On fixed points of a generalized multidimensional affine recursion

Let $G$ be a multiplicative subsemigroup of the general linear group $\Gl(\mathbb{R}^d)$ which consists of matrices with positive entries such that every column and every row contains a strictly positive element. Given a $G$--valued random matrix $A$, we consider the following generalized multidimensional affine equation R\stackrel{\mathcal{D}}{=}\sum_{i=1}^N A_iR_i+B, where $N\ge2$ is a fixed natural number, $A_1,...,A_N$ are independent copies of $A$, $B\in\mathbb{R}^d$ is a random vector with positive entries, and $R_1,...,R_N$ are independent copies of $R\in\mathbb{R}^d$, which have also positive entries. Moreover, all of them are mutually independent and $\stackrel{\mathcal{D}}{=}$ stands for the equality in distribution. We will show with the aid of spectral theory developed by Guivarc'h and Le Page and Kesten's renewal theorem, that under appropriate conditions, there exists $\chi>0$ such that $\P(\{>t\})\asymp t^{-\chi},$ as $t\to\8$, for every unit vector $u\in\mathbb{S}^{d-1}$ with positive entries.


Introduction and statement of the results
We consider the Euclidean space R d endowed with the scalar product x, y = d i=1 x i y i , the norm |x| = x, x , and its Borel σ-field Bor(R d ). We say that R d ∋ x = (x 1 , . . . , x d ) ≥ 0 is positive (resp. R d ∋ x = (x 1 , . . . , x d ) > 0 is strictly positive), when x n ≥ 0, (resp. x n > 0) for every 1 ≤ n ≤ d. By R d + we denote the set of all positive vectors, and we define the set S + = R d + ∩ S d−1 of all positive vectors on the unit sphere S d−1 = {x ∈ R d : |x| = 1} with the distance being the restriction of the Euclidean norm to S + . Given x ∈ R d we denote its projection on S d−1 by x = x |x| . Let Gl(R d ) be the group of d×d invertible matrices on R d , with the operator norm · associated with the Euclidean norm | · | on R d , i.e. a = sup x∈S d−1 |ax| for every a ∈ Gl(R d ).
Suppose that G is a multiplicative subsemigroup of Gl(R d ) which consists of matrices with positive entries such that every column and every row contains a strictly positive element. By G • we denote the multiplicative subsemigroup of G composed of matrices with strictly positive entries. It is easy to see that G provides a projective action on S + which is given by Let A be a G-valued random matrix distributed according to a probability measure µ on G, and B be a random vector independent of A, taking its values in R d + . Let A 1 , . . . , A N and B 0 be independent random variables, where N ≥ 2 is a fixed natural number, A 1 , . . . , A N are independent copies of A, and B 0 is an independent copy of B.
This research project was partially supported by MNiSW grant N N201 392337.
The aim of this paper is to find a random vector R ∈ R d + , independent of A and B, which solves in law D = a generalized multidimensional affine equation i.e.
In order to avoid repetitions in the sequel, and shorten article we have decided to state all necessary definitions and notations in the introduction, and formulate our main results as general as it is possible.
Let M 1 (G) denotes the set of all probability measures on G endowed with the weak topology. We denote by suppµ the support of the measure µ ∈ M 1 (G). If E ⊆ G, let [E] be the subsemigroup of G generated by the set E. For n ∈ N let S n = A n · . . . · A 1 ∈ G, where A 1 , A 2 , . . . ∈ G is a sequence of independent copies of G-valued random matrix A distributed according to µ.
A subsemigroup [suppµ] of G is called contractive if [suppµ] ∩ G • = ∅. In other words, The condition (1.3) was considered by Hennion [9], Hennion and Hervé [10] in the context of limit theorems for the products of positive random matrices.
An element a ∈ Gl(R d ) is proximal if there exists a unique eigenvalue λ a (the dominant eigenvalue) of a, such that r(a) = lim n→∞ a n 1/n = |λ a |.
According to the Perron-Frobenius theorem [11] every a ∈ G • is proximal. Moreover, for every a ∈ G • and its adjoint a * ∈ G • it is possible to choose v a , w a ∈ R d + such that v a > 0, w a > 0 and av a = λ a v a , a * w a = λ a w a , v a , w a = 1, |w a | = 1.
The eigenvector v a determined by these relations will be called the dominant eigenvector of a ∈ G • . This means that we can write R d = R · v a ⊕ v ⊥ a , and the spectral radius of a, restricted to v ⊥ a = {x ∈ R d : x, v a = 0} is strictly less than |λ a |. Furthermore, by the preceding relations we have lim n→∞ a n r(a) n = v a ⊗ w a , (1.4) where v a ⊗ w a is the matrix projector on R · v a . Since v a ⊗ w a x = x, w a v a for every x ∈ R d , (1.4) immediately yields lim n→∞ a n · x = v a ⊗ w a x |v a ⊗ w a x| = v a |v a | = v a ∈ S + , for every x ∈ R d . (1.5) A subsemigroup Γ ⊆ Gl(R d ) is strongly irreducible if there does not exist a finite number (k ∈ N) of proper linear subspaces V 1 , . . . , V k of R d such that If E ⊆ Gl(R d ) we denote by E prox the set of all proximal elements of E. A subsemigroup Γ ⊆ Gl(R d ) is said to satisfy condition (i − p) if Γ is strongly irreducible and Γ prox = ∅. This condition was widely investigated by Guivarc'h and Le Page [5] and [6], see also [7,8,3] and the references given there.
