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M. Misiurewicz, W. Szlenk (1980)
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ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL by MICHAL MISIUREWICZ* o. Introduction Dynamical properties of mappings of an interval into itself are intensely studied from both " theoretical " and " experimental " (numerical experiments) points of view. One of the most important theoretical problems (but also closely related to the problem of reliability of numerical results) is to establish for which mappings there exist invariant probabilistic measures, absolutely continuous with respect to the Lebesgue measure, how many of them, and what are their ergodic properties. The aim of this paper is to answer these questions for a certain class of mappings. They are essentially the piecewise monotone mappings with non-positive Schwarzian derivative, no sinks and trajectories of critical points staying far from critical points (the exact conditions can be found in section 3: conditions (i)-(vi)). For a slightly sinfilar class of mappings M. Jakobson [3] proved the existence of an absolutely continuous invariant measure. Our technique is quite different and it enables us to obtain much information about our measules. After the preliminary results of Sections 1-5, we prove the main theorems (theo- rems (6.2) and (6.3)) in Section 6. For a mapping from our class, there exist a finite number (but at least one) of ergodic invariant probabilistic measures, absolutely conti- nuous with respect to the Lebesgue measure. Their densities are continuous on an open dense set. Images of every finite measure, absolutely continuous with respect to the Lebesgue measure, under nk-th iterations of the mapping (for a certain k), converge strongly (as n--~oo) to a linear combination of those measures. The mapping with every one of those measures is a skew product of a permutation of a finite set (in the base) and an exact transformation. In Section 7 we show that for most widely considered one-parameter families of mappings (like xF-*4o~x(i--x)) , our conditions ((i)-(vi)) are satisfied for a set of parameters of power the continuum. The question, whether the measure of this set of parameters is zero or positive, remains open. However, there is some evidence that * This paper was written during the visit to the Institut des Hautes t~tudes Scientifiques. The author gratefully acknowledges the hospitality of I.H.E.S. and the financial support of the Stiftung Volkswagenwerk for the visit. 17 x8 MICHAL MISIUREWICZ it is zero for generic families. Namely, it can be decomposed into a countable union of subsets, each of them arising from a point " running through " a Cantor set, which also depends on a parameter (one of these subsets is considered in Section 7). This Cantor set has always measure zero and there are reasons to believe that our point " runs too fast " to stay in this Cantor set for a time of positive measure (in a generic case) (1). In sections 8 and 9 we study the problem; when is the measure-theoretical entropy (computable in numerical experiments as a characteristic exponent) equal to the topo- logical entropy, which measures complexity of dynamics from the topological point of view ? The answer for the maps of our class is: almost never (we obtain infinitely many independent necessary conditions). The proof of the main results bases on the properties of the Schwarzian derivative. f,,, "'"'lJ-} ~ and the idea of applying it to mappings of an interval It is defined as Sf-- f, 3_ 2\f'] belongs to D. Singer [io]. Two main properties are: (I) S(fog)=(g')2.(Sfog)+Sg, and consequently if Sf<o and Sg<o then S(fog)<o. In particular, if Sf<o then S(f~)<o for all n>o. (2) If Sf<o then -- is convex on the components of the complement of V]f'l the set of critical points (see (3- I)) and consequently, [f't has no positive strict local minima. One can find examples of mappings with negative Schwarzian derivative for example in [io] and [5], and more properties of these mappings in [2]. With some additional work, it is possible to generalize the result of the paper, replacing the hypotheses C 3 and Sf< o by C 1 with f' Lipschitz and ~ convex on the components of the complement of the set of critical points off (i.e. on the components of I\A). Throughout the whole paper we denote by k the Lebesgue measure, by f" the n-th iterate off, and by E the closure of E. I would like to acknowledge very helpful discussions with J. Guckenheimer and Z. Nitecki. x. Stretching far from critical points In this section, I will be a closed interval, U and V two open subsets of I consisting of a finite number of intervals each, and such that U contains the endpoints of I and Ut3V=I, and f: V-+I will be a continuous mapping. (t) Recently Jakobson proved that the absolutely continuous invariant measures exist for a set of parameters of positive measure (" Absolutely continuous invariant measures for one-parameter families of one-dimensional maps ", to appear in Gommun. Math. Phys.). His set of parameters is much larger and his arguments seem to confirm our conjecture. 18 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 19 We shall call an open interval J C V a homterval if for all positive integers n, f" maps homeomorphically J onto its image. We shall say that f has no sinks if there does not exist an interval J c V and a positive integer n such that f" maps J homeomorphically into J. Lemma ( t . 9 ). -- If f has no sinks and J is a homterval, then the intervals f"J, n=o, I, 2, ..., are pairwise disjoint. Proof. -- Suppose that for some n>o and k>o, f"(J) andf"+k(J) are not disjoint. Then for every p>o, f,+pk(j) and f,+(p+l)k(j) are also not disjoint. Therefore, the set K=p[J=0f"+pk(J) is an interval. For every p the map :, ,.,,, =(:.+<,+,,,[,)o(:.+,, ,), is a homeomorphism, and thus fk K is a homeomorphism. But fk(K) C K and hence f has sinks. 9 Notice that any image of a homterval is also a homterval. The following theorem may be considered as a generalization of the Denjoy theorem for a circle. It was proved by Z. Nitecki. Theorem (1.2). -- Let f have no sinks, be of class (31, f'(x)~eo for all x~V and log If'[ be a Lipschitz function on components of V. Then for every homterval J there exists m>_o such that f"(J) c U. Proof. ~ Denote the Lipschitz constant of log ]f'[ by y. We shall show first: (x. x) If K is an interval such that f" is defined on K then: sup [(if)'[ ,-1 log K <y Z X(fk(K)). infl(f~)'[- k=0 Indeed, if a, b aK then we have: n--1 9 I(f")'(a)l Z [log]f'(fk(a))l--log[f'(fk(b))[]. l~ ](~7(b)[ --k=o Since the setfk(K) is an interval and is contained in V, it is contained in some component of V. Hence: log [f'(fk(a))[--log ]f'(fk(b))[<y]a- b ] < yX(fk(K)), and we obtain (I. I), Since the sets I'\U and I\V are compact and disjoint, Set ~ =dist(I\U, I\V). we have ~>o. 19 20 MICHAL MISIUREWICZ Suppose that there exists a homterval J such that f'(J)\U4=O for every n>o. We claim that: 0.2) there exists n o such that for each n>n o every homterval containingf'(J) is disjoint from U. Suppose that (1.2) is false. Then there exist n>o, k>o and homtervals K and L such that f~(J) c K, f,+k(j) < L and both K and L contain the same endpoint of some component of U and a piece of U adjacent to this endpoint. Then the interval K to L is a homterval and fk(K toL) intersects K to L. This contradicts Lemma (I. I). Hence (1.2) is true. Let M be a maximal homterval containingf"~ By (1.2) we have: (I "a) For each n>o every homterval containingfn(M) is disjoint from U. Now take an open interval L containing M, not equal to M and such that: X(L) ~ -- y(13 + X(I)) AU I by induction that for every k: We shall prove X(fk(L)) < ~ +I and fk a homeomorphism. X(fk(M)) X(I) ~ is For k =o it is obvious. Suppose that (I "5) holds for k=o, I, ..., n--I. We then have X(fk(L)\fk(M))<X~X(fk(M))~. From this, (1.3) and the definition of it follows that fk(L) C V for k = o, ..., n-- i. Hence (since L is an interval), f" L is a homeomorphism. By (I. I) and Lemma (I. I) we obtain: ~'t--1 X(f"(L)) i -- X(f"(L\M)) < X(L\M) e v k=o1: X(I.