Journal of Inequalities in Pure and
Applied Mathematics
THE PRESSURE OF FUNCTIONS OVER (G, τ )–EXTENSIONS
MOHD. SALMI MD. NOORANI
School of Mathematical Sciences
Universiti Kebangsaan Malaysia
43600 Bangi, Selangor
Malaysia.
volume 7, issue 3, article 95,
2006.
Received 28 June, 2005;
accepted 31 March, 2006.
Communicated by: S.S. Dragomir
EMail: msn@pkrisc.cc.ukm.my
Abstract
Contents
JJ
J
II
I
Home Page
Go Back
Close
c
2000
Victoria University
ISSN (electronic): 1443-5756
197-05
Quit
Abstract
Let T : X → X be a (free) (G, τ)–extension of S : Y → Y . Moreover let
fX , fY , fG ≥ 0 be continuous functions defined on X, Y and G respectively.
In this paper we obtain some inequalities for the pressure of fX over the transformation T in relation to the pressure of fY over the transformation S and of
fG over τ.
2000 Mathematics Subject Classification: 37B40.
Key words: (G, τ )–extensions, Topological pressure.
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Contents
Title Page
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3
General Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
4
(G, τ )−Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
5
Final Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
References
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 2 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
1.
Introduction
Let T : X → X be a continuous map of a compact metric space X and τ : G →
G be an automorphism of a compact metric group G. Suppose G acts continuously and freely on the right of T so that the equation T (xg) = T (x)τ (g)
holds true ∀x ∈ X, g ∈ G. Moreover let Y be the G–orbit space and S is the
natural map on Y defined by S(xG) = (T x)G, ∀x ∈ X. Then T is called a
(G, τ )–extension of S.
Bowen [1] studied the topological entropy of the aforementioned extension
system and, amongst other things, showed that the following formula holds:
h(T ) = h(S) + h(τ ),
where h(·) is the topological entropy of the appropriate maps.
In this paper, we are interested in the pressure analogue of Bowen’s formula,
i.e., we consider the pressure of functions defined on the respective dynamical
systems instead of topological entropy. Unfortunately the main result of this
paper (i.e., the analogue of Bowen’s formula, see Corollary 4.6) is somewhat
short of an equality. Our examples indicate that equality holds but we are unable
to prove this in general. The proofs of the results arrived at in this paper are of
course modelled along the lines of Bowen.
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 3 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
2.
Pressure
We shall recall some elementary facts about pressure which are relevant to us
in proving our results. The references for this section are [2] and [4].
As before let T : X → X be a continuous map of a compact metric space
(X, d). Throughout this paper we shall assume that T have finite topological
entropy. Let K be a compact subset of X. A subset F of X is said to be an
(n, )-spanning set for K if for given k ∈ K then there exists x ∈ F such that
d(T i (k), T i (x)) ≤ , ∀0 ≤ i ≤ n − 1. Now let f be a continuous real-valued
function defined on X and consider the set defined by:
(
)
X
Qn (T, f, , K) = inf
eSn f (x) : F (n, )-spans K .
x∈F
(Here we have used the standard notation: Sn f (x) := f (x) + f (T x) + · · · +
f (T n−1 x).) Then it is easy to see that Qn (T, f, , K) ≤ ||eSn f (x) ||rn (T, , K)
where rn (T, , K) is the cardinality of an (n, )-spanning set for K with a minimum number of elements. In particular, by virtue of compactness and continuity, we have 0 < Qn (T, f, , K) < ∞. Now define:
1
Q(T, f, , K) = lim sup log Qn (T, f, , K)
n→∞ n
Lemma 2.1. Q(T, f, , K) < ∞
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 4 of 20
Proof. We know that
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
Qn (T, f, , K) ≤ ||eSn f (x) ||rn (T, , K) < ∞.
http://jipam.vu.edu.au
Hence, since ||eSn f (x) || ≤ en||f (x)|| , we have
Qn (T, f, , K) ≤ en||f (x)|| rn (T, , K).
Thus
1
1
log Qn (T, f, , K) ≤ ||f || + log rn (T, , K).
n
n
In particular we have
Q(T, f, , K) ≤ ||f || + lim sup
n→∞
1
log rn (T, , K).
n
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
It is well known that
the proof.
lim supn→∞ n1
log rn (T, , K) < ∞. Hence this completes
Title Page
We are now ready to define the pressure: The pressure of f with respect to
the subset K of X over the map T : X → X is defined by the quantity:
P (T, f, K) = lim Q(T, f, , K).
