a variational principle for the pressure of continuous transformations

A Variational Principle for the Pressure of Continuous TransformationsAuthor(s): Peter WaltersSource: American Journal of Mathematics, Vol. 97, No. 4 (Winter, 1975), pp. 937-971Published by: The Johns Hopkins University PressStable URL: http://www.jstor.org/stable/2373682 .

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A VARIATIONAL PRINCIPLE FOR THE PRESSURE OF CONTINUOUS TRANSFORMATIONS.

By PETER WALTERS.

Introduction. Ruelle ([17]) has defined the concept of pressure for a continuous Zn action on a compact metric space and proved a variational principle when the action is expansive and satisfies the specification condition. We prove this variational principle, for the case of a Z + action, without these assumptions (Theorem 4.1). The method of proof extends that used by Dina- burg ([5]) and Goodman ([7]) to prove special cases of the result. Several definitions of pressure are given in Sec. 1 and properties of pressure are proved in Sec. 2. Part of the variational principle is proved in Sec. 3 and the proof is completed in Sec. 4. Some remarks on other results appear in Sec. 5.

1. Definitions of Pressure. Throughout the paper X will denote a compact metric space with metric d, C(X,R) the Banach space of real-valued continuous functions on X equipped with the supremum norm, and T: X-> X a continuous map of X into itself. Ba (x) will denote the open ball in X with center x and radius 6. B will denote the closure of a subset B of X. If B CX, d (x, B) = inf{ d (x, y)I y E B }. Logarithms will be to the natural base e. Z + will denote the non-negative integers.

If O < n E Z + and e >O we say a finite subset E of X is (n,E) separated if whenever x, y E E, x7# y, there exists some i with 0 ,e. For qp E C (X, R) define

n-I Pn ( T, qp, e) = sup E exp E qp ( Tx) I E is (n,ce) separated}.

x E=E i=O

Since the exponential function is positive it suffices to take the supremum over maximal (n, E) separated sets (i.e. those that cannot be enlarged to an (n,,E) separated set). Put P(T,q9,E) =lim. sup.'n-,(1/n)logPn(Tq,,E). Then E1 K <2

implies P (T, q9, c1) > P (T, q9,c(2) so that P (T, q4) = lim., OP (T, q9,,e) exists or is xc. The map P (T,.): C (X, R )R->R U { x ) is called the pressure of T. This definition

Manuscript received April 4, 1973.

American Jourmal of Mathematics, Vol. 97, No. 4, pp. 937-971

Copyright C 1976 by Johns Hopkins University Press.

937

938 PETER WALTERS.

was given by Ruelle [17]. We shall see later (Theorem 2.1 (iii)) that if P(T, q)= o for some qp then P(T,q)) is always oc. If 0 denotes the constant function with value 0 then, by Bowen's work [3], P (T, 0) = h (T), the topological entropy of T. It is straightforward to show that P(T, q) does not depend on the metric d. (This will also follow from Theorem 1.2.)

We now give another definition of pressure using spanning sets, which corresponds to the other definition of h(T) given by Bowen [3]. A finite subset F of X is (n, e) spanning if for each x E X there exists y E F with d (Tx, Tiy) < e

for all i with OSiSn-1. Let

( n-I

Qn ( T5 q 5,E)x= inf expi0( Tx)IFis (n,E) spanning}. xEF i=OJ

It clearly suffices to take the infimum over minimal (n, E) spanning sets. If Q (T, q,E) = lim SUPn- n,(l/n)log Qn(T, 5q,e) then Q (T,qg,ce) increases as e de- creases and therefore Q(T, q)=lim.,,0Q(T, p,E) exists or is 00.

THEOREM 1.1.

Q (T5n) P (T5n)

Proof. A maximal (n, e) separated set is also (n, e) spanning. Hence Qn(Tq5,E) S Pn(T5q,c) and therefore Q(T,cp) < P(T,5T).

To prove the converse, let 8 > 0 be given and choose e > 0 so that d(x,y)<e/2 implies jqq(x)-q(qy)j<8. Let O<nEZ+ and X>O. Choose an (n, c) separated set E such that

n-I

Pn(Tqp,E)-X< E exp E 9g (Tkx) xeE i=O

and choose an (n, E/2) spanning set F such that

n-i

Eexp E I(p ( Tx) - < Qn ( T, q9p,E>/2). xEF i=O

Define a map a: E-F by choosing, for each x E E, a point a(x) E F with d (Tx, Tu(x)) < E/2 for all i with 0S i < n - 1. a is injective since E is (n,E)

PRESSURE OF CONTINUOUS TRANSFORMATIONS. 939

separated. We have

n-i n-I

exp E q)(Ty) E exp E q)(T y) yeF i=O yEa(E) i=O

n-i n-I

exp E qp (Tx) exp E qp (Tx) xeE i=O xeE i=O

n-I

min. exp q) [(p(T(x)) - q)(Tix) xeE i=O

> en.

Therefore

n-i

Qn T, qq,,E2)) > exp 1]q (p(Tiy) - yeF i=O

n-I

> - na E exp E qp(Tx)- X xEE i=O

> e - n [Pn ( T, X X.]

Hence Q (T, q,E/2) > e-n8P (T,qq,E) so that

Q ( T, q) > Q ( T, ,2) > - + P (T,q,,E).

We can let E-O and get Q(T,p) > -+P(T,9p). Since this holds for all 6 0 we have Q(T, p) > P(T, p). U

We shall now give two definitions of pressure using open covers. These generalise the definition of topological entropy using open covers [1]. If a, is an open cover of X the diameter of (f is defined by diam(dP ) =

SUpA ESUpxyeA d(x,y). T' d is the cover T-'AIAEde}. If 6 and $ are covers then (T V 93) is the cover (A n B IA E ?, B ( 98 }. d/on will be used to denote Vtn-T-idT. For an open cover ci of X put

n-I

Qn (T, p, d,) = inf| E inf exp E qg (Tx) Ia is a finite subcover of (Tn- Let Q(T,qp, dB)=inf n )xeA

Aea i0O

Let Q (T,q9, C ) =liMsp SP-,c (1 /n)log Qn(T, q, di)



940 PETER WALTERS.

THEOREM 1.2.

P(T, q)=supQ(T, q, d)j? is an open cover of X }

Proof. Let c& be an open cover of X and let 6 be a Lebesgue number for &i. Let F be (n,6/2) spanning. For z E F choose AO(z),...,An- 1(z))E d such

that B,5/2(Tkz) cAk (z) for 0 < k < n-1. Let

n-I

C ((Z) n TkAk(Z) E& (0 k=O

We have X= UZEFC(Z) and

n-i n-I

inf exp E qg(Tx)<exp E qg(Tz). xEC(z) i=O i=O

Therefore

Qn ( T, 'p, d-) < Qn (T T) (, 8 /2)

and so

Q(T,9p, d?) < Q(T, q,6/2) < P(T, q)

by Theorem 1.1. Hence sup Q (T,q, dp ) 0 and choose an open cover d? with diam( d?,) < 6. Let E > 0 be

chosen so that d(x,y)<6 implies Jkp(x)-q( y)I<E. If E is (n,6) separated then each element of n - 1 contains at most one member of E. We have

n-I n- - inf exp E qp (Tx) > exp 4q (Tiy)-nE if y E A E (1.

xeA [i o XA i = O i= O

Therefore if a is any finite subcover of (do1

n-I n-I

inf expE ( Tix)>e- n E exp E q(T'y) AEa i=O yEE i=O

and so

Therefore

sup Q (T,p, )>Q(T,cp, L) >-E + P(T, p,8). C

As 80 we can take E->0 sosupQ(T,q ,()>P(T,P(q). U

THEOREM 1.3. Q (T, qp, )- >P (T, qp) as diam(d,)--O. Therefore for each 8 > O

P ( T,q) = sup{ Q ( T, qp, d) Id is an open cover of X with diam( dB)<3 }.

Proof. Suppose P(T, q)<x. Let X>0. Choose 8>0 so that P(T,q, 8) > P (T, qp)-X/2 and so that d (x, y) < 8 implies I qp (x)- qp ( y)I <X /2. Then by the proof of Theorem 1.2, if diam(df) <6 we have Q(T, q, 6l) - -/22+ P(T,(p,8), and hence Q (T, qp, dT ) > - + P (T, qp). Therefore Q (T, (p, f)-- P (T, qp) as diam( f )->0. The case P (T, qp) = oc is proved similarly. U

If 6/ and (C are covers of X then 6/ < (C if every C E (C is a subset of some member of d(.

LEMMA 1.4. If 6d, (C are open covers of X then 6/ < (C implies

Qn(T, q, d? ) < Qn(T, q, C) and hence Q(T,cq, dl ) < Q(T,cq, (C).

