The Frobenius{Perron Operator

(p.75) CHPTER 4 The Frobenius{Perron Operator The hero of this book is the Frobenius{Perron operator. With this powerful tool we shall study absolutely continuous invariant measures, their existence and properties. This operator was rst introducedby [Kuzmin, 98ab] and describes the e ect of the transformation ona probability density function. 4.Motivation Let X be a random variable on the space = [a;b] having the probability density function f. Then, for any measurable set ½, ProbfX g = fd ; where is the normalized Lebesgue measure on. Let :! be a transformation. Then (X) is also a random variable and it is reasonable to ask: What is the probability density function of (X)? We write Probf (X) g = ProbfX ()g = fd : To obtain a probability density function for (X), we have to write this last integral as Ád ; for some function Á. Obviously, if such a Á exists, it will depend both on f and on the transformation. Let us assume that is non-singular and de ne ¹() = fd ; where f L and is an arbitrary measurable set. Since is nonsingular, () = 0 implies ( ) = 0, which in turn implies that ¹() = 0.

76 4 The Frobenius{Perron Operator Hence ¹ <<. Then, by the Radon{Nikodym Theorem, there exists a Á L such that for all measurable sets, ¹() = Ád : Á is unique a.e., and depends on and f. Set P f = Á. Thus, the probability density function f has been transformed to a new probability density function P f. P obviously depends on the transformation (hence the subscript) and is an operator from the space of probability density functions on into itself. P is referred to as the Frobenius{ Perron operator associated with. t is the major tool used in this book and will make it possible to prove the existence of absolutely continuous invariant measures and to establish many useful properties of these measures and their densities. Since f L ;P f L. Hence P : L! L is a well-de ned operator. f we let = [a;x] ½, we have x P fd = fd : a [a;x] On di erentiating both sides with respect to x, we obtain P f(x) = d fd : dx [a;x] Example 4... Let = [0;] and let :! be de ned by t is easy to see that sin(¼x); 0 x : ([0;x]) = [0; ¼ sin x] [ [ ¼ sin x; ]; 0 x : Hence, for any f L, fd = [0;x] 0 ¼ sin x fd + fd ; 0 x : ¼ sin x Using Leibniz's rule, we obtain P f = d fd dx [0;x] = ¼ p x [f( ¼ sin x) + f( ¼ sin x)]:

4. Motivation 77 Example 4... Let = [0;] and let :! be de ned by ½ x; 0 x ; 4 3 x + 5 3 ; < x ; as shown in Figure 4... t is easy to see that Figure4.. and ([0;x]) = [0; x]; if x < 3 ([0; x]) = [0; x] [ [5 4 3 x;]; if x 4 3 : Thus, we have ([0;x]) = [0; x] [ f[5 4 3 x;] \ Bg; 0 x ; 4 where B = [ ;]. Hence, for any f L, x fd = fd + fâ B d ; 0 x : 5 [0;x] 0 4 3 4 x Using Leibniz's rule, we obtain P f(x) = f(x ) + 3 4 f(5 4 3 4 x)â J(x); (4..)

78 4 The Frobenius{Perron Operator where J = (B) = [ ;]. n Section 4.3, we shall derive a formula for 3 P f which generalizes (4..). n Chapter 5 we shall use the Frobenius{Perron operator to establish the existence of absolutely continuous invariant measures for a large and important class of transformations. Now we present some properties of the operator. 4. Properties of the Frobenius{Perron Operator n this section we de ne and present the basic properties of the Frobenius{Perron operator. We do it formally on an interval = [a;b], but all the ensuing results can be easily extended to a general measure space case. De nition 4... Let = [a; b], B be the Borel ¾-algebra of subsets of and let denote the normalized Lebesgue measure on. Let :! be a nonsingular transformation. We de ne the Frobenius{Perron operator P : L! L as follows: For any f L, P f is the unique (up to a.e. equivalence) function in L such that P fd = fd for any B. () The validity of this de nition, i.e., the existence and the uniqueness of P f, follows by the Radon{Nikodym Theorem. P : L! L is a linear opera- Proposition 4... (Linearity) tor. Proof. Let ½ be measurable and let ; be constants. Then, if f;g L, P ( f + g)d = ( f + g)d = fd + gd = P fd + P gd = ( P f + P g)d :

