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1 Introduction.
In the dimension theory of dynamical systems, and in particular in the study of the
Hausdorff dimension of invariant sets of hyperbolic dynamics, the theory is only developed
to full satisfaction in the case of conformal dynamical systems (both invertible
and non-invertible ). Roughly speaking, these are dynamical systems for which at each
point the rate of contraction and expansion are the same in every direction. Bowen
0
02000 Mathematics Subject classification: Primary 37D35; Secondary 37C45.
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三思论文代写网
[3] was the first to express the Hausdorff dimension of an invariant set as a solution
of an equation involving topological pressure. Ruelle [13] refined Bowen抯 method and
obtained the following result. Assume that f is a C1+?conformal expanding map, ?is
an isolated compact invariant set and fj?is topologically mixing, then the Hausdorff
dimension of ? dimH ?is given by the unique solution ?of the equation
P(fj?‘ log kDxfk) = 0 (1.1)
where P(fj? ? is the topological pressure functional. The smoothness C1+?was recently
relaxed to C1 [10].
For non-conformal dynamical systems there exists only partial results. For example,
the Hausdorff dimension of hyperbolic invariant sets was only computed in some special
cases. Hu [12] gave an estimate of dimension of non-conformal repeller for C2 map.
Falconer [7, 8] computed the Hausdorff dimension of a class of non-conformal repellers.
Related ideas were applied by Simon and Solomyak [15] to compute the Hausdorff
dimension of a class of non-conformal horseshoes in R3.
For C1 non-conformal repellers, in [17], the author uses singular values of the derivative
Dxfn for all n 2 Z+, to define a new equation which involves the limit of a sequence
of topological pressure. Then he shows that the unique solution of the equation is an
upper bounds of Hausdorff dimension of repeller. In [1], the same problem is considered.
The author bases on the non-additive thermodynamic formalism which was
introduced in [2] and singular value of the derivative Dxfn for all n 2 Z+, and gives
an upper bounds of box dimension of repeller under the additional assumptions for
which the map is C1+?and ?bunched. This automatically implies that for Hausdorff
dimension. In [9], the author defines topological pressure of sub-additive potential under
the condition k(Dxf)?k2kDxfk < 1, which means that f is 1unched. They
also obtain an upper bounds of Hausdorff dimension of repeller. In [4], the first named
author prove that the upper bounded of Hausdorff dimension for C1 non-conformal
repeller obtained in [1, 9, 17] are same and it is the unique root of Bowen equation for
sub-additive topological pressure.
In this paper, we introduce the notion of average conformal repeller. Using thermodynamic
formalism for sub-additive potential defined in [5], we prove that Hausdorff
dimension and box dimension of average conformal repellers is the unique root of Bowen
equation for subadditive topological pressure. The map f is only needed C1, without
additional condition. Meanwhile, we introduce sup-additive potential topological pressure
and prove that for special potentials, sub-additive and sup-additive topological
pressures are same. In [2, 11], the authors introduce the concepts of quasi-conformal
and asymptotically conformal repeller by using Markov construction and prove that
its dimension is the unique root of the equation obtained by non-additive topological
pressure. It is obvious that quasi-conformal and asymptotically conformal repeller are
average conformal repellers, but reverse is not true. Therefore our result is a generalization
of the results in [2, 11].
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First we recall some basic definitions and notations.
Let f : X ! X be a continuous map. A set E ?X is called (n; ? separated set
with respect to f if x; y 2 E then dn(x; y) = max0穒穘? d(fix; fiy) > ? For x 2 X
and r > 0, define
Bn(x; r) = fy 2 X : fiy 2 B(fix; r); for all i = 0; ???; n ?1g:
If ?is a real continuous function on X and n 2 Z+, let
Sn?x) =
n? Xi=0
?fi(x)):
We define
Pn(? ; ? = supfXx2E
exp Sn?x) : E is a (n; ? ?separated subset of Xg:
Then the topological pressure of ?is given by
P(f; ? = lim
?0
lim sup
n!1
1
n
log Pn(? ?:
Next we give some properties of P(f; ? : C(M;R) ! R [ f1g.
