ON THE DEFINITION OF RELATIVE PRESSURE FOR FACTOR MAPS ON SHIFTS OF FINITE TYPE. 1. Introduction

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1 ON THE DEFINITION OF RELATIVE PRESSURE FOR FACTOR MAPS ON SHIFTS OF FINITE TYPE KARL PETERSEN AND SUJIN SHIN Abstract. We show that two natural definitions of the relative pressure function for a locally constant potential function and a factor map from a shift of finite type coincide almost everywhere with respect to every invariant measure. With a suitable extension of one of the definitions, the same holds true for any continuous potential function.. Introduction The introduction to the paper 4 included, for factor maps between subshifts and the identically zero potential, a finite-range definition of the relative pressure function that is different from the standard one, 8, which involves complete bisequences. While this variation in the definition had no bearing on the results of that paper, it does seem useful to clarify the extent to which the two definitions differ; in particular, in some situations one definition may be easier to use than the other. We show in this note that for a factor map π : X Y, where X is a shift of finite type and Y is a subshift, the two relative pressure functions for the identically zero potential can be different, but they coincide almost everywhere with respect to every invariant measure on Y. Therefore, for each ergodic invariant measure ν on Y, the two definitions lead to the same value of the maximal possible relative entropy h µ (X Y ) of any invariant measure µ on X over Y. More generally, for any locally constant potential function (one that depends on only finitely many coordinates), the analogously defined two relative pressure functions coincide almost everywhere with respect to every invariant measure on Y. Finally, we show that this statement continues to hold for an arbitrary continuous potential function with a suitably generalized definition of the finite-range relative pressure function. Several useful ideas for the study of the relative entropy function were developed in 6 in connection with questions about the existence of compensation functions, and we adopt and extend them for our purposes here. For general background on symbolic dynamics, entropy, and pressure, see 2, 3, Mathematics Subject Classification. Primary: 37A35, 37B40; Secondary: 37B0.

2 2 KARL PETERSEN AND SUJIN SHIN If (X, S) is a topological dynamical system, then M(X) will denote the set of all S-invariant Borel probability measures on X with its weak* topology. Given x X, let n µ x = lim δ S n i x M(X) if it exists, in which case we call x a generic point. Denote by G(X) the set of all generic points in X. If X is a shift space, then for each n, B n (X) denotes the set of n-blocks in X, and B(X) = n B n (X). Given b b n B n (X), n, we define rb b n r+n = {x X : x r = b,..., x r+n = b n }, and we abbreviate 0 b b n n by b b n. We denote the shift transformation by σ and the usual metric for a subshift by ρ. Let S : X X and T : Y Y be continuous maps of compact metrizable spaces and π : X Y a factor map. i.e., a continuous surjection with π S = T π. For a given compact subset K of X, for n and δ > 0, denote by n,δ (K) the set of (n, δ)-separated sets of X contained in K. Let f C(X). Fix δ > 0 and n. For each y Y, let { P n (π, f, δ)(y) = sup Define P (π, f) : Y R by x E ( n exp f(s i x)) E n,δ ( π {y} )}. P (π, f)(y) = lim lim sup δ 0 n ln P n(π, f, δ)(y). The function P (π, f) is called the relative pressure function associated with f. It is Borel measurable and T -invariant. For ν M(Y ), let M(ν) = π (ν) denote the set of measures in M(X) that project to ν under π. Given f C(X), the function P (π, f) : Y R satisfies the relative variational principle : for each ν M(Y ), { } P (π, f)dν = sup h(µ) + fdµ µ M(ν) h(ν). In particular, for a fixed ν M(Y ), sup{h µ (X Y ) : µ M(ν)} = sup{h(µ) h(ν) : µ M(ν)} = Y P (π, 0) dν. 2. Two different definitions of relative pressure Hereinafter, let X be a shift of finite type and Y a subshift, on finite alphabets A(X) and A(Y ), respectively. Let π : X Y be a factor map, so that Y is a sofic system (see 2). We treat the 2-sided case, the -sided case being very similar. Let f C(X) be a locally constant function, i.e. one that depends on only finitely many coordinates x m x m. By passing to a higher block representation if necessary, we may assume that π is represented