{L)} X(f"(M)) X(f~(M)) X(M) < } e-,{~ + x(I}} e'~glo (x@I~ +*) x{/"{M)' < ~ eY [(x@I) + 1) t(1) - 13 - t(I)] -- X(I) -- X(I) X(I)" This ends the proof of (I.5). Sincef"I is a homeomorphism for each n>o, L is a homterval. This contradicts the maximality of M. 9 Theorem (x.3). -- Let f have no sinks, f be of class C 3, f'(x)eeo for all xsV and Sf<o. Then there exists m>I such that, if fJ(x)r for j--o, ..., m--I, one has [(fm)'(x)l > i. 20 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 2i Proof. -- Suppose that for every n>I there exists x n such that: (i. 6) l(fn) ' (x.)I<I and (I.7) f~(xn) r for j=o,...,n--I. Take the maximal open interval Jn containing x n and such that f~ is defined on Jn. The point x~ divides Jn into two subintervals. Since S(fn)<o, on one of these subintervals we have [(fn)'l~I. Denote this subinterval by L n. By the maximality of J~, there exists k(n)<n such that (I.8) fk(~)(Ln) has a common endpoint with some component of V. We claim that: (I.9) X(L.)-+o as n-~ oe. n oo If not, there exists a sequence (~)~=1, x0eI and ,>o such that n;-+oo and Xni-+X o as i-+o% X(L,i)~r for every i, and all intervals L~ are on the same side of xni. For every j we have f~(x,i)6U if n~2>j and hence fJ(Xo)6U. But one of the inter- vals (x 0- r x0) and (Xo, x0+ ~) is a homterval, which contradicts Theolem (1.2). This proves (I.9). By (I.6) we have X(fn(L,))<X(L~), and therefore: (I. IO) X(fn- k(~)(fk(~l(L,))) -~o as n~oo. In view of (I .8) and the fact that fk(~)(x~)$U, fk(n)(L~) contains some component of the set U c~V. Hence, there exists a component K of I5 c~V and a sequence (ni)i~ 1 such that nl--->oo as i-+c~ and f~(ni)(L~i)DK for every i. From (I.lO) we get X(f~i-kC~i)(K))-+o as i-+m and therefore: (I.II) ni -- k (?zi) ----> oo as i----> o0. Let a be some point of condensation of the sequence (fk(~/)(x,~.))~= 1 . Since fJ(fk(~i)(x~i))~U for nl--k(ni)>j, we obtain in view of (1.11): (I.I2) fJ(a)q!U for every j. Let M be the minimal open interval such that Kw{a}CM. Then, by (i.ii), M is a homterval. This contradicts (I. I2) and Theorem (i.2). 9 ~. Estimates I We want to obtain estimates, which enable us to prove the convergence of the images of the Lebesgue measure to an absolutely continuous measure. In particular, we want to estimate measures of inverse images of neighbourhoods of the set of critical values off". First we shall deal with the part of estimates which can be done in the context similar to that of Section i. 21 22 MICHAL MISIUREWICZ In this section U is an open subset of I consisting of a finite number of intervals and such that the endpoints of I belong to U. We denote by f: I\U-+I a map of class C 1 such that: (2.x) If'l>~>i and the function log If'l is Lipschitz on the components of I\U, and by B a subset of I\U such that f(B)CB and: (2. =) ~ = dist(B, U)>o. Define En={xeI :fk(x)$U for k=o, ..., n--I} (notice that E, is a domain off"). Now fix an element a of B and a number ~e[o, I) and define: for for for x<a, or for x~ a. Proposition (2. x ). --- There exist constants (independent on a and on whether supp ? is to the right or to the left of a) ~e(o, I), ~e(o, I), 0>o such that for every n_~o: (2.3) x(E.) < ~"~(I), (2.4) f,. ~dx<(n+~)OK". Proof. ~ We can, instead of U, take a slightly smaller set and extend f onto the complement of this set in such a way that all hypotheses still hold (perhaps ~ will be slightly smaller and the Lipschitz constant of log [f'i slightly larger) and also the image of every component of the domain off is the whole intelval. The new sets E n will be perhaps slightly larger, but at least not smaller. Therefore we may simply assume that the image of every component of I\U under f is the whole interval I. We shall filSt prove (2.3). Let K be a component of En, let Y be a Lipschitz constant for log If'l. By the same asguments as in the proof of Theorem (i. 2), (I. I) holds. By (2.1) we have k(I)~X(ff(K))>0~"-kX(fk(K)), and thus: n--I n--i oo Y, X(fk(K))<X(I) Y, ~-"<X(I) Y, ~-~= X(I). k=O k=O j=l 0~--I 1"),(1) Set 8=e =-1. In such a way we obtain: sup I(f")'I K <8. (o,.3) inf l(f-), I -- m(x)={~x-al-r x>a,x<__a, 9(x)={~x--a]-r ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 0 3 We have f"(K)=I and f,(K\E.+,)=U. Therefore X(I)>X(K)iKnf[(f")' [ and x(U)<_x(K\E.+JsupI(f")' f. By (a.5)we get: X(K\E.+~) X(U) inf[,(ff)' [ x(U) x(K) > I + > X(K r~ E~+,)-- X(K) -- I + ~ ] (f~ 2 I -~- ~-~ " Set ~q---- Clearly o < ~ < I. We have X(KmE,+I)<~X(K) for every compo- x(u)" I-~--- ~x(~) nent K of E,. Summing over components of En, we get x(En+l)<~q.X(E,). Hence, by induction we obtain (2.3)- Now we shall prove (2.4). Since f(B)CB, we have B CE, for all n. For k=o, i, 2, ..., denote by Jk the component of Ekc~supp q~ containing a. One of the endpoints of Jk is a; denote the other one by a k. We have: n--1 (2.6) E, nsupp q~ =k__UoH k uJ, where Hk=E,n(Jk\Jk+l ). The endpoints offk(Jk+l) are fk(a)~B and fk(ak+l)~d. From this, using (2.2) (which is clearly true also for subintervals of K instead of K) and (2.5), we obtain: I a- ak+, I > sup [(fk),[ > sup I(fk)' [ -> 8 inf [(fu)' I " ak+l ak ak Consequently, since fk(Hk)Cfk(E,)CE,_k, we have: X(ff(Hk)) (2.7) ~ q0 dx < X(Hk) sup qo k -- ~k -- <infl(fk) '1 ta-a~§ ltk < X(E. k) ~-;8~<inf[ (if)' I) ~-* < ~"-~x(I) ~-~a~(~-*) k. Since we have: (2.8) fa ~dX fla-~tx ~ I dx= la--a.ll-~< (x(I))x-~ =J0 - i-~ i-~ (~-1).. Set ~=max(~q, ~-1) ~Lnd O=max~X(I)~ -;~, i_~ ]" Clearly, o < ~ < x and 02>0. Now (2.4) follows from (2.6), (2.7) and (2.8). 9 3" Estimates H In this section (and the next ones) I will be a closed interval, A a finite subset of I, containing its endpoints, and fl I\A-+I a continuous map, strictly monotone on components of I\A. 28 MICHAL MISIUREWICZ ~4 Sometimes (especially in later sections) we shall pretend that f is defined on the whole of I. When speaking aboutf"(x) for an x such that f" is in fact not defined at x, we shall mean a one-sided limit (it is usually clear which one). However, in order to be more rigorous, we introduce (and sometimes use) the following notations: = I � --}\{(left endpoint of I, --), (right endpoint of I, +)}, +)=x+, ag: I--.I given by f(x, e)=(x', g) /right if ~:+} and e'=~ if and where x' is a limit off(y) asy tends to x from the {left if /right if ~:~} neighbourhood ofx. If xeI only iff preserves the orientation in a (left if then Y is its first coordinate (i.e. (y+)V=(y_)V=y). We set A=Ic~(Ax{+,--}). If the reader becomes confused about the use of the above notations, he can'always think about the case off continuous and omit all --'s and v's. Now we make further assumptions on f. (i) fis of class C 3, (ii) f' + o, (iii) Sf<o. Consider f on a component of I\A. Set g We have: v'lf'l" I J'tv g' =-- ~g ~--v, ,, i[ I [f,,'~2 f'"f'--(f")q I +g j=-:gSf. Hence: : ]"' Sf 2 /l ,j This means that (iii) is equivalent to the following condition: (iii') -- is convex on components of I\A. ~/If'l From this it follows that one-sided limits off' at the elements of A exist. We shall denote them asf' at the corresponding points. Clearlyf' exists also outside of A. (iv) If fP(x)=x then (.f')'(x)>I. 2d ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL ~o 5 From (iv) it easily follows that f has no sinks. Indeed, if J is a subinterval of I and f" maps homeomorphically J into itself, then there exists a point xeJ� with f2"(x)=x and (f2n)'(X)<I. (V) There exists a neighbourhood U of A such that for every ae.~ and n>o, (f"(a)) v sa u (I\U). (vi) For every aeA there exist constants ~,~,o~>o and u>o such that: ~[x--~]"< lf'(x)l < ~lx-~[" ((a', Y+ a) if a=g+, for every xe/(~-- a , ~) if a=~_. Taking smaller e and larger ~ we can obtain the above inequalities on the whole corresponding component of I\A. Clearly (vi) is satisfied iff has non-zero one-sided derivatives (first, second or higher) at all elements of A. We make also two additional assumptions: (vii) ]f'l>i on I\U, (viii) if ae.