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volume 7, issue 3, article 95, 2006.

Received 28 June, 2005;

accepted 31 March, 2006.

Communicated by:S.S. Dragomir

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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.

EMail:msn@pkrisc.cc.ukm.my

2000c Victoria University ISSN (electronic): 1443-5756 197-05

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The Pressure Of Functions Over (G, τ)–Extensions Mohd. Salmi Md. Noorani

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J. Ineq. Pure and Appl. Math. 7(3) Art. 95, 2006

http://jipam.vu.edu.au

Abstract

Let T:X → X be a (free) (G, τ)–extension of S:Y → Y. Moreover let fX, fY, fG ≥ 0be continuous functions defined onX,Y andG respectively.

In this paper we obtain some inequalities for the pressure off_X over the trans- formationT in relation to the pressure off_Y over the transformationSand of fGoverτ.

2000 Mathematics Subject Classification:37B40.

Key words:(G, τ)–extensions, Topological pressure.

1. Introduction

LetT: X →Xbe a continuous map of a compact metric spaceXandτ: G→ G be an automorphism of a compact metric groupG. Suppose Gacts 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 letY be theG–orbit space andSis the natural map on Y defined by S(xG) = (T x)G, ∀x ∈ X. ThenT is called a (G, τ)–extension ofS.

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),

whereh(·)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.

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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 letT: 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 ofX. A subsetF of X is said to be an (n, )-spanning set for K if for givenk ∈ K then there existsx ∈ F such that d(Tⁱ(k), Tⁱ(x)) ≤ , ∀0 ≤ i ≤ n−1. Now letf be a continuous real-valued function defined onX and consider the set defined by:

Q_n(T, f, , K) = inf (

X

x∈F

e^Sⁿ^f(x): F (n, )-spansK )

.

(Here we have used the standard notation: S_nf(x) := f(x) +f(T x) +· · ·+ f(Tⁿ⁻¹x).) Then it is easy to see thatQ_n(T, f, , K) ≤ ||e^Sⁿ^f^(x)||r_n(T, , K) wherer_n(T, , K)is the cardinality of an(n, )-spanning set forK with a minimum number of elements. In particular, by virtue of compactness and continuity, we have0< Q_n(T, f, , K)<∞. Now define:

Q(T, f, , K) = lim sup

n→∞

1

n logQn(T, f, , K) Lemma 2.1. Q(T, f, , K)<∞

Proof. We know that

Qn(T, f, , K)≤ ||e^Sⁿ^f(x)||rn(T, , K)<∞.

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Hence, since||e^Sⁿ^f(x)|| ≤e^n||f(x)||, we have

Q_n(T, f, , K)≤e^n||f^(x)||r_n(T, , K).

Thus 1

nlogQ_n(T, f, , K)≤ ||f||+ 1

nlogr_n(T, , K).

In particular we have

Q(T, f, , K)≤ ||f||+ lim sup

n→∞

1

nlogrn(T, , K).

It is well known thatlim sup_n→∞ ¹_nlogr_n(T, , K)<∞. Hence this completes the proof.

We are now ready to define the pressure: The pressure off with respect to the subsetK ofX over the mapT: X →Xis defined by the quantity:

P(T, f, K) = lim

→0Q(T, f, , K).

Remark 1.

1. As is well-known, the metric onX can be arbitrarily chosen as long as it induces the same topology onX.

2. WhenK = X we obtain the usual definition of the pressure, P(T, f), of the functionf over the mapT: X →X.

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3. Recall thatE is an(n, )-separated set ofK ⊂ X if for any two distinct pointsx, y ∈Ethere exists some0≤i < nsuch thatd(Tⁱ(x), Tⁱ(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

Pn(T, f, , K) = sup (

X

x∈E

e^Sⁿ^f^(x): E (n, )-separatesK )

.

As is well known, the next step is to define P(T, f, K) = lim

→0lim inf

n→∞

1

n logP_n(T, f, , K).

The following two results are straight forward consequences of the definition of pressure.

Proposition 2.2.

P(T, f, K)≤P(T, f) wheneverK ⊂X

Proof. LetF (n, )-spanX. ThenF also(n, )-spansK ⊂X. Hence

inf (

X

x∈F

e^Sⁿ^f^(x): F (n, )-spansK )

≤inf (

X

x∈F

e^Sⁿ^f(x): F (n, )-spansX )

.

ThusQ_n(T, f, , K)≤Q_n(T, f, ). The result follows by taking the appropriate logarithms and limits.

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Lemma 2.3. Lets_n(T,4, X)denote the cardinality of a(n,4)-separated set ofXwith maximum number of elements. Then

P_n(T, f,8, X)≤e^n||f||s_n(T,4, X).