→0
Remark 1.
1. As is well-known, the metric on X can be arbitrarily chosen as long as it
induces the same topology on X.
2. When K = X we obtain the usual definition of the pressure, P (T, f ), of
the function f over the map T : X → X.
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 5 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
3. Recall that E is an (n, )-separated set of K ⊂ X if for any two distinct
points x, y ∈ E there exists some 0 ≤ i < n such that d(T i (x), T i (y)) > .
It can be checked that the above definition can also be arrived at by using
separating sets. In this case we shall be concerned with the quantity
(
)
X
Pn (T, f, , K) = sup
eSn f (x) : E (n, )-separates K .
x∈E
As is well known, the next step is to define
P (T, f, K) = lim lim inf
→0 n→∞
1
log Pn (T, f, , K).
n
The following two results are straight forward consequences of the definition
of pressure.
Proposition 2.2.
P (T, f, K) ≤ P (T, f )
whenever K ⊂ X
Proof. Let F (n, )-span X. Then F also (n, )-spans K ⊂ X. Hence
(
)
(
)
X
X
Sn f (x)
Sn f (x)
inf
e
: F (n, )-spans K ≤ inf
e
: F (n, )-spans X .
x∈F
x∈F
Thus Qn (T, f, , K) ≤ Qn (T, f, ). The result follows by taking the appropriate
logarithms and limits.
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 6 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
Lemma 2.3. Let sn (T, 4, X) denote the cardinality of a (n, 4)-separated set
of X with maximum number of elements. Then
Pn (T, f, 8, X) ≤ en||f || sn (T, 4, X).
Proof. For any > 0, it is not difficult to check that
Pn (T, f, 2, X) ≤ Qn (T, f, , X)
and
rn (T, , X) ≤ sn (T, , X).
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Hence since
Qn (T, f, , X) ≤ ekSn f k rn (T, , X)
we have
Title Page
Contents
Pn (T, f, 8, X) ≤ Qn (T, f, 4, X)
kSn f k
≤e
rn (T, 4, X)
≤ enkf k sn (T, 4, X).
JJ
J
II
I
Go Back
Close
Quit
Page 7 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
3.
General Extensions
In this section we shall start off with a straightforward modification of a crucial
estimate of Bowen and later show how this estimate is used when dealing with
pressure. But first recall the following definition:
Let T : X → X and S : Y → Y be continuous maps of compact metric spaces
X and Y . Moreover let π be a continuous surjective map from X to Y such that
π ◦ T = S ◦ π. Then as is well-known T is called an extension of S.
With respect to this extension system, we have the following result which is
essentially due to Bowen [1].
Lemma 3.1. Let > 0, α > 0 and integer n > 0 be arbitrary. Also let fX be
a continuous positive function defined on X. Then there exists some δ > 0 such
that if Yn is an (n, δ)-spanning set for Y with minimum cardinality then for any
(n, 4)-separating set F of X we have
(a+α)(n+M )
Card. F ≤ Card. Yn · e
,
where a = supy∈Y P (T, fX , π −1 (y)) and M is some finite positive real number.
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Title Page
Contents
JJ
J
II
I
Proof. Let a = supy∈Y P (T, fX , π −1 (y)). For each y ∈ Y , choose an integer
m(y) > 0 such that
Go Back
a + α ≥ P (T, fX , π −1 y) + α
1
≥
log Qm(y) (T, fX , , π −1 (y)).
m(y)
Quit
(*)
Close
Page 8 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
Also for each y ∈ Y let Ey be a (m(y), )-spanning set of π −1 (y). Now consider
the open neighborhood of π −1 (y)
m(y)−1
Uy =
[
\
z∈Ey
k=0
T −k B2 (T k z).
Then it is clear that
(X/Uy ) ∩
\
π −1 (Bγ (y)) = ∅,
γ>0
where Bγ (y) is the open-ball centered at y with radius γ. Since X is compact,
the finite intersection property for compact sets then implies there exists some
γ = γ(y) > 0 such that π −1 (Bγ (y)) ⊂ Uy . In particular, since X is compact,
there exists y1 , y2 , . . . , yr ∈ Y such that Y is covered by the open balls Bγ (yi ),
i = 1, 2, . . . , r and
π −1 (Bγ (yi )) ⊂ Uyi ,
where
m(yi )−1
Uyi =
[
\
z∈Eyi
k=0
T −k B2 (T k z)
and Eyi is an (m(yi ), )-spanning set of π −1 (yi ).