~ is a periodic point forg~ then it is a fixed point forJ~ Lemma (3-x ). -- If f satisfies conditions (i)-(vi) then some iterate satisfies conditions (i)-(viii) (perhaps with a different set A). m--1 Pro@ -- Let m>I, 3~=f ", A = [.J f-k(A). It is easy to see that (i)-(vi) are k=O m--1 also satisfied by J~ A instead off, A. In (v) we take U= [.J f-k(U) instead of U. k=0 In (vi) we use a simple computation showing ;:hat if o~x~i<g$(x)<%x ~i for i-~I, 2, this is also true for i=3 with some 0~a, c%, u 3 where ga=glog2 (we can take: u3 = (U1-1- I) (u2-}- I)-- I ) . It is also clear that iff satisfies also (vii) (resp. (viii)) then so does 3~ Now it remains to show that iff satisfies (i)-(vi) then some iterate satisfies (vii) and some (perhaps an other) one satisfies (viii). But the first fact follows from Theorem .(I-3) and the second one is trivial (notice that an image of a periodic point is periodic). 9 ,, o~ Set: Ai={aeA:f(a)=a}, A~=A\A1, c,=Ulf~(X), C= _lLJfi(A), B=~. For a measurable function q0 on I we denote by q~X the measure which is absolutely continuous with respect to X and with the density (i.e. Radon-Nikodym derivative) % For a measure ~x and a map g, g*(~t) denotes the image of ~ under g, i.e. a measure such that for every measurable set E: (E) = (3.2) 4 26 MICHAL MISIUREWICZ For an absolutely continuous map g, g, denotes its Perron-Frobenius operator, i.e. for a measurable function V: (g,(~)) .X=g*(~.X). Notice that by (3.2), fo_l(~.)gdX=f~g,(9)dX. It is easy to check that we have (if g is differentiable): Proposition (3.2). -- If f satisfies (i)-(iii) then for every x~ I\~,: X(I) . < (3.3) f, (I) (x)_ dist (x, I~,)" n--1 Proof. -- Let J be a component of the set on which f" is defined (i.e. I\kU0f-k(A)).= The setf"(J) is an interval; denote it by (a, b). Set g =f" a" The map g is a homeo- morphism and Sg> o. By the formula for the Schwarzian derivative of a composition wehave o=S(id)=S(gog-1)=S(g-1)+((g-1)')~(Sgog-J). Hence S(g-~)>o. We have I I g,(~)_ Ig'l~ -t _ [(g-i),[. By (3.i) applied for g-l, the function W/~ is concave. Therefore the function A/](g-l) ' ] is convex, and consequently g,(i) is also convex (a reader not convinced by this argument may look at Lemma (4.3))- Thus: I fc X(g-t(a, x)) 9 /, g,(I)dX-- x--a x(J) g,(I)(X) ~ or < -- dist (x, ~,) I (T X(g--l(X, b)) ,~g*(I)dk-- b--x (because a, b~,). .-1 Summing over all components of I\kO0f-k(A),= we obtain (3-3). 9 Lemma (3.3). -- Let g :,(b, c)-~R be a map of class 121, g'>o. Let ~, co>o and u>o be such that o~(x--b)"<_g'(x)<co(x--b)" for every x~(b, c). Let o<~<I and let a function q~ : (b, c) ~R be given by a formula eO(x ) -= (x-- b) -~. Then for some constants &'>o, o<4<1 we have g,(9)(y)<3(y--g(b))-~ for every ye(g(b),g(c)). q~(x) ~ b) -~ =--~I (x-- b) -~-~ But: Proof. -- We have g,(9)(g(x))=g-~<. g(x)--g(b)= g'(t)dt co(t--b)"dt=r Pdt=--(x--b) ~+i, f/ r'-' ,o dO u n t- I o~(x--b)"(x-- ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 27 < I iu (g(x)--g(y))) i~-g. g*(qg)(Y)--~ t 2r-Icai \--~+u Thus we put ~ ~-r 1(~)~ =- . Clearly o<~<I, ~>o. 9 Now we assume that f satisfies (i)-(viii). By (v), BkA is disjoint from U. Since Sf<o and by the continuity off', there exist open intervals Ua, aeA, such that [JU~ aeA /left if a=a'+] endpoint is disjoint from B\A, ]f'l>c,>~ on I\,eAU^U, and dis the tright if a=Yj of U~. It is easy to see that we can also have lf'[~>I on U a if aEA 1. Lemma (3.4). -- For every aeA~ there exist constants y, ~>o and ~, ~e[o, I) such that: (3."t) f."(i)(x)<yIx--~l -~ for every n>o and xeU~, (3.5) (f u,).(f."(i))(/)<~ly--(f(~))tl -~ for every n>o, yef(Ua). Proof. -- By (viii) and the definition of A S we can assume that A~={al, ..., at} and no iterate off takes al to aj for j<_i. Then we use induction, proving first (3-4) for al, then (3.5) for al, then (3.4) for as, then (3.5) for as, etc. To prove (3.4) for a i we writef.n(I) in the form: =(: o). + (: where G= U Ua s, T={j:f(@=a,}. Then we obtain an estimation of the first jet summand from (3.5) for jeT, and the second one from Proposition (3.2) and the fact that by the definition ofUa, dist(B\A, U^U~)>o. We get a finite sum of expressions aEA of the form vlx- ,l (notice that a constant is also of this form for ~=o), and the sum is not greater than some function of the same form. Thus we obtain (3.4), but in general only for x from some semi-neighbourhood of ~ smaller than Uoi. Using once more Proposition (3-2) we obtain an estimation by a constant on the rest of U a . Again, the sum of a function of the form y ix--~ I -; and a constant is not greater than some function of the form v lx-a' l (3.5) for a i follows from (3.4) for a i and Lemma (3.3). 9 Then Lemma (3-5)-- Let HCI and Hk={x:fi(x)EH for i=o, ...,k--I}. for every s, m we have: 27 28 MICHAL MISIUREWICZ Pro@ -- Let Gk=f-~(Hk)\H. We have f-i(Hk)=GkwH1,+, , and by induction we get: s+m--1 f-m(H,)= U .f-'-'~+l+'~(Gk)uH,,+,~,. k=s Hence: fH s+m--1 fy(i)dX=X(f-m(H,))< Y~ X(f-'-'~+~+k(Gk))+X(U,+m) $ k~8 (s o). (j: + k = s dl(C,-k) < ~ f sup(f i\tt) (ft,'(l))d),.--~-),,(Hs+m). II --k= sdH k n>_O Lemma (3.6). -- For every aEA 1 and z>o there exists a neighbourhood W of ~ such that: J,~["~u f,n(I)dX<~ for every n2o. Proof. -- Let aeA 1. There exists a constant ~>i such that: (3.6) If(x)--~l>~lx--~] for all xeU a. The set Vk={x :fi(x)~U, for i=o, ..., k--i} is a neighbourhood of h" in U,, and by (3.6) we have: x(I) (3.7) x(v~)< ~k 9 Therefore it is enough to prove that: (3.8) supf f,'*(I)dX~ o as s~oo. m>OdV s First we write, as in the proof of Lemma (3.4): (f i\Ua),(f*n(I))=( f G),(fin(I))+( f i\(OUUa)), (f*n(I))' where G= U Ub, R={b~X\{a}:f(b)=a). The same arguments as in the proof of b~R Lemma (3.4) show that: sup(f[ ~ (f:(~))(y)<_~ly-al -~ n>Ok [ I\Ua]* for some constants 8>0, o~<i (notice that RCA2, so we can use (3.5)). In view of Lemma (3.5) (for H=U,; then Hk=Vk) and (3.7) we get: oo f;,(II X(I) supf f','(I)dx<kY~ ~ -~t-~dt+ ~,+----~ ,,,>_oJv, -- = .~o 5",-- | ' /X(I)~l-~-kX(I) ~ ~ (~_l)~t_. X(I) '--&O as s~oe. This proves (13.8). 9 28 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 29 Lemma (3.7). -- For every ~>o there exists a neighbourhood W of B\A such that: n~o. fwf.*(i)dX<~ for every Proof. -- If we put Aw O^U~ instead of U and B\A instead of B, then the hypo- aCA theses of Section 2 are satisfied. Let :E~ be defined as in Section 2. Clearly, En is a neighbourhood of B\A. Hence, it is enough to prove that: (3-9) sup~ f.m(I) d X---> o as s--> o. m~OdE s By Proposition (2.1) and Lemma (3-4) there exist constants ~, ~e(o, I) and 0>o such that X(E,)<~X(I) for all s>o and: f sup(f U U,)(f,n(I))d~.~(S-~-I)0~s for all s>o. JE Hence, by Lemma (3.5) (for H = I\ U^U~, H k = Ek) we obtain: affA supf f,m(I)dk~ ~ (s-]-I)0~8+k(I)~qs-->o re>odE s k=s as S~+ o0. This proves (3-9). 9 Proposition (3.8).- If f satisfies (i)- (vi) then for every ,>o there exists 8>o such that if GCI and X(G)<8 then fGf,~(I)dX<, for all n. Proof. -- Suppose first that f satisfies (i)-(viii). Since a function of the form q0(x)=ylx-h'[-~ (o<_~<I) is integrable at h', from Lemma (3.4) it follows that the statement of Lemma (3-6) holds also for all aeA2. This and Lemmata (3.6) and (3-7) imply that for every ~>o there exists a neighbourhood W of B such that: f f2( )dx< for all n. Now suppose that f satisfies (i)-(vi). By Lemma (3.i), there exists rn>i such thatf m satisfies (i)-(viii). In order to distinguish the sets A and B defined for different iterates off we shall use the symbols A(f k) and B(f k) for those sets defined forf k. There exists q>i such that the set on whichf k is defined (i.e. I\A(fk)) consists of at most q components for k---- o, I, ..., m-- 1. Fix ,>o. Since fm satisfies (i)-(viii), there exists an open neighbourhood W of B(f ~) such that: ff~n(i)dX<~ for all n2o. (3.I0) Jw Take: dist(B(fm), I\W) (3 ii) = 3x(I) 29 MICHAL MISIUREWICZ 3o Clearly ~q>o. There exist neighbourhoods U k of A(fk), k = o, I, ..., m-- i such that: (3. ~2) x(uD< ~. Take: (a.,a) ~-----~ min infl(fk)'[. qO<k<m--1 I\U k By OiL also a>o. Suppose now that GCI and X(G)<& Fix ke[o, m--l]. We want to estimate faf.