ON LINEAR RECURRENCES WITH POSITIVE VARIABLE COEFFICIENTS IN BANACH SPACES

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1 ON LINEAR RECURRENCES WITH POSITIVE VARIABLE COEFFICIENTS IN BANACH SPACES ALEXANDRU MIHAIL The purpose of the paper is to study the convergence of a linear recurrence with positive variable coefficients with elements from a Banach space to estimate the speed of the convergence. The conditions depend of the sum of the coefficients the minimum of the coefficients. In this paper we study the case when the sequence of the sum of the coefficients is convergent to 1. The results are first proved for recurrences of real numbers. These results are extended later for Banach spaces. AMS 2000 Subject Classification: 40A05. Key words: linear recurrence, homogeneous linear recurrence, Banach space. 1. INTRODUCTION AND PRELIMINARIES The aim of the paper is the study of linear recurrences with positive variable coefficients in Banach spaces see Definition 1.1 below. The paper extends the results obtained in [2], where the convergence or the divergence of the linear recurrence was studied. Here we want to estimate the speed of the convergence of the linear recurrence. The Banach spaces will be supposed to be real Banach spaces although this is not necessary. The results are also valid for complex Banach spaces because every complex Banach space can be seen as a real Banach space. For a Banach space X, 0 X denotes the identity element of X for a set A X, A denotes the Banach subspace of X generated by A conv A denotes the convex closure of the set A, that is { conva = a i x i x 1, x 2,..., x n A, a 1, a 2,..., a n [0, 1], } a i = 1. i=1,n i=1,n n, m, k, l, i, j, p denotes natural numbers if we do not say otherwise, δj i = { 1 if i = j = 0 if i j. MATH. REPORTS 1262, , 9 29

2 10 Alexru Mihail 2 Definition 1.1. Let X be a real normed space k 1 be fixed. Let an = a 1 n, a 2 n,..., a k n n k+1 be a sequence of elements from Rk b n n k+1 be a sequence of elements from X. The sequence x n given by x n+k = a 1 n+k x n+k 1 + a 2 n+k x n+k a k n+k x n + b n+k for n 1 is called the linear recurrence of order k with coefficients a n n k+1, free terms b n n k+1 initial values x 1, x 2,..., x k X. The sequence is called homogeneous if b n = 0 for every n k + 1. For two sequences of real numbers a n b n, a n b n means that a n is convergent if only if b n is convergent. The infinite series products are denoted by. Notation. a l = 0 a l = 1 if m > n. l=m,n l=m,n The following lemma is well-known see [1] or [3]. Lemma 1.1. For a sequence of real positive numbers a n we have a 1 + a n a n ; b 1 a n > 0 a n < if a n < 1 for every n N. We recall, from [2], the following important remark on homogenous real linear recurrences with variable coefficients: Remark 1.1. Let a n n k+1 be a fixed sequence with a n R k + x n be the homogeneous linear recurrence of order k associated with the sequence a n n k+1 with initial values x 1, x 2,..., x k > 0 that is x n+k = a 1 n+k x n+k 1 + a 2 n+k x n+k a k n+k x n for n 1. Then a If there exist initial values x 1, x 2,..., x k > 0 such that lim x n = 0, n lim x n = respectively x n is bounded, then for every initial values n x 1, x 2,..., x k not necessary greater than 0 in the first the third case lim x n = 0, lim x n = respectively x n is bounded. n n b If there exists initial values x 1, x 2,..., x k > 0 such that lim x n n 0,, then lim a j n = 1. n This remark suggested us to divide the problem when x n is real in the cases lim x n = 0, lim x n = x n is a bounded sequence, in particular when lim x n 0,. These cases depend on the behavior of the n n n

3 3 On linear recurrences with positive variable coefficients in Banach spaces 11 sequence a j n. In Banach spaces we have similar cases. When a n k+1 j n is smaller than 1, x n tends to be convergent to 0, for example when lim sup a j n < 1 or when < 1 = 0. When n lim inf n a j n n k+1 a j n a j n is greater than 1, x n tends to be divergent, for example when a j n > 1 or when a j n > 1 n k+1 a j n =. In particular if x 1, x 2,..., x k are in a cone then x n tends to be divergent to. Also when a j n is convergent to 1 quickly enough, the sequence x n is bounded if the coefficients are not to small then it is convergent to a limit which is different from 0 in general. These cases are also valid for inhomogeneous linear recurrences. In this paper we will study the last case when the series 1 is convergent. This case seems to be difficult in the n k+1 a j n sense that the calculations are longer than in the other cases. The paper is divided into 6 sections. The first section is the introduction the second one gives some particular cases in which is possible to calculate the general term. These cases give an idea of what happens in the general case. The third section contains the study of the real homogenous case when the sum of the coefficients is 1. The next section contains the study of the general homogenous case when the sum of the coefficients is 1. The fifth section contains the results obtained in the general inhomogeneous case when the sum of the coefficients is 1. The last section contains the general case. 2. PARTICULAR CASES In this part we will give some particular cases when it is possible to find the general term of the recurrence. We study first the case k = 1. Let a n b n be sequences of real numbers with a n 0 let x n n 0 be the sequence given by the linear recurrence x n+1 = a n x n + b n for n 0. An inductive calculation shows us that x n = a l x 0 + j=l+1,n a j b l.