3 ON THE DEFINITION OF RELATIVE PRESSURE 3 by a one-block map from B (X) to B (Y ), which we denote again by π, and that f is a function of the two coordinates x 0 x. For B = b b n B n (X), π(b) means the n-block π(b ) π(b n ) of Y ; given v B n (Y ), π (v) denotes the set of n-blocks of X that project to v by the block map π. Given y Y, for each n, let D n (y) consist of one point from each nonempty set π (y) x 0 x x n. The potential function f determines a block map F : B 2 (X) R by F (b 0 b ) = exp(f(x)) for any x b 0 b. For a block B = b b n B(X), put s f (B) = F (b b 2 )F (b 2 b 3 ) F (b n b n ), and for a block w B(Y ), put S f (w) = B s f(b) where the sum is taken over all B B(X) that are mapped to w by π. Then for each y Y, ( P (π, f)(y) = lim sup n ln n ) exp f(σ i x) () = lim sup n ln (see 8, Theorem 4.6). In particular, for all y Y, P (π, 0)(y) = lim sup s f (x 0 x x n ) n ln Dn (y) (with f 0). Define another Borel-measurable function Φ f : Y R by Φ f (y) = lim sup n ln S f(y 0 y y n ) for y Y. It can be shown that Φ f (y) Φ f (σy) for all y Y. Also P (π, f)(y) Φ f (y) for all y Y. Given f C(X), one may have P (π, f)(y) < Φ f (y) for some y Y, as seen in the following example (which in 5 and 6 was shown to be a factor map for which there exists no saturated compensation function). Example. Let X, Y be the shifts of finite type determined by allowing the transitions marked on the figure and the one-block factor code π : X Y map to, and 2, 3, 4, 5 to 2. X π 2 Y

4 4 KARL PETERSEN AND SUJIN SHIN For each k, let a k = 2 k +, and define y Y by y = a 2 a 2 2 a 3 (so that y i = 2 for all i < 0). Whenever πx = y, then x 0, ) = y 0, ) = 2 a 2 a 2 2 a 3, since each a k is odd and so π (2 a k ) = {2 a k}. Thus Dn (y) = for all n which implies that P (π, 0)(y) = 0. Meanwhile, fix k and set n k = 2 k+ + 2k 2. Then π y 0 y y nk = π 2 a 2 a 2 2 a k = π 2 a k = 2 2 k +. Consider f 0 C(X). Then Φ f (y) lim sup ln π y 0 y y nk k n k ( ln 2 2 k + ) = lim k 2 k+ + 2k 2 = ln 2 > 0 = P (π, 0)(y) Essential agreement of the two definitions for locally constant potential functions Our goal is to prove the following. Theorem. Let X be an irreducible shift of finite type, Y a subshift, and π : X Y a factor map. Let f C(X) be a function which depends on just two coordinates, x 0 x, of each point x X. Then for each ν M(Y ), we have P (π, f)(y) = Φ f (y) a.e. dν(y), and hence P (π, f)dν = Φ f dν. Remark. We may assume that f 0. For if f takes some negative values, choose a constant c such that f + c 0 and use the equations P (π, f + c) = P (π, f) + c, Φ f+c = Φ f + c. Thus we have F M for some constant M. For notational convenience, set s = s f, S = S f and Φ = Φ f, and let T (y) = P (π, f)(y) for y Y. The map R : M(X) R + defined by R(µ) = h(µ) h(πµ) is upper semicontinuous in the weak topology 8, Lemma 2.2. Using this one can easily prove the following 6. Lemma. For any f C(X) the affine map L : M(Y ) R given by L(ν) = T dν is upper semicontinuous. For q, let P q (Y ) = {y Y σ q (y) = y} and P (Y ) = q P q(y ). Lemma 2. Let y P q (Y ), q. Then (2) Φ(y) = lim nq ln S(y 0y y nq ). Also Φ(y) = T (y).