4. Properties ofthe Frobenius{Perron Operator 79 Since this is true for any measurable set, P ( f + g) = P f + P g Proposition 4... (Positivity) Let f L and assume f 0. Then P f 0. a.e. Proof. For B, P fd = fd 0: Since B is arbitrary, P f 0. Proposition 4..3. (Preservation of ntegrals) P fd = fd : Proof. Since P fd = () fd = fd ; the result follows. Proposition 4..4. (Contraction Property) P : L! L is a contraction, i.e., kp fk kfk for any f L. Proof. Let f L. Let f + = max(f;0) and f = min(0; f). Then f + ;f L ;f = f + f and jfj = f + + f. By the linearity of P f, we have P f = P (f + f ) = P f + P f : Hence, and jp fj jp f + j + jp f j = P f + + P f = P jfj; kp fk = jp fjd P jfjd = jfjd = kfk ; where we have used Proposition 4..3. t follows from this result that P : L! L is continuous with respect to the norm topology since kp f P gk kf gk :

80 4 The Frobenius{Perron Operator Proposition 4..5. (Composition Property) Let :! and ¾ :! be nonsingular. Then P o¾ f = P ± P ¾ f. n particular, P nf = P n f. Proof. Since and ¾ are nonsingular their composition ± ¾ is also nonsingular. Let f L and B: P o¾ fd = fd = fd and P (P ¾ f)d = ( ±¾) P ¾ fd = ¾ ( ) ¾ ( ) fd : Hence P o¾ f = P P ¾ f a.e. By induction, it follows that P nf = P n f a.e. Recall that the Koopman operator U : L! L is de ned by U g = g ± and that for f L ; g L, we denote R fgd by hf; gi. Proposition 4..6. (djoint Property) f f L and g L, then hp f;gi = hf;u gi, i.e., (P f) gd = f U gd : (4..) Proof. Let be a measurable subset of and let g = Â. Then the left hand side of (4..) is P fd = fd and the right hand side is f ( ± )d = f  d = fd : Hence (4..) is veri ed for characteristic functions. Since the linear combinations of characteristic functions are dense in L, (4..) holds for all f L. The following proposition says that a density f is a xed point of P if and only if it is the density of a -invariant measure ¹, absolutely continuous with respect to a measure.

4. Properties ofthe Frobenius{Perron Operator 8 Proposition 4..7. Let :! be nonsingular. Then P f = f a.e., if and only if the measure ¹ = f, de ned by ¹() = R f d, is -invariant, i.e., if and only if ¹( ) = ¹() for all measurable sets, where f 0; f L and kf k =. Proof. ssume ¹( ) = ¹() for any measurable set. Then f d = f d and therefore P f d = f d : Since B is arbitrary, P f = f a.e. ssume P f = f a.e. Then P f d = f d = ¹() By de nition, P f d = f d = ¹( ) and so ¹( ) = ¹(). Let D(X;B;¹) denote the probability density functions on the measure space (X;B;¹). When we wish to emphasize the underlying measure in the Frobenius{Perron operator, we shall write P ;¹. P ;¹ acts on D(X;B;¹), while P ;º acts on D(X;B; º). Suppose ¹ << º and º << ¹, i.e., ¹ is equivalent to º. Then ¹ << º << º << ¹: The following result presents a relation between P ;¹ and P ;º. Proposition 4..8. Let ¹ be equivalent to º. Then ¹ = fº, where f D(X; B;º), and for any g L (X; B;¹), Proof. For any B, (P ;¹ g)d¹ = P ;¹ g = P ;º(f g) : (4..) f gd¹;

8 4 The Frobenius{Perron Operator and P ;º (fg) d¹ = P ;º (fg)dº = fgdº = gd¹: f Since B is arbitrary, the result is proved. Now we will prove some properties of P = P ;¹, where ¹ is a - invariant measure. Proposition 4..9. Let :! and let ¹ be a -invariant measure. Then (P f) ± = E(fj (B)) a.e. Proof. Since (P f) ± is obviously B-measurable, it is enough to prove that (P f) ± satis es the condition of Theorem.4.. Let = (B);B B. Then (P f) ± d¹ = (P f) ± d¹ B = P fd¹ = fd¹ = fd¹: B B Corollary 4... f :! and ¹ is a -invariant measure, then P is a contraction on any space L p ; p +. Proof. Let p < +. Then (kp fk p ) p = jp fj p d¹ = j(p f) ± j p d¹ X X = je(fj B)j p d¹ X E(jfj p j B)d¹ = jfj p d¹ = (kfk p ) p and kp fk p kfk p. Let p = +. Then X kp fk = ess supjp fj = ess supj(p f) ± j = ess supje(fj (B))j ess supjfj: X