4 12 Alexru Mihail 4 j=l+1,n We remark that x n n 0 is convergent for every x 0 if only if the product a l is convergent the sequence b l a j is also convergent. Let us suppose that the series a n 1 is convergent this imply that the product a l is convergent different from 0. Because the convergence of the product a l of the sequence b l a j does not change j=l+1,n when we change a finite number of terms of the sequence a n maintaining the condition a n 0 we can suppose also that in this case we have a n 1 ε, 1 + ε for an ε > 0. If b n 0 the sequence x n n 0 is convergent for every x 0 if only if the series b l is convergent. To see this let us remark that 0 < 1 a n 1 a j 1 + a n 1 < +. If the j=l+1,n series b l is convergent then x n n 0 is also convergent for every x 0. If the series b l is divergent there exists a sequence a n such that x n n 0 is divergent. We now study the case k = 2 when the sum of the coefficients is 1. Lemma 2.1. Let x n n 0 be the sequence of real numbers given by the linear recurrence x n+1 = 1 a n x n + a n x n 1 for n 1 where 0 < a n < 1. If x 1 x 0 the sequence x n is convergent if only if the product a j is convergent to 0. Proof. We have j 1 x n+1 x n = a n x n x n 1, x n+1 x n = 1 n x 1 x 0 x n+1 = x 1 x l j=1,l a l a j + x 0. If the sequence x n n 0 is convergent then the sequence x n+1 x n n 0 is convergent to 0 then a j is also convergent to 0. j 1

5 5 On linear recurrences with positive variable coefficients in Banach spaces 13 If a j is convergent to 0, because 0 < a n < 1, the sequence a l j 1 is decreasing to 0 the series 1 l a j is convergent by the Leib- nitz corollary for alternate series. j=1,l Lemma 2.2. Let x n n 0 be the sequence given by the linear recurrence x n+1 = 1 a n x n + a n x n 1 + b n for n 1, where 0 < a n < 1 b n is a sequence of real numbers. Then a If the sequence x n n 0 is convergent for every x 0 every x 1 then the product a j is convergent to 0, the sequence 1 l a j b l j 1 is convergent the sequence 1 l b l 1 j j=l,n also convergent. b If the product a j is convergent to 0 the sequence j 1 1 l b l 1 j j=l,n i=l+1,ja i j=l+1,n i=l+1,ja i is convergent then the sequence x n n 0 is convergent for every x 0 every x 1. c If the product j is convergent to 0, the series j 1a b j is convergent the sequence is bounded then the j 1 sequence j n i=n+2,j+1a i x n n 0 is also convergent for every x 0 every x 1. d If there is an ε > 0 such that a n 1 ε for n 1 the series b j a i are convergent, then the sequence x n is also j 1 j n+1 i=n,j convergent for every x 0 every x 1. Proof. Let x n n 0 be the sequence given by the linear recurrence x n+1 = 1 a n x n + a n x n 1 + b n for n 0 where 0 < a n < 1. Then x n+1 x n = a n x n 1 x n + b n = a n x n x n 1 + b n. If we denote x n+1 x n by y n then the sequence y n n 0 is defined by y n = a n y n 1 + b n y 0 = x 1 x 0. is

6 14 Alexru Mihail 6 [ We have y n = 1 n a l x 1 x l Then = [ 1 j j=1,n l=1,j = x 1 x 0 1 j j=1,n x n = x 0 + x j x j 1 = j=1,n a l x 1 x 0 + l=1,j = x 1 x 0 1 j j=1,n l=1,j a l + j=1,n a l + l=1,j l=1,j 1 l i=l+1,j 1 l+j j=l+1,n a j b l ]. a i b l ] + x 0 = i=l+1,j 1 l b l 1 j j=l,n a i b l + x 0 = i=l+1,j a i + x 0. a If the sequence x n n 0 is convergent for every x 0 every x 1 then the sequence y n n 0 is convergent for every y 0 from the case k = 1 it follows that the series 1 l a j b l is convergent the product j=l+1,n a j is also convergent. Because 0<a n <1 we have a j =0. Taking x 0 = j 1 j 1 i x 1 =0 we obtain that the sequence 1 l b l 1 j j=l,n i=l+1,ja is also convergent. b If the product a j is convergent to 0, then the sequence j 1 1 j j=1,n l=1,ja l is also convergent. Taking account that the sequence 1 l b l j=l,n i=l+1,ja i is convergent, we obtain that the sequence x n n 0 is convergent for every x 0 every x 1. i c Let the sequence be bounded by M > 0. We j n i=n+2,j+1a i have to prove that the sequence 1 l b l 1 j is j=l,n i=l,ja