Proposition 1.1. Let f : M ! M be a continuous transformation of a compact
metrisable space M. If '1; '2 2 C(X;R), then the followings are true:
(1) P(f; 0) = htop(f):
(2) jP(f; '1) ?P(f; '2)j ?k'1 ?'2k.
(3) '1 ?'2 implies that P(f; '1) ?P(f; '2).
Proof. See Walters book [16].
Corollary 1. Let f : M ! M be a continuous transformation of a compact metrisable
space M. If ' 2 C(M;R) and ' < 0 then function P(? = P(f; ?) is continuous and
strictly decreasing in ?
Proof. It easily follows from Proposition 1.1.
The paper is organized as follows. In Section 2, we develop variational principal
for sub-additive potential. In Section 3, we introduce the definition of average conformal
repeller and give related results and the main theorem. In section 4, we develop
sup-additive thermodynamics formalism and variational principal for sup-additive potential.
In section 5, we give the proof of main result.
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2 A sub-additive thermodynamics formalism
Let f : X ! X be a continuous map. A set E ?X is called (n; ? separated set with
respect to f if x; y 2 E then dn(x; y) = max0穒穘? d(fix; fiy) > ? A sub-additive
valuation on X is a sequence of functions 羘 : M ! R such that
羗+n(x) ?羘(x) + 羗(fn(x));
we denote it by F = f羘g.
In the following we will define the topological pressure of F = f羘g with respect to
f. We define
Pn(F; ? = supfXx2E
exp 羘(x) : E is a (n; ? ?separated subset of Xg:
Then the topological pressure of F is given by
P(f;F) = lim
?0
lim sup
n!1
1
n
log Pn(F; ?:
Let M(X) be the space of all Borel probability measures endowed with the weak*
topology. Let M(X; f) denote the subspace of M(X) consisting of all f-invariant
measures. For ?2M(X; f), let h?f) denote the entropy of f with respect to ? and
let F?? denote the following limit
F?? = lim
n!1
1
n Z 羘d?
The existence of the above limit follows from a sub-additive argument. We call F??
the Lyapunov exponent of F with respect to ?since it describes the exponentially
increasing speed of 羘 with respect to ?
In [5], authors proved that the following variational principal
Theorem 2.1. [5] Under the above general setting, we have
P(f;F) = supfh?T) + F?? : ?2M(X; f)g:
3 Average conformal repeller
Let M be a C1 Riemann manifold, dimM = m. Let U be an open subset of M and let
f : U ! M be a C1 map. Suppose ??U is a compact invariant set, that is, f?= ?
and there is k > 1 such that for all x 2 ?and v 2 TxM,
kDxfvk ?kkvk;
where k:k is the norm induced by an adapted Riemannian metric. Let M(fj?; E(f)
denote the all f invariant measures and the all ergodic invariant measure supported
on ?respectively. By the Oseledec multiplicative ergodic theorem, for any ?2 E(f),
we can define Lyapunov exponents ?(? ??(? ?????竛(?; n = dimM.
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Definition 3.1. An invariant repeller is called average conformal if for any ?2 E(f),
?(? = ?(? = ???= 竛(? > 0.
It is obvious that a conformal repeller is an average conformal repeller, but reverse
isn抰 true.
Next we will give main theorem.
Theorem 3.1. (Main Theorem) Let f be C1 dynamical system and ?be an average
conformal repeller, then the Hausdorff dimension of ?is zero t0 of t 7! P(F), where
F = flog(m(Dxfn); x 2 ? n 2 Ng: (3.2)
where m(A) = kA?k?
The proof will be given in section 5.
Theorem 3.2. If ?be an average conformal repeller, then
lim
n!1
1
n
(log kDfn(x)k ?logm(dfn(x))) = 0
uniformly on ?
