount of ∆t in
formula (5.16) allows us to move 7a towards the endpoint by at least that
much. To this end, we again consider the interval (5a′ , 3a′), but this time the
point within we single out, is point e determined by
(e) = p5a,3a(7a) + ∆t. (5.20)
From (5.14), using Proposition 2.1 with points 0, 3a′ and e in place of 1, 0
and x respectively, or by a direct check, one gets
pe,0(3a′) − p7a,0(3a) > ∆ā− ∆t. (5.21)
Doing the homogeneous mapping, we have
ph(e),0(h(3a′)) − ph(7a). 0(h(3a)) > ∆ā− ∆t. (5.22)
Again, we consider an interval (f, d), where d has same meaning as above,
and point f is defined by
pf,1(d) − p8a,1(4a) = ∆ā− ∆t. (5.23)
From (5.18) and (5.23), it follows by Proposition 2.1 that pd,−∞(c)−pd,−∞(f) =
∆t, and again by this same proposition pc,0(d) − pf,0(d) = ∆t. Hence, by
(5.19), we have pf,0(d) > p8a,0(4a). This, and (5.22) lead to
(p5a,3a (7a)+∆t)),0
(4a′) > p8a,0(4a). (5.24)
We can describe what we have found so far in the following way. We
move the parameter up, from ā to ā′. In the odd family, we see 3a moving
to 3′a by more than ∆ā. Consequently, the non-euclidean coordinate of 5a
vary, within its dynamically determined base interval, by at least ∆t, plus
an additional increment which is sufficient to make up for the increased –
due to larger nonlinearity of the new new domain intervals – strength of
the non-euclidean push backwards. In the next odd return we do not let 7a
move all the way to its new position 7a′ at once. Instead, we first only add
∆t to its Poincaré coordinate. This corresponds to starting from the point
e in the already fully enlarged domain (5a′ , 3a′), rather than from 7a′. We
have just seen that not only is the nonlinearity of (e, 3a′) larger than that
of (7a, 3a), but the nonlinearity of (ê, 4a′) is larger than that of (8a, 4a) also.
Here ê is the dynamical successor of e on the even side. This latter estimate
from the below on the the nonlinearity, turns out to be fundamental for the
prospective iterates.
Recall we defined e by (5.20) so as to have p5
(e) = p5a,3a(7a) + ∆t.
The same way we derived (5.11) from (5.10) we also get
(ê) − p6a,4a(8a) > ∆t. (5.25)
This will be needed, when it comes to definite growth in both odd and even
family. But now, for points e and ê we have stronger input: in both cases,
we know that the nonlinearity of the remaining part of the base interval in-
creased. Therefore, we will now be able to proceed pretty much like in the
initial step, that led from (1, 3a′) to (2a′ , 4a′), rather than use the earlier de-
scribed shortcut. Similarly to that initial step, we again want to know that
the surplus exceeding ∆t in (5.25) will make up for possible loss, inflicted by
increased nonlinearity of (5a′ , 7a′), upon next return to (5a′ , 7a′). However,
we have to overcome a serious obstacle. Formula (4.7) we previously used to
that goal, holds true only so long as the critical point is the endpoint corre-
sponding to non-euclidean +∞. This is of course not the case for (5a, 3a),
nor for all other intervals in our odd and even families, except for the initial
ones. For intervals not bounded by the critical point, we only know mono-
tonicity of the strength of non-eucliean push and this, in general, does not
give control over an amount of the gain in Poincaré coordinates discrepancy.
Composing h mappings over two arbitrary, successive domains, yet worsens
the the situation. Fortunately, all this can be fixed with (5.21) and (5.24) in
place. Increased nonlinearity of that part of a domain interval which bounds
us away from the endpoint, provides an effective replacement for the critical
endpoint. In particular, we will see that the gain in the non-euclidean coor-
dinates discrepancy is even better than that in formula (4.7) . This is why
we have striven for those nonlinearity inequalities. As soon as we are over
with the part which takes 7a to e, the remaining part, in which we move e
further to 7a′ , will require only an easy estimate. All the above holds true for
even successors, ê and eventually 8a′ , as well. With one extra observation
to make, we will be able to do arbitrarily long iterates, preserving the ∆t
discrepancy all along the way.