Proof. Let y be a finite subcover of C . For each C Ey choose an A E don 'such that C C A and let a denote the collection of these sets. Then

n-i n-i

inf exp E (q ( Tx) < inf exp E (q (Tx) A xEA i=O ~~xeC i=O

Aea XA 0cey

so that

Qn (T, P, /) < Qn (T,p,).

We now give a fourth definition of pressure. If (T is an open cover of X let

P/ (T,g, c)n= inf sup exp E n p( ix)Ia is a finite subcover of (fo- AEaXeA i=O

LEMMA 1.5. P(T,9qP, a,)=limn Oo(l/n)logPn(T,q, d?) exists and equals

infn(1/n)1ogPn(T,q), (Bi).

Proof. If CElR we have Pn(T, p+C,dB)=encPn(Tq,dp,() so we may as well suppose qp > 0.

942 PETER WALTERS.

We shall use the result that if (a,,} is a sequence of positive real numbers with a, <a, + am for all n,m then lim(l/n)an exists and equals inf(l/n)an. We have to show

Pn + m ( T, qp, ('i) < Pn ( T, qp, (Bi) Pm ( T, qp, CT).

Let a be a finite subcover of on- 1 and y a finite subcover of Gfom-1. Then

n+m-1

sup exp E qg ( Tx) Be=aVT--y xEB i=O

< 0 E sup exp E n p(TTx) [ sup exp E (Tx)] [AEaXeA i=O J y xEC i=O

so that

Pn + m ( T, 95, di) S Pn ( T, 95, d?) Pm(T,d). T

THEOREM 1.6. P (T, qP) = lim6,O[sup.{P (T, 'P, di )Idiam( ( ) <6 }]. In fact P(T,p, d)-*P(T, p) as diam(d,)->O.

Proof. Let 8--0. Let d( be an open cover with diam(d?) <6. Let be a Lebesgue number for d-. Let E > 0 be such that d (x, y) < 8 implies I qp (x) - q ( y) <,E. Let F be (n,T/2) spanning. For z E F choose A, E (f such that B,12(Tiz)

cA, (0S jn n-1). If C(z)= nnI'T -AE A 'on then X= UZEFC(Z). Forz EF

n-1 n-1 sup exp E qg ( Tix) exp E q ( Tiz) + nE

x e C (z) i=O i=o

so after summing over z E F we see that

Pn (T T) s, (T-) < en - Qn ( T, (, IT/2) .

Therefore

P(T, p, C?) < Q (T,P,T/2) +cE

< P ( T, qp) +,E.

by Theorem 1.2. This gives sup(P(T,q, d,)diam(dB)<6) < P(T,9))+ E and hence lim,6O[sup(P(T,qp, dB)ldiam(d))<6)] < P(T,q). On the other hand, it is clear that Q (T, 9p, d ) 0.

We now show the second statement of the theorem. Let X >0. Choose 8 > 0 such that d (x, y) < 8 implies qg (x) - ( y) < ; and (using Theorem 1.3) so that diam(6d))<6 implies Q(T,p, ,ci)>P(T,qp)-X if P(T,p)< oo. Then, by the above, if diam(d, )< 8 then P(T, q, di) S P(T, q)+X and P(T, q, Ci) > Q (T, p, d) > P (T, T))-A. The case P (T, q)= oc is proved similarly. U

Remark. In some cases sup{P(T,q, CTi)j is an open cover of X} can be strictly greater than P (T, q,).

LEMMA 1.7. If T:X--X is sur/ective then Q(T,T'6i)Q(T,q,di) and P(T,p, T ' )=P(T,q, ( ).

Proof.

Qn ( T, 99, d?,)

Qn(T,p, Td)

n

inf exp E p(Tix) AEia j1

< sup n-1 a is a finite subcover of T 160

inf exp E i p ( Tx)

If a is any finite subcover of T - - 1 then

n n inf exp E q(Tix) inf exp 9 p(T1x)

ACa XA _x EA

S max n-1

infep q p( Tx) A Ea inf exp i= q (Tx) AEa i=O xEA i=O

< max sup [exp(q (Tnx)- qp (x))] AEa xEA

< exp(211p11)

Therefore Qn (T, p, 6) S e211IIQn (T, , T' - ) so that

Q (T, qp, (Bf) < Q(T, 9, T - 1 ).

Similarly Qn (T, 9p, T1 -d) < e211llQn (T, p, & ) and hence

QT(The po fr( ))i=sQ(T,s,mTil

The proof for P (T, qp, d) is similar. N

944 PETER WALTERS.

LEMMA 1.8. If T:X--X is surjective and k>O then Q(T,q), do) =Q(T,qg,d ) and P(T,p,,0k))=P(T,qq,d ). If T is a-homeomorphism and m > O, k > O then Q (T, q), dm)= Q(T, qp, ?i) and P (T,, m) = P (T,qp, di).

Proof. Since

[ ~~n-i Qn (T, q, dok)=inf. inf exp E qg (Tx) Ia is a subcover of /4+n-1

we have

Qn +k ( T, qp, d,) e 11 SQn ( T, qp, (B) Qn +k ( T, qp, d,) e

Hence

ak )(p i) = Q ( T, qp, aB

If T is a homeomorphism

Q (T,T(p, m)= Q (T,(p,6+m) by Lemma 1.7

=Q ( T, (p,i) by the above.

The results for P are proved similarly. U

A homeomorphism T: X->X is expansive if 36 >0 such that x# y implies d (Tx, T'y)> 6 for some i E Z. 8 is called an expansive constant for T. T is

expansive iff there is an open cover i so that n __ T is at most one

point whenever Ai E c [13]. Such a cover is called a generator. Anosov diffeomorphisms ([18]) are expansive and any subshift of the shift modelled on a

finite number of symbols is expansive. A continuous map T;X--X is positively

expansive if 3 8 > 0 such that x # y implies d (Tx, Ty) > 8 for some i > 0. T is

positively expansive iff there is an open cover 6, with n ' T - nA at most one

point whenever Ai E c6 [13]. c6 is then called a one-sided generator. Expanding maps ([18]) and one-sided subshifts of the 1-sided shift on a finite number of

symbols are positively expansive.

THEOREM 1.9. If T is an expansive homeomorphism then P (T, p) = P (T, p 68) = Q (T, p 68) if 28 is smaller than an expansive constant and P (T, p) = P ( T, p, 6() = Q (T, p, 6() if 6, is a generator. The corresponding results

are true for positively expansive maps.



Proof. If &T is a generator then diam( 61 - )--O as n-xoo [13]. By Theorem 1.3

P(T,cp)= lim Q(T,9, in 1)

=Q (T,T,(i) by Lemma 1.8.

By Theorem 1.6

P(T,cp)= lim P(T,T(p, cd1)

=P(T,q),pd) byLemma8.

Now let 281 > 0 be less than an expansive constant 28. Let C be the collection of all open 8-balls of X. Let F be an (n, 61) spanning set. For z E F choose

AO(z),...,A (z) C such that B0(T,k)= Ak(z) and put C (z)

= n- T kAk(z)e Crn- Then X= UzEFC(Z) and

n-1 n-1

inf exp , (p ( Tx) < exp E (p (Tz). C (z) i-O i=O

Therefore

Qn ( T, cp, ) , Qn ( T, (p, 81)

so that

Q (T, , (C) < Q (T, Ip, 1)

However, (C is a generator so that

P(T,p) > Q (T,(p,81) > Q (T,p, ()= P(T,cp)

and so P (T, (p) = Q (T, , 8).

Since P(T,p,8) > Q(T,cp,6) we get P(T,5)= P(T,cp,8). The corresponding results for positively expansive maps are proved similarly. U

2. Properties of P(T, p). Theorem 2.1 gives the dependence of P(T, T)

on (p and Theorem 2.2 describes how pressure behaves relative to the various

natural operations on dynamical systems. Some other properties are given in

Corollary 4.12.

THEOREM 2.1. Let T:X->X be continuous and (pEC(X,R). Then

(i) P(T,p) < P(T, )if (p ,

(ii) P (T,0) = h(T).

946 PETER WALTERS.

(iii) h(T)+infcp < P(T,cp) < h(T)+supT(. .P(T,p) = oo'?*h(T)= ox). (iv) V E>OP(T, *,E): C(X,IR)-I1R is convex (and hence so is P(T, ): C(X,R)

-4R unless it is infinite). (v) Vc>OjP(T,q,c)-P(T,4,c)jAjlp-4fl (if P(T,.) is finite then

jP(T,(p)-P(T,-+)j< I1p--II). (vi) P(T,cp+4,oT--2)=P(T,p))V4iEC(X,R) (the case 42= p gives

P(T,cp o T) = P(T,cp)). (vii) If c is a constant P (T, cp + c) = P (T, p) + c.