Proof. Let
Fn(x) = log kDfn(x)k ?logm(dfn(x)); n 2 N; x 2 ?
It is obviously that the sequence fFn(x)g is a non-negative subadditive function sequence.
That is say
Fn+m(x) ?Fn(x) + Fm(fn(x)); x 2 ?
Suppose (3.2) is not true, then there exists ? > 0, for any k 2 N, there exits nk ?k
and xnk 2 ?such that
1
nk
Fnk(xnk) ??:
Define measures
筺k =
1
nk
nk? Xi=0
眆i(xnk ):
Compactness of P(f) implies there exists a subsequence of 筺k that converges to measure
? Without loss of generality, we suppose that 筺k ! ? It is well known that ?
is f -invariant. Therefore ?2M(f).
For a fixed m, we have
lim
k!1ZM
1
m
Fm(xnk)d筺k = ZM
1
m
Fm(xnk)d?
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It implies
lim
k!1
1
nk
nk? Xi=0
1
m
Fm(fi(xnk)) = ZM
1
m
Fm(x)d?
For a fixed m, let nk = ms + l; 0 ?l < m. The sub-additivity of fFng implies that for
j = 0; ?? ?1,
Fnk(xnk) ?Fj(xnk) + Fm(fj(xnk) + ???+ Fm(fm(s?)fj(xnk))
+Fm+l(fm(s?)fj(xnk))
Summing j from 0 to m ?1, we get
Fnk(xnk) ?
1
m
m? Xj=0
s? Xi=0
Fm(fim+j(xnk))
+
1
m
m? Xj=0
[Fj(xnk) + Fm+l(f(s?)m+j(xnk ))]
Let C1 = maxi=1;ⅱ?;2m? maxx2?Fi(x).
Fnk(x) ?
(sm+l)? Xj=0
1
m
Fm(fj(x)) ?
1
m
sm? X j=(s?)m
Fm(fj(x)) + 2C1
?
nk? Xj=0
1
m
Fm(fj(x)) + 4C1:
Hence we have
lim
k!1
1
nk
Fnk(x) ?lim
k!1
1
nk
nk? Xi=0
1
m
Fm(fi(x)) = ZM
1
m
Fm(x)d?
The arbitrariness of m 2 N implies that
lim
k!1
1
nk
Fnk(x) ?
1
m ZM
Fm(x)d? 8m 2 N:
Hence
lim
m!1
1
m ZM
Fm(x)d??? > 0:
Then ergodic decomposition theorem [16] implies that there exists 樄 2 E(f) such that
lim
m!1
1
m ZM
Fm(x)d樄 ?? > 0:
On the other hand, from Oseledec theorem and Kingman抯 subadditive ergodic theorem,
we have lim
m!1
1
m RM log kDfn(x)kd樄 = 竛(樄) and lim
m!1
1
m RM logm(fn(x))d樄 =
?(樄). Therefore
竛(樄) ??(樄) ??:
This gives a contradiction to assumption of average conformal.
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4 Sup-additive variational principal
In this section, we first give the definition of sup-additive topological principal. Then
we prove the variational principal for special sup-additive potential.
Let f : X ! X be a continuous map. A set E ?X is called (n; ? separated set with
respect to f if x; y 2 E then dn(x; y) = max0穒穘? d(fix; fiy) > ? A sup-additive
valuation on X is a sequence of functions 'n : M ! R such that
'm+n(x) ?'n(x) + 'm(fn(x));
we denote it by F = f'ng.
In the following we will define the topological pressure of F = f'ng with respect
to f. We define
P?n(F; ? = supfXx2E
exp 'n(x) : E is a (n; ? ?separated subset of Xg:
Then the topological pressure of F is given by
P?f;F) = lim
?0
lim sup
n!1
1
n
log Pn(F; ?:
For every ?2M(X; f), let F?? denote the following limit
F?? = lim
n!1
1
n Z 'nd?
The existence of the above limit follows from a sup-additive argument. We call F??
the Lyapunov exponent of F with respect to ?since it describes the exponentially
increasing speed of 'n with respect to ?