4. Properties ofthe Frobenius{Perron Operator 83 Proposition 4..0. Let :! and ¹ be -invariant measure. Let denote the constant function equal to everywhere. Then, (a) is ergodic () for any f D(X;B; ¹) n X P ;¹ k n f! ; k=0 weakly in L as n! +. (b) is weakly mixing () for any f D(X; B;¹) n X jp k n ;¹f j! 0; k=0 weakly in L as n! +. (c) is mixing () for any f D(X;B;¹) weakly in L as n! +. P n ;¹ f! ; Proof. ll statements are direct consequences of properties (a), (b), and (c) of Theorem 3.4. and the djoint Property (Proposition 4..6). Proposition 4... Let :! and ¹ be a -invariant measure. Then is exact () for any f D(X;B;¹); as n! in the L -norm. P n ;¹f! ; Proof. ssume is exact. The ¾-algebras n (B) form a decreasing sequence of ¾-algebras. Since is exact, the ¾-algebra B T = T n n (B) consists of sets of measure 0 or. By Proposition 4..5 and Proposition 4..9, (P n f ± n ) = (P nf) ± n = E(fj n (B))! E(fjB T ) in L (X; B;¹) as n!. Since B T consists of sets of ¹-measure 0 or, E(fjB T ) = R fd¹ =. Thus, we have X j(p n f) ± n jd¹! 0 as n! +: X

84 4 The Frobenius{Perron Operator But X j(p n f) ± n jd¹ = = X X j(p n f) ± n ± n jd¹ jp n f jd¹: Thus P n f! in L as n!. Now let P n f! as n! in L. Let B and assume ¹() > 0. We will show that as n! ¹( n )! : Let f = ¹() Â. Then R X f d¹ = and as n!. We have ¹( n ()) = = n () n () n () v n : v n kp n f k! 0 d¹ P n f d¹ n () P n f d¹ v n = (P n f )d¹ n ( n ) f d¹ v n Since v n! 0 as n!, we have the result. Recall that the transformation : M()! M() is de ned by º() = º( ). Let º = f ¹. Then n º() = º( n ) = fd¹ = P nfd¹ n = (P nf)â d¹ = fâ ( n )d¹: (4..3) Let M () ½ M() denote the space of probability measures. Proposition 4... Let :! be strongly mixing on the normalized measure space (;B;¹). Let º M () be absolutely continuous with respect to ¹. Then on any set B; n º! ¹ as n! +. Proof. Since º << ¹, there exists f D(¹) such that º() = fd¹:

4.3 Representation of the Frobenius{Perron Operator 85 Then by (4..3), n º() = f  ( n )d¹: Since is strongly mixing, we have f  ( n )d¹! fd¹  d¹ = ¹() as n!. We already know that P is continuous with respect to the norm topology on L (; B; ). The nal property of P establishes the fact that P is also continuous in the weak topology of L (;B; ). Proposition 4..3. Let (;B; ¹) be a normalized measure space and let :! be nonsingular. Then P : L! L is continuous in the weak topology on L. Proof. Let f n! f weakly in L as n!. We want to prove that P f n! P f weakly in L as n!, i.e., for all g L, (P f n )gd¹! (P f)gd¹: Now, by Proposition 4..6, (P f n )gd¹ = f n (g ± )d¹: Since g ± L and f n! f weakly, we have f n (g ± )d¹! f(g ± )d¹ = (P f)gd¹: Thus, (P f n )gd¹! (P f)gd¹ as n!, i.e., P f n! P f weakly in L. 4.3 Representation of the Frobenius{Perron Operator n this section we derive an extremely useful representation for the Frobenius{Perron operator for a large class of one-dimensional transformations. These transformations, which are piecewise monotonic functions on an interval into itself, contain many of the transformations of interest in one-dimensional dynamical modeling and analysis.