~"+k(I)dX. Let J1, .-.,Jp be the components of I\A(fk). By the definition of q, we have p<q. We have: (3.x4) f.~"(I)dX dl-kia) ~fwf*mn(') d~" "~-fUk\Wf, mn(I) d~, AC ~ ~ fmn(I)d~.. i ffi 1 J l-k( o) ~ Ji\ (Uk k) W) By (3.II), (3.I~) and Proposition (3.2): ?,(I) (3" IS) fU f*~"(I)dX< k(Uk)" kxw -- dist (I\W, B(f=)) < - --3" By (3.II), (3.I2), (3.I3) and Proposition (3.2): (3. I6) x(I) i_~1 s j,\(vku w)f.~"(I)dx<p.x(f-k(G)~J,\Uk), dist (I\W, B(f'~)) X(G) X(I) q.3 X(I) --<q" inf I(f~)'l " dist(I\W, B(f~)) < inf (fk)' I dist(I\W, B(fm)) --3 i\Uk -~. Jikgk Now from (3-I4), (3- IO), (3. I5) and (3. I6) we obtain fGf,~"+~(,)ux<~. 9 Lemma (3.9).- If f satisfies (i)-(vi) then X(B)=o. Proof. -- By Lemma (3.i), there exists m~i such that f" satisfies (i)- (viii) . m--1 We may assume that A(fm)= U f-i(A(f)). Then: i=0 m m--1 C(f) c U f'(J,(f))= u Uof'(c(fm))..= m ra--'l The setiUtfi(A(f)) is finite, and hence it is enough to prove that X( U (f~(C(f")))V)=o. i=0 But: m--t m--1 U (f~(C(f")))vc U f~(B(f"))w{(f~(a))V:aeX(f), o<i<m}. i=0 i =0 In--J. We have X( U fi(B(fm)))=o because by Lemma (3.7) (for n=o), X(B(f"))=o. The set {(f~(a))V: aeA(f),o<i<m} is finite. 9 30 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 3 ~ 4" Functional spaces If f satisfies (i)-(vi) then the existence of an invariant probabilistic measure, absolutely continuous with respect to X, follows already from Proposition (3.8). It is [i.-I \o~ onou h to ta e we*" ofa sub que e of " owowr, we want n=1 to know something about ergodic properties of such measures, their densities and how many of them there exist. Thus we are going to establish several auxilliary facts. Let J be an open interval. We denote by 9,(J) the set consisting of all C" positive functions z on J such that ~ is concave, and the function o. Lemma (4. x ). -- An integrable non-zero function "~ belongs to ~2(J)/f and only if "~ =g,(I) for some C 3 diffeomorphism g:J*~J with non-positive Schwarzian derivative. Proof. -- Let J* be an open interval, g :J*-+J a C a diffeomorphism, v=g.(I). As in the proof of Proposition (3.2) we deduce that is concave if and only if Sg < o. This proves that if Sg<o then xe~(J). ~ Now assume that xe~2(J) , v+o and z is integrable. Let "~(x)=f:v(t)dt, where a is a left-hand endpoint of J. Set J*=~(J), g=~-l. Clearly, we have z=g.(I), g is a C 3 diffeomorphism and Sg~o. 9 Lemma (4.2).- If v, +~r(J) then -~++e~,(J). Proof. -- Take [a,b]CJ, t~(o, i), c=ta+(i--t)b. We want to show that (if ., +,o): i t i--t > -~ (4"x) %/v(c)++(c)--~c/;(al++(a) %/-~,(b) + +(b) Take affine functions p, o such that: I I I I (4.2) p(a)= %/x~' or(a) = %/qb(a) ' e(b) = ~v/+(b ) . P(b)= ~ ~/~(b ) ' Since % ~e~r(J), we have: I I "(c) <-V+(c)" ~(C)<V-(6, i.e.: Hence v(c)+~(c)<(~(c)).-~+(~(c)) -~, (4.3) ~(c)<_/,(~1 + +(~), where q(x)-- ~/(~(x))_2 + (~(x)) -2 31 3 2 MICHAL MISIUREWICZ Since p and a are affine functions, their derivatives are constants. Denote these derivatives by e and ~ respectively. We have: #(x) = - -~ (~(x))~. (- 2)(~,(p (x))-~+ 13(,,(x))-~) = (~(x)) ~ (~(~ (x))-~ + [X,,(x))- ~), r = 3 (~(x)) ~. (~o(x)) ~. (o~(p(x))-~+ t3(~(x))-~) ~ + (~ (x))~ (- 3) (~,~ (P (x))- ~+ t3%,(x)) - ~) ~,(p (x))-, + ~,(,,(x),)-, 1 = 3(~(x))~[ (<p(x))-% ~(~,(x))-~) ~ (~(x)) ~ J =- 3(~(x)? (~,(x))-~ (,Xx))-~(p;) ~x))~< o, i.e. the function q~ is concave on [a, b]. Hence, ?(x)>t~(a)+(i--t)~(b). By (4.2), I I ~(a)-v',(~) + +(~)' ~(b)= Together with (4.3), this gives (4. i). 9 %/v(b)++(b) Lemma (4-3). -- Every element of No(J) is a convex function. Pro@- Take [a,b]CJ, t~(o,i), c--ta+(I--t)b, zeN0(J)\{o}. We have: I t i--t I [(i ,) where q~(x) = ~(a) x/z~b): x + ~ " We have: ~0'(x) =--2(q~(x))3/2.( i ~) \Vz(a) %/ ' - 2 I I r = (,(dx)) __= >o (V'~(a) v~10):)~ and therefore the function q0 is convex. Thus: x(c)~(t)~t~(I)@(I--t)q~(o):tv(a)@(I--t)~(b). 9 Now we consider again a mapping f which satisfies (i)-(vi). Denote by ~r the set of all functions z on I\B such that v e~r(J) for all componentsJ of I\B. We shall consider on N 0 the topology of uniform convergence on compact subsets of I\B (shortly u.c.s, topology). Clearly, ~s N 0 for all r. 32 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 33 Lemma (4.4). -- The set ~o is closed in the space of all continuous functions on I\B with the u.c.s, topology. Proof. -- Let J be a component of I\B, %~N0(J) for n=I, 2,... and ~,-+~ uniformly on compact subsets of J. Suppose that x, yeJ, v(x)=o, "~(y)>o. Then I I I ~-+ 0% ~- -< + 0% and from the concavity of I %/%(y) ,V/~(y ) ~ it follows that -- ~ -- oo for all z belonging to the component of J\{y} which does not contain x. Since __ >o, this is impossible. Therefore either v = o on the whole of J or z I I i is positive and uniformly on compact subsets of J. But then ~vv is concave, i.e.-:e~o(J). ~-~-~ If we have a sequence of elements of ~0 convergent in the u.c.s, topology, then we apply the above argument to every component J of I\B and conclude that the limit function is also an element of ~0- 9 Proposition (4-5). -- Let p be a C ~ real function on I. Let H be the closure of the convex hull of the set {f,"(p)}~~ 0 in the u.c.s, topology. Then: (a) /f HC o; (b) every element of H is continuous on I\B; (c) H is compact; (d) H C LI(X) ; (e) the L 1 topology and the u.c.s, topology coincide on H. Proof. -- Suppose first that pe~ 2. By Lemmata (4. I) and (4.2), all functionsf,~(p) belong to ~2, and therefore to ~0. Hence, their convex combinations belong to ~0. Now (a) follows from Lemma (4.4). By Lemma (4.3), all functions f,"(p), and therefore also their convex combinations, are convex on the components of I\B. By Proposition (3.2), they are all bounded x(I) by the function ,(x)=supi P'dist(x, B)" Hence they are equicontinuous on compact subsets of I\B (cf. [I]). By Arzel~t-Ascoli's theorem, from every sequence of these functions we can choose a u.c.s, convergent subsequence. This proves (c) in the case Let now p be an arbitrary C 2 real function on I. If for a C 2 positive function -: on I we have -~" 3(v')2>o then =-- v -3/2 <o, and consequently zeN s. Therefore ~2 contains an open cone in the space of C 2 functions on I. This cone is non-empty, for instance because z(x) = ---, where a is a point to the left of I, x--a 88 MICHAL MISIUREWICZ belongs to it. Hence, there exist two C 2 functions on I, % +, which also belong to ~2 and such that ~--+--p. Thus, by (a), every convex combination of functions of the formf.n(p) is a difference of two elements of ~0, and therefore is continuous on I\B. Now (b) follows from the fact that the u.c.s, limit of continuous functions is continuous. If we have a sequence of convex combinations of functions of the form f.n(p), then we can write every function as a difference of convex combinations of functions of the form f."(q0) and f.n(~b), respectively. Since we know already that (c) holds for elements of ~2, we can find a convergent subsequence of our sequence. This proves (c). The function 9 is bounded on I and hence p~Ll(X). We have: (4.4) f [f,"(o)lgx=f uet~-"C x) [(f")'(Y)l p(y) dX(x) --dyer- (x)[(f")' (Y)[ and hence all functions f~"(P) (and consequently also their convex combinations) belong to LI(?,). Let now (%)n~~ be a sequence of convex combinations of functions of the formf.k(p), and let %-+a 0 in the u.c.s, topology. We shall show that %-+a0 also in the L 1 topo- logy. By Proposition (3.8) and Lemma (3.9), for every r there exists an open neighbourhood W of B such that: 3 sup [ P l for all n > o. (4-5) fwf:(I) ex< Since I\W is a compact subset of I\B, there exists n o such that: [e.--%ldX< -~ for all n>n o. (4.6) ,Ji \w 3 From (4.5), in view of Fatou's lemma, it follows that: (4"7) fw[~o[dX<lim~inffw[%] dx< ~-. -- 3 If n~no, then from (4-5), (4 .6) and (4.7) we obtain fl(~,~--%[d3.