Theorem 4.1. Let f be C1 dynamical system and ?be an average conformal repeller,
and F = f'n(x)g = f log kDfn(x)kg for t ?0 be a sup-additive function sequence.
Then we have
P?f;F) = supfh?T) + F?? : ?2M(X; f)g:
Proof. First we prove that for any m 2 N
P?f;F) ?P(f;
'm
m
):
For a fixed m, let n = ms + l; 0 ?l < m. From the sup-additivity of f'ng, we have
'n(x) ?
1
m
m? Xj=0
s? Xi=0
'm(fim+j(x)) +
1
m
m? Xj=0
['j(x) + 'm+l(f(s?)m+j(x))]:
Let C1 = mini=1;ⅱ?;2m? minx2X 'i(x). Then it has
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'n(x) ?
(sm+l)? Xj=0
1
m
'm(fj(x)) ?
1
m
sm? X j=(s?)m
'm(fj(x)) + 2C1
?
n? Xj=0
1
m
'm(fj(x)) + 4C1:
Hence we have
exp('n(x)) ?exp(
n? Xj=0
1
m
'm(fj(x)) + 4C1):
Thus
P?n(F; ? = supfXx2E
exp 'n(x) : E is a (n; ? ?separated subset of Xg
?Pn(
1
m
'm; ? ?exp(4C1):
It implies
P?f;F) ?P(f;
1
m
'm) :
From the arbitrary of m 2 Z+, we have
P?f;F) ?P(f;
1
m
'm); for all m 2 Z+:
By the variational principal in [16], for every ?2M(f), we have
P?f;F) ?P(f;
1
m
'm) ?h?f) + ZM
1
m
'n(x)d? 8m 2 N:
Hence we have for every ?2M(f)
P?f;F) ?h?f) + lim
m!1ZM
1
m
'n(x)d?
Therefore
P?f;F) ?supfh?f) + lim
m!1ZM
1
m
'n(x)d? ?2M(f)g
Let ﹏(x) =  logm(Dfn(x)) for t ?0. Then it is sub-additive. By the theorem
in [5], we have
P(f; f﹏g) = supfh?f) + lim
m!1ZM
1
m
﹏(x)d? ?2M(f)g
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By the definitions,  logm(Dfn(x)) ? log kDfn(x)k for t ?0 implies that
P?f;F) ?P(f; f﹏g):
Theorem 3.2 implies that for any ?2M(f), it has
lim
m!1ZM
1
m
﹏(x)d?= lim
m!1ZM
1
m
'n(x)d?
Therefore
P?f;F) = supfh?f) + lim
m!1ZM
1
m
﹏(x)d? ?2M(f)g:
This completes the proof of theorem.
5 The proof of main theorem
In this section, we will give the proof of main theorem. First we state some known
results.
In [1], Barreira prove the following theorem.
Theorem 5.1. If f is a C1 expanding map and ?is a repeller, then
s1 ?dimH??dimB??dimB??t1
where s1 and t1 are the unique roots of the Bowen抯 equations P(f; log kDf(x)k) = 0
and P(f; logm(Df(x))) = 0 respectively.
Since ?is f-invariant, it is fn-invariant. Hence we have the following corollary.
Corollary 2. If f is a C1 expanding map and ?is a repeller, then
sn ?dimH??dimB??dimB??tn
where sn and tn are the unique roots of the Bowen抯 equations P(fn; log kDfn(x)k) =
0 and P(fn; logm(Dfn(x))) = 0 respectively.
Next we prove that the sequences ft2kg and fs2kg are monotone.
Theorem 5.2. The sequence fs2kg is monotone, and
lim
k!1
s2k = s?
Then we have s?is the unique root of equation P?f;flog kDfn(x)kg) = 0.