7 7 On linear recurrences with positive variable coefficients in Banach spaces 15 convergent. Let u n = u n+1 u n = 1 n l b l 1 l j=l,n n+1 u n u = l b n l 1 This imply that the sequence convergent. d results from c. 1 j i=l+1,n+1 +1 i=l+1,n+1 i=l+1,j a i b l i=l+1,n+1 a i. Then +1 a i b l = a i M l. b 1 l b l j=l,n i=l+1,n+1 i=l+1,ja i a i b l is 3. THE REAL HOMOGENOUS CASE WHEN THE SUM OF THE COEFFICIENTS IS 1 In this part we study real homogenous linear recurrence with positive coefficients when the sum of the coefficients is one. For a real linear recurrence of order k as in Definition 1.1 let y n = min{x n, x n 1,..., x n k+1 }, n k, z n = max{x n, x n 1,..., x n k+1 }, n k, d n = z n y n, m n = min{a 1 n, a 2 n,..., a k n}, n k + 1, m l n = min{m n, m n 1,..., m n l+1 }, n k + l m n =m k 1 n, n 2k 1. Lemma 3.1. Let a n n k+1 be a fixed sequence with a n R k + such that c n = a j n = 1. Let x n be the homogeneous linear recurrence of order k associated to the sequence a n n k+1 with initial values x 1, x 2,..., x k, that is x n+k = a 1 n+k x n+k 1 + a 2 n+k x n+k a k n+k x n. Then d n+k 1 d n 1 m n+k 1 for every n k. Proof. Let y n = min{x n, x n 1,..., x n k+1 } z n = max{x n, x n 1,..., x n k+1 } for n k. It is clear that y n x n+1 z n. It follows that z n+1 z n, y n+1 y n d n+1 d n.

8 16 Alexru Mihail 8 where If m n+k 1 = 0 the result is obvious. So, we can suppose m n+k 1 > 0. We want to estimate d n+k 1 = z n+k 1 y n+k 1 = x i x j, max i,j=n,n+k 1 x n+i = a 1 n+ix n+i 1 + a 2 n+ix n+i a k n+ix n+i k, i = 1, k 1, under the conditions a 1 n+i + a 2 n+i + + a k n+i = 1, i = 1, k 1, m n+i a 1 n+i, m n+i a 2 n+i,..., m n+i a k n+i, i = 1, k 1, y n x n+1 i z n, i = 1, k. Let us consider the function d n+k 1 : d n+k 1 t 1,..., t k 1, t k = where x i : k i=1 k [y n, z n ] R defined by i=1 max x i t 1,..., t k 1, t k x j t 1,..., t k 1, t k, i,j=n,n+k 1 [y n, z n ] R for i n k + 1 are the functions defined such that x i t 1,..., t k 1, t k is the value of the ith term of the linear recurrence if x n+1 i = t i for i = 1, k that is x i are defined inductively by by x i t 1,..., t k 1, t k = a 1 i x i 1 t 1,..., t k 1, t k + +a 2 i x i 2 t 1,... t k 1, t k + + a k i x i k t 1,..., t k 1, t k, i n + 1, x i t 1,..., t k 1, t k = t n+1 i, i = n k + 1, n. It is easy to see that x i are linear functions. Because x = maxx, x, d n+k 1 is the maximum of a finite family of linear functions defined on a compact convex set. After the linear programming theory it results that the maximum is taken into an extreme point of the convex set. So, we can suppose that x n+1 i = z n or x n+1 i = y n for i = 1, k. If m n+k 1 > 0 then m n > 0 y n < x n+1 < z n. Also we have y n y n+i 1 < x n+i < z n+i 1 z n for i = 2, k 1. It follows that d n+k 1 < d n. Let l be such that d n+l < d n d n+l 1 = d n. This means that z n+l 1 = z n+l 2 = = z n y n+l 1 = y n+l 2 = = y n. Then z n+l < z n or y n+l > y n. Let us suppose that z n+l < z n. In this case, for j = 1, l x n+j = a 1 n+jx n+j 1 + a 2 n+jx n+j a k n+jx n+j k 1 m n+j z n+j 1 + m n+j y n+j 1 = 1 m n+j z n + m n+j y n = = z n m n+j d n z n m n+k 1 d n.