Proof. For any >0, it is not difficult to check that P_n(T, f,2, X)≤Q_n(T, f, , X) and

rn(T, , X)≤sn(T, , X).

Hence since

Q_n(T, f, , X)≤e^kSⁿ^f^kr_n(T, , X) we have

P_n(T, f,8, X)≤Q_n(T, f,4, X)

≤e^kSⁿ^fkrn(T,4, X)

≤e^nkf^ks_n(T,4, X).

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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:

LetT: X →X andS: Y →Y be continuous maps of compact metric spaces XandY. Moreover letπbe a continuous surjective map fromXtoY such that π◦T =S◦π. Then as is well-knownT is called an extension ofS.

With respect to this extension system, we have the following result which is essentially due to Bowen [1].

Lemma 3.1. Let > 0,α > 0and integern > 0be arbitrary. Also let f_X be a continuous positive function defined onX. Then there exists someδ >0such that ifY_nis an(n, δ)-spanning set forY with minimum cardinality then for any (n,4)-separating setF ofX we have

Card.F ≤Card.Y_n·e^(a+α)(n+M⁾,

wherea= sup_y∈Y P(T, f_X, π⁻¹(y))andM is some finite positive real number.

Proof. Let a = sup_y∈Y P(T, f_X, π⁻¹(y)). For eachy ∈ Y, choose an integer m(y)>0such that

a+α≥P(T, f_X, π⁻¹y) +α

≥ 1

m(y)logQm(y)(T, fX, , π⁻¹(y)).

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Also for eachy∈Y letE_ybe a(m(y), )-spanning set ofπ⁻¹(y). Now consider the open neighborhood ofπ⁻¹(y)

U_y = [

z∈Ey

m(y)−1

\

k=0

T^−kB₂(T^kz).

Then it is clear that

(X/U_y)∩ \

γ>0

π⁻¹(B_γ(y)) =∅,

whereB_γ(y)is the open-ball centered atywith radiusγ. SinceX is compact, the finite intersection property for compact sets then implies there exists some γ = γ(y) > 0such thatπ⁻¹(B_γ(y)) ⊂ U_y. In particular, sinceX is compact, there existsy₁, y₂, . . . , y_r ∈ Y such thatY is covered by the open ballsB_γ(y_i), i= 1,2, . . . , r and

π⁻¹(B_γ(y_i))⊂U_y_i, where

U_y_i = [

z∈E_yi

m(yi)−1

\

k=0

T^−kB₂(T^kz)

andEyi is an(m(yi), )-spanning set ofπ⁻¹(yi).

Now letδ > 0 be the Lebesgue number of this cover ofY and let Yn be a (n, δ)-spanning set forY with minimum cardinality. Hence for each y ∈ Y_n we can define c_j(y)as the element in {y₁, y₂, . . . , y_r}satisfyingB_δ(S^j(y)) ⊂ Bγ(ci(y))for eachj = 0,1, . . . , n−1.

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Next define recursively the positive integers t₀(y) = 0,

t_s(y) =

s−1

X

r=0

m c_t_r_(y)(y)

for eachs= 1,2, . . . , q,whereq=q(y)having the propertytq+1(y)≥n. Now for eachq+ 1-ple

(x₀, x₁, . . . , x_q)∈E_c_t

0(y)(y)×E_c_t

1(y)(y)× · · · ×E_c_tq_(y)_(y) define the set

V(y; (x₀, x₁, . . . , x_q)) =

x∈X: d T^t+t^s^(y)(x), T^t(x_s)

<2

∀0≤t < m c_t_s_(y)(y)

& 0≤s≤q}. Then it is easy to check that

[

(y;(x0,x1,...,xq))

V(y; (x₀, x₁, . . . , x_q)) =X

and ifF is an(n,4)-separated subset ofX, then

(3.1) Card.(F ∩V(y; (x₀, x₁, . . . , x_q))) = 0or1

for eachq+ 2-ple(y; (x₀, x₁, . . . , x_q)),wherey ∈ Y_n andx_s ∈ E_c_ts(y)_(y),s = 0,1, . . . , q.

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To complete the proof of this lemma, we shall now obtain an estimate for the numberNy of theq+ 1-ple’s(x0, x1, . . . , xq(y)). Since

N_y =

q(y)

Y

s=0

r_m(^cts(y)(y)) T, , π⁻¹ c_t_s_(y)(y)

we have by virtue of(∗)andfX ≥0

logN_y ≤

q(y)

X

s=0

logr_m(^cts(y)(y)) T, , π⁻¹ c_t_s_(y)(y)

≤

q(y)

X

s=0

logQ_m(^cts(y))(T, f_X, , π⁻¹(c_t_s_(y)))

≤

q(y)

X

s=0

m c_t_s_(y)(y)

(a+α).