To carry out the above described strategy, we recall that in Proposition
2.2 of [9] we gave an explicite formula for the strength of non-euclidean push,
which turns out to be
|pr,s(ϕ(x)) − pp,q(x)| = |ϕ
x,q + ϕ
p,x| (5.26)
We also noticed there, that for the homogeneous mapping h restricted to
some interval, the quantities h
and h
depend solely on the nonlinearity of
that domain interval. By monotonicity of the strength of the non-euclidean
push as a function of the nonlineatity of the domain, also the limit values, h
and h
, behave monotonically. By all the above, taking (5.21) into account,
we have
(ê) − p6a,4a(8a) = h5a′ ,3a′ (p5a,3a(7a) + ∆t) − p6a,4a(8a) =
),h(3
)(h(e)) − ph(5a),h(3a)(h(7a)) > ∆t + (h
,e − h
5a,7a
) (5.27)
The sign at the superscript of h in (5.27) depends only on an orientation
of the domain, so (5.27) provides a better estimate than we could derive
from (4.7), if the endpoint 3a′ coincided with the critical point. Doing the
successive h-map step on the even side, because of (5.24), we get in the same
),h(4
)(h(ê))−ph(6a),h(4a)(h(8a)) > ∆t+ (h
5a,7a
) + (h
,ê−h
6a,8a
(5.28)
These are formulas analogous to (5.7) and (5.8), and what we want now, is
a similar estimate where the input is 7a′ and 8a′ , rather than e and ê. To
move from e to 7a′ we could simply invoke Lemma 2.4 of [9]. However, there
is no generalization of that lemma which could be used over two unrelated
domains. We need to be a bit more careful, and use the dynamical relation
between an interval and its image. Doing the mapping h, by homogeneity
and (4.7) we have
),0(h(7a′)) − ph(5
),0(h(e)) = (p5
,0(7a′) − p5
,0(e)) + (h
(5.29)
Before we do another mapping h, we take the image over onto (1,−∞), so
that h(3a′) goes onto (4a′), and cut off at 0. Because of this truncation
,0(8a′) − p6
,0(ê) > (p5
,0(7a′) − p5
,0(e)) + (h
,e). (5.30)
Now, acting by homogeneous map, we get
),0(h(8a′)) − ph(6
),0(h(ê)) > (p5
,0(7a′) − p5
,0(e)) +
,e) + (h
,ê) (5.31)
We neglect a positive summand (p5
,0(7a′)− p5
,0(e)) and truncate at h(4a′)
to arrive at
),h(4
)(h(8a′)) − ph(6
),h(4
)(h(ê)) > (h
,e) + (h
(5.32)
Similarly, neglecting a positive summand at (5.29), followed by cutting off at
h(3a′) leads to
),h(3
)(h(7a′)) − ph(5
),h(3
)(h(e)) > (h
,e). (5.33)
The inequalities (5.27) and (5.33), in conjunction with (5.28) and (5.32), give
),h(3
)(h(7a′)) − ph(5a),h(3a)(h(7a)) > ∆t + (h
5a,7a
) (5.34)
and also
),h(4
)(h(8a′))−ph(6a),h(4a)(h(8a)) > ∆t+(h
5a,7a
6a,8a
(5.35)
which are the desired estimates. Now, by the same argument which led from
(5.7) and (5.8), through (5.9) to (5.10), we can see that (5.34) and (5.35)
imply
(11a′) − p9a,7a(11a) > ∆t. (5.36)
In the same way we obtained (5.11) from (5.10), we can also derive
(12a′) − p10a,8a(12a) > ∆t. (5.37)
We have completed the second cycle. Those were necessary to initialize the
inductive procedure. We are now in a position to do the final argument,
which can be used repeatedly. We believe that, because all the elaborate
notation of the first two cycles is already in place, it will be more instructive
to present this argument in detail as the next cycle, rather than in general
terms. It will be obvious that what we do, is tantamount to the inductive
step.
We pick a point ǫ ∈ (9a′ , 7a′), such that
(ǫ) = p9a,7a(11a) + ∆t. (5.38)
We will prove that the nonlinearity of (ǫ, 7a′) is larger than that of (11a, 7a),
and simultaneously the nonlinearity of (ǫ̂, 8a′) is larger than that of (12a, 8a).
As before, ǫ̂ stands for the dynamical successor of ǫ on the even side. This
will permit to bypass the non-critical endpoint obstacle in the next cycle,
the way we did earlier, with e and ê. To show this nonlinearity increase, we
proceed in several steps. First, in (5a′ , e) we find a point β = p
,e(p5a,7a(9a)).