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Proof. First we prove that the sequence fs2ng is monotone increasing. Let 'n =
og k(Dfn(x)k and F = f'ng: Then it is a sup-additive function sequence. For a
fixed k 2 N,
Pk(? ? = supfXx2E
exp Sn?x) : E is a (n; ? ?separated subset of Xg:
For 8?> 0; by the uniformly continuity of f, there exists ?> 0 such that if E ?M
is an (n; ? separated set of f2k+1, then E is an (2n; ? separated set of f2k and ?! 0
when ?! 0. Using the subadditivity of 'n, the Birkhoff sum Sn?k+1 of '2k+1 with
respect to f2k+1 has the following property:
Sn'2k+1(x) = '2k+1(x) + '2k+1(f2k+1
x) + ???+ '2k+1(f2k+1(n?)x)
?'2k(x) + '2k(f2k
x) + '2k(f2k+1
x) + '2k(f2k+1
f2k
x)
+ ???+ '2k(f2k+1(n?)x) + '2k(f2k+1(n?)f2k
x)
= S2n'2k(x)
where S2n'2k(x) is the Birkhoff sum of '2k with respect to f2k .
Thus
Pn(f2k+1
; '2k+1; ? ?P2n(f2k
; '2k ; ?:
Hence
P(f2k+1
; '2k+1) ?2P(f2k
; '2k):
Therefore if s2k+1 is the unique root of Bowen抯 equation P(t'2k+1) = 0, then we
have
0 = P(f2k+1
; s2k+1'2k+1) ?2P(f2k
; s2k+1'2k):
The monotone decreasing of the function P(f2k ; t?k) implies that s2k ?s2k+1.
The arbitrariness of k implies that the sequence fs2kg monotone decreasing.
Next we prove that
P?f;F) ?
1
k
P(fk; 'k) 8k 2 N:
For a fixed k 2 N, let n = km + r, 0 ?r < k, and let C = minx2M max1穒穔羒(x).
For 8?> 0; by the uniformly continuity of f, there exists ?> 0 such that if E ?M is
an (n; ? separated set of f, then E is an (m; ? separated set of fk and ?! 0 when
?! 0. Using the sup-additivity of 'n, we have
'n(x) ?'k(x) + 'k(fk(x)) + ???+ 'k(f(m?)k(x)) + 'r(fmk(x)):
Thus
P?n(f;F; ? ?Pm(fk; 'k; ? ?e:
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Hence
P?f;F; ? ?
1
k
P(fk; 'k; ?:
It gives that
P?f;F) ?
1
k
P(fk; 'k):
Therefore
P?f;F) ?
1
2k P(f2k
; ?k) 8k 2 N:
Let tF = ft羘(x)g. Then we have
P?f; s2kF) ?
1
2k P(f2k
; s2k?k) = 0 8k 2 N:
The monotone decreasing of P?f; tF) with respect to t implies that the unique root
s?of the equation
P?f; tF) = 0
satisfies
s??s2k 8k 2 N:
Thus
s??s = lim
k!+1
s2k :
Next we want to prove that
s ?s?
For a fixed m,
1
2mP(f2m
; s2m'2m) = 0
using the variational principle, for any ?2M(f) 組(f2m), it has
h?f) +
1
2ms2m ZM
'2md?=
1
2m(h?f2m
) + s2m ZM
'2md? ?0:
Let m ! 1, we
h?f) + s lim
m!1ZM
1
2m'2md??0:
Using sup-additive variational principle, we have
P?f; sf'ng) ?0:
Since P(f; tf'ng) is strictly monotone decreasing with respect to t, we have
s??s:
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Lemma 5.1. If 羘(x) is a subadditive sequence, then
lim
k!1
1
2k P(f2k
; ?k) ?lim
m!1
P(f;
?m
2m ):
Proof. For a fixed k 2 N. It is well known that if E ?M is an (n; ? separated set of
f2k , then E is an (n2k; ? separated set of f. By the definition
P(f2k
; ?k) = lim
?1
lim sup
n!1
1
n
log supfXx2E
exp( ?Sn?k(x))
: E is a (n; ? separated set of f2kg;
where
( ?Sn?k(x)) = ?k(x) + ?k(f2k
x) + ???+ ?k(f(n?)2k
x):
Hence for a fixed m < k, let 2k = 2mq + r and C = maxx2M maxi=1;ⅱ?;2m 羒(x), the
subadditivity of 羘 implies that
?k(x) ?