Now let δ > 0 be the Lebesgue number of this cover of Y and let Yn be a
(n, δ)-spanning set for Y with minimum cardinality. Hence for each y ∈ Yn
we can define cj (y) as the element in {y1 , y2 , . . . , yr } satisfying Bδ (S j (y)) ⊂
Bγ (ci (y)) for each j = 0, 1, . . . , n − 1.
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 9 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
Next define recursively the positive integers
t0 (y) = 0,
ts (y) =
s−1
X
m ctr (y) (y)
r=0
for each s = 1, 2, . . . , q, where q = q(y) having the property tq+1 (y) ≥ n. Now
for each q + 1-ple
(x0 , x1 , . . . , xq ) ∈ Ect0 (y) (y) × Ect1 (y) (y) × · · · × Ectq (y) (y)
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
define the set
V (y; (x0 , x1 , . . . , xq )) = x ∈ X : d T t+ts (y) (x), T t (xs ) < 2
∀0 ≤ t < m cts (y) (y) & 0 ≤ s ≤ q} .
Then it is easy to check that
[
Title Page
Contents
JJ
J
V (y; (x0 , x1 , . . . , xq )) = X
II
I
Go Back
(y;(x0 ,x1 ,...,xq ))
Close
and if F is an (n, 4)-separated subset of X, then
(3.1)
Card.(F ∩ V (y; (x0 , x1 , . . . , xq ))) = 0 or 1
for each q + 2-ple (y; (x0 , x1 , . . . , xq )), where y ∈ Yn and xs ∈ Ects (y) (y) , s =
0, 1, . . . , q.
Quit
Page 10 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
To complete the proof of this lemma, we shall now obtain an estimate for the
number Ny of the q + 1-ple’s (x0 , x1 , . . . , xq(y) ). Since
Ny =
q(y)
Y
−1
T,
,
π
c
(y)
t
(y)
s
s (y) (y))
rm(ct
s=0
we have by virtue of (∗) and fX ≥ 0
log Ny ≤
q(y)
X
log rm(ct
s (y) (y)
) T, , π
−1
cts (y) (y)
s=0
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
q(y)
≤
X
log Qm(ct
s (y)
−1
) (T, fX , , π (cts (y) ))
s=0
≤
q(y)
X
Title Page
Contents
m cts (y) (y) (a + α).
s=0
Recall that
tq+1 (y) =
q(y)
X
m cts (y) (y) .
r=0
Also tq+1 (y) = tq (y) + m ctq (y) (y) . Therefore
tq+1 (y) ≤ n − 1 + m ctq (y) (y)
≤ n + M,
JJ
J
II
I
Go Back
Close
Quit
Page 11 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
where M = max{m(y1 ), m(y2 ), . . . , m(yr )}.
Hence
log Ny ≤ (n + M )(a + α)
so that Ny ≤ e(n+M )(a+α) . In particular, (3.1) now implies
Card. F ≤ Card. Yn · Ny
≤ Card. Yn · e(n+M )(a+α)
and this completes the proof of this lemma.
The Pressure Of Functions Over
(G, τ )–Extensions
Some remarks are in order:
Mohd. Salmi Md. Noorani
1. Apart from the choices of integers m(y), the rest of the proof of the above
lemma is an exact copy of Bowen’s theorem [1, Theorem 17].
Title Page
2. When F is a (n, 4)-separating set of X with maximum cardinality, then
Contents
(n+M )(a+α)
sn (T, 4, X) ≤ rn (S, δ, Y )e
where as before sn (T, 4, X) is the cardinality of such F and rn (S, δ, Y )
is the cardinality of Yn .
JJ
J
II
I
Go Back
The following theorem is the pressure analogue of Bowen’s Theorem [1,
Theorem 17] (see also [3]).
Close
Theorem 3.2. Let fX and fY be continuous real-valued functions defined on X
and Y respectively such that fX ≥ 0 and fY ≥ 0. Then
Page 12 of 20
P (T, fX ) ≤ ||fX || + P (S, fY ) + sup P (T, fX , π −1 y).
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
y∈Y
Quit
http://jipam.vu.edu.au
Proof. Let > 0, α > 0, integer n > 0 be arbitrary and a = supy P (T, fX , π −1 y).
Also let sn (T, 4, X) denote the cardinality of a (n, 4)-separated set of X with
maximum cardinality. Then Lemmas 2.3 and 3.1 give us
Pn (T, fX , 8, X) ≤ enkfX k sn (T, 4, X)
≤ enkfX k rn (S, δ, Y )e(n+M )(a+α)
≤ enkfX k Qn (S, fY , δ, Y )e(n+M )(a+α) ,
where the last line follows from the positivity of fY .