9 9 On linear recurrences with positive variable coefficients in Banach spaces 17 Also from the assumption that x n+1 i = z n or x n+1 i = y n for i = 1, k it follows that x n+1 i = y n for i = 1, k l. If there is an i {1, 2,..., k l} such that x n+1 i = z n we should have z n+l = z n. Then z n+l z n m n+k 1 d n. Finally, d n+k 1 d n+l = z n+l y n+l z n+l y n z n m n+k 1 d n y n = d n 1 m n+k 1. Proposition 3.1. Let a n n k+1 be a fixed sequence with a n R k + x n be the homogeneous linear recurrence of order k associated to the sequence a n n k+1 with initial values x 1, x 2,..., x k, that is x n+k = a 1 n+k x n+k 1 +a 2 n+k x n+k a k n+k x n such that c n = a j n = 1. If m n = then the sequence x n is convergent to a finite limit l x n l d k 1 mlk 1+p, l=2,[ n p where p N is fixed n p + 2k 2. n k+1 Proof. It is clear that y n x n+1 = a 1 n+k x n+k 1 + a 2 n+k x n+k a k n+k x n z n. It follows that z n+1 z n, y n+1 y n d n+1 d n. To prove the convergence it is enough to prove that d n = z n y n 0 when n because [y n+1, z n+1 ] [y n, z n ]. From Lemma 3.1, d n+k 1 d n 1 m n+k 1 for every n k. It follows for n 2 that d nk 1+p d k+p 1 1 mlk 1+p dk 1 mlk 1+p. l=2,n l=2,n Because + = m n = k m lk 1+p there is a p {1, 2,..., k} p=1 such that m lk 1+p = +. It follows that d n = z n y n 0 when n. Let l = lim x n. Then x n l d n d n [ n p x n l d k 1 mlk 1+p. l=2,[ n p k 1+p 4. THE GENERAL HOMOGENOUS CASE WHEN THE SUM OF THE COEFFICIENTS IS 1 In this part we study real homogenous linear recurrence in Banach space with positive coefficients when the sum of the coefficients is one.

10 18 Alexru Mihail 10 Theorem 4.1. Let X, be a normed space. Let a n n k+1 be a fixed sequence with a n R k + x n X be the homogeneous linear recurrence of order k associated to the sequence a n n k+1 with initial values x 1, x 2,..., x k X, such that c n = a j n = 1. If m n = then x n is convergent to a limit l X x n l d k p N is fixed, n p + 2k 2 d k = max i, l=2,[ n p x i x j. 1 mlk 1+p, where Proof. Let us suppose first that X is finite dimensional. Let ϕ : X R be a linear function. Then ϕx n is the homogeneous linear recurrence of order k associated to the sequence a n n k+1 with initial values ϕx 1, ϕx 2,..., ϕx k R. From Proposition 3.1 ϕx n is convergent to a finite limit l ϕ ϕx n l ϕ max ϕx i ϕx j 1 mlk 1+p. i, l=2,[ n p Since max ϕx i ϕx j ϕ max x i x j = ϕ d k we have i, i, ϕx n l ϕ ϕ d k 1 mlk 1+p. l=2,[ n p Since X is finite dimensional, x n is convergent to a limit l X ϕl = l ϕ for every linear function ϕ : X R. It follows that x n l d k 1 mlk 1+p. l=2,[ n p This ends the proof in the case of finite dimensional spaces. The case when X is not finite dimensional can be reduced to the case when X is finite dimensional because x n x 1, x 2,..., x k for every n 1. Corollary 4.1. Let X, be a normed space. Let a n n k+1 be a fixed sequence with a n R k + x n X be the homogeneous linear recurrence of order k associated to the sequence a n n k+1 with initial values x 1, x 2,..., x k X, such that c n = a j n = 1. Let K n = conv {x n, x n 1,..., x n k+1 } for n k l be the limit of the sequence x n. Then K n+1 K n K n = {l}. In particular, we have l max{ x 1, x 2,..., x k }.

11 11 On linear recurrences with positive variable coefficients in Banach spaces 19 Proof. We have x n+1 K n so K n+1 K n. The rest result from the convergence of the sequence x n. 5. THE GENERAL INHOMOGENEOUS CASE WHEN THE SUM OF THE COEFFICIENTS IS 1 The following two lemmas are technical results that give us the possibility to reduce the inhomogeneous case to the homogenous one. Lemma 5.1. Let X, be a Banach space. Let f : N N be an increasing function such that lim fn =. Let xm n X, l m m 1 X for n m N be sequences of vectors d n,m for m N, t m m 0 be sequences of positive numbers such that 1 xm n is convergent to l m; 2 m 0 l m is convergent; 3 xm n l m d n,m t m for m fn; 4 m 0t m is convergent; 5 d n,m 0 when n for every m N; 6 there exists a M > 0 such that d n,m < M for every n, m N with m fn. Then the sequence y n = xl n is convergent to m 0l m. l=0,fn Proof. Let ε > 0 be fixed. From 2 4 there exists m 0 N m 1 N such that m 0 m 1, t m < ε 3M l m < ε 3. There m m 0 m m 1 also exists n ε m such that fn ε m for every n n ε, ε d n,m 3 for m = 0, m 0. Then for n n ε we have m 0 tm y n l m m 0 m fn+1 m fn+1 l m + l m + m=0,fn m=0,fn xm n l m d n,m t m l m + d n,m t m + d n,m t m m fn+1 m=0,m 0 m=m 0 +1,fn l m + + M t m m m 1 m=0,m 0 d n,m m=0,m 0 t m m=m 0 +1,fn