(viii) P(T, + 1) < P(T,p)+ P(T,4). (ix) P(T,ccp) < cP(T,) if c > I and P(T,cp) > cP(T,p) if c < 1. (x) IP (T, (p) O, a + =1. By Holder's inequality

n-1 n-1

?Iexpa E (p (Tx)IIexpI , 4'(Tx) xEL i=O J L i=O ]

n-1 n-1 i B

< E exp E (p (Tix) E exp E 42 (Tx) xEE i=O xEE i=O

so that Pn (T, a(p + f34e) < (Pn ( T, (p, r))a (Pn ( T, 4,, c)) . Therefore P ( T, a(p + fC:) < a?P (T, , r) + /3P (T, A,E).

(v) n-1

| exp E (p (Tx)

< SUp E SU XeP n- 1 E is (n, C) separated P~~(T,4,,c) ex p ya- 4(Tx)

xEE 1=0

exp E p(Tx)

?sup max. =

E is (n, c) separated

1=0

? e 'im -.1

Therefore P (T, ,) - P (T, ,,) < I(p - 'i and by symmetry

(vi) Since

Pn (T + (P + 4 ? T- )=suP eXp( 2 P( Tx) + 4 (Tnx) x EE i=o

-4(x)) IE is (n, c) separated}

we have

e 211A'lPn (T, p, E) S Pn (T, p + 4 o T- 41, c) S Pn (T, p, c) e2u'4"e .

Therefore

P (T, + 4o T- ,c) = P (T,q,c).

(vii), (viii), (ix) and (x) are clear from the first definition of P (T, T). In (ix) one uses the inequalities 2 jaic < 1 if c > 1 and 'En= jai = 1 ai > 0, and 'En= la' > 1 if c?1 and S1jai =1 ai>O. U

Remark. If c E R then P (T, cp) is not equal to cP (T, p) in general.

If T: X-> X is surjective then we can define a compact metric space X* and a homeomorphism T* of X* as follows. Let X* = {{xn )'ITxn+1 = xnn > 0), and T*(xO, xl, x2,2.) = (Tx0, xO, x1 x2, ...). We suppose X* is equipped with the metric

d* defined by d*({ Ynt ) fzn) = od (y nzn)/2 . (X*, T*) is called the inverse limit system of (X, T). T* is a homeomorphism and if r:X-X is defined by

7J({ xn}) = xo then rT* = Tv. If (Xl, d1), (X2, d2) are metric spaces we shall suppose X1 x X2 is equipped

with the metric defined by

d ((x, y), (u, v)) = max[ dj(x, u), d2( y, v)]

THEOREM 2.2. Let T: X-X be continuous and p EE C (X, R). Then

(i) P(Tm,?i-- lzpo TTi)=mP(T, p) if m>O.

(ii) P (T -1, p) = P (T, (p) if T is a homeomorphism. (iii) If Y is a closed subset of X and TY c Y then P (T,5p) > P (IT y, 1i y). (iv) P (T5 q)) = P (TIx,, qPIx.) where XO = n o0TnX. (v) If X1,X2,..., XP are closed subsets of X with TXi c Xi for each i and

X=U f1Xi then P(T,5p) =maxi. P(T1x,<qx). (vi) If : X-- Y is continuous and T, is a continuous map of the compact

metric space Y with 7TT= T, then PV(TC o 7T) > P(T, ,+) Vd C (Y,IR).

948 PETER WALTERS.

(vii) If T: X--X is sujective and (X*, T*) is the inverse limit of (X, T) then P(T*,cp*)=P(T,cp) where p*peC(X*,R) is defined by *(xO,xl,x2,...) = T(xo).

(viii) If Ti: Xi-Xi is a continuous map of the compact metric space Xi and cipE C(Xi,R) (i= 1,2) then P(Tlx T2,'p1x (P2)=P(T1,(p1)+ P(T2, P2) where P1 X P2 E C (X1 X X2, R) is defined by (T1 x p2)(X, y) = (Pl(x) +

'P2(Y)

(ix) If h: X-Y is a homeomorphism then P(T,p) = P(hTh ,p 0 h).

Proof.

(i) If F is (nm, c) spanning for T then F is (n, c) spanning for Tm. Hence

m-1 Qn Tm, fO (poTi) ? Qnm(T,(p5T)

i =O

and so

m-1 P Tm, m T poT) ?mP(T,cp).

i =o

Now let c>0. Choose 8 >0 so that d(x,y)<a implies d(Tix,Tiy)<c for all with 0 < j m -1. If F is (n, 8) spanning for Tm then F is (nm, c) spanning for T. Hence Qn ( 7Tm5 mi=-ol(p ? T', 8) > Qnm (T, (p, E) and so

m-1 m-1 \

P Tm, po ) T i > Q Tm, 9O ? Ti',) > mQ (T, (p5c) i =o i =o

(ii) If E is (n, c) separated for T then Tn'- E is (n, c) separated for T and conversely. Hence

n-1 n-1

exp E (p (Tx) exp E p(T-ix) xEE i=O yE Tn- E i=O

and

Pn (T,q,p) = Pn (T 1,(p,).

Therefore

P (T, (p) = P (T -1'P)

(iii) is clear from the first definition of pressure.

(iv) By (iii) it suffices to show P (TO, (po) > P (T, (p) where To =Tlx and

(PO=-(PlX" Let e >0 be given. Choose 8>0 so that diam(6C )<68=P(T,p,6i )

> P(T,p)- c, diam(iO) < 83=P(TO,0o, ) < P)(TO,o)+ e and so that d(x,y) < 8

=I(p(X) - CO( y) < .

Let C, be a finite open cover of X with diam(&6) < 6 and let d?0 be the cover of XO induced by (i.e. = {A n XolA E 6i }). For each member Ao of c0 cho-6se a member A of iT with AO= A n X0 and let i1 denote the collection of these sets A. Choose k so that P (TO, (po, 6io) > (1/ k) logPk ( TO, To' 6io) - c. Choose N so that for every subcover of VZ-'To- 6i0 TNX is contained in the union of the corresponding sets in V -o'T - i 1. Then

-logPk (TO (PO, 6i?0) ) -logPk (T,?o TN, T-N6C) - E

so that

P(To,(Po) > P(To,(Po, 6i 0) - >-klogPk (To,(po, 6iO) -2c

> klogPk(T, o TN, T-Nd)-3c

? P(T,TpoTN,T NI)-3c by Lemma 1.5

= P (T, p, 6i)-3c

> P(T,cp)-4c.

Therefore P (T, To(po) = P ( T, (p).

(v) Put Ti = T1x and (Pi = (pTx. By (iii) it suffices to show maxiP(Ti, p() > P(T,cp).

If Fi(n,c) spans Xi with respect to Ti then u 1F11(n,c) spans X with respect to T. Therefore Qn (T,cp,re) <EP= jQ( ( Pi,cp ,) o>o so that (l/n1)logQn (T,cp,c)-,Q(T,cp,) and so that

maxi Q, (Ti,(pi, ) is achieved for the same value of i (say io). Then

Q (T,(p,c) < Q (Ti,(pi.,c).

Now choose ck->O so that the corresponding io is the same, say i', for each Ck'

Then P (T,9) S, P ( Tj, 99).

(vi) Let c >0. Choose 8 >0 so that d (xl, x2) < 6 implies p(x1l, rx2) < c where p is the metric on Y. If F is (n,8) spanning for T then srF is (n,c)

950 PETER WALTERS.

spanning for LT and

n-I n-I

exp 2 AP (7Tx) > 2 exp ( Ti(L y) xEF i=O yE rF a=O

> Qn ( Tr, I, E)

Therefore

P(T,?4 r) > Q (T,I v 3I)> Q( Tv,4)

and so

P (T, ? r) > P Tlff)

(vii) By (vi) it suffices to show P(T*,cp*) < P(T,). Let E* be (n,e) separated for T,. Choose N so that X'=Ndiam(X)/2i < E/

2. If d*({y1},{,z})>E then X=Od(y1,z)/2'>e/2 so that d(y,z,)>?E/4 for some j with 0< j<N-1. If {y},{z,) EE* there exists 0<i?n-1 with

d*(T*j { y}, T*j {z;}) > E and so d (Tiyo, Tizo) > E/4 for some j with 0 j< n-I or d(yi,zi)>c/4 for some i with 0< i<N-l. Hence G={ YN-il{ y}E-E*} is (N+ n, E/4) separated for T. Therefore

N+n-l

PN+n (T,T,E/4)> , exp E p(Tix) xEG i=O

N-2 N+n-l

- Zexp E p (Tx) + , (Tx) x?G i=O N-1

N+n-1 > e-l J E exp Y (p(TTx)

x G N-1

n

-e - (N- I)T11 exp p * (T* ly) y E=E* j-O

and so PN+ n(T,p,E/4) > e-(N-l)JI Pn + I(T*,p*,E). Hence

P(T,T) > P(T*,T*).