86 4 The Frobenius{Perron Operator De nition 4.3.. Let = [a;b]. The transformation :! is called piecewise monotonic if there exists a partition of ; a = a 0 < a < ::: < a q = b; and a number r such that () j(ai ;a i) is a C r function, i = ; :::;q which can be extended to a C r function on [a i ;a i ]; i = ;:::;q, and () j 0 (x)j > 0 on (a i ;a i ); i = ; :::;q. f, in addition to (), j 0 (x)j > wherever the derivative exists, then is called piecewise monotonic and expanding. Note that () implies that is monotonic on each (a i ;a i ). n example of such a transformation is shown in Figure 4.3.. Figure4.3. We now proceed to nd P for piecewise monotonic. By the de nition of P, we have P fd = fd ; (4.3.) for any Borel set in. Since is monotonic on each (a i ;a i );i = ; :::; q, we can de ne an inverse function for each j(ai ;a i ). Let Á i = j Bi ; where B i = ([a i ;a i ]). Then Á i : B i! [a i ;a i ] and () = [ q i= Á i(b i \ ); (4.3.)

4.3 Representation of the Frobenius{Perron Operator 87 T where the sets fá i (B i )g q i= are mutually disjoint. See Figure 4.3.. T Note also that, depending on ;Á i (B i ) may be empty. On substituting (4.3.) into (4.3.), we obtain qx P fd = fd = i= qx i= Á i (B i \) B i\ f(á i (x))j Á i 0 (x)j d ; where we have used the change of variable formula for each i. Now qx P fd = f(á i (x))já 0 i (x)jâ Bi (x)d Since is arbitrary, i= = P f(x) = qx i= qx i= f( i (x)) j 0 ( i (x))j  (a i ;a i ) (x)d : f( i (x)) j 0 ( i (x))j  (a i ; a i ) (x) (4.3.3) for any f L. There is a more compact way of writing (4.3.3): X f(z) P f(x) = j 0 (z)j : (4.3.4) zf (x)g For any x, the set f (x)g consists of at most q points; if z is one of these points, i.e., z (a i ;a i ) for some i, the corresponding term f(z) will appear on the right hand side of (4.3.4). j 0 (z)j Remark 4.3.. The operator P is not -to-. To see this, let us consider, the symmetric triangle transformation on [0; ]. Let f = on [0; ) and on [ ;]. Then, P f = 0 a.e. Thus P is not a -to- operator. Example 4.3.. Let : [0; ]! [0;] be de ned by rxe bx, where r and b are such that is well-de ned (i.e., b > and r be). The graph of such a is shown in Figure 4.3.. Then, P f(x) = f( (x)) f( + (x)) j 0 ( (x))j j 0 ( (x))jâ [re b ;] (x); where and are the two monotonic components of.

88 4 The Frobenius{Perron Operator Problems forchapter 4 b = 5, r = 5e Figure4.3. Problem 4... Let : [0;]! [0; ] be de ned by rx( x), where 0 < r 4. Find P f. Problem 4... Let = [0;], B =Borel ¾-algebra on andlet be Lebesgue measure on. Let :! be de ned by px mod, where p is a positive integer greater than or equal to. Find P f. Then, prove that is exact. Problem 4..3. For and f as in Figures 4.4. and 4.4. respectively, nd P f. Problem 4..4. For ½ x; 0 x ; x + 3 ; x ; nd P. Let ½ 0; 0 x < f(x) = ; g(x); x ; where g is symmetric with respect to the line x = 3 4. Show that P f = f. Problem 4..5. Let be an interval of the real line. Let and be measurable, nonsingular transformations from!. Show that P ± = P ± P.

Problemsfor Chapter 4 89 Figure4.4. Figure4.4. Problem 4..6. Let be an interval of the real line and :! be a measurable and nonsingular transformation. Show that where n = ± ± :: :. P n = P n ; Problem 4..7. Let n! uniformly and let f n be the invariant density associated with n, i.e., P n f n = f n. f f n! f weakly in L, show that P f = f. Problem 4..8. Let f L ; g L or the other way around. Prove that P ((f ± ) g) = f P (g);a:e: Problem 4.3.. Let 4x( x) on [0;]: Show that f(x) = ¼ p x( x) is a xed point of P f. Problem 4.3.. Let ½ x; 0 x ; 4 3 x + 5 3 ; x : Find V 0 P f, where (a) f(x) = x (b) f(x) = sinx.