12 20 Alexru Mihail 12 ε l m + 3 m m 1 m 0 Notation. l=m,n t m a l = 0 m=0,m 0 t m + M l=m,n m m 0 +1 a l = 1 if m > n. t m ε 3 + ε 3 + ε 3 = ε. Lemma 5.2. Let X, be a Banach space. Let f : N N be an increasing function such that lim fn = fn n. Let xm n X n for m N, l m m 0 X be sequences of vectors d n,m for m N, t m m 0, c m m 1 s m m 0 be sequences of positive numbers such that 1 xm n is convergent to l m; 2 m 0s m is convergent; 3 l m s m t m s m ; 4 xm n l m d n,m t m for m fn; 5 d n,m = c l for m fn 1; l=m+1,fn 6 c n 0, 1]. Let y n be the sequence is convergent l=0,fn xl n. Then the series l m m 0 y n m 0l m min m 0 =0,fn min m 0 =0,fn min m 0 =0,fn m m 0 +1 m m 0 +1 m m 0 +1 s m + l=m 0 +1,fn s m + l=m 0 +1,fn s m + c l c l l=m 0 +1,fn m=0,m 0 m=0,m 0 t m t m cl l=m+1,m 0 c l t m. m 0 Proof. Since l m s m the series m is convergent it m 0 m 0 m 0s results that the series m 0l m is convergent. Let n be fixed m 0 be a natural number such that m 0 fn. As in the proof of Lemma 5.1, we have y n l m m 0 m fn+1 l m + m=0,m 0 d n,m t m + m=m 0 +1,fn d n,m t m.

13 13 On linear recurrences with positive variable coefficients in Banach spaces 21 Then y n l m m 0 m m 0 +1 m fn+1 s m + m m 0 +1 s m + m m 0 +1 m=0,m 0 l=m 0 +1,fn s m + s m + l=m+1,fn c l l=m 0 +1,fn m=0,m 0 c l c l t m + t m cl l=m+1,m 0 m=0,m 0 t m l=m 0 +1,fnc l m 0 t m. m=m 0 +1,fn s m Because m 0 is an arbitrary natural number such that m 0 fn we obtain the desired conclusion. Remark 5.1. If the conditions from Lemma 5.2 are fulfilled c l = 0 then the conditions from Lemma 5.1 are also fulfilled. The following lemma is obvious. Lemma 5.3. Let a n n k+1 be a fixed sequence with a n R k + x n be the homogeneous linear recurrence of order k of real numbers associated to the sequence a n n k+1 with initial values x 1, x 2,..., x k R, that is x n+k = a 1 n+k x n+k 1 + a 2 n+k x n+k a k n+k x n such that c n = a j n = 1. Then x n [min{x 1, x 2,..., x k }, max{x 1, x 2,..., x k }]. Lemma 5.4. Let X, be a Banach space. Let a n n k+1 be a fixed sequence with a n R k + b n n k+1 be a sequence of elements from X x n be the linear recurrence of order k associated to the sequence a n with initial values x 1 = x 2 = = x k = 0, that is x n+k = a 1 n+k x n+k 1 + a 2 n+k x n+k 2 + +a k n+k x n +b n. We suppose that c n = a j n = 1 b n = 0 for n k + 1. Then sup x i x j b j. i,j 1 Proof. Let us consider the following linear recurrences xl n for l {1, 2,..., k} defined by xl 1 = 0, xl 2 = 0,..., xl k = 0 for l {1, 2,..., k} xl n+k = a 1 n+k xl n+k 1 + a 2 n+k xl n+k a k n+k xl n + b n+k δ l n+k. Then x n = x1 n + x2 n + + xk n.