Hence
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
log Pn (T, fX , 8, X) ≤ nkfX k + (n + M )(a + α) + log Qn (S, fY , δ, Y ).
In particular, on dividing by n and taking limit superior, we have as → 0
Title Page
Contents
P (T, fX ) ≤ kfX k + P (S, fY ) + a + α.
The result then follows since α is arbitrary.
Corollary 3.3. Let X and Y be compact metric spaces and T : X → X and
π : X → Y be continuous such that π ◦ T = π. Moreover let f ≥ 0 be
continuous on X.Then
JJ
J
II
I
Go Back
Close
Quit
sup P (T, f, π −1 y) ≤ P (T, f ) ≤ ||f || + sup P (T, f, π −1 y).
y∈Y
y∈Y
Proof. The first inequality follows from Prop. 2.2. Then in the above theorem
take S = Id, fY = 0 and fX = f .
Page 13 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
By taking f = 0 in the above corollary, we have
Corollary 3.4. Let X and Y be compact metric spaces and T : X → X and
π : X → Y be continuous such that π ◦ T = π. Then
h(T ) = sup h(T, π −1 y).
y∈Y
The last corollary is contained in [1, Corollary 18].
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 14 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
4.
(G, τ )−Extensions
As in the introduction let T : X → X be a (G, τ ) - extension of S : Y → Y .
For the rest of this paper let fY and fG be positive real-valued functions defined
on Y and G respectively. Now define the function f : X → R as f (xg) =
fY (xG) + fG (g) so that f is also positive and continuous.
Lemma 4.1. The function f is well-defined.
Proof. Let xg = zg 0 where x, z ∈ X and g, g 0 ∈ G. Consider the projection
map π : X → Y . Then π(xg) = π(zg 0 ) implies xG = zG so that x = zg for
some g ∈ G. Hence f (xg) = f (zgg) = fY (zG) + fG (gg). Also f (zg 0 ) =
fY (zG) + fG (g 0 ). And this implies f (xg) = f (zg 0 ) if fG (gg) = fG (g 0 ). But
this is true since x = zg implies zgg = zg 0 and in turn by virtue of free acting
this implies gg = g 0 . In other words fG (gg) = fG (g 0 ) and this completes the
proof.
Let y ∈ Y . Then recall that Qn (T, f, , π −1 y) is defined as
(
)
X
Qn (T, f, , π −1 y) = inf
eSn f (x) : F (n, ) − spans π −1 y .
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Title Page
Contents
JJ
J
II
I
Go Back
x∈F
Close
We have
Quit
Proposition 4.2. Given y ∈ Y , integer n ≥ 1 and > 0,
−1
n||fY ||
Qn (T, f, , π y) ≤ e
Page 15 of 20
Qn (τ, fG , δ)
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
for some δ > 0.
http://jipam.vu.edu.au
Proof. Let d and d0 be the metrics associated with X and G respectively. Then
since G acts continuously on X, we have by uniform continuity, there exists
δ > 0 such that d(xg, xg 0 ) ≤ whenever d0 (g, g 0 ) ≤ δ. Now let x ∈ π −1 y
and En be a (n, δ)-spanning set for G. Then it is easy to check that xEn is a
(n, )-spanning set for π −1 y. Observe that by commutativity of T and S (via π)
and the relation T (xg) = T (x)τ (g) we have
eSn f (xg) = eSn fY (xG) eSn fG (g)
with respect to the appropriate maps T , S and τ , so that
X
X
eSn f (xg) = eSn fY (xG)
eSn fG (g)
g∈En
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
g∈En
Title Page
or
X
eSn f (xg) = eSn fY (zG)
xg∈zEn
X
eSn fG (g) .
g∈En
Therefore
Qn (T, f, , π −1 y) ≤ eSn fY (xG) Qn (τ, fG , δ).
Note that the above manipulation is independent of x ∈ π −1 y since if x0 ∈
π −1 y then x0 = xg for some g ∈ G so that x0 G = xG which in turn implies
Sn fY (x0 G) = Sn fY (xG). The result follows since eSn fY (xG) ≤ enkfY k .
Theorem 4.3. Given y ∈ Y , we have
P (T, f, π −1 y) ≤ kfY k + P (τ, fG ).