For s ≥ 0 we write where µ * n is the n-th convolution power of µ ∈ M 1 (G). The limit above exists and it is equal to inf n∈N G a s µ * n (da) 1 n , because u n (s) = G a s µ * n (da) is submultiplicative, i.e. u m+n (s) ≤ u m (s)u n (s) for every m, n ∈ N. Moreover, Let s ∞ = sup {s ≥ 0 : κ µ (s) < ∞} ∈ R + ∪ {∞}, then by the Hölder inequality I µ = [0, s ∞ ) or I µ = [0, s ∞ ]. For technical reasons we have to assume that there is s 1 ≤ 1 2 such that E( A s1 ) ≤ 1 N . Our "existence" result is the following Theorem 1.7. Assume that A is a G-valued random matrix distributed according to a probability measure µ on G, and B is a random vector independent of A, taking its values in R d + , such that P({B > 0}) > 0. Let A 1 , . . . , A N and B 0 be independent random variables as in (1.1), where N ≥ 2 is a fixed natural number, A 1 , . . . , A N are independent copies of A, and B 0 is an independent copy of B. Suppose further that [suppµ] ⊆ G satisfies condition (C) and there exist s 1 ∈ (0, 1/2], and Then there exists a unique vector R ∈ R d + and its independent copies Section 3 contains a detailed proof of Theorem 1.7, which is similar in spirit to that of [12]. However, the multidimensional framework, we consider, provides some difficulties which do not appear in the one dimensional case. Namely, the method developed in [12], which gives finiteness of appropriate moments for the solution of (1.1), breaks down in higher dimensions. This problem will be dealt with the help of condition (C).
Let λ d be the Lebesgue measure on R d . If ν is a probability measure on R d , then by ν = ν a + ν s we denote its Lebesgue decomposition with respect to λ d , where ν a is the absolutely continuous part with respect to λ d , i.e. ν a ≪ λ d , and ν s is the singular part with respect to λ d , i.e. ν s ⊥ λ d . We have also ν a ⊥ ν s . Since ν is positive then its total variation ν = ν(R d ) = 1. We say that the measure ν is singular if ν s = 1, otherwise ν is nonsingular, i.e. ν s < 1. Now we can state our main "tail" result. Theorem 1.8. Fix a natural number N ≥ 2, a G-valued random matrix A distributed according to µ, and a random vector B with law η, independent of A, taking its values in R d + , such that P({B > 0}) > 0.
As we mentioned before, the proof is based on concepts of [3] with considerable complications determined by the structure of equation (1.1). The most important tool which allows us to establish relation (1.9) is Kesten's renewal theorem [16].We need to check that its assumptions are satisfied (see Section 4). This is the most difficult part of the paper and requires the spectral theory of transfer operators developed by Guivarc'h and Le Page ( [5], [6] and [3]), which is summarized in Section 2. But we touch only a few aspects of their theory and restrict our attention to the results which will be used in Sections 3 and 4. Guivarc'h and Le Page approach significantly simplifies and clarifies proofs developed by Kesten in [15], and what is most important for us, it is applicable to our situation.

Transfer operators
Let C(S + ) be the space of continuous functions on S + with the supremum norm | · | ∞ .
Given a closed subset V of S + , M 1 (V ) denotes the set of all probability measures on V , endowed with the weak topology. We say that U ⊆ S + is a subspace of S + , if U = V ∩ S + for some subspace V ⊆ R d . A measure ν ∈ M 1 (S + ) is said to be proper if ν(U ) = 0 for every subspace U S + . Here and subsequently, Λ(Γ) = {v a ∈ S + : v a is the dominant eigenvector of a ∈ Γ prox }, where Γ is a subsemigroup of G such that Γ prox = ∅.
The following Proposition 2.1 due to Guivarc'h and Raugi [7] (see also [8]) contains the relevant properties of (i − p) semigroups which will be used in the sequel.
The main purpose of this section is to summarize a number of properties of operators P s , P s * , see Theorem 2.3 below. x, y s ν s * (dy), for x ∈ S + .
(iii) I µ ∋ s → e s ∈ C(S + ) is continuous in the uniform topology.