14 22 Alexru Mihail 14 Also xl n = yl n b l, where yl n are the real linear recurrences defined by yl 1 = 0, yl 2 = 0,..., yl k = 0 for l {1, 2,..., k} yl n+k = a 1 n+k yl n+k 1 + a 2 n+k yl n+k a k n+k yl n + δn+k l. It follows that x n = yl n b l. From Lemma 5.3, yl n [0, 1]. The rest is obvious. l=1,k Definition 5.1. We denote by A the set of all sequences a n with a n R k + c n = a j n = 1 for n k + 1. Definition 5.2. Let X, be a Banach space. Let b 1, b 2,..., b k be elements from X. Let a n be an arbitrary sequence with a n R k + x n be the linear recurrence of order k associated to the sequence a n with initial values x 1 = x 2 = = x k = 0, defined by x n+k = a 1 n+k x n+k 1 + a 2 n+k x n+k 2+ +a k n+k x n+δnb 1 1 +δnb δnb k k such that c n = a j n = 1 for n k. x n depends on the sequence a n n so we can write x n = x n an n. Let v m b 1, b 2,..., b k = sup max x i an n xj an n a n n A i,j=k+1,m for m N m k + 1. Lemma 5.5. With the notations from the above Definition 5.2 we have 1 [ ] v m b 1, b 2,..., b k = sup max yi l, an n yj l, an n bl, a n n A i,j=k+1,m l=1,k where y m l, a n n are the real linear recurrences defined by m 1 for l {1, 2,..., k} y 1 l, am m = y 2 l, am m = = yk l, am m = 0 y n+k l, am m = a 1 n+k y n+k 1 l, am m + + a 2 n+k y n+k 2 l, am m + + a k n+k y n l, am m + δ l n+k ; 2 v m b 1, b 2,..., b k b j ; 3 v m b 1 + e 1, b 2 + e 2,..., b k + e k v m b 1, b 2,..., b k + e j ; 4 v m b 1, b 2,..., b k v 2k b 1, b 2,..., b k for m 2k. Proof. 1 It is easy to see that x i an n = y i l, an n bl. l=1,k

15 15 On linear recurrences with positive variable coefficients in Banach spaces results from the fact that y i l, an n [0, 1] see Lemma For m 2k the recurrence is homogenous we have x m+p conv{x m, x m 1,..., x m k+1 } conv{x 2k, x 2k 1,..., x k } if p N. Lemma 5.6. Let X, be a Banach space. Let a n n k+1 be a fixed sequence with a n R k + b n n k+1 be a sequence of elements from X. Let x n be the linear recurrence of order k associated to the sequence a n n k+1 with initial values x 1, x 2,..., x k, that is x n+k = a 1 n+k x n+k a k n+k x n + b n+k such that c n = a j n = 1. Let d n = sup x i x j. i,j=1,n Then d k + b l d k+p, where p N. l=k+1,k+p Proof. It is enough to give the proof in the case p = 1. The general case results by induction. We have d k+1 = max d k, sup x n+1 x i i=1,n x n+1 x i a 1 n+1 x n x i + a 2 n+1 x n 1 x i + + +a k n+1 x 1 x i + b n+1 d k + b n+1. It follows that d k+1 d k + b n+1. The following theorem extends the results of the homogenous case to the inhomogeneous case when the sum of the coefficients of the linear recurrence is 1 it is the main result of this section. Theorem 5.1. Let X, be a Banach space. Let a n n k+1 be a fixed sequence with a n R k +, b n n k+1 be a sequence of elements from X x n be the linear recurrence of order k associated to the sequences a n n k+1 b n n k+1 with initial values x 1, x 2,..., x k, that is x n+k = a 1 n+k x n+k 1 + a 2 n+k x n+k a k n+k x n + b n+k, such that c n = a j n = 1. If m n = b n < then x n is convergent to a limit n 2k 1 a X x n a [ n p min m 0 =0 [ n p min m 0 =0 n k+1 l m 0 k 1+p+1 l m 0 k 1+p+1 b l + d k + [ n p r l,p b l + d k + l k+1 [ n p b l l=m 0 +1 l=m mlk 1+p 1 mlk 1+p,

16 24 Alexru Mihail 16 where r l,p = v 2k 0X, b lk 1+p+1, b lk 1+p+2,..., b lk 1+p+k 1, p N is fixed, n p + 2k 2 d k = max x i x j. i, Proof. The idea of the proof is to decompose the inhomogeneous linear recurrence x n into a countable sum of homogenous linear recurrences. Then using Theorem 4.1 we can estimate the speed of the convergence of these recurrences. From Lemma 5.1 we obtain the convergence from Lemma 5.2 we can estimate the speed of the convergence of the inhomogeneous linear recurrence x n. It is enough to make the proof only in the case p = 1. For the general case let us consider the linear recurrence of order k, x n, associated to the sequences a n n k+1 b n n k+1 with initial values x 1, x 2,..., x k, that is x n+k = a 1 n+k x n+k 1+a 2 n+k x n+k 2+ +a k n+k x n+ b n+k, where x n = x n+p 1, a j n = a j n+p 1 b n = b n+p 1. We have c n = a j n = c n+p 1 = 1. Applying the results of the Theorem to the linear recurrence x n for p = 1 taking account that d k + b l d k+p 1 = sup x i x j i,+p 1 l=k+1,k+p 1 from Lemma 5.6 we can obtain the results of the Theorem for the linear recurrence x n for a general p. Let us consider the sequences xl n for l N defined by x0 n+k = a 1 n+k x0 n+k 1 + a 2 n+k x0 n+k a k n+k x0 n, x0 1 = x 1, x0 2 = x 2,..., x0 k = x k for l = 0 xl n+k = a 1 n+k xl n+k 1 + a 2 n+k xl n+k a k n+k xl n+ +b n+k δ lk 1+2 n+k + δ lk 1+3 n+k + + δ l+1k 1+1 n+k, xl 1 = 0, xl 2 = 0,..., xl k = 0 for l > 0. [ Then x n = x0 n + x1 n + + x n 2 ] k 1. n From Theorem 4.1 Corollary 4.1 the sequences xl n for l N are convergent to the limits al such that a0 x j, i=1,k