(viii) If Fi (n, E) spans Xi with respect to Ti then F1>X F2(n, c) spans X> XX2

with respect to T1 X T2. Also

n-1

exp 1, (q) 1X T2) ((Tj XT2)'(x, y)) (x,y)EF1xF2 i=O

n-I ~~~n-I

exp E qI(T1ix))( 2exp E (p2(T2Y)) xEF1 i=O yEF2 i?O

so that Qn (TIX T2, T1 X T92,c) S Qn (T1, p1,c) Qn (T2,cp2,E). Therefore P(T1X T2,T1 X 92) < P(T,9I) + P(T2,92).

Now let 6 >0 and suppose di is an open cover of Xi with diam(6ij) < 8. Let E > 0 be such that di(x, y) < 8 implies IPiT(x)- Pi( y)I < E (i = 1,2) and let T, be a Lebesgue number for di. Let Si be a maximal (n,iJ/2) separated set for Ti. Then Si (n,j/2) spans with respect to Ti so by the proof of Theorem 1.6

Pn (Ti,cpi, G/j) < ene. Qn (Ti, qqjji/2).

Since

Qn ( Ti, qgi, Ti /2) < Pn ( Ti, qPi, Tj/2)

we have

Pn ( Ti, qgi, (i) < e nfPn ( Ti, qqjj t/2).

If T = min(TI, T2) then S1 x S2 is (n, /2) separated with respect to T1 x T2 so that

Pn (T1 X T2, T91 X 92, T/2) > Pn (T1, TP1, 1/2) Pn (T2, T2, 2/2)

> e - Mn ( TI, qP1, g/ 1) Pn ( T2, (P21 -2)

Therefore

P (T1 X T2, zQl XT92j/2) > lim sup [-2E+ IlogPn (TI,pqI, CT) n-oo n

+ Ilog Pn ( T2 92, Cd-2)]

=-2c+P(T1,cp1,CTI)+P(T2,592C2) byLemmal.5

Hence

P (T1 X T2,51 X 92) >-2E + P(T1,cp1, 61) + P(T2,cp2 52)

952 PETER WALTERS.

and using Theorem 1.6 and the fact that E->O as 8 >O, we get

P (T1 X T2,T91 X T92) > P (T1I,91) + P (T2,T92).

(ix) This follows from Theorem 1.2 since h maps open covers of X into open covers of Y. U

Remark. It is easy to see from the definitions that P (id, p) = supxex,(x). By Theorem 2.2 (i) if Tm = id (m >0) then

P (T, p) = msup [T(x)+99(x)+* +T(Tmlx)].

The idea of representing a continuous map T: X--X by subshifts of Hilbert cubes helps us to solve the variational problems in the later sections. The idea is due to Goodwyn [9]. We now describe these ideas.

Let 1q be the closed unit cube in RK, let Yq 11q)Z+ and letqSq:YqYq be the shift mapping i.e. Sq (xO, x1, x2,... ) = (x, x2, ...). Consider Iq equipped with the metric p(a,b)=maxi aia-biI where a=(al,...,aq), b=(bi,...,bq). The metric on Yq will be d ( xn}, t Yn }) = Y, ??= op (xn,y Y) /2n. A one-sided representation of (X, T) is a continuous map : X-Yq for some q such that 7TT=Sq7T. ST (X) is then a closed subset of Yq and let T, denote Sq 1,(x). T7:r T (X) -ST(X) is called a subshift of Sq: Yq -Yq. If T is surjective then so is T,

Let Wq = (Iq)Z and let Uq: Wq -> Wq be the shift defined by Uq (X *1 0x1x2. .)=(. .X_XOXx2. ) Let Wq have the metric d(xn}, ( Yn}) = OOP(Xn, Yn)/2I I. If T: X->X is a homeomorphism then a two-sided representation of (X,T) is a continuous map v :X--Wq for some q such that 7TT= Uq 7 and T, = Uq 71(x) is a homeomorphism. If Y is a closed subset of Wq and Uq Y= Y then (Y, Uq I Y) is called an invertible subshift of (Wq, Uq). For p9eC(X,R) let RI(T,9p) denote the set of all one-sided representations ST of

(X,T) for which T9=9P9,o for some p(PeC(7T(X), R1). Similarly if T is a homeomorphism let R2(T,9p) be the collection of all two-sided representations of (X,T) for which Tp= p7r0 T for some Tp E C(7T(X), 1R).

THEOREM 2.3. If T:X--X is continuous and p& C(X,R) P(T,T9)

=sup,gERi(T,cp) P(T,,q,,) where 9=9p9,ov. If T is a homeomorphism P(T,cp)

=SupIT E R2(T,qm) P (Tp X ).

Proof. Suppose T is a homeomorphism. Assume firstly that P (T, 9p) < xo. By Theorem 2.2(vi) it suffices to show P (T, 9p) < supN eR2(T, ) P (T, 9Xn). Let 8 > O. Choose e > O so that

P(T,9) < P(T,Tp,c)+ 6.

Then P(T,cp) < (l/n) log P (T,p,E)+28 for infinitely many n. For each of these infinitely many n there is a (n,E) separated set E with P(T,T) < (1/n) log (2TxE

exp 1-' qp(Tx))+38. Choose b>O so that b+q)(x)>OVxeX and let (l/c) =maxx[Tg(x)+b]>O. Cover X by E/4 balls Bl,...,Bk and let fi: X-[O, 1] be defined by fi (x) = d (x,Bi)/diam(X) 1 i < k. Define v: X- Wk+l by

c [9(Tnx) + b 3

7T(X)n= fi( rfTx) neZ.

fk(Tnx)

We have vrT= Uk+l1r. Let 4: q(X)-1R be defined by yn( Yn}) = ( yo0/c) - b where

Yoi

Yo= 0 402 elk+l.

Yok +Y1

Then = o vT and so v E R2(T, 9).

If E is (n,c) separated for T and x, yEE3j with OS j<n-1 so that d(Tix,Tiy)>E. TixEBi2 for some i, so fiI (x)=O but fiL(T'y)=d(T'y,Bj,)/ diam(X) > E/2 diam (X). Hence <(E) is (n, E/2 diam (X)) separated for

TX = Uk+lI17T(X). 2xeE exp 2' -O1 9p(Tx) =2 yeE exp 21} (T,y) so that

n-1

P(T,59) S-log exp E A(T,y) +38 for infinitely many n n

yeGE i=O

< 1 logPn (T7r A,' /2 diam(X )) + 38. n

Therefore

P(T,T) < P(TT,45,c/2 diam(X)) +38

< P(T,,2) +38.

Hence

P(T,9)) O we could choose E > O so that N < P ( T, 9, E)

954 PETER WALTERS.

and proceed as above to construct 7 E R2(T, p) with N < P (T, q,,). Hence

P (T, p) = sup P (T, T99P) if T is a homeomorphism. so E R2(T T,)

If T is not a homeomorphism the proof is similar to the above with S: X-* Yk+1

defined by

c[cp(Tnx) + b]

f1( T'x) , nZ +. 7V(X)n= Lni X E + E

fk (Tnx)

If T: X-- X is continuous M (T) will denote the collection of all T-invariant Borel probability measures on X. M (T) is a convex set which is compact in the weak topology, and its extreme points are exactly those ergodic measures for T. E (T) will denote the ergodic members of M (T). If pi E M (T), h, (T) will denote the entropy of T calculated with the measure tt. [16]. If (p E C (X, R ) }(9p) will denote the integral fJx pdp. A cover dC of X is tt -disjoint if tt( U (U n VI U, V E 6, U# V}) = 0. Our variational principle will relate P(T,Tp) and h,,(T) + v}). The following will be useful.

THEOREM 2.4. If T: X-X is continuous, m E C (X,R) and (E M (T) then h, (T) + (9) = supN eR,(T, ) [hA. , - l(T1) + a o l@ - 1(99,)] where =, o qT = 9. If T is a homeomorphism we can take the supremum over R2(T, (p).

Proof. ,o-l(p,)= ,u(p) so the result reduces to hi,(T) =SUPgER1(Tq9)

ht, ,-1(T,). As in Theorem 3 of [1] we can construct a sequence (Ym) of pt-disjoint closed covers of X with h11(T)=limm,O h11(T,ym). Fix m and let

Ym=(Fl5 .Fk}. Define fi:X->[0,1] by Lj(x)=d(x,Fi)/diam(X) (1<i<k). Choose b so that 9p(x) + b > OV x e X and let (I/c) = maxx [Tp(x) + b]. Define

c [ p ((Tx) + b 3

X--- XYk+ 1by 7(X)n= f(Tx) ,ne Z +.

fk (Tnx)

(If T is a homeomorphism one would define 7: X-- Wk+l in a similar way in order to prove the second part of the theorem). ST E RI( T, p). Let Vi = t t yn } E

Yk+i JYo,i=0} i=2, .,k+ 1 where

Yoi1

Yo

Yok +1

Let -ym denote the cover of v (X) defined by the closed sets (X) n Vi (2 < i < k + 1). Then ST - lYr = Ymn and ym is t o?T' -1 disjoint. Also

h,, (TX Ym) = h,,. I1( Tv , Ym) < h, T,

and so

h (T)< sup ho I1(T)T X

7rER1(T,)

3. An Inequality. In this section we shall prove half of Theorem 4.1; namely,

THEOREM 3.1. If T:X--X is continuous and pE&C(X,R) then SUP,EM(T) [h (T) + A}(p)] < P(T,cp).