1
2m
2m? Xj=0
q? Xi=0
?m(fi2m+j(x)) +
1
2m
2m? Xj=0
[羓(x) + ?m+l(f(q?)2m+j(x))]
?
2k? Xi=0
1
2m?m(fi(x)) + 4C:
Thus for 1 ?j ?n ?1, we have
?k(f2kj(x)) ?
2k? Xi=0
1
2m?m(fi(f2kj(x)) + 4C:
Hence
?Sn?k(x) = ?k(x) + ?k(f2k
x) + ???+ ?k(f(n?)2k
x)
?
n2k? Xi=0
1
2m?m(fi(x)) + 4nC
= Sn2k(
1
2m?m)(x) + 4nC:
It gives that
Pn(f2k
; ?k ; ? ?Pn2k(f;
1
2m?m; ? ?e4nC:
Thus
P(f2k
; ?k) ?2kP(f;
1
2m?m) + lim
n!1
1
n
log e4nC
= 2kP(f;
1
2m?m) + 4C:
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Therefore
lim
k!1
1
2k P(f2k
; ?k) ?P(f;
1
2m?m) for all m 2 Z+:
Hence
lim
k!1
1
2k P(f2k
; ?k) ?lim
m!1
P(f;
1
2m?m):
Lemma 5.2.
lim
n!1
P(f;
?k
2k ) ?P(f;F):
Proof. Since f : ?! ?is expanding map, h?f) is an upper-semi continuous function
from M(fj? to R. From variational principal of topological pressure [16], we have
that for every k 2 Z+ there exists ?k 2M(fj? such that
P(fj?
1
2k ?k) = h?k (f) + Z?
1
2k ?kd?k :
Since M(fj? is compact, it implies that ?k has a subsequence which convergence to
?2M(fj?. Without loss of generality, suppose that ?k convergence to ? Using the
subadditivity and invariant of ?k , then we have for every k 2 N
h?k (f) + Z?
?k(x)
2k d?k ?h?k (f) + Z?
?(x)d?k :
Furthermore for fixed s 2 N. If k > s, from the subadditivity and invariance of ?k , it
has
h?k (f) + Z?
?k(x)
2k d?k ?h?k (f) + Z?
?s(x)
2s d?k :
Since h?f) is a upper-semi continuous function, we have
lim
k!1
P(f;
?k
2k ) = lim
k!1
(h?k (f) + Z?
?k(x)
2k d?k)
?lim
k!1
(h?k (f) + Z?
?s(x)
2s d?k)
?h?f) + Z?
?s(x)
2s d?
Since sequence fR?羘(x)d筭 is sub-additive sequence, it has
lim
n!1Z?
羘(x)
n
d?= inf
n?fZ?
羘(x)
n
d筭:
The arbitrariness of s 2 N implies that
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lim
k!1
P(f;
?k
2k ) ?h?f) + lim
s!1Z?
?s
2s (x)d?
Hence by variational principal of the sub-additive topological pressure in [5], we have
lim
k!1
P(f;
?k
2k ) ?h?f) + lim
s!1Z?
?s
2s (x)d??P(f;F):
This completes the proof of lemma.
Theorem 5.3. The sequence ftng is monotone, and
lim
n!1
tn = t?
where t?is the unique root of equation P(f;flogm(Dfn(x))g) = 0.