17 17 On linear recurrences with positive variable coefficients in Banach spaces 25 al x0 n a0 d k j=lk 1+2,l+1k 1+1 j=2,[ n 1 xl n al v 2k 0X, b lk 1+2,..., b lk 1+k b j for l > 0, 1 mjk 1+1 for l = 0 j=l+1,[ n 1 1 mjk 1+1 for l > 0 n l + 1k We can apply Lemma 5.2 with x n as y n, xm n as xm n, am as l m, r l,1 = v 2k 0X, b lk 1+2, b lk 1+3,..., b l+1k 1+1 as tl for l 1, 1 m lk 1+1 as cl, s l = b j as s l for l 1, x j as s 0, j=lk 1+2,l+1k 1+1 d k as t 0 [ ] n 1 k 1 for fn. It results that the series x n a min m 0 =0,[ n 1 where a = n 0an. l m 0 +1 s l + d k + r l,1 n 0 l=m 0 +1,[ n 1 i=1,k an is convergent 1 mlk 1+1, The last inequality follows from the fact that see Lemma v 2k 0X, b lk 1+2, b lk 1+3,..., b l+1k 1+1 b j. j=lk 1+2,l+1k 1+1 It remains us to prove the convergence of the sequence x n. Since + = m n = k m lk 1+p there is a p {1, 2,..., k} such that n k p=1 m lk 1+p = +, so 1 mlk 1+p = 0 for every m0. From l m 0 +1 Remark 5.1 conditions from Lemma 5.1 are also fulfilled. Therefore the convergence results from Lemma THE GENERAL INHOMOGENEOUS CASE The next result is obvious. Lemma 6.1. Let n N, ε > 0 a 1, a 2,..., a n, a be such that, nε 1, a i ε for i 1, n, a 0 a 1 + a a n = 1 + a. Then there exists

18 26 Alexru Mihail 18 b 1, b 2,..., b n such that b 1 + b b n = 1 b i a i ε if a < 0 or a i b i ε if a > 0. Theorem 6.1. Let X, be a Banach space. Let a n n k+1 be a fixed sequence with a n R k +, b n n k+1 be a sequence of elements from X x n be the linear recurrence of order k associated to the sequences a n n k+1 b n n k+1 with initial values x 1, x 2,..., x k X, that is x n+k = a 1 n+k x n+k 1 + a 2 n+k x n+k a k n+k x n + b n+k. We suppose that c n = a j n = 1 + a n a n <. If m n = b n < n k+1 n k+1 then x n is convergent to a limit l X x n l A m 0 p + B p min m 0 =0,[ n p min A m 0 p + d k + m 0 =0,[ n p l k+1 where p N is fixed, n p + 2k 2, A m 0 p = M = j k+1 B p = l=m 0,[ n p bl + M a l l m 0 k 1+p+1 n k+1 1 mlk 1+p l=m 0,[ n p bl + M a l, d k+p 1 + rl,p b + Ms l,p, 1 mlk 1+p, max{1, c j } max x ij + 1 b l, m n = min i=1,k k, m n, r l,p b = v 2k 0X, b lk 1+p+1, b lk 1+p+2,..., b lk 1+p+k 1, d k+p 1 = s l,p = sup x i x j i,+p 1 j=lk 1+p+1,l+1k 1+p a j. Proof. As in the proof of Theorem 5.1 we can make the proof of the inequality only in the case p = 1. We prove first that x n is bounded. Let us suppose that X = R x n R +. Let y n = min{x n, x n 1,..., x n k+1 } for n k z n = max{x n, x n 1,..., x n k+1 } for n k.