This generalizes the result of Goodwyn [9] which corresponds to the case = 0. The proof will depend on the following lemmas. For any cover 6C. of X

(not necessarily an open cover) define

inf ~ sup expn-1 Pn ( T, , i)5 = inftE sup exp E (p( Tx) Ia is a finite subcover of 6Pn tAEa xEA i=O

and

P(T,cp, CT) = lim sup IlogPn (T,q, i9) n- oo n

= lim I

logPn ( T,9, i )

as Lemma 1.5 is true for any cover 6C. This coincides with our previous definitions when Cd is an open cover.

LEMMA 3.2. (c.f. Proposition 2 of [9]). For a closed finite cover e P ( T, 5, 6 ) < P (T,5 p) + logp (d ) where p (d) is

the largest number of members of e having non-empty intersection.

956 PETER WALTERS.

Proof. Choose 8 > 0 so that any closed 8-ball is met by at most p( 6(C) sets of d . If B;(w,n)= (xId(Tx,Tw) < 8 0? i S n-I) then at most p( d)' sets of dOn meet B6 (w, n). Let E = SUPd (x y) <?68 | (x )-Tp ( y) I. We know U w E F B6 (w, n) = X for a (n, a) spanning set F. Let A E dr'1 SUPXEA exp : i2-O =cp( Tx) S exp

(Ii= p (Ttw) + nE) for some w E F chosen so that xA E B,6 (w, n) where XA is a point of A where the supremum is attained. Since each B6 (w, n) can meet at most p(c )n sets of cd - 1 we have

E supexp I (Tx)< p()fln exp I( (Tw)+nE). A En' xeA i=O wEF i=O

Hence

Pn ( T,, (g) O and

P (T, g, (g) < logp (6P) + P (T,p)

LEMMA 3.3.. If 0 < k E Z + is fixed and ql,.. ., qk are given then the maximum of pi( (-log pi + qi) subject to pi?, > pl=i is log ik_ e1q

and is attained when pi = e i/ 1 eqi.

Proof. This is an easy result that can be proved using the Lagrange multiplier method. U

LEMMA 3.4. If 6B is a tt-disjoint closed cover of X, ,u E M (T), then

h,,(T, CT) + A (99) 0 and let a be any finite subcover of dTonj

n-1 ~~~~n-1

nj (z A( 9 z T ) < i t(A)sup E 9p(Tx). i=O Aea xeA i=O

Hence

1y@ 1)yg=nH()yg I H cTn-I H , (A ) [-o+ A)+spEg(Tx

n 0n [ ~~~n-i1

< - 1i A) og i () SUP E cp (Tx) nE x EA i=0 A E a ~ ~ ~ A =

n-1

< -log E exp sup E 9p (Tx) by Lemma 3.3. A ea xEA i=O

Therefore

I HI, ( cTn)

-I (g n

n-1i- < inf| -log sup exp , p(Tix)la is a finite subcover of c |

An a xEA i=O

so that h. (T, 4S ) +t(9p) < P (T, 9p,). U

LEMMA 3.5. [9]. Let v be a finite Borel measure on Iq. There is a sequence C1 < C62< 6,3< ... of closed v-disjoint covers of Iq with diam( (n )-*O and P((2n) < q + 1 Vn.

This result is proved in [9] p. 683. The following is a well known result about entropy.

LEMMA 3.6. ([2] p. 87). Let T:X- X be continuous and 1i,EM(T). If C < 2 < dT3 < ... is a sequence of tt disjoint closed covers and if the a- algebra generated by (T -T Ck I k > 1 i > 0) is the a-algebra of all Borel sets then h, ( T) = liMk oo hi (T, CTk)

LEMMA 3.7. It suffices to prove Theorem 3.1 for subshifts of the shift on (Iq)Z+ for all q > 0.

Proof. Suppose Theorem 3.1 is true for subshifts of the shift on (Iq)Z+ for all q >0. Suppose h,, (T) < oc. Let E >0. By Theorem 2.4 choose 7 E R1(T, p) with

hA,(T) + ,u(q) < h?) + tL o 77T- '(99) + E.

Then

h,,(T)+p(cp) < P(Tq,qp) + E

< P ( T, p) + E by Theorem 2.2 (vi).

958 PETER WALTERS.

Therefore h,, (T) + ,u(9) < P (T, ). The proof is similar if h,, (T) = x U

Proof of Theorem 3.1. By Lemma 3.7 it suffices to consider (X, T) as a subshift of the shift on (Iq)Z+. We only need consider the case P (T, (p) < xc. Let y C M(T). Define i on Iq by (A) = q(r -'(A) n X) where 7ro: (Iq)Z+_Iq is the projection onto the 0-th coordinate. Let 61 < 82 < 63< ... be a sequence of v-disjoint closed covers of 1q determined by Lemma 3.5. Let (6k = 1 (Ck) n X. 6'k is a closed n-disjoint cover of X and el< 6d'2< ?35 < . Since diam (C - )->0 the hypothesis of Lemma 3.6 is satisfied and so

htt ( T ) = limm h,, ( T, 60,n).

Hence

ht,,(T) + ,u (9) = lim [ hl,,(T, e9n) + Itp)

< lim P ( T, q, dn) by Lemma 3.4 n-i-oo

< P(T,p) +log(q+ 1) by Lemma 3.2.

Now let m > 0. Consider (X, TM). Th.e natural homeomorphism h: (Iq)Z+

_(Inqm)Z+ given by

Xm- X2m - 1

conjugates (X, TM) with a subshift (hX,hTmhl) of the shift on (Iqm)Z+ We have

m-1

mh,(T)+mA((p)=h,(TM )+P E (po Ti i =o

= h lh1(hTmh-) + o h-( E (po h(hTil) i =O

m-1

< P hTmh 1, E 99o?h -l(hTih-1) + log(mq +1) i =o

by the above m-1

=P T m 1 (p o T) + log(mq + 1) by Theorem 2.2 (ix) i =o

=mP ( T, 9p) + log(mq + 1) by Theorem 2.2 (i)

Therefore

h, ( T) + y q).

4. A Variational Principle. In this section we wish to prove the following result which generalizes the work of Dinaburg [5], Goodman [7] and [7'] (some oversights in [7] are corrected in [7']) and Ruelle [17]. Ruelle proved the result when T is an expansive homeomorphism satisfying the specification property.

THEOREM 4.1. If T: X->X is continuous and qp E C(X,R) then

P(T,T)= sup. [h, (T)+ aEM(T)

We shall need some lemmas before giving the proof of this theorem. The first lemma can be found in [10] and is an application of the Hahn-Banach theorem.

LEMMA 4.2. Let T: X-*X, S: Y-* Y be continuous maps of compact metric spaces and let T X-> Y be a sur/ective continuous map with VT= S7r. For each v E M (S) there exists ,u E M (T) with o 7T-1P= P.

LEMMA 4.3. It suffices to prove Theorem 4.1 for homeomorphisms.

Proof. Suppose Theorem 4.1 is true for homeomorphism. Let T: X->X be surjective, and let (X*, T*) be the inverse limit of (X, T). Since T* is a homeomorphism we have

P ( T., qg) = sup hv (T*) + P ((p*) P E M (T*)

where Tp*(Xo,xl, )=q(p(xo). By Theorem 2.2 (vii) P(T,) = P(T*,T*). Also every P E- M (T*) is the inverse limit of some ,u E M (T) and h,, (T) = h?,(T*) ([16] Sec. 9.9) and P(Tp*)= ((p). Therefore P(T,(p)=SUP,EM(T) [h,(T)+A(9))] and Theorem 4.1 is true for surjective maps. If T: X->X is any continuous map let X O= O TnX and To= TIXO. TO:Xo->Xo is surjective and so P(To(po) = SUPveM(Tl) [h,(T0)+ (@p0)] where (po=(pIx0 By Theorem 2.2 (iv) P(T,9) =P(To,(po). Also each tEM(T) satisfies [t(X-X0)=O and so can be con- sidered a member of M (To). Therefore P (T, Tp) = SUp, EM(T) [h, (T) + ,A(T)]. U

960 PETER WALTERS.

LEMMA 4.4. It suffices to prove Theorem 4.1 for all invertible subshifts of the shift on (Iq)Z for all q > 0.