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 16 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
Proof. From Proposition 4.2 we have
Qn (T, f, , π −1 y) ≤ en||fY || Qn (τ, fG , δ).
Therefore
lim sup
1
1
log Qn (T, f, , π −1 y) ≤ kfY k + lim sup log Qn (τ, fG , δ),
n
n
that is
Q(T, f, , π −1 y) ≤ ||fY || + Q(τ, fG , δ).
The result follows by taking → 0 and δ → 0.
Combining Theorem 3.2 and Theorem 4.3 and taking f = fX , we have
Proposition 4.4.
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Title Page
Contents
P (T, f ) ≤ kf k + kfY k + P (S, fY ) + P (τ, fG )
We also have:
JJ
J
II
I
Go Back
Proposition 4.5.
P (T, f ) ≥ P (S, fY ) + P (τ, fG )
Close
Proof. Let > 0 and let d00 be the metric on Y . Then by the uniform continuity
of π and the fact that G-acts freely on X there exists some δ > 0 such that
a. d00 (π(x), π(z)) ≤ when d(x, z) ≤ δ
b. d(xg, xg 0 ) > δ when x ∈ X and d(g, g 0 ) > .
Now let Gn ⊂ G be (n, )-separated and Yn ⊂ Y be (n, )-separated and choose
Quit
Page 17 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
Xn ⊂ X so that π|Xn : Xn → Yn is a bijection. Then Xn Gn is (n, δ)-separated.
Thus
X
Pn (T, f, δ) ≥
e(Sn f )(xg) = Pn (S, fY , ) · P (τ, fG , ).
xg∈Xn Gn
In particular
P (T, f, δ) ≥ P (S, fY , ) + P (τ, fG , ).
The result follows by taking → ∞ and δ → ∞.
The Pressure Of Functions Over
(G, τ )–Extensions
Corollary 4.6.
Mohd. Salmi Md. Noorani
P (S, fY ) + P (τ, fG ) ≤ P (T, f ) ≤ kf k + kfY k + P (S, fY ) + P (τ, fG ).
And by taking fY ≡ 0 ≡ fG , we recover Bowen’s formula
Corollary 4.7.
h(T ) = h(S) + h(τ ).
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 18 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au
5.
Final Remarks
When fY and fG are both constant, by using the variational formula for pressure
and Bowen’s formula, it is easy to deduce that P (T, f ) = P (S, fY ) + P (τ, fG ),
i.e., equality holds in this trivial case.
Perhaps, a non-trivial example supporting the equality is as follows:
Example 5.1. Let Y = {−1, 1}Z and σ : Y → Y be the full two-shift. Consider
the group extension given by
σ̂ : Y × Z3 → Y × Z3
(y, g) 7→ (σy, (g + 2y0 ) mod 3).
Of course, in this case τ = Id. Also, let f (y, g) = fY (y) + fG (g), where
fY (y) = 0 if y0 = −1, fY (y) = 1 if y0 = 1 and fG = 2, constant. Then one
can easily check that P (σ, fY ) = log(1 + e) and P (Id, fG ) = 2. Moreover
it is not difficult to see that P (σ̂, f ) = log(e2 + e3 ). In particular, we have
P (σ̂, f ) = P (σ, fY ) + P (Id, fG ).
We end with the following conjecture:
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Title Page
Contents
JJ
J
II
I
Go Back
Conjecture 5.1. Let T : X → X be a (free) (G, τ )–extension of S : Y → Y
such that T has finite topological entropy. Also let f : G → R be defined as
f (xg) = fY (xG) + fG (g) where fY , fG are positive real-valued functions on
Y and G respectively. Then
Page 19 of 20
P (T, f ) = P (S, fY ) + P (τ, fG ).
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
Close
Quit
http://jipam.vu.edu.au
References
[1] R. BOWEN, Entropy for group endormorphisms and homogeneous spaces,
Trans. Amer. Math. Soc., 153 (1971), 401–414.
[2] A. KATOK AND B. HASSELBLATT, Introduction to the Modern Theory
of Dynamical Systems, Cambridge University Press, Cambridge, 1995.
[3] H.B. KEYNES, Lifting topological entropy, Proc. Amer. Math. Soc., 24
(1970), 440–445.
[4] P. WALTERS, An Introduction To Ergodic Theory, GTM 79, SpringerVerlag, Berlin, 1982.
The Pressure Of Functions Over
(G, τ )–Extensions
Mohd. Salmi Md. Noorani
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 20 of 20
J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006
http://jipam.vu.edu.au