Recall that

t_q+1(y) =

q(y)

X

r=0

m c_t_s_(y)(y) . Alsot_q+1(y) = t_q(y) +m c_t_q_(y)(y)

. Therefore

t_q+1(y)≤n−1 +m c_t_q_(y)(y)

≤n+M,

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whereM = max{m(y₁), m(y₂), . . . , m(y_r)}.

Hence

logN_y ≤(n+M)(a+α) so thatNy ≤e^(n+M^)(a+α). In particular, (3.1) now implies

Card.F ≤Card.Y_n·N_y

≤Card.Y_n·e^(n+M)(a+α) and this completes the proof of this lemma.

Some remarks are in order:

1. Apart from the choices of integersm(y), the rest of the proof of the above lemma is an exact copy of Bowen’s theorem [1, Theorem 17].

2. WhenF is a(n,4)-separating set ofX with maximum cardinality, then s_n(T,4, X)≤r_n(S, δ, Y)e^(n+M)(a+α)

where as befores_n(T,4, X) is the cardinality of suchF andr_n(S, δ, Y) is the cardinality ofYn.

The following theorem is the pressure analogue of Bowen’s Theorem [1, Theorem 17] (see also [3]).

Theorem 3.2. Letf_X andf_Y be continuous real-valued functions defined onX andY respectively such thatf_X ≥0andf_Y ≥0. Then

P(T, f_X)≤ ||f_X||+P(S, f_Y) + sup

y∈Y

P(T, f_X, π⁻¹y).

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Proof. Let >0,α >0, integern >0be arbitrary anda= sup_yP(T, f_X, π⁻¹y).

Also letsn(T,4, X)denote the cardinality of a(n,4)-separated set ofX with maximum cardinality. Then Lemmas2.3and3.1give us

Pn(T, fX,8, X)≤e^nkf^X^ksn(T,4, X)

≤e^nkf^X^kr_n(S, δ, Y)e^(n+M)(a+α)

≤e^nkf^X^kQ_n(S, f_Y, δ, Y)e^(n+M^)(a+α), where the last line follows from the positivity offY.

Hence

logPn(T, fX,8, X)≤nkfXk+ (n+M)(a+α) + logQn(S, fY, δ, Y).

In particular, on dividing bynand taking limit superior, we have as→0 P(T, f_X)≤ kf_Xk+P(S, f_Y) +a+α.

The result then follows sinceαis arbitrary.

Corollary 3.3. Let X andY 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 onX.Then

sup

y∈Y

P(T, f, π⁻¹y)≤P(T, f)≤ ||f||+ sup

y∈Y

P(T, f, π⁻¹y).

Proof. The first inequality follows from Prop. 2.2. Then in the above theorem takeS=Id,fY = 0andfX =f.

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By takingf = 0in the above corollary, we have

Corollary 3.4. Let X andY be compact metric spaces and T: X → X and π: X →Y be continuous such thatπ◦T =π. Then

h(T) = sup

y∈Y

h(T, π⁻¹y).

The last corollary is contained in [1, Corollary 18].

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4. (G, τ )−Extensions

As in the introduction let T: X → X be a (G, τ) - extension ofS: Y → Y. For the rest of this paper letfY andfGbe positive real-valued functions defined on Y and G respectively. Now define the function f: X → R as f(xg) = f_Y(xG) +f_G(g)so thatf is also positive and continuous.

Lemma 4.1. The functionf is well-defined.

Proof. Let xg = zg⁰ wherex, z ∈ X and g, g⁰ ∈ G. Consider the projection map π: X → Y. Then π(xg) = π(zg⁰)impliesxG = zGso that x = zg for some g ∈ G. Hence f(xg) = f(zgg) = f_Y(zG) +f_G(gg). Also f(zg⁰) = f_Y(zG) +f_G(g⁰). And this impliesf(xg) = f(zg⁰)if f_G(gg) = f_G(g⁰). But this is true since x= zg implieszgg = zg⁰ and in turn by virtue of free acting this implies gg = g⁰. In other wordsf_G(gg) = f_G(g⁰)and this completes the proof.

Lety∈Y. Then recall thatQ_n(T, f, , π⁻¹y)is defined as

Q_n(T, f, , π⁻¹y) =inf (

X

x∈F

e^Sⁿ^f^(x): F (n, )−spansπ⁻¹y )

.

We have

Proposition 4.2. Giveny∈Y, integern ≥1and >0, Q_n(T, f, , π⁻¹y)≤e^n||f^Y^||Q_n(τ, f_G, δ) for someδ >0.