<~. Hence (~n---> (~0 in L x. Now (d) follows from (4-4) whereas (e) follows from the fact that the identity mapping from H with the u.c.s, topology onto H with the L ~ topology is continuous, and from (c). 9 Theorem (4.6). -- If f satisfies (i)-(vi), then there are no homtervats for f. Proof. -- Suppose that J c I is a homterval. Take a C 2 non-negative function p on I such that supppCJ, fpdk>o. By Proposition (4.5), there exists a sub- 3~ ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 35 fl i o~ U(X). Then sequence (2 (P)),=l convergent to some function ~ in f~dX=f~dX>o. But f has no sinks and hence by Lemma (i. i) all intervals f"~(J) are pairwise disjoint. Hence f, ni(p) converges to o pointwise--a contradiction. 9 Then there exists n > o such Lemma (4.7). -- Let UEI be an open non-empty set. that f"(U) contains some component of I\B. Pro@ -- By Theorem (4.6), there exists an interval (x,y) C U and k, m> o such that fk(x), fro(y) EA. Let n = max(k, m) + I. Take the maximal interval (x, z) disjoint n--1 from A(f ")= Uf-l(A). Clearly z~y and hence (x,z) CU. Butf"(x,z) is an i=0 interval and its endpoints belong to B. 9 5. Spectral Decomposition In this section we still assume that f satisfies (i)-(vi). We shall decompose the possible support of absolutely continuous invariant measures in a way in some sense similar to the spectral decomposition for Axiom A diffeomorphisms. However, the reader should remember that we are interested in absolutely continuous measures. Therefore a part of the non-wandering set (a Cantor set of measure o) can remain outside our set (i.e. the set U~'). Denote the set of all components ofI\B byof. Since f"(B) ff B, if for some J, LsJ the set fn(j) intersects L, then it contains L. Hence the reader can think about our system as a topological Markov chain with a countable number of states or as a walking on an oriented graph with a countable number of vertices. We define two relations on of: J~,L if and only if there exists n>o such that fn(j) DL, J g L if and only if there exists n> o such that f"(J) DJ u L. Let ~r be the set of all elements Jsof such that, for every Lsof, if J~L then L~J. Lemma (5. i). -- The relations ,~ and ~ are equivalence relations on ~. Proof. ~ io. f0(j)DJ=JuJ. 2 ~ . If J~L then L~J by the definition ofd/g. Let J~L. Then there exist k,n>_o such that fk(J) DJuL, f"(L) DJ. Hence f"+k(L) DLwJ, i.e. E~J. 3 ~ . If J~L and L~M then clearly J~M. Let J~L and J~M. Then there exist k,m,n~o such that f~(J) DJuL, f"(J) DJuM, f"(L) DJ. Hence f~+"+"(L) DLuM, i.e. L~M. 9 Lemma (5.2). -- The number of equivalence classes of the relation ,-~ is not greater than Card A- 2. 86 MICHAL MISIUREWICZ Proof. -- Let NCJ4 ~ be an equivalence class of the relation ~-~. We shall show that the set K---- [J ~ contains in its interior an element of A which is not an endpoint of I. Suppose that K does not contain such a point. Let J be an open interval contained in K\A. Since f is a homeomorphism on intervals disjoint from A and maps them onto intervals, and since f(K\A) C A, we obtain by induction that f"(J) is contained in K\A and f" a is a homeomorphism for every n>o. This contradicts Theorem (4.6). 9 Set 0f={ [J N : N is an equivalence class of ~ in ~,o}. We shall use the nota- tion fk(K) instead of the more precise (fk(~)) ~. Lemma (5.3). -- The set 2/~ is finite, and f maps elements of o,~ onto elements of ~. Moreover, for every KeJC there exists n>i such that fn(K)=K. Proof. -- Let J, L, M, Reef', J ~ L, f(J) 3 M, f(L) ~ R. Then there exist k, m>o such that fk(j) DjuL and fro(M) DJ. Hence, fk+"+l(M) DMuR, i.e. M~R. This shows that f maps elements of ~ into elements of ~. Let Ke~. In view of the definition of W there exists n> I such that f'(K) C K. Then [J fk(K) must be a closure of a union of all elements of some equivalence class of ~. Hence, f"(K)=K and alsofk(K), k=i, ..., n--i, are all elements of ~('. By Lemma (5.2), -~ff is finite. 9 Then there exists n>o such that Lemma (5.4)- -- Let KeJi" and let JeJ, JCK. f"(J) ~ K. Proof. -- By Theorem (I '3), there exists m_>I such that if L%I and the sets L, f(L),...,f"-l(L) are disjoint from A, then X(fm(L))>X(L). Using this argument repeatedly, we see that for n=o, i, 2, ..., the setf"m(J) is a union of a finite number of intervals, each not shorter than a=min/x(J), min X(fk(M))~>o. Mt~A~O 0 <k<ra Since JeJ and JCK, also JeW. Hence there exists k>I such that 9(J) 3J. If r>l then fk,,(,-t)(j)3j and thus 9m'(J)~fk~t(j). Hence (fk""(J))~__ 0 is an ascending sequence of sets. If L%~', L CK, then there exists n>o such that f"(J) DJwL and therefi)re also f,m~(j)D L. Consequently, we obtain the following situation: K is the closure of the union of an ascending sequence of sets (f~m'(J))~'=0, and every term of this sequence is the union of a finite number of intervals, each not shorter than ~. Since inside I there is enough room only for a finite number of such intervals, after some n o their number must stabilize (although a priori it can be smaller for fkm~(j)). As n(>n0) tends to infinity, these intervals can only become longer and n~0 36 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 37 longer. Denote the limit intervals by K1, ..., K,. If xn~K\fkm-(J) and x,-+~, then ~ must be an endpoint of a certain I~.. Let aeI be an endpoint of some K~ (a--h'+ if it is a left endpoint and a=~_ if it is a right one). We shall show that there exists a point b~I such that b is an interior point of a certain Kj, and fkm"(b)=a for a certain n>o. Suppose that this is not true. Then a is an element of a periodic fkm-orbit consisting of endpoints of Kj's and every element of this orbit has only one preimage (under f~m). Together with the fact that a is repelling, this implies that there exists a (one-sided) neighbourhood U of h" and an open non-empty set V C K such that fkm"(V) is disjoint from U for every n2o. But by Lemma (4.7), s~ contains an element of J. Since fk"(K) C K, this element is contained in K. Hence there exists r>o such that fk"~(V)D J, and consequently fkmn(V) intersects U for a certain n--a contradiction. Consequently, there exists ~(a)>o such that h'eKirafk"t(~)(J). Now if we take ~=max{t(a)'a is an endpoint of Ki; i=i,...,s}, we have K=fk"t(J). 9 Corollary (5.5). -- Each Keg( is a union of afinite number of intervals. Remark (5.6). -- Iff is also continuous then each Ke~C is an interval. Notice that in the case off continuous, the proof of Lemma (5.4) becomes much easier (each fkmn(j) is an interval). From Lemmata (4-7) and (5.4) it follows that: Proposition (5.7). -- If KE~{" and fk(K)=K, then for every open non-empty set UC i=0 ~ f-r there exists n2o such that f"~(U) DK. In particular, fk Ii is topologically exact. 6. Absolutely Continuous Invariant Measures Now we shall apply the results of Sections 4 and 5 to obtain the main results of the paper. As usual, f is a mapping satisfying (i)-(vi). By I[" [I we shall denote the norm in the space LI(X), and by Xe the characteristic function of a set M. Notice that by (4.4), the L 1 norm of the operator f, is not greater than i. Lemma (6. i). -- Let Keg(, fk(K)= K, M = 0 f-"k(K). Let ~ be a non-negative n=O continuous function on I\B such that o<f~0dX<+ oo, supp q~CK, J,~(~)=% Then for fM p dX every C3 function p on I, lirn zMf,"k(9)= q~dX?j in La(X) and u.c.s. 37 38 MICHAL MISIUREWICZ Proof. -- Let p be a C 2 function on I. Suppose that z~f,"ik(p)~ as i~oo, in the L 1 topology 9 Set: = --~ and +:~ --% +, z.s:, (o)_f.oex f.oex f~dx f~dx Since f-"ik(M)=M, we have: f+i dX =f f.