5 ON THE DEFINITION OF RELATIVE PRESSURE 5 Proof. For a block w B(Y ) with ww B(Y ), we have S f (w n+m ) M S f (w n )S f (w m ). Thus (/n) ln S f (w n ) converges as n (see 3, p. 240), and hence Φ(y) = lim nq ln S(y 0 y nq ). To show that Φ(y) = T (y), set w = y 0 y y q B q (Y ), so that y = w.ww = y q.y 0 y y q y 0 y. Let π (w) = l and define an l l, 0- matrix A = (A uv ) by A uv = if and only if uv B 2q (X), where u, v π (w), that is, A is the transition matrix between blocks in the inverse image of the repeating word w that forms y. Note that some blocks in π (w) may be preceded or followed only by allowable q-blocks in X that do not map to w. So we let B be the reduction of A obtained by excluding all the zero columns and rows together with their corresponding rows and columns. Then by (2), Φ(y) = lim nq ln u i π (w) Since B is essential, we have s(u u n+ u n+2 )A u u 2 A un+ u n+2 u i π (w) u i π (w) l 2 M 2q l 2 M 2q s(u u n )A u u 2 A un u n. s(u u n )B u u 2 B un u n l 2 M 2q x D nq(y) u i π (w) s(x 0 x x nq ) s(u u n )A u u 2 A un u n. Now we take logarithms, divide, and take the limit on n to get Φ(y) = T (y). For the proof of the following result, we refer to 6. Lemma 3. Let y G(Y ) and let y (s) P ls (Y ), l s s, for each s. If there is N such that y (s) 0,l s N = y 0,l s N for all s large enough, then µ y (s) µ y as s. Let A now denote the alphabet of X. Fix y Y. Given b, c A and n, let Γ n y(b, c) = u S(buc),

6 6 KARL PETERSEN AND SUJIN SHIN where the sum is taken over all u s in B n (X) such that π(buc) = y 0 y y n. (If Γ n y(b, c), then πb = y 0 and πc = y n.) Then Γ n y(b, c) = S(y 0 y y n ). b,c A It is not difficult to check the following. Lemma 4. Let y P q (Y ), q, and b A. Then for k, Γ q y (b, b) k Γ qk y (b, b). Lemma 5. Let y G(Y ). Then there is a sequence {y (s) } s= P (Y ) such that µ y (s) as s and Φ(y) lim s Φ(y (s) ). µ y Proof. Observe first that there exist a symbol, say a, of A(Y ) and a strictly increasing sequence {m k } k=0 Z+ such that y mk = a for all k 0 and (3) Φ(y) = lim ln S(y 0 y mk ). m k m k Let y = σ m 0 y Y and for each k 0 put n k = m k m 0 0. Then yi = a if i = n k for some k 0 (n 0 = 0). Let C = π (a). For k, choose b k, c k C so that Γ n k y (b k, c k ) = max b,c C Γn k y (b, c) (so π(b k ) = y 0 and π(c k ) = y n k ). Since F M, it follows that Thus by (3), (4) Φ(y) lim k M 2 S(y 0 y n k ) Γ n k y (b k, c k ) S(y 0 y n k ). = lim k ln M m0 S(y m0 y mk ) m k ln S(y0 yn n k ) = lim k k n k ln Γ n k y (b k, c k ). Notice that there exist b, c C such that b k = b and c k = c for initely many k s, say k s s, where k s as s. Since X is irreducible, there is w B d (X) for some d m 0 such that c wb B d+2 (X). Let a a d = π(w) B d (Y ). Fix s and put l s = n ks + + d. Define y (s) P ls (Y ) by y (s) = a d.y 0y y n ks a a d y 0y (y 0... y n ks is the image under π of a word w = w 0 w n ks in X with w 0 = b k = b and w n ks = c k = c, so that w.w ww w is a legitimate point x (s) X, and y (s) = π(x (s) )). Since y (s) 0,l s d = y 0,l s d and l s m ks k s s, it follows from Lemma 3 that µ y (s) µ y