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Proof. Letdandd⁰ be the metrics associated withXandGrespectively. Then since G acts continuously on X, we have by uniform continuity, there exists δ > 0 such that d(xg, xg⁰) ≤ whenever d⁰(g, g⁰) ≤ δ. Now let x ∈ π⁻¹y and E_n be a(n, δ)-spanning set forG. Then it is easy to check that xE_n is a (n, )-spanning set forπ⁻¹y. Observe that by commutativity ofT andS(viaπ) and the relationT(xg) =T(x)τ(g)we have

e^Sⁿ^f(xg)=e^Sⁿ^f^Y^(xG)e^Sⁿ^f^G^(g) with respect to the appropriate mapsT,Sandτ, so that

X

g∈En

e^Sⁿ^f(xg) =e^Sⁿ^f^Y^(xG) X

g∈En

e^Sⁿ^f^G^(g)

or

X

xg∈zEn

e^Sⁿ^f(xg)=e^Sⁿ^f^Y^(zG) X

g∈En

e^Sⁿ^f^G^(g). Therefore

Q_n(T, f, , π⁻¹y)≤e^Sⁿ^f^Y^(xG)Q_n(τ, f_G, δ).

Note that the above manipulation is independent of x ∈ π⁻¹y since if x⁰ ∈ π⁻¹y thenx⁰ = xg for some g ∈ Gso thatx⁰G = xG which in turn implies S_nf_Y(x⁰G) = S_nf_Y(xG). The result follows sincee^Sⁿ^f^Y^(xG) ≤e^nkf^Y^k.

Theorem 4.3. Giveny∈Y, we have

P(T, f, π⁻¹y)≤ kf_Yk+P(τ, f_G).

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Proof. From Proposition4.2we have

Q_n(T, f, , π⁻¹y)≤e^n||f^Y^||Q_n(τ, f_G, δ).

Therefore lim sup1

nlogQ_n(T, f, , π⁻¹y)≤ kf_Yk+ lim sup 1

nlogQ_n(τ, f_G, δ), that is

Q(T, f, , π⁻¹y)≤ ||f_Y||+Q(τ, f_G, δ).

The result follows by taking→0andδ →0.

Combining Theorem3.2and Theorem4.3and takingf =f_X , we have Proposition 4.4.

P(T, f)≤ kfk+kf_Yk+P(S, f_Y) +P(τ, f_G) We also have:

Proposition 4.5.

P(T, f)≥P(S, f_Y) +P(τ, f_G)

Proof. Let >0and letd⁰⁰be the metric onY. Then by the uniform continuity ofπand the fact thatG-acts freely onX there exists someδ >0such that a. d⁰⁰(π(x), π(z))≤whend(x, z)≤δ

b. d(xg, xg⁰)> δwhenx∈X andd(g, g⁰)> .

Now letG_n⊂Gbe(n, )-separated andY_n⊂Y be(n, )-separated and choose

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X_n ⊂Xso thatπ|_Xn: X_n →Y_nis a bijection. ThenX_nG_nis(n, δ)-separated.

Thus

P_n(T, f, δ)≥ X

xg∈XnGn

e^(Sⁿ^f^)(xg) =P_n(S, f_Y, )·P(τ, f_G, ).

In particular

P(T, f, δ)≥P(S, fY, ) +P(τ, fG, ).

The result follows by taking→ ∞andδ → ∞.

Corollary 4.6.

P(S, f_Y) +P(τ, f_G)≤P(T, f)≤ kfk+kf_Yk+P(S, f_Y) +P(τ, f_G).

And by takingf_Y ≡0≡f_G, we recover Bowen’s formula Corollary 4.7.

h(T) =h(S) +h(τ).

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5. Final Remarks

Whenf_Y andf_Gare both constant, by using the variational formula for pressure and Bowen’s formula, it is easy to deduce thatP(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. LetY ={−1,1}^Zandσ:Y →Y be the full two-shift. Consider the group extension given by

ˆ

σ: Y ×Z³ →Y ×Z³ (y, g)7→(σy,(g+ 2y0)mod3).

Of course, in this case τ = Id. Also, let f(y, g) = f_Y(y) + f_G(g), where f_Y(y) = 0 ify₀ = −1, f_Y(y) = 1 ify₀ = 1 andf_G = 2, constant. Then one can easily check that P(σ, f_Y) = log(1 + e) and P(Id, f_G) = 2. Moreover it is not difficult to see that P(ˆσ, f) = log(e² +e³). In particular, we have P(ˆσ, f) =P(σ, f_Y) +P(Id, f_G).

We end with the following conjecture:

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) = f_Y(xG) +f_G(g)where f_Y, f_G are positive real-valued functions on Y andGrespectively. Then

P(T, f) = P(S, f_Y) +P(τ, f_G).

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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, Springer- Verlag, Berlin, 1982.