~'k(p)dX q~ dx Jq) If nj--ni=l>o , then f, tk+i=+j. Clearly +i-++ in the L 1 topology. Suppose that + is not equal a.e. to o. By Proposition (4.5) we can assume that + is continuous on I\B. Then there are open non-empty sets U, V C M such that + u>O, + v<O. By Proposition (5.7), there exists to0>o such that fe'k(U)DK and ff.k(V) 3K. Hence, for every xaK\B and to>Q we have If, ek(+)(x)l<f.e~(l+l)(x ). Denote ~=ll+ll-IIf/~ . Since X(K)>o, we have: = f l+ldx-- f If, t~ = f (f,t~ (I + l) --I f,t~ (+)1) dx >o. If t>to o then I If~(+)[I = I lfY-t')~(f/'~(+))ll <_ I If'~(+)ll, and thus: (6.i) II+II-IIAtk(+)ll_>~ for all t~ . Henc e: But there exist i, j such that t=nj--ni>go, I[+,-+t[<- ~ and [I +5--+ll <-~o 2 2 l[ + l[- I I f, tk(+)II < l[ +-f, tk(+)I ] <_ ]l + - +~]l + I] +j-f, tk(+) ] I <11+- +~l[ + IIA'%- +)tl <~. This contradicts (6 i). Hence, +=% and consequently, zMf,~k(p) JMPdt 9 -+ q~ in the ? dX L 1 topology. Now the u.c.s, convergence follows from Proposition (4-5). 9 Theorem (6.2). -- Let f satisJ) (i)-(vi), and let KeJ~U. Then there exists kZI and a probability measure ~ZK, absolutely continuous with respect to the Lebesgue measure andfk-invariant such that: (a) fk(K) = K and the interiors of K and f~(K) are disjoint for i = I, . 9 k-- I, (b) supp ~K----- K, (4 ~-~ ~ ~o, (d) inf d~K>O, K\B d), 38 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 39 (e) /f M= 6 f-"k(K) and Z~.peLX(X) then lim 7~.f,"~(O)= pdX .-d~ in L*(X); n=O ~ --->- oO if additionally p is continuous on M then the convergence is also u.c.s., (f) the system . (K,f k K' ~'K) is exact, (g) ~K is the unique probability measure on K, absolutely continuous with respect to X and fk-invariant. Proof. -- Let k be the smallest positive integer such that fk(K)=K. By Lemma (5.3), such an integer exists. Now (a) follows from Lemma (5.3). By Proposition (4.5), the closed convex hull of {f, kn(I)}~=0 is compact in LI(X) and hence, by Markov-Kakutani theorem, it contains a fixed point ? off, k (if the reader does not want to use strong theorems, he can take instead a point of condensation of the sequence ~0f*k'(I) . We set pts=~-----.X. Clearly ~(I)= I. Since / ) *= n=l JK q~ dX if(K) = K, we have i.e. ~K is fk-invariant. .x- f?(v).x= fK dx From the definition of ~K it follows that supp aK C K. Proposition (4.5) implies (c). Since ~K(K) =I, inaf~xK>o for a certain component J of K\B. By Proposition (5.7), is bounded and consequently there exists n>o such that f"k(J) DK. But [f'] inf d~K>o. t"k(s) dX This proves (d), and also ends the proof of (b). and let p be a function on I such that We shall prove (e). Let M = 6 f-"~(K), n=0 --~ L[p--~]dX< ~ By 7~. o~LX(X) 9 Let ~?>o. There exists a C 2 function + such that Lemma (6. i) there exists n o such that for all n> n o we have: ~.f,~k(+)_ +dx dX <-'3 If n>n o then: % [I z~.f,"~(v-- +))l] + [[ This proves the convergence in LI(X). 89 MICHAL MISIUREWICZ 4o Let now p be also continuous. Let ~>o and let E be a compact subset of M~B. db~K Set ~ = sup Clearly ~<-k oo. Hence there exists a C 2 function + such that E --~-" sup [ p -- ~b [ < ~ where 8 -- By Lemma (6. I) there exists n o such that if 2(X(M) q- I)~" n > n o then: f?n(i ) d~K (f.~) d~K --X(M) dZ ~i on E and f, kn(+)__ +d)~ d~k ~ on n. If n>n o then: [f,~(~__+) [ ~ f, km([ ~__+ l)--~ ~f?n( I ) ~(t(M)~--~K-~ I)~__~ (x(M)~ +~)a on E, and consequently: dX--fp dX d~K JM dX dx +f~ < (X(M) ~ -k I ) ~ -}- a --~ 8X(M) ~ = ~. This proves the u.c.s, convergence. We shall prove (f). Suppose that the system , (K,f k K' btK) is not exact. Then there existsaset ECK such that E=Knf-"k(f"k(g)) for all n>o and o<~tK(E)<~. Then for every n>o we have f,"k(7~)--=o outsidef"k(E) and hence: A"k(ZE) - x(L) = X(E) b~x(K\f "k (E)) = X(E) ~-K(K\E)> o. This contradicts (e). Hence, the system , (K,f k K' ~t is exact. At last, (g) follows from (e). 9 From Theorem (6.2), Lemma (5.3) and Lemma (5.2) we obtain easily Theorem (6.3). --- Let f satisfy (i)-(vi). Then there exist probability f-invariant measures ~1, ..., ~8, absolutely continuous with respect to the Lebesgue measure, and a positive integer k such that: (a) supp Exi=[.J~ for certain equivalence classes ~ of the relation ~, i=i, ..., s, (b) supp ~t i n supp ~j is a finite set if i W-j, (c) I<s<CardA--2, (d) ~t~ /s ergodic, i=I, ..., s, (e) ~dNe~o, i=I, ...,s, #0 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 4t (f) mf over the set supp ~i\B is positive, i = I, ..., s, (g) /f peLl(X)then ,lim~co E ~=l 1~162 in Ll(X), where ~i= r-"(~.pp~) p dX; n=0 if moreover p is continuous, then the convergence is also u.c.s., (h) for every finite Borel measure v, absolutely continuous with respect to X and f-invariant, one has ~ = ~ ~i~ti, where ,r 6 f-"(supp ix,)). i=1 n=0 7" Examples We are going to show that for a large class of one-parameter families of mappings (looking like 4ex(I--X)) conditions (i)-(vi) are satisfied for a set of parameters of power the continuum. Let F: [o, i]� be a continuous mapping. For ee[o, I] we denote by f~ : I->I the mapping given by f~(x)=F(cq x). Let I =[a0, ai]. We assume that: (a) f~(ao)=f~(al)=a o for all c~e[o, i], (b) fo(x)=ao for all xeI, (c) for every 0~e(o, i] there exists c~eI such that f~ is strictly increasing on [a0, c~] and strictly decreasing on [c~, al], (d) fi(q) = ai. Lemma (7.I).- The mapping c~c~ is continuous. -~ >o such that Proof. -- Fix ee(o, I] and ~>o. By (c) there exists (7.x) if f~(x)>f~(c~)--~ then Ix--c~l<~. The mapping F is continuous and [o, I]� is compact. Therefore F is uniformly continuous. Hence, there exists 8>0 such that: for all xeI. (7.2) if Icr ~[<~ then Is <- Let By (7.2) we have: and by (7.i), [c~--c~]<,. 9 Now define %=sup{o~:f~(c~,)<_c~,}. Lemma (7. m ). -- % < I . Proof. J By (a) and (d), we have q<a 1. The mapping a~f,(c~)--c~ is conti- nuous in view of Lemma (7. I). Hence the set {~ :f~(c~)>c~} is open and contains I. I f*~+:(P)=~__ 42 MICHAL MISIUREWICZ Lemma (7.3)- -- For every .E(.o, I] there exists exactly one b~(c,, at) and exactly one b" ~(ao, c~) such that f~(b')~f,(b~)=b~. Proof. -- Let "~(~o, I]. Then f~(c~)--c~>o, f~(al)--al<o and therefore there exists b~(c~, al) such that f~(b~)=b~. It is unique becausef~ is decreasing on (Q, al). Now f~(ao)<b~, f~(c~)>f~(b~)~b~ and therefore there exists b'(ao, c~) such that f,(b'~)=b~. It is unique because f~ is strictly monotone on (a0, c~). 9 Now we must make a new assumption on f: (e) there exists a neighbourhood U of the set {(e, c~) : eE(o, i]} in (o, i] � I such that for every ee(o, I], U, ={x: (~, x)EU} is an interval and f~ satisfies the Lipschitz condition with the constant "V/2 on U~. Lemma (7-4:). -- There exists ~>~o such that f~(c~)> b'. Pro@ -- By (e), there exists ~>o such that if ]e--%l<z thenf~ satisfies the Lipschitz condition with the constant -V/2 on the interval [c~--z, Q-t-z]. By Lem- mata (7.i) and (7.2), there exists e>% such that [e--%]<z and Thenf~ satisfies the Lipschitz condition with the constant ~r on the interval [c~,f~(Q)] and b~, belongs to this interval. Hence: (7.3) ~ [f~(c~)--b~l _~-<'V~ lf~(c~)--b,l _< b~--c~,. If b~-- c~ > c~ -- b~, then o < c~ -- b" <f~ (c~)-- c~ < z and f~ satisfies the Lipschitz condition with the constant ~ also on the interval [b'~, c~], and hence also: (7-4) -lf~(c,)--b~t<_c~, -b'. But if b~--c~<c~--b'~ then (7.4) follows from (7.3). Summing (7 9 3) and (7.4) we get If~(c~)--b~[<b~--b'~, and therefore f~2(c~)L> b~.' 