Proof. First we prove that the sequence ft2ng is monotone decreasing. Let 羘 =
ogm(Dfn(x)): For a fixed k 2 N,
Pk(? ; ? = supfXx2E
exp Sn?x) : E is a (n; ? ?separated subset of Xg:
For 8?> 0; by the uniformly continuity of f, there exists ?> 0 such that if E ?M
is an (n; ? separated set of f2k+1, then E is an (2n; ? separated set of f2k and ?! 0
when ?! 0. Using the subadditivity of 羘, the Birkhoff sum Sn?k+1 of ?k+1 with
respect to f2k+1 has the following property:
Sn?k+1(x) = ?k+1(x) + ?k+1(f2k+1
x) + ???+ ?k+1(f2k+1(n?)x)
??k(x) + ?k(f2k
x) + ?k(f2k+1
x) + ?k(f2k+1
f2k
x)
+ ???+ ?k(f2k+1(n?)x) + ?k(f2k+1(n?)f2k
x)
= S2n?k(x)
where S2n?k(x) is the Birkhoff sum of ?k with respect to f2k .
Thus
Pn(f2k+1
; ?k+1; ? ?P2n(f2k
; ?k ; ?:
Hence
P(f2k+1
; ?k+1) ?2P(f2k
; ?k):
Therefore if t2k+1 is the unique root of Bowen抯 equation P(t?k+1) = 0, then we
have
0 = P(f2k+1
; t2k+1?k+1) ?2P(f2k
; t2k+1?k):
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The monotone decreasing of the function P(f2k ; t?k) implies that t2k ?t2k+1.
The arbitrariness of k implies that the sequence ft2kg monotone decreasing. Hence
limit exists and we denote the limit of this sequence by t. From the proof as above, we
have
P(f2k+1; ?k+1)
2k+1 ?
P(f2k ; ?k)
2k ?????
P(f2; ?)
2 ?P(f; ?:
Next we prove that
P(f;F) ?
1
k
P(fk; 羕) 8k 2 N:
For a fixed k 2 N, let n = km + r, 0 ?r < k, and let C = maxx2M max1穒穔羒(x).
For 8?> 0; by the uniformly continuity of f, there exists ?> 0 such that if E ?M is
an (n; ? separated set of f, then E is an (m; ? separated set of fk and ?! 0 when
?! 0. Using the subadditivity of 羘, we have
羘(x) ?羕(x) + 羕(fk(x)) + ???+ 羕(f(m?)k(x)) + 羠(fmk(x)):
Thus
Pn(f;F; ? ?Pm(fk; 羕; ? ?eC:
Hence
P(f;F; ? ?
1
k
P(fk; 羕; ?:
It gives that
P(f;F) ?
1
k
P(fk; 羕):
Therefore
P(f;F) ?
1
2k P(f2k
; ?k) 8k 2 N: (5.3)
Let tF = ft羘(x)g. Then we have
P(f; t2kF) ?
1
2k P(f2k
; t2k?k) = 0 8k 2 N:
Therefore the unique root t?of the equation
P(f; tF) = 0
satisfies
t??t2k 8k 2 N:
Thus
t??t = lim
k!+1
t2k :
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Next we want to prove that
t ?t?
From 5.3 and lemma 5.1, 5.2, we have the sequence f 1
2kP(f2k ; ?k)g is monotone decreasing
and it converges to P(f;F). By the definition, it is easy to prove that
0 ?
P(f2k ; t?k)
2k ?
P(f2k ; t2k?k)
2k ?jt ?t2k jC; 8k 2 N;
where C = maxx2M j?(x)j: Let k ! 1, we have
P(f; tF) = 0:
Hence it has,
t = t?
Theorem 5.4. t?= s?
Proof. From theorems as above, we have functions
P(f;flogm(Dfn(x))g)
and
P(f;flog kDfn(x)kg)
coincide and both of them have unique zero points. Therefore
t?= s?
The proof of main theorem:
From Corollary 2 and theorems 5.4 as above, we have
dimH?= dimB?= dimB = s?= t?
This completes the proof of main theorem.
Acknowledgement. Author would like to thank Prof.Dejun Feng and Prof.Marcelo
Viana for their discussions and suggestions. This work is partially supported by
NSFC(10571130), NCET, and SRFDP of China.
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