Then, in (β, e) we find ε, such that pβ,e(ε) = p9a,7a(11a) + ∆t. We use the
the fact that Poincaré coordinate of the point e, compared to that of 7a, is
already moved by ∆t towards the endpoint, to ascertain that the nonlinearity
of (ε, e) is larger than nonlinearity of (11a, 7a). This is so, because according
to Proposition 2.5, pβ,3
(e) = p9a,3a(7a) + ∆t, and the nonlinearity of (e, 3a′)
is larger than nonlinearity of (7a, 3a); we have
|(β, e)|
|(e, 3a′)|
= (exp ∆t)
|(9a, 7a)|
|(7a, 3a)|
> (exp ∆t)
|(β, δ)|
|(δ, e)|
, (5.39)
where δ ∈ (β, e) is a point such that
|(δ,e)|
|(e,3
|(11a,7a)|
|(7a,3a)|
. Thus, pe,β(ε) >
pe,β(δ), and consequently pε,3
(e) > pδ,3
(e) or, in other words,
|(ε,e)|
|(e,3
|(δ,e)|
|(e,3
. Since (e, 3a′) has larger nonlinearity than (7a, 3a), the nonlinearity
of (ε, e) must be larger than nonlinearity of (11a, 7a). Next, we do the map-
ping h and consider the situation on the even side. The interval (ε, 3a′) has
larger nonlinearity than (11a, 3a) and pε,3
(e) > p11a,3a(7a), so by principles
of monotonicity of the strength of non-euclidean push in nonlinearty of the
domain, as well as in the coordinate of the point, the action of hε,3
makes
pε̂,4
(ê) > p12a,4a(8a),
where ε̂ is the dynamic successor of ε. Because (ê, 4a′) has larger nonlinearity
than (8a, 4a), it follows that the nonlinearity of (ε̂, ê) is also larger than that
of (12a, 8a). By the same two principles applied to pβ,e(ε), we get pβ̂,ê(ε̂) >
p10a,8a(12a) + ∆t, but because of the proved nonlinearity increases, we can
also claim that
ph(β),h(e)(h(ε)) − ph(9a),h(7a)(h(11a)) > ∆t + (h
β,ε − h
9a,11a
) and
h(β̂),h(ê)
(h(ε̂))− ph(10a),h(8a)(h(12a)) > ∆t+ (h
β,ε − h
9a,11a
) + (h
β̂,ε̂− h
10a,12a
We are through with the first part of the inductive step. Now, our immediate
plan is to move e to 7a′ , then β up to 9a′ , and to replace ε by ǫ, keeping all the
above gains untouched, both on the odd and on the even side. Having done
all that, we will easily be able to move ǫ to 11a′, to complete the procedure.
Denote by λ the point determined by p5
(λ) = p5a,7a(9a), and let
τ ∈ (λ, 7a′) be such, that pλ,7
(τ) = p9a,7a(11a) + ∆t. The interval (5a′, 7a′)
has larger nonlinearity than (5a′ , e), so consequently (λ, 7a′) has larger non-
linearity than (β, e), and (τ, 7a′) has larger nonlinearity than (ε, e). Thus
(h(τ), h(7a′)) has larger nonlinearity than (h(ε), h(e)). The distance of 8a′
to the critical point 0 is smaller than similar distance for the point ê, so
the truncation argument after cutting of at 0, implies that (τ̂ , 8a′) has larger
nonlinearity than (ε̂, ê) and, in turn, larger than (12a, 8a). Again, we in-
crease the intervals in question, choosing 9a′ in place of λ, and replacing τ
by ǫ. Then, of course, (ǫ, 7a′) has yet larger nonlinearity, so (h(ǫ), h(7a′))
has larger nonlinearity than (h(τ), h(7a′)) and , after truncation, (ǫ̂, 8a′) has
larger nonlinearity than (τ̂ , 8a′). It immediately implies
),h(7
)(h(ǫ)) − ph(9a),h(7a)(h(11a)) > ∆t + (h
,ǫ − h
9a,11a
), (5.40)
ph(10
),h(8
)(h(ǫ̂))−ph(10a),h(8a)(h(12a)) > ∆t+(h
9a,11a
β̂,ε̂−h
10a,12a
(5.41)
This is what we aimed at. By the same argument that earlier let us replace e
by 7a′ and ê by 8a′ to derive formulas (5.34) and (5.35), we can now replace
ǫ by 11a′ and ǫ̂ by 12a′ , arriving at
),h(7
)(h(11a′)) − ph(9a),h(7a)(h(11a)) > ∆t + (h
9a,11a
), (5.42)
ph(10
),h(8
)(h(12a′)) − ph(10a),h(8a)(h(12a)) >
∆t + (h
9a,11a
) + (h
10a,12a
). (5.43)
Similarly to (5.36) and (5.37), we also get
(15a′)−p13a ,11a(15a) > ∆t, and p14a′ ,12a′ (16a′)−p14a ,12a(16a) > ∆t.
This completes the inductive step. The claim of the theorem follows imme-
diately. �
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	Introduction
	Notation and preliminaries
	The period 4 super-sink
	The period 8 super-sink
	RLRLRLRL…