9 Now we define el=sup{e" 2 .f~(Q)>b~}. By Lemma (7.4), we have ~t>~o. n oo Let ~=(~( )),=0 be a o-i sequence. Let D o [b'~,c~], D~=[c~,b,] (for ~>~o)" Lemma (7.5). -- There exists a descending sequence (Pk)k~o of closed intervals contained in [el, I], such that: (7.5) f~+2"(c,)ED~(") for n=o, i, ...,k and for every eePk, (7.6) there exist ~(k), v(k)EPk such that 3+2k f~(k) (c~(k)) = c~(k) f~ck~ (%kl)= bv(k) /f ~(k)=o 3+2k { ' b,(k) /3" ~(k)=i. 42 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 43 Proof. -- We shall use induction. The functions e~b'~, 0~c~, e~b~ and ~/~(c~) are continuous on [el, i] f~(c~) = b~, f~a(q)< b 1. Hence it is obvious that there exists and we have b'~<c~<b~, ~ an interval P0 satisfying (7-5) and such that its endpoints (as ~(o) and y(o)) satisfy (7.6). Suppose now that we have found already Pm-t such that (7.5) and (7.6) hold for k=m--~. The function o~f~+~m(c~) is continuous. We have by (7.6): f~+~_'~l(c.:(,,_ll)=b(,,_~l and f~+ z'~)(c~(m_ll )=f~m_l)(c~(m_ll)< b;(m_~l. Hence, as before, there exists an interval PmCPm_I satisfying (7.5) for k=m and such that its endpoints (as ~(m) and y(m)) satisfy (7.6) for k=m. [] From Lemma (7.5) it follows immediately that: Proposition (7.6). There exists ~(~)e[0~, I] such that c3+2~r~ ~r~(,~l for n~o, I~ 2, ... Proposition (7-7). -- There is no homterval joining c~(~) with a periodic point of f~(~). Pro@ -- Denote f=f~(~l, c = c~(~), b = b~(~l , b' -- b'~(;l. Suppose that J is a homterval joining c and p and f"(p) =p. Then f2,(p) =p and f2,, preserves an orienta- tion at p (and hence on J). We may assume that 2n> 4 (otherwise take 4 instead of 2n). Then, by the definition of ~(~), f2"(c)ef[b', b], and hence f2"(c)>b. Clearly p+c. If p>c then p =f2"(p)>f2"(c)2b, and consequently beJ. Then b =f2"(b)>f2"(c)2b- a contradiction. If p<c then c<b~f2"(c), i.e. cef2"(J). This contradicts the assumption that J is a homterval. [] Proposition (7.8). -- The point c~(~) does not belong to the closure of the set {f~)(%r ~ Proof. ~ We use the same notation as in the preceeding proof. If ~(~)----el then "n C oo f3(c)=b and {J ()},=l={f(c), b', b}~c. Consider the case e(~)>~l. We have f2(c)<b', and hence there exists an open interval V~c such that if xeV then f2(x)<b'. Thus f"(c)r for n odd. For n even and greater than 2 we have f~'(c)ef([b', b])C [b, al]. For n----2, f"(c)<b'. Conse- quently, since VC(b',b), fn(c)r for all n>i. [] Theorem (7-9). -- Let the family {f~,}r e [o, 11 of mappings of I = [ao, al] into itself satisfy. the following conditions: (I) (~, x),f~(x) is continuous, (2) f~ is of class C3 for every c~e[o, i], (3) (~,x)~f2(x) is continuous, (4) f2'(x)<o for every ~(o, I], xeI, (5) Sf~<o for every 0~e[o, i], MICHAL MISIUREWICZ (6) f~(ao)=fe(al)= a o for every q, (7) supfo = ao, (8) sup fl =a 1. Then there exists a set 0 C [o, I] of power the continuum such that f~ satisfies conditions (i)-(vi) for every ~0. Proof. -- Clearly, our family satisfies conditions (a)-(d). Fix ~e(o, I]. By (3), there exists ~>o such that if ] e-- ~ { < ~ and [x-- c~ [ < ~ then [f~'(x){ < %/~. But there exists 8 >o such that if [ 0~-- } [ < 8 then [c=-- c~ [ < ~. Hence, if ] e-- }[ < min(8, ~) and Ix-c l< then This proves (e). Now we take a sequence ~ and e(~) obtained in Proposition (7.6). We shall prove that f~(~) satisfies (i)-(vi). As before, we denote f=f~(~), c =c=(~). Conditions (i)-(iii) are satisfied by (2), (4) and (5) (notice that (ii) means that f'(x)#o at xEI\{a0, al, c}). Suppose that (iv) is not satisfied. Then there exists a periodic point p of period n such that ] (fn), (p) l< x" If p=a 0 then by (4), f(x)<x for all x>a o. By (5), [(f~")'] has no positive strict local minima and hence there exists an open interval J such that p is one of the endpoints of J and o< ](f2,),[ <I. We can take as J a maximal such interval. Then the other endpoint of J is either an endpoint of I or a point at which (f2,), is o. In the first case we obtain a contradiction, because f2,(j)cj and consequently I(f2k")'(x)[<_I for every k at the endpoint x of J. But since f'(a0)>I , no image of x is %. In the second case, some image of J is a homterval joining c with a periodic point. This contradicts Pioposition (7.7). Hence (iv) holds. Condition (v) follows from Proposition (7.8). Condition (vi) follows from the fact that f'(ao)#O , f'(al)#o , and f"(c)#o (we take u=o foI a 0 and al and u=I for c). m 8. Entropy We still assume that f satisfies (i)-(vi). Iff is also continuous then the topological entropy off is equal to the topological entropy of the corresponding symbolic system (see e.g. [6]). Here we use for coding the partition into components of I\A. There- fore, in the case off not necessarily continuous we can define h(f) simply as the topo- logical entropy of the corresponding symbolic system (cf. [8]). It is easy to see that both systems are conjugate to each other after removing a countable number of points from both spaces. Hence there is a one-to-one correspondence between probabilistic invariant non-atomic measures, and thus the topological entropy so defined is equal to the supremum of metric entropies. 44 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 45 The goat of this section is to give necessary conditions for an absolutely continuous measure to be a measure with maximal entropy. The first reason why an absolutely continuous measure may be not a measure with maximal entropy, is that it often happens that h(f,u,,]<h(f)',us This reason was pointed out by J. Guckenheimer. I k--1 Notice that if Kegg', fk(K)=K and ~,_~0f*(~K).= is a measure with maximal entropy for f, then ~: is a measure with maximal entropy for fk. Theorem (8. I). -- Let f satisfy (i)-(vi), KeJ{', fk(K)=K. If the absolutely continuous measure ~K is a measure with maximal entropy for f k K' then for every periodic point x~K\B of period n for fk, ](f,~k),(x)]=~3,, where h(f k K)=log [L Proof. -- In view of Corollary (5-5), K is a union of a finite number of intervals. Since K is fk-invariant, we can " cut out " the gaps between them and we obtain a piecewise continuous mapping of an interval onto itself. By Proposition (5- 7), f K is strongly transitive (see [8]). By Theorem (6.3)(f), h(f k K)> h~K(f k K)>O. Hence, by the theorem of Parry [8], fk I is conjugate to a piecewise linear mapping g such that Ig'h--~ (in the case off continuous, this follows also from [4]). Denote this conjugacy by e. Clearly, g satisfies (i)-(vi), and it has only one set in its " spectral decompo- sition "--the whole interval. Therefore there exists a unique g-invariant probabilistic d~ measure v, absolutely continuous with respect to X. Clearly ~ o ~ is continuous on K\B. Denote by � the measure (~-~)*(,~), and by + and q~ the measure theoretical jacobians of ~z K and � respectively (see [7]). We have: (8. x) d? -= \ dX of . l(fk)'l d~XK dZ (dv ) d,Joaof k (8.2) d,, o =13 --O(Y dx dx For every probabilistic f~-invariant measure ~ on K we have, in view of (8.2): 45 46 MICHAL MISIUREWIGZ In view of Theorem (4.6), our systems have one-sided generators, and hence (see [7]) the measure-theoretic entropy is equal to the integral of the logarithm of the measure theoretic jacobian. Thus, by (8.3) we get: (8.4) h~(fk)--h~.(f~)=flog ~ d~K--flog q0d� =(f iog + dv. :--log +(log log v log If a function 0 belongs to LI(~zK) then for the conditional expectation of p with respect to the inverse image of the whole ~-field under f~, we have the formula: E~(~ {f-~(N))(x) = Z 0(Y) e t-kltk( )) + (y)" Hence: = 51 ! = o. Assume now that VtK is a measure with maximal entropy forfk I and that x~K\B,, / K' f'n(x)=x. By (8.4) and (8.5), we have q~-~+ ~tl~-almost everywhere. The whole trajectory of x (under f ~) is disjoint from B and hence, in view of (8.i), (8.2) and Theorem (6.2) (c), q~ and + are continuous in some neighboulhood of the trajectory of x. Thus, they are equal on the whole trajectory of x. Therefore by (8. I) and (8.2) we obtain: n--1 n--1 n--1 I(f"~)'(x)[ = 1-I I(f~)'(fr = [| +(fi~(x))=.II ~?(f'k(x))=~". 