7 ON THE DEFINITION OF RELATIVE PRESSURE 7 or equivalently µ y (s) µ y as s. To see that Φ(y) lim s Φ(y (s) ), fix s. By Lemma 4, Φ(y (s) ) = lim ln S(y (s) 0 y (s) p p l p l s ) lim sup ln Γ p ls (b s p p l y (s), b ) s lim sup p l s ln Γ ls y (s) (b, b ) = l s ln Γ ls y (s) (b, b ). It is clear that Γ n ks y (b, c ) Γ ls (b y (s), b ). Thus from (4), Φ(y) lim s lim s which completes the proof. ln Γ n ks y n (b k s, c ks ) = lim ks s ln Γ ls (b l y (s), b ) lim s s Φ(y(s) ), n ks ln Γ n ks y (b, c ) Let E = {y G(Y ) T dµ y = T (y)}. Then ν(e) = for every ergodic invariant measure ν on Y, and hence ν(e) = for every ν M(Y ). For y E, let {y (s) } s= P (Y ) be a sequence obtained from Lemma 5 so that µ y (s) µ y as s and Φ(y) lim s Φ(y (s) ). By Lemma, lim sup s T (y (s) ) = lim sup s T dµ y (s) T dµ y = T (y). It follows from Lemma 2 that T (y (s) ) = Φ(y (s) ) for all s. Thus Φ(y) lim s Φ(y(s) ) = lim s T (y(s) ) T (y) Φ(y). Hence T (y) = Φ(y) for all y E. Let ν M(Y ). Since ν(e) =, we have T dν = T dν = Φdν = Φdν, which completes the proof of Theorem. E E 4. A finite-range relative pressure function for general continuous potentials To extend Theorem to the case of an arbitrary potential f C(X), we need to define a suitable function corresponding to Φ f. Let f C(X). Fix n. For each i = 0,,, n, define a block map F i n : B n (X) R by F i n(b b n ) = exp(f(x)) σ i x b b n

8 8 KARL PETERSEN AND SUJIN SHIN (so the imum is taken over all x in the cylinder set i b b n n i ). For each n-block B B n (X), put and for a block C B n (Y ) put Define Ψ f : Y R by Ψ f (y) = lim sup n s f (B) = Fn(B), i S f (C) = π(b)=c s f (B). n ln S f(y 0 y y n ) for y Y. Now, given n, for each i = 0,,, n define F i n : B n (X) R by F i n(b b n ) = sup exp(f(x)). σ i x b b n Using F i n, we similarly define s f, S f and Ψ f as follows. For each B B n (X), and for C B n (Y ), s f (B) = S f (C) = n π(b)=c F i n(b), s f (B). Define Ψ f : Y R by Ψ f (y) = lim sup n ln S f (y 0 y y n ) for y Y. It can be easily shown that if f depends on just two coordinates, x 0 x, of each point x X, then Ψ f and Ψ f both are equivalent to Φ f, and hence Theorem would apply. In the current more general situation, we still have the analogous result. Theorem 2. Let X be an irreducible shift of finite type, Y a subshift, π : X Y a factor map, and f C(X). For each ν M(Y ), we have P (π, f)(y) = Ψ f (y) = Ψ f (y) a.e. dν(y), and hence P (π, f)dν = Ψ f dν = Ψ f dν.

9 ON THE DEFINITION OF RELATIVE PRESSURE 9 Proof. As before, we may assume that f 0. Thus we have a constant M > 0 such that exp(f) M and hence Fn i F n i M for all n and for i = 0,,, n. Let T = P (π, f) as before. Set s = s f, S = S f, and Ψ = Ψ f. For y Y, define θ(y) = lim sup n ln s(x 0 x x n ). We show that T (y) = θ(y) for all y Y. Note first that if x 0 x n B n (X), n, then Thus from (), given y Y, s(x 0 x n ) = θ(y) = lim sup lim sup n σ i z x 0 x n exp(f(z)) n = exp(f(σ i z)). z x 0 x n n ln n ln n Suppose that there exist y Y and ɛ > 0 for which θ(y) = T (y) 2ɛ. z x 0 x n exp(f(σ i z)) n exp ( f(σ i x) ) = T (y). Since f is uniformly continuous, there is p 0 such that whenever ρ(z, x) < 2 p or, equivalently, z p,p = x p,p, then f(z) f(x) < ɛ. Fix x X and n > 2p. For each i = p, p +,, n p, F i n(x 0 x x n ) = z x 0 x n exp(f(σi z)) exp(f(σ i z)) exp z X ρ(σ i z,σ i x)<2 p Since F i n for each i and exp(f) M, we have s(x 0 x x n ) n p i=p F i n(x 0 x x n ) n p i=p n M 2p e (n 2p)ɛ exp(f(σ i x)). ( f(σ i x) ɛ ). exp ( f(σ i x) ɛ )