Proof. Suppose Theorem 4.1 is true for invertible subshifts of the shift on (Iq)Z for all q >0. By Lemma 4.3 we only need to show Theorem 4.1 for homeomorphisms. Let T: X->X be a homeomorphism and P (T, q) < oc. Let E > 0. By Theorem 2.3 choose STER2(T, (p) with P (TX, qP) > P (T, (p)-E. Choose

EM(T,) with h?, (T,, ) + p ((p,,) > P ( T, p,))- E

> P(T,cp) -2E.

By Lemma 4.2 choose tiE M (T) with ,u o7r P1= Then (<p) = P((p) and so

h,,(T) + ,()> h^,(T,r ) + P (9') > P(T,cp)-22.

Therefore SUP, eM(T) [h,(T)+ji(q)] >P(T,cp) and the converse is true by Theorem 3.1.

The case where P ( T, (p) = ox is proved similarly by choosing for each N > 0 7rER2(T, (p) with P (T,, ,,) > N. N

The next lemma is proved in the proof of the theorem in [8].

LEMMA 4.5. If T: X->X is an expansive homeomorphism the map M (T) ->R given by u->h,,(T is upper semi continuous.

The two sided shift on s symbols is the homeomorphism of Z (s) = (1,2,.. .,s} defined by T(. x*1 x1 .)=(. xlxoxlx2 ). If Y is a closed subset of Z (s) with TY= Y then (Y, Tl y) is called an invertible subshift of (Z (s), T). We have already mentioned that an invertible subshift is expansive. If (YT1y) is an invertible subshift then .4EC(Y,R) is called a cylinder function if there exists N>O and a function X:(1,2,.. .,s}2N+l 1*R so that

#({xln})=X(x_N,...,xN). An invertible subshift (YT1y) of (Z(s),T) is of finite type (or an intrinsic Markov chain [14]) if there exists N>O such that no { xf } H X \ Y has the property that for each n c Z there exists y c Y with

(Xn .. Xnf+N) = ( Yn . . , Yn+ N). The least such N is called the order of (Y, TI y) If

(Y1Ty) is a subshift of (Z(s),T) of finite type and order 1 the associated structure matrix X=(aij) is the sXs matrix defined by aij=1 if there exists y= { y(}EY with yo=i y =i, and %i =0 if no such y E Y exists.

We shall need the Perron-Frobenius theory of non-negative matrices ([6] pp. 50) which we now describe. Let A be a p x p matrix with non-negative entries (i.e. A > 0) which is irreducible (i.e. for every pair (i, j) there exists q > 0

for which A q has strictly positive (i, j)-th element.) Then A has a simple eigenvalue X such that the right and left eigenvalues corresponding to it have all their entries positive and also X) Izi for all other eigenvalues z of A. We shall use the following result, recalling that the spectral radius of a square matrix A is the maximum absolute value of the eigenvalues of A.

LEMMA 4.6. If A > 0 is a square matrix and X is strictly greater than the spectral radius of A then (X -A)- .> O.

Proof. Consider A as a linear operator on R P where A is p x p. The requirement A > 0 is equivalent to the property R P D u >0 implies Au > 0. (1/X)A has spectral radius strictly less than 1 and so X- E'X0(1/Xn)An converges and has sum (X-A)-l. Hence RPDu>O implies (X-A)-O. Therefore (X-A) -O. U

We shall let 1 denote the norm on Rn given by Juj=X%jjuj if u = (ul,... , un). Some of the ideas for the proof of the following lemma are from [14]. The lemma appears in physics literature.

LEMMA 4.7. Let (X, T) be an invertible subshift of the shift on Z (s) = { ,...,s}Z with finite type of order 1, let 2 be its structure matrix and let

a,, ... ., as C R. If p EC (X, R) is defined by (p ({ x})= ax and

ea, e

A=2 eA a2

O e qs

then P (T, p)=logX where X is the largest eigenvalue of A. Also there exists ,u E M (T) with P (T, ) = h,, (T) + (cp). (The proof will show there exists a real eigenvalue X such that Izi < X for all other eigenvalues z of A.)

Proof. We can suppose each state { 1,.. ,s } occurs in X by reducing s if necessary. For n > 0 let

n-1 (Di =E ex E p(Tix) = Eexp(ax.+ axl+ -+ an_l+ ai)

i=O

where the summation is over all x= {x)} E X with xn = i (1 S i < s). We have

on 1 e n(l n a(2i + '''+ n(Dsn ]s

so if (D = (?D1 D25 . on) E R s then D|1 = ID A. Therefore (D = IDOAn. If D

= {G1, . . ., Gs} denotes the natural generator for X (Gi= {{xn} E X I xo = i}) then

962 PETER WALTERS.

by Theorem 1.9

P(T, p) P(T,cp,)

1 n-1 = lim - log E exp E (p(Tix) by Lemma 1.5

n--on ACDn-i i=O i0

1 SE D = lim l logI4nJ where Ij= E on n

n--3oonf =

We shall now calculate P (T, p). Define an equivalence relation on { 1,. . ., s ) by i- j if there exist

{x,},{yn}EX and k, m )O with xo=i, xk=j and yO=', Ym=i. Equivalently i -j iff there exist k,m> 0 such that Ek has non-zero (i, j)-th entry and Em has non-zero (, ji)-th entry. Let S, .. ., St denote the equivalence classes, numbered so that if 1 O and all i E S E Sk. Then E has the form

Fl, ... 0

F21 F22

Fti . Ftt

and hence A has a similar form

E- ... 0

E22 E22 A=

Et Et

By the definition of the equivalence relation for each k, 1 < k < t, Ekk iS

irreducible and so each Ekk has a simple eigenvalue Xk with Xk > IZl for all other eigenvalues z of Ekk and the corresponding left and right eigenvectors for Ekk

have all entries positive. Let X be the largest of the Xk's.

Suppose ko is the smallest so that Ek.k. has X as an eigenvalue. We seek a left eigenvector u E R 1 for X. Decompose R s = R Sl + . + R S* into a direct sum

of t subspaces corresponding to the decomposition of A (Ei is si X si) and then we seek u = (u, . .. , ut) such that (u1, ... , ut)A = X(ul, ... , ut). Put Uk0+ 1 = = Ut =0 (if ko < t). Then uk Ekk. = Xuk so we can choose uk R sko to have all

coordinates strictly positive (i.e. uk0>O). Lemma 4.6 now allows us to choose ur ER ,1 < r < k0, with non-negative coordinates. Therefore u >, Uk0>O and Uk0+l = = Ut=0. We now find a right eigenvector v corresponding to X. Decompose Rs=Rs"+ +R* as above and we seek v=(vl, ,vt) with Av =Xv. Put v= =VkO1 0 (if ko > 1). Then EkokoVko =

XVk0 so we can choose

Vkk ERSko with vk >0. Lemma 4.6 then allows us to solve for vr ERs(ko< r < t) with vr>O. Hence v>, Vko >0 and vl,..., Vk 0=O. The inner product of u

and v in Rs, u.v, equals b1c,+.. +bsko c>kO0> where u=(A . bSko ERsko V = (C1, ... . CSk )ER sko Normalize u and v so that u v = 1.

Choose 08 >0 so that (ea',...,e,)>(u/8). Then 4>n=(ea1,...,e a)An

> uAn/6 and therefore (Dn > nu / Hence (1/n) log I(nl > (1/n) log (XnIU/ 8 )->log X. However ODn = I?oAn so that I'DnI < 10OH1A n where IAn Iis the norm of the linear map An. Therefore IDn 1/n = - %11/nIAn11/n-A since X is the spectral radius of A. Hence (1/n) log nI-->log X and so P(T,(p) =log X.

Let Y denote the closed subset of X defined by { yn) }E Y iff yn E SkoVn E

Z. TY= Y. Let y denote the Markov measure on Y defined by the transition matrix pi2 = cjeij/Xci 1 < i, j < sk0, where v = (cl, ..., csk ) and ei2 is the (i, j)th entry

of Ek 0k, and the invariant initial distribution p =( p . P50) where pi = bi . ci. We can consider y as a measure on X. Then t E M?(T) and h, (T) - l Pi p0 log pi0 [2]. An easy calculation shows h, (T) = log X - (q) and hence

P(T,Tp)=h,(T)+ M(z). U

COROLLARY 4.8. Let (X, T) be an invertible subshift of the shift on

Z(s)={1,2,...,s)z of finite type and let cpEC(X,R) be a cylinder function. Then there exists ,u E M (T) with P ( T, p) = h, (T) + i (p).