9 9- Entropy for quadratic maps The best known (and easiest for computations) family of maps satisfying the hypotheses of Theorem (7.9) is the family of maps f~ : [o, I] --~ [o, i] given by the quadratic polynomials f~(x)=4ex(I--X ). Comparison of the graphs of topological entropy [4] and the characteristic exponents [9] of these maps suggests that the only case when the metric and topological entropies are equal is e = i. In this case they are clearly equal since fl is smoothly conjugate to the piecewise linear map with slope i 2. It often happens that the topological entropy on the support of the absolutely continuous measure is smaller that the topological etttropy on the whole interval. As J. Guckenheimer pointed out, this effect takes place e.g. when there is an interval J c~(~I-~ invariant underf~ and with o-th, ~-st and 2-nd images disjoint. containing \ 21 46 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 47 Then the absolutely continuous measure is supported by Jt3f~(J)uj~(J) (or even a smaller set). Butwe have then h(f~ a)<~ l~176 --2 For our family this happens for ee , o.9642oo... (of course, we consider only those e, for which an absolutely continuous invariant measure exists). This effect is unavoidable for such families. However, the question remains, what happens in the case when the support of an absolutely continuous measure is an interval. I { ,(i)}o Then the probabi- We are interested in the maps f, for which ~ r f~ ~ ,=1 listic invariant absolutely continuous measure is unique (since Card A=3). We shall denote this measure by [x~. We shall use the notations of Section 7 (remember I ) that c~=- for all 0~ . Proposition (9. I ). -- The support of ~(~) is an interval. bl t Pro@ -- Let f=f,(~), b = b~l~) , = b~(~), ~ = ~-(~1" In the proof of Propo- sition (7.8) we can take as V the maximal interval with the given properties. We have VC supp ~x and hence f3(V)C supp Ix. But f2(V)~b' and hence f~(V)~b. Since there are no homtervals, there exists n such that f~+3(V)~I. Therefore: 2 I . I But this interval is invariant, and hence it must be equal to supp ix. 9 We are going to show that ~(r is not a measure with maximal entropy for an uncountable set of sequences ~. To simplify the computations we will work first with the equivalent family of maps g~, where g~(x)=x2--~, ~e[o, 2]. It is easy to check [ I -}-'V/I -}-4~ I + ~/I--@~] that g0 maps the interval , - into itself. Lemma (9.2). -- Let o < y < ~ < 2 be such that gv has no periodic point of prime period 3 but g8 does. Then u Proof. -- Denote W~(x)=g~(x)--x=x2--x--~. It is easy to check that: g~(x) - x =W~(x). pdx) (9.I) where P~(x) =W~(x). (W~(x) + 2x + ~)2 + 2x(W.(x) + 2x + t) + (9.2) = x6 + x5 + (I-- 3{~)x4 + (I-- 2L3)x3 + (I-- 3L3 + 3L3~)x ~ 47 48 MICHAL MISIUREWICZ From (9.2) it follows that if W~(x) = o then: P~(x)= 2x(2x--~ I ) + I= 4Xz @ 2x @ I = 3X2 @ (x-~- I )2>o. Therefore x is a periodic point of prime peiiod 3 if and only if P~(x) = o. i -t-'V/i -t-4~ then g~(x)=x and hence P~(x)>o. If x= then P~(x) -g~(x)-x -- -- I >0. Hence, there exists ~e(y, 3] and: g~ (x) - x yG such that P~(y)=o and P'~(y)=o. We also have P~(g~(y))----P~(g~(y))---- o and the points y, &,(y), g~(y) are distinct. We have: (g~)'(y)---I = (g~(x) -- X)' x=y = (W~, P~)' (y) = o and hence also (g~)' (&,(y) )=(g~)' (g~( y) )= :. Thus: (W,. P',)(&,(y)) = (g~)' (g~(y) )-- :--(W'. P,)(&,(y))= o, and analogically (W~.P'~)(g~(y))=o. Consequently, P'~(g,(y))----P~(g~(y))' 2 =o. Hence, the polynomial P, has three double zeros. Since its degree is 6, it is a square of a certain polynomial Q of degree 3. The coefficient of x 6 in P~ is i and therefore we may assume that also the coefficient of x 3 in Q is i. Let: Q(x) = x 3 -k- ~2x 2 + ~:x § ~0. Comparing coefficients of x 5, x 4, x 3 and x in P~ and O~ we obtain: 2~ 2 = i \ 2~: 2 I ~.0~=I -2~+~ ~ ) I 3 3 ~2 5 : From the first three equations we obtain successively: ~2 = ~, ~l = g-- 2 e' 16 4 Then from the fourth equation we get ~ 74 +49~4=o, i.e. ~=-'47 9 Proposition (9.3). -- g~ has a periodic point of prime period 3 if and only if ~ > 7. X 7 -- I Proof. -- We get from (9.e): Po(x)=xG+xS+x4+x3+x~+x+I - Since X--I P0 has no real zeros, go has no periodic points of prime period 3. If I 1= i, then 2 Re(z2)=(2 Re z)2--2 =g2(2 Re z). Hence, if we take as z a complex root of : of degree 7, then 2 Re z will be a periodic point of g2 of prime period 3. ~8 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL 49 Now, if ~>7 then g~ has a periodic point of prime period 3 by Lemma (9.2) applied to y = ~, ~ = 2; if ~ < 7 then g~ has no periodic point of prime period 3 by Lemma (9.2) applied to y=o, 8=~. 9 Since g~ is (linearily) conjugate to f~, where e= x--(I-}-%/~), we obtain Corollary (9.4). -- f ~ has a periodic point of prime period 3 if and only if ~> ~ ( ~ + 2 ~v/ 2 ) . Lemma (9.5)-- If ~(~)<4(~-}-2%/~) then h~(~)(f~(~t)<log Pro@ -- Let f=f~(~), ~----~(~). By Proposition (9-~), supp ~ is an interval. Hence (see [8]), fl~uv,~ is conjugate to a piecewise linear map f~with Iffl =[~, where h(fls~ ~v) =log ~. It is easy to see that ?is linearily conjugate to the map g given by {~x if x<i g(x)= [~(2--x) if x>I Suppose that h~(f)> log -- Then on some subinterval of [o, 2], containing i. I -}- "~/-5 22 222 is a periodic trajectory of prime ~> log-- The trajectory 2 I+~ 3' I+~ 3' I+~ 3 2~ and period 3. In order to see this, it is enough to notice that o< ---2-~ ~<I I+~ ~ I+~ ..... 2~ 3 This i < ~ < ~ = g(i). Hence f has also a periodic trajectory of prime period 3- contradicts Corollary (9.4). 9 Theorem (9.6). -- For the family {f~}~e[0,1], f~(x)-=-4ex(I--x), f~ : [o, i] "--> [O, I], there exists a set AC 13 of power the continuum such that if eeA then the absolutely continuous measure is not a measure with maximal entropy. Proof. -- Let A be the set of all e(~) such that there exists k with the property that all blocks of i's appearing in ~ are shorter than k. ( i) b'=b~ remember that -=c~ . Since Let etA, f=f~, ~=~, b=b~, f3+2n(b)eD~ t for all n, Suppose that tz is a measure with maximal then b does not belong to fn ~=1" We entropy. Then, by Theorem (8.I), we have if'(b)[=~, where log ~=h~(f). have 4eb(i--b)=b, b+o and thus 4~(I--b)=I, i.e. b=i----. Then 4 e (9.3) ~= [f'(b)[ =14e--abe 1=14~-8~+2[=4~-2. 49 MICHAL MISIUREWICZ 5 ~ I-J-~; Suppose first that c~<I(i +2A/i). By Lemma (9-5) we have ~<--, and hence by (9.3): (9.4) ~< 5-k%/5. We have c~ = c~(~)2 ~ and f~l = b'~. Since: b'~ = I -- b~ -- and f =l (-0 = - 11, 4~ 5+%/5 Since el is a zero of a polynomial Q(x)=i6x4-i6x3+i. By (9-4), 51<---- Q(5 +8~-) - 7- 32 5%/5 < ~ Q(I) we must have =o and Q'(x)=-i6x2(4x-3), i i el=-. But =- and hence -<s 0. Consequently, ~Xl~Z 0. This contradicts 2 2 2-- Lemma (7.4). Suppose now that c~2I(i+2%/]). Then, by (9.3): (9.5) ~__~ 2%/2-- I. [(:)] [i] We have log ~=h~(f)<h(f). Take the intervals Jl= f2 , b' , J2= b',-~ , r 1 r / \~ L A L ~ /J is contained in {o}ui__01J i. The mapf is monotone for 7=I, 2, 3, 4, and therefore li h(f) is equal to the topological entropy of the symbolic system obtained by coding with respect to {J~}~=t (see e.g. [6]). We have f(J1)cJ=wJ~, f(J=)c J4, f(J~)c J,, f(J4) cJ1uJ~uJ3. Hence h(f) is not larger than the topological entropy of the topo- (i I I IO;t logical Markov chain with the transition matrix o o . The characteristic poly- O O I I nomial of this matrix is x.P(x), where P(x)=x3--2x--2. Hence ~ is not larger than the largest zero of P. We have: and 60 ABSOLUTELY CONTINUOUS MEASURES FOR CERTAIN MAPS OF AN INTERVAL By (9"5), ~-~2"~/~--I>J~" Hence P(2,V/2--I)<O. But: P(2%/~-- I) = 18"V/~-- 25 >o, a contradiction. 9 REFERENCES [I] N. BOURBAKI, Fonctions d'une variable rdelle (Livre IV), Paris, Hermann, x958 (chap. I, w 4, exerc. Ia). [2] J. GUCKENHEI~mR, Sensitive dependence to initial conditions for one dimensional maps, preprint I.H.E.S. (x979). [3] M. JAKOBSON, Topological and metric properties of one-dlmensional endomorphisms, Dokl. Akad. Nauh SSSR, 9.43 (I978), 866-869 (in Russian). [4] J. MILNOR, W. THURSTON, On iterated maps of the interval, preprint. [5] M. MISlUR~WlCZ, Structure of mappings of an interval with zero entropy, Publ. Math. LH.E.S., 53 (I98I)~ 000-000. [6] M. MISIUREWIGZ~ W. SZLENK, Entropy of piecewise monotone mappings, Ast~risque, 50 (I977) , 299-3io (full version will appear in Studia xkIath., 6'/). [7] W. PARRY, Entropy and generators in ergodic theory, New York, Benjamin, I969. [8] W. PARRY, Symbolic dynamics and transformations of the unit interval, Trans. Amer. Math. Soc., 19.2 (1966), 368-378. [9] R. S~Aw, Strange attractors, chaotic behavior and information flow, preprint, Santa Cruz (I978). [IO] D. SINO~R, Stable orbits and bifurcation of maps of the interval, SIAM J. Appl. Math., 85 (1978), 26o-267. Instytut Matematyki, Uniwersytet Warszawski, PKiN IX p, oo-9oi Warszawa, Poland Manuscrit refu le 3 ~ juin I979.
Publications mathématiques de l'IHÉS – Springer Journals
Published: Aug 30, 2007
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