10 0 KARL PETERSEN AND SUJIN SHIN It follows that θ(y) lim sup = lim sup n ln M 2p e (n 2p)ɛ n ln n exp(f(σ i x)) n exp(f(σ i x)) ɛ = T (y) ɛ, which is a contradiction. Therefore T (y) = θ(y). Observe next that for each y Y, θ(y) = lim sup lim sup n ln n ln s(x 0 x x n ) s(x 0 x x n ) = Ψ(y), π(x 0 x n ) =y 0 y n since the summation for Ψ(y) is taken over the larger set than the one for θ(y). As we have seen before, it is possible to have θ(y) < Ψ(y) for some y Y. We will prove that if y Y is periodic, then θ(y) = Ψ(y), or, equivalently T (y) = Ψ(y). Fix y Y and for each n, let τ n (y) = s(x 0 x x n ). Assuming that D n (y) D n+ (y) for each n (without loss of generality), we have τ n (y) = n exp(f(σ i z)) z x 0,n n x D n+ (y) n z x 0,n exp(f(σ i z)) = τ n+ (y). exp(f(σ i z)) z x 0,n Hence τ n (y) increases as n increases. An easy computation shows that given an increasing sequence {a n }, for any q, lim sup a n /n = lim sup a nq /(nq). Letting a n = ln τ n (y) proves that for any q, (5) θ(y) = lim sup nq ln s(x 0 x x nq ). x D nq(y)

11 Let y P q (Y ), q. We claim that ON THE DEFINITION OF RELATIVE PRESSURE (6) Ψ(y) = lim sup = lim sup nq ln S(y 0y y nq ) nq ln s(x 0 x x nq ) π(x 0 x nq ) =y 0 y nq. Then, using (5) and (6), we can proceed as in Lemma 2 to show that Ψ(y) θ(y) and therefore Ψ(y) = T (y). To verify (6), let ɛ > 0 be given and (using uniform continuity of f) choose p 0 so that whenever ρ(z, x) < 2 p or z p,p = x p,p, then f(z) f(x) < ɛ. Given an integer m 2p + q, there is n N such that 2p < nq m < (n + )q. Put k = m nq 0 and v = y 0 y y nq. Then (7) S(y 0 y... y m ) = π(u)=y 0 y m m z u m π(u)=v w B k (X) uw B(X) π(u)=v w B k (X) uw B(X) z uw exp(f(σ i z)) nq p M k+2p exp(f(σ i z)) i=p z uw exp(f(σ i z)). Let u B nq (X) and w B k (X) be any blocks such that uw B(X). Let p i nq p. Note that if z uw and z u, then ρ(σ i z, σ i z) < 2 p so that f(σ i z) f(σ i z) < ɛ. Thus From this inequality and (7), we have z uw exp(f(σi z)) z u exp ( f(σ i z) + ɛ ). S(y 0 y... y m ) M k+2p π(u)=v w B k (X) uw B(X) M k+2p Bk (X) π(u)=v nq p i=p nq exp ( f(σ i z) + ɛ ) z u exp ( f(σ i z) + ɛ ) z u = M k+2p Bk (X) e nqɛ S(y 0 y... y nq ).