19 19 On linear recurrences with positive variable coefficients in Banach spaces 27 Then x n+k = a 1 n+k x n+k 1 + a 2 n+k x n+k a k n+k x n + b n+k a 1 n+k z n+k 1+ a 2 n+k z n+k 1+ + a k n+k z n+k 1+ b n+k = c n+k z n+k 1 + b n+k z n+k max{x n+k, z n+k 1 } max{c n+k z n+k 1 + b n+k, z n+k 1 } z n+k 1 max{1, c n+k } + b n+k. Now, it is easy to see that z n z k max{1, c j } + j=k+1,n z k + b l max j=k+1,n i=k+1,n max{1, c j } b i+1 j=i+1,n z k + b l max{1, c j } j k+1 max{1, c j }, where max{1, c j } = 1. j=n+1,n This proves that x n is bounded by z k + b l max{1, c j } j k+1 when X = R x n R +. If X = R x n R, let x n be the homogeneous linear recurrence of order k associated to the sequences a n n k+1 R k b n n k+1 R with initial values x 1 = x 1, x 2 = x 2,..., x k = x k defined by x n+k = a 1 n+k x n+k 1 + a 2 n+k x n+k a k n+k x n + b n+k. It follows that x n x n max j + x b l max{1, c j }. j k+1 Let X be a Banach space let ϕ : X R be a linear continuous function with ϕ 1. Then ϕx n is the homogeneous linear recurrence of order k associated to the sequence a n n k+1 with initial values ϕx 1, ϕx 2,..., ϕx k R; that is ϕx n+k = a 1 n+k ϕx n+k 1 + a 2 n+k ϕx n+k a k n+k ϕx n + ϕb n+k. From the previous, ϕx n is bounded by ϕxj + ϕb l max{1, c j } max j k xj + b l max{1, c j } = M. From the Hahn-Banach theorem, it follows that the sequence x n is bounded in norm by M. Indeed, if there exists x m such that x m > M there j k

20 28 Alexru Mihail 20 also exists ϕ : X R a linear continuous function such that ϕ = 1 ϕx m = x m > M, which is a contradiction. Using Lemma 6.1 when a n+k 0 with m n+k = min 1 k, m n+k for ε, k for n, a 1 n+k, a2 n+k,..., ak n+k, a n+k for a 1, a 2,..., a n, a we obtain e 1 n+k, e2 n+k,..., ek n+k such that e 1 n+k + e2 n+k + + ek n+k = 1 ei n+k ai n+k m n+k if a n+k < 0 or a i n+k ei n+k m n+k if a n+k > 0. Then where x n+k = a 1 n+k x n+k 1 + a 2 n+k x n+k a k n+k x n + b n+k = = e 1 n+k x n+k 1 + e 2 n+k x n+k e k n+k x n + d n+k, d n+k = b n+k + a 1 n+k e1 n+k xn+k a k n+k ek n+k xn. If a n+k = 0 we take a i n+k = ei n+k d n+k = b n+k. In this way we have founded two sequences e n n k+1 with e n R k + d n n k+1 with d n X such that e j n = 1, e j n m n = min 1 k, m n, dn+k b n+k M a n+k m n n 2k 1 x n+k = e 1 n+k x n+k 1 + e 2 n+k x n+k e k n+k x n + d n+k. We also have m n+k = min 1 k, m n+k = min mn+k, m n+k 1,..., m n+1. The new recurrence fulfills the conditions from Theorem 5.1. Indeed = if only if m n = d n < since n 2k 1 n k+1 b n < a n+k <. n k+1 Let r l b = r l,1 b = v 2k 0 X, b lk 1+2, b lk 1+3,..., b l+1k 1+1, s l = s l,1 = a j r l d = r l,1 d = v 2k d lk 1+2, d lk 1+3,..., j=lk 1+2,lk 1+k d lk 1+k. From Lemma we have r l d r l b + Ms l from Lemma we have r l d max d j max b j + Ms l. j=lk 1+2,l+1k 1+1 j=lk 1+2,l+1k 1+1

21 21 On linear recurrences with positive variable coefficients in Banach spaces 29 From the Theorem 5.1 the sequence x n is convergent to a limit l X x n l min d l + d k + [ n p l d 1 mlk 1+1 r min m 0 =0,[ n 1 m 0 =0,[ n 1 l m 0 k 1+2 min m 0 =0,[ n 1 l m 0 k d k + rl b + Ms l A m d k + l k+1 l=m 0 +1,[ n 1 l=m 0 +1 bl + M a l + bl +M a l 1 mlk 1+1 l=m 0 +1,[ n 1 1 mlk 1+1. REFERENCES [1] Tom Apostol, Mathematical Analysis. Addison-Wesley Publishing, Massachusetts, USA, [2] Alexru Mihail Alexru-Petre Tache, On linear recurrence with variable coefficients. Anal. Univ. Bucureşti Mat , [3] Gheorghe Siretchi, Calcul diferential şi integral. Vol. 1, 2. Editura Ştiinţifică şi Enciclopedică, Bucureşti, Received 21 October 2008 University of Bucharest Faculty of Mathematics Computer Science Str. Academiei Bucharest, Romania mihail alex@yahoo.com