Proof. Let (X, T) be of order N' and let (p depend only on the coordinates between - M and M. We can choose N so that 2N+ 1 ) max(N', 2M+ 1). The homeomorphism h:{ 1, 2, ... .,s} Z ->{1, 2, ..., 5S2N+ 1Z defined by {h(x))} =f((XlN, ...5X. +N)) where f:{1, 2,...,s)2N+1- >{1 2,... 52N+1) is a fixed bijection, conjugates (X, T) with a subshift of { 1, 2,... s2N+ 1Z of finite type with order 1 and 9p corresponds to a cylinder function which depends only on the 0-th coordinate in {1, 2,...,S 2N+1)Z. The result then follows by Lemma 4.7 and Theorem 2.2 (ix). M

LEMMA 4.9. If (Y,S) is an invertible subshift of the shift on Z(s)= { 1, 2, ... ,s)Zand # E C (Y, R) there exists ,u E M (S) with P (S,.4) = h,, (S) + ,L(#).

Proof. We first show it suffices to prove the lemma for cylinder functions. Suppose the result is true when 41 is a cylinder function. Let 4 E C (Y, R) and

964 PETER WALTERS.

for each n >0 choose a cylinder function 4'n with I1- nII <1/n. There is

tUn E M (S) with P (S,ljn) = hv, (S) + in (4J'). Choose a convergent subsequence t-n E M (S). By Theorem 2.1 (v)

P (S,4)= uM P (S,4E)

= lim [hr , (S ) + nn (-ln)

n-- [ no < lim [hv}( S ) + An (WJ +n

h,, (S )+ 1) by Lemma 4.5

So by Theorem 3.1 P(S, 2)=h (S)+ (v). Suppose, therefore, that 4 is a cylinder function depending only on the

coordinates between times - k and k. We shall employ a method used by Parry [15] and Goodwyn [10]. For m > 0 define

Ym{= xE{ ,S...s}ZIVfnlEZ3yEYwithxn+i=yn+j OS<iSm}

Ym is closed, n = 1 Ym = Y and TYm = Ym where T is the shift on Z (s). For m > 2k+ 1 we can define 42m E C(Ym,R) by 'm(x) =4 ( y) for any y E Y with Xi= yo,-k < i < k. If Sm :Ym -Ym denotes the shift then Sm is of finite type and so by Corollary 4.8 there exists Itm E M (Sm) with P (Sm, 'm) = h4 (Sm) + Am ('Pm)

Considering each Jim as a member of M (T) choose a convergent subsequence I1mj I1- E M (T). Since characteristic functions of cylinder sets are continuous we have imj(C) -i (C) for every cylinder set C in Z (s). If C is a cylinder set and C n Y= 0 then ,m, (C) = 0 so ,u(C) = 0. Therefore ji is concentrated on Y and we can consider ji as a member of M (S). We have

P ( Sns 5 '4m) = h,4,j ( Sn, ) + iltm(/m )

= ht,(T) + iis ('mPr) for a fixed io and i > io

so that

lim sup P (Sm, 5'mj) < h4 ( T) + P (mj by Lemma 4.5

h= h(S ) + A (1).

Using Theorem 3.1 it suffices to show P (S7,,',)--P(S,') Let ={G1,.. Gs} be the natural generator for T. i.e. Gi={{xn}Ixo=i}. Then tnYm={Gln Ym5 ... , G n Ym} is a generator for Sm and D n Y is a generator for S. Let E > 0. Choose N >0 such that (1/N) log PN(S545 n Y) < P(S;P) +cE. Clearly if m

> N+2k+ 1 then PN(Sm,4im, fl n Ym) = PN(S,41, 4n Y). Therefore by Lemma 1.5 and Theorem 1.9

P ( Sm,4m) < N logPN(Sm, 4m, t n Ymi)

-= logPN(S,41,n Y) if m>N+2k+1

<P(S,41)+E.

Therefore P (Sm, m) - P (S, 41) and the proof is complete. U

LEMMA 4.10. If T:X--X is a homeomorphism, n>O and aEM(T') then m= (1/n) Ei`d a (T -) E M(T) and hm(T') > ha(Tn).

Proof. Clearly m E M (T). It follows from the convexity of the function x log x that the map -- ,h, (S) from M (S) to R is concave for any continuous S. Hence hm(Tn) > (1/n) Yi.n- ha.T_.(Tn) and since h (Tn)= hoT-t(Tn) we have hm(Tn) we have hm(Tn)> h,(Tn). U

LEMMA 4.11. [7']. If 6T = {A1,... ,An} is an open cover of X there is a Borel partition q = { B1, . ... B} of X with BiCAi(1 < i < n).

Proof. We use induction in n. The result is obvious for n= 1. Suppose the result is true for any metric space and any open cover by n - 1 sets. Let

(-{A1, ... An} be an open cover of X. Put C=X-_ Un-1 Ai and choose an open set D with C cD cD cAn. Let Y= X-D. {A1ln Y, ... ,Anln Y} is an open cover of Y so by the inductive assumption there is a Borel partition {B1...,Bn-1} of Y with BicAin Y (1 < i < n-1). Then the Borel partition { B1 ... . Bn_ 1, D } satisfies the requirements for n. U

If C is an open cover of X and E C (X,R) then var p=supcEe supx yecl(x) - 9( Y) I

Proof of Theorem 4.1. By Lemma 4.4 it suffices to consider (X, T) to be an invertible subshift of the shift on (1q)Z. Let , = {E1, . . . EJ be finite open cover of X and let d={ C1, . . ., C} be a closed cover of X with Ci c Ei < i < s). Let Z (s)= {1,2, ...,s}z and put Y= {{cn} E Z(s)l n n? T-CnC }. Y is a closed shift-invariant subset of Z (s) and let S: Y-- Y be the restriction of the shift to Y. Put Q=((x5y)XxYxEn' ?_ooT-nCy. if y={ Yn }} 2 is a closed subset of X x Y and we can define U: 2-42 by U(x, y) = (Tx, Sy). The natural projections P1: 2 y-X P2: 2 Y satisfy P1 U = Tp1, P2 U = SP2.

Suppose 41 E C (Y, R) satisfies 4p2 > pp1. By Lemma 4.9 there exists t E M (S) with P (S,4) = h,, (s) + i(141). By Lemma 4.2 choose v E M (U) with

966 PETER WALTERS.

vop27=I[. Put a=vop 0 EM(T). Let D-{GI,...,Gj} be the natural generator for S. (Gi= {{y} E YI yo =i}). ' is an open cover of Y and partitions Y. We have

Q(T,qp,dC)? Q(S,4,) since 4p2> p1andp - (Gi)cp7 '(Ci)

haT,7)-r(u71) >hr(,t)-H ('/7')([16] p25)

- h, (S,)-Hv( 7

= h,(S) - H1(D/71)

since D is a measure-theoretic generator for S with respect to every [Le M (S). Hence h.(T)>Q(T, p,(e)- -(4)-HV( '//-)H By Theorem 1.3 if P(T,i)<x there exists 8 > 0 such that if C is an open cover of X with diam ( C) < 8 then

var<e (q) < 2 and Q (T, qp, C ) > P (T, qp)-a . Since I q has dimension q we can choose an open cover 6D {Dl,... ,D8} of it with diam (6D ) <K8/3 and so that each point of Iq lies in at most q+1 members of 6D. Put 6i ={El,...,Es} where Ei= {(xeXjx0ED where x= {xj}}. Choose a closed cover C

-{ Cl, * * C.} of X with Ci c Ei. Each x E X lies in at most q + 1 members of CT so we can find an open neighbourhood Ox of x which intersects at most q + 1 members of RB. Take a finite subcover {F1,.. .,F,} of {OxIxeX} and let

= {,... ,3Bt) be a Borel partition of X with Bi C Fi(1 < i < t) (Lemma 4.11). If

P1 = t'= { Gi,..., G.} and p7 -q = 71'= {B', ... .,B'}, then (x, y) e Gi' n B,' implies x e Ci n B, so each element B,' of 7q' intersects at most q + 1 elements of D'.

Hence H( /'/7')< log (q + 1). Choose N >0 so that diam (NN) < 8. Then Q(T,q, 5(N) > P(T,5)-

and if

4'(y)= sup p(x) N

xE n T-nCy -N

then 4 E C ( Y R ), 0p2 > Tp1, and 014'p2-qTP 11 <a Therefore

ha(T) + a(q() > Q (T599, () - () Hv (D /'') + a(qp)

=Q (T,cp,,PN)-jt(4)-Hv (t'/'') +a(q) byLemmal.8

P(T, - 2-log(q+1)+(qa)-pjp)

> P (T5 ,q)-log(q + 1)-1

since a(9)-9() = v(0p2)-v pp1) < 1140P2 -PII < 21 So for an invertible subshift (X,T) of the shift on (Iq)Z and cpEC(X,R) there is a aE-M(T) with

ha (T) + a (q) > P (T, )-log(q + 1)-1. Let n > 0 and consider (X, T'). The natural homeomorphism h :(Iq)z__(Iqn)z given by

xon nO1X 1 h( xixoxi ... (.j(

X1Xn-l X2n-1J

conjugates (X, Tn) with an invertible subshift (hX, hTnh -) of the shift on (1qn)Z. By the above there exists a EM (Tn) such that

n-1 n-1

h17(Tn) + Y, 9 o T' >P T,n , I ?oT' - log(nq +1) - i=o i =o

= nP (T, )-log(nq + 1)-1 by Theorem 2.2 (i).

Put m=(l/n) yn-1 (T-')EM(T). By Lemma 4.10 hm(Tn)> ha(Tn) so nhm(T)+nm(cp)>nP(T,q)-log(nq+1)-1. Therefore for each n>O we have constructed m E M (T) vvith hm (T) + m(q) > P (T,q) -(1 + log(nq + 1)) / n. Therefore SUpmeM(T) [hm(T) + m(q9)] > P (T, q9) if P (T, q)) < oo.

When P(T,q)-= oo, for each M >O we can choose 6 so that diam (C) < 8

implies var(qp) M. The above proof yields a E M (T) so that ha(T) + a(q9) > M-log (q + 1). Hence SupmeM(T) [hm(T) + m (p)]= Uo.

Theorem 4.1 implies the following results, some of which can be proved in a straightforward way without using Theorem 4.1.

COROLLARY 4.12. Let T:X--X be continuous and EEC(X,R). Then

(i) If { Xi I i E I } is any family of closed sets with TXi c Xi for each i CE

and X= U iE Xi then P(T,cp)=supi:e P(TIy,q p1 ). (ii) If 5 denotes the non-wandering set of T then

P(T,p) = P(Tjg, p91).

(iii) Let {Tt)}tER be a 1-parameter group of homeomorphisms of X and

9pEC(X). Define qptEC(X) by cpt(x)=f p(T'x)ds. If M({Tt}) denotes

those Borel probability measures invariant for the flow { Tt} (i.e.

968 PETER WALTERS.

M ({ Tt}) = ntER M (Tt)) then

P (Tt,qTt) = SUp [ (Tt) + (ut)p IeM({T.})

= sup [hit(Tt) + gt] ,ZeM({T,})

Also P (Tt, 'Pt) t Ip (T,, ,P). (iv) P (Tn, np) > nP (T,q 9) (n > 0). (v) If 4 E C (X, R) inf, e M (T) It () < P (T, + 4)-P (T, ) < sup E M (T)/()

(vi) h(T) = 0 implies P (T,q ) = SUP4EM(T) ItJ(T)

(vii) If T is uniquely ergodic (i.e. M (T) has only one member, jt say) then P (T T,p) = h,, ( T) + ,u (qp) = h ( T) + ti (p)).

Part (i) is a strengthening of theorem 2.2 (v) and part (v) is a strengthening of Theorem 2.1 (iii) and (v).

Proof.

(i) By Theorem 2.2 (iii) it suffices to show P (T, 9p) < supi e g P (Ti, Pqi) where Ti = TIx and (Pi =Pglx. Let E > 0. Choose 1i E M (T) with P (T, 9p) < h,, (T) + p(T) + E. Let I = JE(T) ddT be the ergodic decomposition of ji into ergodic measures [12]. Then P (T,9 9) < f E (T) [ h, (T) + v (qp)] dT + ,, and since each ergodic measure v is concentrated on some Xi we have

h,(T)+ v(p) < supP(Tii) VPE(T)

and therefore

P(T,cp) sup [h,(Tt)+1'(q)t)] M ({ T.})

Let E(Tt) denote the ergodic members of M(Tt). E(Tt) is the collection of

extreme points of M (Tt) and therefore P (Tt, qt) = SUPE(Tj[h,(Tt) + t(q9t)]. Let

[tE EE(Tt). For each sEllR [?oT, is also a member of E(Tt) and h,,(Tt) h,,(T )(Tt). Let j = (1/t)fJt (T-)ds E M({ Ts}). By 9.8 of [16] h(Tt)=(1/t)

fth (T)( Tt)ds = h (Tt). Clearly tj2 (q) = (q(t) = 2(qt) SO that

P (Tt,qgt) = SUp [ hy( Tt )+ 1 (qt) ] M({T.})

=sup [h1(Tt)+ tlA(q9)1 M({T.})

Let t >0.

P (Ttqt) = SUp [ h,(Tt )+1'(qt) ] M ({ T.})

=t sup [ h (T,)+ lu)] M ({ T.})

tP (T,, P1).

If t <0 then

P(ITt,qt)=P(T_t,qu-t) by Theorem 2.1 (vi)

=-tP(T_1,qu-1) by the above

=-tP (T1,, P1).

(iv)

P (Tn np)= SUp [h, (Tn ) + pI(nP)] E M (Tn)

> sup [hm(Tn)+m(nrp)] mEM(T)

=n sup [hm(T)+m(q()1 mEM(T)

= nP(T, P)

(v)

P (T, 9 +4)= su [ h, (T) +p) + )] 1 EM(T)

 sup ) h,,(T) +

K(9) ] + if

1(4T)

P (T, (p) + inf Tp) -L E M (T)

970 PETER WALTERS.

(vi) Since 0 < h,(T) < h(T), t E M (T) we have

h, (T)=OtV E M(T)soP(T,qm)= sup tt(p) 1 EM(T)

(vii) When M(T)={14 we have, by Theorem 4.1, h,(T)+ t(q)=P(T,q) and h,(T)=h(T).

5. Remarks. If the map M(T)--R given by t-*>h,(T) is upper semi continuous (which is true if T is expansive), the last part of the proof of Theorem 5.1 of [17] shows that for jtEM(T)

h, (T) = inf [P(T,q)-M,(g)]. c E C (X, R)

It is not hard to see that the converse is true. Clearly, when 1i--h, (T) is upper semi continuous there exists a jt E M (T)

with h, (T) + jt (q) = P (T, m) since M (T) is compact. Such a jt is called on equilibrium state for (T, p). This is not true in general as Gurevic has shown for the case qi = 0 [11]. Bowen has shown that if T is an expansive homeomorphism satisfying the specification condition and qi belongs to a certain linear subspace of C (X, R) there is a unique jt, E M (T) with h,, (T) + j,(q) = P(T, m) [4].

If P (T, p) < ox then any finite Borel measure jt on X satisfying j(t) < P (T, 9)V9 E C (X, R) is a member of M (T) (see [17] Prop. 6.3). Conversely any member of M (T) has this property.

UNIVERSITY OF WARWICK.

REFERENCES.

[1] R. L. Adler, A. G. Konheim and M. H. McAndrew, "Topological entropy," Trans. Amer. Math. Soc., 114 (1965) 309-319.

[2] P. Billingsley, Ergodic Theory and Information, Wiley (1965). [3] R. Bowen, "Entropy for group endomorphisms and homogeneous spaces," Trans. Amer.

Math. Soc., 153 (1971) 401-414. [4] , "Some systems with unique equilibrium states, Math. Systems Theory, 8 (1975),

193-202. [5] E. I. Dinaburg, "The relation between topological entropy and metric entropy," Soviet

Math., vol. II, No. 1 (1970) 13-16. [6] F. R. Gantmacher, The Theory of Matrices, vol. 2, Chelsea (1964). [7] T. N. T. Goodman, "Relating topological entropy and measure entropy," Bull. London Math&.

Soc., 3 (1971) 176-180.


[7'] , "Topological entropy, topological sequence entropy and closed generators," Ph.D. thesis, University of Sussex (1972).

[8] "Maximal measures for expansive homeomorphisms," J. London Maths. Soc., (2) 5 (1972).

[9] L. W. Goodwyn, "Topological entropy bounds measure-theoretic entropy," Proc. Amer. Math. Soc., 23, 3, (1969) 679-688.

[10] , "Comparing topological entropy with measure-theoretic entropy," Amer. J. Math., 94 (1972), 366-388.

[11] B. M. Gurevic, "Topological entropy of denumerable Markov chains," Soviet Math. Dokl., 10, No. 4 (1969) 911-915.

[12] K. Jacobs, "Ergodic decompositions of the Kolmogorov-Sinai invariant," Proc. Symposium on Ergodic Theory (Ed. F. Wright), Academic Press (1963).

[13] H. B. Keynes and J. B. Robertson, "Generators for topological entropy and expansiveness," Math. Systems Theory, 3 (1969) 51-59.

[14] W. Parry, "Intrinsic Markov chains," Trans. Amer. Math. Soc., 112 (1964) 55-66. [15] "Symbolic dynamics and transformations of the unit interval," Trans. Amer. Math.

Soc., 122 (1966) 368-378. [16] V. A. Rohlin, "Lectures on the entropy theory of transformations with invariant measure,"

Russian Maths. Surveys, 22 (1967) 1-52. [17] D. Ruelle, "Statistical mechanics on a compact set with ZV action satisfying expansiveness

and specification," Trans. Amer. Math. Soc., 185 (1973), 237-252. [18] S. Smale, "Differentiable dynamical systems," Bull. Amer. Math. Soc., 73 (1967) 747-817.



a variational principle for the pressure of continuous transformations

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