90 4 The Frobenius{Perron Operator Problem 4.3.3. Let ½ x; 0 x ; ( x); < x : Let S consist of all functions f of the form f = Â [0; ] + Â ( ;] where ; [0;]. Let f = (f ; f ) where f = Â [0; ] and f = Â ( that P f = (f ;f )µ. Problem 4.3.4. Let : [0;]! [0;] be de ned by (x)â [0; 4 )(x) + (x)â [ 4 ; ) (x) + 3 (x)â [ ;](x); ;]. Show where (x) = 4x, (x) = 3 x, 3(x) = x. Let S denote the class of all functions f : [0;]! [0;], where f = Â [0; 4 )+ Â [ 4 ; ) + 3 Â [ ;] and ; ; 3 [0; ]. For any f : [0;]! [0;], let f (f ;f ;f 3 ) where f = fâ [0; ), f = fâ 4 [ ), and f 4 ; 3 = fâ [ ;]. Show that for f S, Find a xed point of P. 0 4 4 4 P f = (f ;f ;f 3 )@ 0 0 : Problem 4.3.5. Let : [0;]! [0;] be de ned by 8 >< x + ; x [0; ]; >: x ; x ( ;]: Find P : and Problem 4.3.6. 3 : [0; ]! [0;] be de ned by 3 = ±, where Find P 3. (x) = rx( x); 0 r 4 8 >< x + ; x [0; (x) = ]; >: x ; x ( ;]: Problem 4.3.7. Let : [0;]! [0; ] be de ned by rxe bx, where r > 0; b > 0. Find P.

Problemsfor Chapter 4 9 Problem 4.3.8. Show that (a) f(x) = is an invariant density for 8 x >< ; 0 x >: x ; x ; (b) is an invariant density for 8 >< >: + 3 8 0; 0 x < >< 4 f(x) = ; 4 x < >: 3 ; x x ; x; 0 x x ; (c) f(x) = 4 ¼ +x is an invariant density for 8 x >< x ; 0 x p x >: x ; p x ; (d) f(x) = (x ) is an invariant density for : [0;]! [0;] given by 8 jx 3 + j3 ; (e) f(x) = x is an invariant density for (+x) 8 x >< x ; 0 x 3 >: x x ; 3 x ; (f) f(x) = px p is an invariant density for 8 x >< ( ) ; 0 x ( p ) p ( x p ) p >: ; ( ) p x ; ( ) p

9 4 The Frobenius{Perron Operator (g) f(x) = cos x is invariant under arctan( tanx) ; ¼ x ¼: Problem 4.3.9. (Di±cult) Let : [0; ]! [0;] be given by ( x p+( p)x ; 0 x q( ) q( )x q q +( q+q )x ; x ; where 0 < p and q > 0 are real numbers. Let and S (x) = p + x p S (x) = q + q q( )x q q + q( )x : Let,, ± be de ned by the equations S (±) = ±, S ( ) = ±, S ( ) =, respectively. Thenverify thataninvariantdensity of is givenas follows: Case (a): 6= ±. Case (b): = = ± 6= 0. Case (c): = = ± = 0. f(x) =j x + f(x) = x + ± (x + ) ; f(x) : Problem 4.3.0. Consider a family of transformations ( ax ( a)x a (x) = ; for 0 x ; x; for < x ; a : Show that the a 's satisfy the assumptions of Problem 4.3.9 ( = ; p = a ; q = ). Find P a -invariant functions f a for a > and show that f = lim a! + f a (pointwise) is a P -invariant nonintegrable function. j;

Problemsfor Chapter 4 93 Problem 4.3.. Let : [0;]! [0;] be nonsingular and let h : [0;]! [0; ] be a di eomorphism. Prove (a) P f = f implies P ¾ g = g, where ¾ = h ± ± h and g = (f ± h ) j(h ) 0 j; (b) if f is a -invariant density, then g is a ¾-invariant density. Problem 4.3.. Let ¾ : [0;]! [0; ] be de ned by ¾ = h± ±h, where h(x) = p x and is as in Problem 4.3.4. Find the ¾-invariant density g. Problem 4.3.3. Let : [0;]! [0;] be piecewise monotonic with respect to the partition P( ) [0; ];[ ;]ª, where sin¼x. Find the partition P( 3 ) with respect to which 3 is piecewise monotonic. Problem 4.3.4. Let : [0;]! [0;] be de ned by 4x( x): Suppose ¹ is a Borel measure on [0;] de ned by d¹ = is Lebesgue measure on [0;]. Find P ;¹. d p x( x), where Problem 4.3.5. Let be the tent transformation (de ned in Problem 4.3.3 and shown in Figure 4.4.). Let ¾(y) = 4 y ( y); y [0; ]. Use the di eomorphism h(x) = sin ( ¼ x) and Problem 4.3. to show that the density g(y) = p is ¾-invariant. ¼ y( y) Problem 4.3.6. Prove that linear homeomorphism h(x) = a x+b conjugates x + x + to ¾(x) = x + c, h( ) = ¾(h), if a = ; b = ; c = + ( ): n particular, 4x( x) on [0;] is conjugated to ¾(x) = x on [ ; ]. Find the density of the ¾-invariant absolutely continuous invariant measure. Problem 4.3.7. a) The transformation : R! R, given by x x, is called Boole's transformation. Show that preserves Lebesgue measure. b) generalized Boole transformation : R! R is given by (x + a 0 + nx i= b i x a i );

94 4 The Frobenius{Perron Operator where n N [ f0g; a i R; for i = 0;;:::; n and b i < 0, for i = ; :::; n: Show that preserves Lebesgue measure. Problem 4.3.8. a) Let : R! R be given by atan(x); x 6= k ¼ ; where k = ; 3; ::: and a >. Show that the density function f(x) = p=¼ p +x is -invariant for p > 0 satisfying equation a tanh(p) = p. b) More generally, let a(tan(bx) + c); x 6= k b ¼ ; k = ; 3;:::. Find the ranges of a;b; c for which we can nd p;q R p=¼ such that the density f(x) = p +(x q) is -invariant. c) Let tanx; x 6= k ¼ ; k = ; 3;:: :: Show that the density f(x) = x is -invariant. Note that tan(0) = 0 and tan 0 (0) =. Thus, 0 is a xed point of at which the slope of is equal to. Such a xed point is called indi erent. is an example of a transformation with an indi erent xed point. Such transformations have in nite ¾- nite invariant measures ([Thaler, 980]). The transformation of case c) has ¾- nite absolutely continuous invariant measure ¹ = x. Hint: Use the formula +X k= s;t R; or its special form (s = 0) +X k= Problem 4.3.9. Let s + (t + k¼) = tanh(s)( + tan (t)) s(tan (t) + tanh (s)) ; (t + k¼) = sin (t) = + tan (t) tan : (t) ( x x ; 0 x < x ; x : Show that f(x) = x is a xed point of P.

Problemsfor Chapter 4 95 Problem 4.3.0. Let be the Gauss transformation de ned in Problem 3..7. a) Show that f(x) = +x is a xed point of P. b) Let n and de ne (n) (x) = ½ n x; for 0 x n (x); for n < x : Show that (n) preserves the same density f. Problem 4.3.. Show that the transformation with countably many branches, de ned by f x xg; where ftg denotes the fractional part of t, preserves the density f(x) = : The transformation x is called the \backward continued fraction transformation" and has interesting connections with geodesic ow on hyperbolic plane (see [dler and Flatto, 984]). Problem 4.3.. Let d be an integer or + and let f : [0;d)! [0; ) be an increasing function with lim x!d f(x) =. We de ne f (x) (mod ). is a piecewise monotonic transformation with d branches. a) Show that the equation for the -invariant density h is given by d X h(x) = h(f(x + k))f 0 (x + k): () k=0 b) Let f (x) = x +x ; x > 0. Show that h (x) = x. c) Let > 0; 6= and f (x) = (x ( + x) + ) ; x > 0. Show that the -invariant density is h = x. Problem 4.3.3. (See [Lasota and Yorke, 98].) Let be a ¹- nonsingular transformation and let P be the Frobenius{Perron operator induced by. function h L (X;B;¹) is called a lower function for P if h 0; R X hd¹ > 0, and lim k(h P n n!+ f) + k = 0 for all f D(X;B; ¹). a) Prove that if h is a lower function, then P h is also a lower function. b) Prove that if h ;h are lower functions, then max(h ;h ) is also a lower function.

96 4 The Frobenius{Perron Operator c) Prove that if h n ; n = ;;::: are lower functions and h n! h; in L (X;B;¹) as! +, then h is also a lower function. d) Prove that if P has a lower function, then P has a lower function h satisfying P h = h. e) Prove that if h is a lower function and P h = h, then h = ( khk )h is also a lower function. f) Prove that if P has a lower function, then has an invariant density g such that P n f! L g; for any f D(X;B;¹), i.e., the dynamical system (X; B; ; g ¹) is exact.