12 2 KARL PETERSEN AND SUJIN SHIN Since k < q and p, q are fixed, it follows that Ψ(y) = lim sup m lim sup m ln S(y 0 y m ) nq ln S(y 0y... y nq ) + ɛ. Taking ɛ arbitrarily small completes the proof of the claim. Next, we will show that Ψ f (y) = T (y) for y P (Y ). It is straightforward to check that T (y) Ψ f (y) for all y Y. Set s = s f, S = S f, and Ψ = Ψ f. Define θ : Y R by θ(y) = lim sup n ln s(x 0 x x n ). Note that T (y) θ(y). Suppose there exist y Y and ɛ > 0 for which θ(y) = T (y) + 2ɛ. Similarly to the foregoing, there is p 0 such that whenever ρ(z, x) < 2 p or equivalently, z p,p = x p,p, then f(z) f(x) < ɛ. Fix x X and n > 2p. If p r n p, then Thus It follows that F r n(x 0 x x n ) = s(x 0 x x n ) = p r=0 n p M 2p n p F n(x r 0 x n ) r=p M 2p e (n 2p)ɛ exp θ(y) lim sup = ɛ + lim sup sup exp(f(z)) σ r z x 0 x n sup exp(f(z)) exp(f(σ r x) + ɛ). ρ(z,σ r x)<2 p r=p F r n(x 0 x n ) n p F n(x r 0 x n ) M 2p ( n p r=p n ln M 2p e (n 2p)ɛ n ln exp r=p n r=n p F r n(x 0 x n ) exp(f(σ r x) + ɛ) ) f(σ r x) M 2p e (n 2p)ɛ exp ( n r=0 ( n r=0 ( n ) exp f(σ r x) f(σ x)) r = ɛ + T (y), r=0 ) f(σ r x).

13 ON THE DEFINITION OF RELATIVE PRESSURE 3 which is a contradiction. Therefore T (y) = θ(y). Let y P q (Y ), q. It is not difficult to see that T (y) = lim sup nq ln s(x 0 x x nq ) = θ(y) and Ψ(y) = lim sup x D nq(y) nq ln S(y 0 y nq ). We can proceed again as in Lemma 2 to show that Ψ(y) = T (y). The remainder of the proof is the same as before. Recall that according to Theorem 4.6 of 8, if for each n =, 2, and y Y we denote by D n (y) a set consisting of exactly one point from each nonempty set x 0 x n π (y), then for each f C(Y ), P (π, f)(y) = lim sup n ln exp ( n f(σ x)) i. From Theorem 2 it now follows that we will obtain the value of P (π, f)(y) a.e. with respect to every invariant measure on Y if we delete from the definition of D n (y) the requirement that x π (y): Corollary. For each n =, 2, and y Y denote by E n (y) a set consisting of exactly one point from each nonempty cylinder x 0 x n π y 0 y n. Then for each f C(Y ), ( P (π, f)(y) = lim sup n ln n ) exp f(σ i x) x E n(y) a.e. with respect to every invariant measure on Y. Acknowledgment. The authors thank the University of Warwick, where much of this research was accomplished, for its outstanding hospitality. References. F. Ledrappier and P. Walters, A relativised variational principle for continuous transformations, J. London Math. Soc. 6 (977), D. Lind and B. Marcus, An Introduction to Symbolic Dynamics and Coding, Cambridge Univ. Press, K. Petersen, Ergodic Theory, Cambridge Univ. Press, K. Petersen, A. Quas, and S. Shin, Measures of maximal relative entropy, Erg. Th. Dyn. Sys. 23 (2003), S. Shin, An example of a factor map without a saturated compensation function, Erg. Th. Dyn. Sys. 2 (200), S. Shin, Relative entropy functions for factor maps between subshifts, preprint.

14 4 KARL PETERSEN AND SUJIN SHIN 7. P. Walters, An Introduction to Ergodic Theory, Springer-Verlag, P. Walters, Relative pressure, relative equilibrium states, compensation functions and many-to-one codes between subshifts, Trans. Amer. Math. Soc. 296 (986), 3. Department of Mathematics, CB 3250, Phillips Hall, University of North Carolina, Chapel Hill, NC USA address: Department of Mathematics, Korea Advanced Institute of Science and Technology, Daejon, , South Korea address: