Iteration of the rational function z 1/z and a Hausdorff moment sequence

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1 Iteration of the rational function z /z and a Hausdorff moment sequence Christian Berg University of Copenhagen, Institute of Mathematical Sciences Universitetsparken 5, DK-200 København Ø, Denmark Antonio J. Durán Universidad de Sevilla, Departamento de Análisis Matemático Apdo (P. O. BOX) 60, E-4080 Sevilla, Spain Abstract In a previous paper we considered a positive function f, uniquely determined for s > 0 by the requirements f() =, log(/f) is convex and the functional equation f(s) = ψ(f(s + )) with ψ(s) = s /s. Denoting ψ (z) = ψ(z), ψ n (z) = ψ(ψ (n ) (z)), n 2, we prove that the meromorphic extension of f to the whole complex plane is given by the formula f(z) = n ψ n ( (+ / ) z ), where the numbers are defined by λ 0 = 0 and the recursion + = (/2)( + λ 2 n + 4). The numbers m n = /+ form a Hausdorff moment sequence of a probability measure µ such that t z dµ(t) = /f(z). Key words: Hausdorff moment sequence, iteration of rational functions MSC 2000: 44A60, 30D05 This work was initiated during a visit of the first author to the University of Sevilla in January 2007 supported by D.G.E.S, ref. BFM C03-0, FQM-262 (Junta de Andalucía). Corresponding author. addresses: berg@math.ku.dk (Christian Berg), duran@us.es (Antonio J. Durán). Preprint submitted to Elsevier 0 April 2008

2 Introduction and main results Hausdorff moment sequences are sequences of the form a n = 0 tn dν(t), n 0, where ν is a positive measure on [0, ]. Hausdorff s characterization of moment sequences was given in [7]. See also Widder s monograph [8]. For information about moment sequences in general see []. In [5] we introduced a non-linear transformation T of the set of Hausdorff moment sequences into itself by the formula: T ((a n )) n = /(a 0 + a + + a n ), n 0. () The corresponding transformation of positive measures on [0, ] is denoted T. We recall from [5] that if ν 0, then T (ν)({0}) = 0 and 0 t z+ t dν(t) t z d T (ν)(t) = for Rz 0. (2) 0 It is clear that if ν is a probability measure, then so is T (ν), and in this way we get a transformation of the convex set of normalized Hausdorff moment sequences (i.e. a 0 = ) as well as a transformation of the set of probability measures on [0, ]. By Tychonoff s extension of Brouwer s fixed point theorem to locally convex topological vector spaces the transformation T has a fixed point, and by () it is clear that a fixed point (m n ) n is uniquely determined by the recursive equation m 0 =, ( + m + + m n )m n =, n. (3) Therefore giving m = + 5, m 2 = 2 m 2 n+ + m n+ m n = 0, (4) ,. 4 In [6] we studied the Hausdorff moment sequence (m n ) n and its associated probability measure µ, called the fixed point measure. It has an increasing and convex density D with respect to Lebesgue measure on ]0, [, and for t we have D(t) / 2π( t). We studied the Bernstein transform f(z) = B(µ)(z) = 0 t z dµ(t), Rz > 0 (5) t 2

3 as well as the Mellin transform F (z) = M(µ)(z) = 0 t z dµ(t), Rz > 0 (6) of µ. These functions are clearly holomorphic in the half-plane Rz > 0 and continuous in Rz 0, the latter because µ({0}) = 0. As a first result we proved: Theorem ([6]) The functions f, F can be extended to meromorphic functions in C and they satisfy f(z + )F (z) =, z C (7) f(z) = f(z + ), z C. (8) f(z + ) They are holomorphic in Rz >. Furthermore z = is a simple pole of f and F. The fixed point measure µ has the properties 0 t x dµ(t) <, x > ; 0 d µ(t) t =. (9) Using the rational function the functional equation (8) can be written ψ(z) = z z, (0) f(z) = ψ(f(z + )), z C. () The function f can be characterized in analogy with the Bohr-Mollerup theorem about the Gamma function, cf. [2]. Using the following notation for iterations of a function ρ ρ (z) = ρ(z), ρ n (z) = ρ(ρ (n ) (z)), n 2, we proved: Theorem 2 ([6]) The Bernstein transform (5) of the fixed point measure is a function f : ]0, [ ]0, [ with the following properties (i) f() =, (ii) log(/f) is convex, (iii) f(s) = ψ(f(s + )), s > 0. 3

4 Conversely, if f :]0, [ ]0, [ satisfies (i)-(iii), then it is equal to f and for 0 < s we have ( ( ) ) f(s) = ψ n mn s n. (2) In particular (2) holds for f. m n The rational function ψ is a two-to-one mapping of C \ {0} onto C with the exception that ψ(z) = ±2i has only one solution z = ±i. Moreover, ψ(0) = ψ( ) = and ψ is a continuous mapping of the Riemann sphere C = C { } onto itself. It is strictly increasing on the half-lines ], 0[ and ]0, [, mapping each of them onto R. The sequence ( ) n is defined in terms of (m n ) n from (3) by m n λ 0 = 0, + = /m n, n 0, (3) i.e. λ =, λ 2 = + 5, λ 3 = 2 By (6) and (7) we clearly have ,. 4 hence by () which can be reformulated to m n = F (n), = f(n), n 0, (4) + = 2 = ψ(+ ), n 0, (5) ( ) + λ 2 n + 4, n 0. (6) The main purpose of this paper is to prove that equation (2) holds for f in the whole complex plane. Theorem 3 Let a n, b n : C C be the entire functions defined by ( ) z ( ) z λn λn+ a n (z) =, n 2, b n (z) =, n. (7) The meromorphic function f from Theorem is given for z C by f(z) = n ψ n (a n (z)) = n ψ n (b n (z)), (8) and the convergence is uniform on compact subsets of C. 4

5 Remark 4 In [6] it was proved that the Julia and Fatou sets of ψ are respectively R and C \ R. For z C \ R we have ψ n (z) for n. Notice that a n (z), b n (z) are close to when n is large because + / according to Lemma 5 below. Also ψ n ( ) = 0 for all n. 2 Proofs For a domain G C we denote by H(G) the space of holomorphic functions on G equipped with the topology of uniform convergence on compact subsets of G. We first recall some easily established properties of the sequence ( ) n, which are needed in the proof of Theorem 3. Lemma 5 ([6]) () n 2n, n 0. (2) ( ) n is an increasing divergent sequence and + / is decreasing with + =. n (3) n (λ 2 n+ λ 2 n) = 2. λ 2 n (4) n n = 2. The following lemma gives monotonicity properties of the sequences in (8). Lemma 6 The sequence (ψ n (a n (s))) n is decreasing for s > 0. The sequence (ψ n (b n (s))) n is increasing for 0 < s < and decreasing for < s, while ψ n (b n ()) = for all n. Proof: Define the function ρ(y) = log(2 sinh y) = log ψ(e y ), y > 0, (9) which is easily seen to be increasing and concave, and positive for y > log λ 2. It satisfies ρ(log + ) = log, n. (20) Let s > 0. The slope of the chord of ρ over the interval [log, log + ] majorizes the slope over [log +, log + + s log(+ / )] by concavity, i.e. ρ(log + ) ρ(log ) ρ(log + + s log(+ / )) ρ(log + ), log + log s log(+ / ) hence using (20) ρ(log(a n+ (s)) log a n (s), 5

6 or ψ(a n+ (s)) a n (s), which shows that the sequence (ψ n (a n (s))) n is decreasing. For s > the slope of the chord of ρ over the interval [log +, log +2 ] majorizes the slope over [log +, log + + s log(+2 /+ )] by concavity, i.e. ρ(log +2 ) ρ(log + ) ρ(log + + s log(+2 /+ )) ρ(log + ), log +2 log + s log(+2 /+ ) hence using (20) or ρ(log(b n+ (s)) log b n (s), ψ(b n+ (s)) b n (s), which shows that the sequence (ψ n (b n (s))) n is decreasing. If 0 < s < the inequality between the slopes is reversed and we get that (ψ n (b n (s))) n is increasing. Proof of Theorem 3. The proof is given in a number of steps. : For any 0 < s < we have n [ψ n (a n (s)) ψ n (b n (s))] = 0. Note that the quantity under the it is positive because a n (s) > b n (s) and ψ is increasing. By the mean value theorem we get for a certain w ]b n (s), a n (s)[ ψ n (a n (s)) ψ n (b n (s)) = (a n (s) b n (s))(ψ n ) (w) = (a n (s) b n (s))ψ (ψ n (w))ψ (ψ n 2 (w)) ψ (w). Since < b n (s) < w < a n (s), we get k < ψ k (b n (s)) < ψ k (w), k = 0,,..., n, hence 6

7 ψ n (a n (s)) ψ n (b n (s)) n a n (s) b n (s) ψ (ψ k (w)) a n (s) b n (s) = (( λn k=0 ( n + ) k=0 λ 2 n k ) s ( ) s ) n ( λn+ k= ( ) s ) n λn+ (( ) s λn = (n + ) (( λn k= ) s ( ) s ) λn+ ) + λ 2 k ( + ) k where we have used k λ k from Lemma 5 part., For < y < x there exists y < ξ < x such that s(x y) if 0 < s x s y s = s(x y)ξ s sx s (x y) if < s, (2) and using / λ 2 for n 2 we get ( ) s ( ) s ( λn λn+ s max(λ s λn 2, ) λ ) n+. By (6) we have + = 2 ( + 4 λ 2 n so the final estimate is + 4 λ 2 n ) = ( 2 ) λ 2 n λ 2 n λ 2 n λ 2 n λ2 n λ 2 n, λ 2 nλ 2 n ψ n (a n (s)) ψ n (b n (s)) max(λ s s(n + ) 2, ) (λ 2 λ 2 n λ 2 n ), n which tends to zero by part 2, 3 and 4 of Lemma 5. 7

8 2 : The two sequences (ψ n (a n (s))) n and (ψ n (b n (s))) n have the same it R(s) for s > 0, and R is a function satisfying the properties of Theorem 2, hence R(s) = f(s). By the monotonicity of Lemma 6 and the inequality ψ n (a n (s)) > ψ n (b n (s)) > ψ n ( ) = 0, it follows that the two sequences converge. By the its are the same, which we denote by R(s). The inequality R(s) > 0 for s > 0 follows for 0 < s from R(s) ψ(b (s)) = ψ(λ s 2) and for s > from ψ n (b n (s)) ψ n (+ ) =. Clearly R() =. In particular we have and using ψ (n+) (a n+ (s)) R(s), ψ n (b n (s + )) R(s + ), ϕ(x) = (/2)(x + x 2 + 4), x R (22) as the continuous inverse of ψ ]0, [ we get ψ n (a n+ (s)) ϕ(r(s)). From the equation a n+ (s) = b n (s + ) we finally get ψ(r(s + )) = R(s). Iterating (9) leads to ρ n (y) = log ψ n (e y ) for all n, and therefore log ψ n (a n (s)) = ρ n (log + s log( / )) is a sequence of concave functions of s > 0. Therefore the it log R(s) is concave. It now follows from the uniqueness part of Theorem 2 that f(s) = R(s) for s > 0. Remark 7 The inequality ψ n (b n (s)) f(s) ψ n (a n (s)), 0 < s, (23) follows from the monotonicity of Lemma 6 because R(s) = f(s). It can however be seen directly from the properties of f without using the uniqueness part of Theorem 2. Combining (23) with the result of, we then see that the sequences (ψ n (a n (s))) n and (ψ n (b n (s))) n converge to f(s) for 0 < s. To see (23), we notice that for any convex function h on ]0, [ we have h(n) h(n ) h(n + s) h(n) s h(n + ) h(n), 0 < s, n 2. 8

9 By taking h = log(/f), we get using f(n) = hence log s log f(n + s) log +, ( ) s ( ) s λn+ λn f(n + s), 0 < s, and (23) follows by applying ψ n. In the following steps we need the quantity ρ n,n = N = N k= + k, n N, (24) where the second equality follows by summing λ j λ j = /λ j from j = n+ N to j = n. We denote D(a, r) = {z C z a < r}. 3 : For N N, 0 < c, n > N ψ(d(, cρ n,n )) D(, cρ n,n ). If z < cρ n,n we get and hence ψ(z) = z ( z ) < cρ n,n ( + z ) z = ( z) z > ρ n,n = N, ) ( N ψ(z) < cρ n,n ( + = c + λ ) n N = cρ n,n. N + k N k= Iterating n N times using ρ N,N = λ N λ 0 = λ N we get 4 : For N n and 0 < c ψ (n N) (D(, cρ n,n )) D(λ N, cλ N ). 9

10 5 : For 0 < c, z cn, N n we have b n (z) D(, cρ n,n ). For a > and z C, z we have the elementary inequality Applying this with a = (+ / ) N we get a z z (a ). (25) b n (z) = ( λ ( n+ ) z c ( λ ) n+ ) N = c( λ N n+ ) ( + ) k = c k=0 + ) ( N λ k n+ = c + c + k= λ k n where we have used the inequalities which are equivalent to N k=0 k+ < c N k=0 N k=0 ( + ) k k = cρ n,n, λ k n+ k+ λ k n, k =,..., N, (26) (k ) log f(n + ) + log f(n k + ) k log f(n), but they hold because R = log f is concave. Combining 4 and 5 we get 6 : For 0 < c, n N, z cn ψ (n N) (b n (z)) D(λ N, cλ N ). In particular, the sequence ψ (n N) (b n (z)), n N of holomorphic functions in the disc D(0, N) is bounded on compact subsets of this disc. 7 : For 0 < s <, n N we have n ψ (n N) (b n (s)) = f(s + N). Since b n (s) > we know that ψ (n N) (b n (s)) > λ N. For each N we see that ψ N ]λ N, [ R is a homeomorphism with inverse ϕ N, where ϕ is given by (22). Since ψ n (b n (s)) f(s) we get ϕ N (ψ n (b n (s)) ϕ N (f(s)) = f(s + N) 0

11 i.e. n ψ (n N) (b n (s)) = f(s + N). 8 : Let N N. For z D(0, N) we have n ψ (n N) (b n (z)) = f(z + N), and the convergence is uniform on compact subsets of D(0, N). By Montel s theorem the sequence ψ (n N) (b n (z)) has accumulation points h in the space H(D(0, N)) of holomorphic functions on D(0, N). By 7 we know that h(s) = f(s + N) for 0 < s < N. From the uniqueness theorem for holomorphic functions, all accumulation points then agree with f(z + N) H(D(0, N)), and the result follows. 9 : For z C we have n ψ n (b n (z)) = f(z), uniformly on compact subsets of C. For a compact subset K C we choose N N such that K D(0, N) and know by 8 that ψ (n N) (b n (z)) converges uniformly to f(z + N) for z K. We next use that ψ N : C C is continuous, hence uniformly continuous with respect to the chordal metric on C, and since ψ N (f(z + N)) = f(z), the result follows. 0 : For z C we have uniformly on compact subsets of C. n ψ n (a n (z)) = f(z), In fact, so ψ n (a n+ (z)) = ψ n (b n (z + )) f(z + ) ψ(ψ n (a n+ (z))) ψ(f(z + )) = f(z). Acknowledgment The authors want to thank Michael Olesen, Nykøbing Katedralskole for useful comments to an early version of this manuscript.

12 References [] Akhiezer, N. I., The classical moment problem, Oliver and Boyd, Edinburgh, 965. [2] Artin, E., The Gamma Function, Holt, Rinehart and Winston, New York, 964. [3] Beardon, A. F., Iteration of rational functions, Graduate Texts in Mathematics vol. 32, Springer-Verlag, Berlin-Heidelberg-New York, 99. [4] Berg, C., Durán, A. J., A transformation from Hausdorff to Stieltjes moment sequences, Ark. Mat. 42 (2004) [5] Berg, C., Durán, A. J., Some transformations of Hausdorff moment sequences and Harmonic numbers, Canad. J. Math. 57 (2005) [6] Berg, C., Durán, A. J., The fixed point for a transformation of Hausdorff moment sequences and iteration of a rational function, Math. Scand. 02 (2008)??. [7] Hausdorff, F., Momentenprobleme für ein endliches Intervall, Math. Z. 6 (923) [8] Widder, D. V., The Laplace Transform, Princeton University Press, Princeton, 94. 2