• Moreover, there exists a unique stationary measure π s ∈ M 1 (S + ), (π s * ∈ M 1 (S + )) for ). This result was proved by Guivarc'h and Le Page and its, quite long and far from being obvious, proof can be found in [5] and [6]. Notice that in view of the cocycle property σ s (x, a 2 a 1 ) = σ s (x, a 1 )σ s (a 1 · x, a 2 ), a 1 , a 2 ∈ G, x ∈ S + of (2.4) σ s (x, a) = |ax| s e s (a · x) e s (x) , the Markov operators Q s and Q s * defined in Theorem 2.3 can be rewritten in the following form n ∈ N, x ∈ S + , a ∈ G and φ is an arbitrary measurable function on S + .

Construction of the solution
Recall that A stands for a G-valued random matrix distributed according to the measure µ ∈ M 1 (G), and B for a random vector taking its values in R d + , independent of A. In this section we construct a solution of the equation (1.1). The idea of the construction goes back to [12]. It is not difficult to imagine that we have to study a sequence of random variables that are obtained by iterating (1.1). Let N ≥ 2 be a fixed natural number and R * 0,1 , . . . , R * 0,N be independent and identically distributed (i.i.d.) copies of the initial random variable R * 0 ∈ R d + . We consider the sequence (R * n ) n≥0 such that A n+1,k R * n,k + B n+1 , for every n ≥ 0, (3.1) where A n+1,1 , . . . , A n+1,N , B n+1 and R * n,1 , . . . , R * n,N , n ≥ 0 are independent. Moreover, for n ≥ 1 R * n,1 , . . . , R * n,N are i.i.d. copies of R * n from the previous iteration. For n ≥ 0 A n+1,1 , . . . , A n+1,N are i.i.d. copies of A and B n+1 is an independent copy of B.
We will look more closely at the sequence (R * n ) n≥0 . Let A = {A i1,...,in : (i 1 , . . . , i n ) ∈ {1, . . . , N } n , n ∈ N} be the set consisting of i.i.d. copies of A, and B = {B i1,...,in : (i 1 , . . . , i n ) ∈ {1, . . . , N } n , n ∈ N} ∪ {B 0 } the set consisting of i.i.d. copies of B independent of A. Additionally we assume that A 0 = Id a.s. and the initial random variable R * 0 is always independent of A, B, A and B. Now and for n ≥ 0 be the partial sum of the sequence (W n ) n≥0 . Then is a candidate for a solution of (1.1). It is not hard to see that W n satisfies where A k and W n−1,k are independent of each other and W n−1,1 , . . . , W n−1,N have the same distribution as W n−1 . In view of the above calculations, R (n) satisfies the recursion To obtain a solution with an initial condition, let . Now we have following Lemma 3.8. Assume now that (R * n ) n≥0 and (R (n) ) n≥0 are the sequences defined in (3.1) and (3.3) respectively, then for every n ∈ N we have Proof. Observe that for n = 1, (3.9) follows from definition. For more details we refer to [12]. Now we have simple, but very useful Proof. We refer to [5].
To take the limit in (3.4) we need an estimate for E (|W n | s ). Suppose for a moment that s ≤ 1. Then, in view of inequality (3.11), we have We would like to show that for an appropriate s > 0, not necessarily less or equal 1, E (|W n | s ) decays exponentially. This is contained in Lemma 3.12. For the sake of computations we have to assume that there exist s 1 ∈ (0, 1/2] such that E( A s1 ) ≤ 1 N . Lemma 3.12. Assume that [suppµ] ⊆ G satisfies condition (C), and there exist s 1 ∈ (0, 1/2], and Then for every s ∈ (s 1 , s 2 ), there exist finite constants K s > 0 and η < 1 such that for every n ∈ N Proof. By Theorem 2.3 κ(s) is strictly log-convex so N κ(s) < 1, for every s ∈ (s 1 , s 2 ) and for s ≤ 1, (3.13) follows from the calculation above. From now we assume that s ∈ (1, s 2 ) and it is fixed.
. . , N } n and n ∈ N. We order the set of indices writing {1, . . . , N } n = {i 1 , . . . , i N n } and we choose p ∈ N and p ≥ 2, such that Observe that by the inequality (3.11), there exist constants Since j i1 , j i2 , . . . , j i N n ∈ {0, 1, . . . , p}, the constants above do not depend on n ∈ N and we may define Therefore, When N n > p, (3.14) also holds with the universal constant c p p,s which does not depend on n ∈ N, but we have to estimate in a more subtle way. Before we do that we need to introduce a portion of necessary definitions. For every r ≤ k, and j 1 ≤ . . . ≤ j k , let L(j 1 , . . . , j k ) = k l 1 , l 2 , . . . , l r , Then it is not difficult to see that for every k ≤ p Let now η = max{η 1 , η 2 , . . . , η p } < 1, where This implies that Hence in this case (3.13) follows with K s = 2 p−1 p!c p p,s E (|B| s ) < ∞ and η < 1.
Proof of Theorem 1.7. First of all we show that E(|R| s ) < ∞ for every s < s 2 . By Lemma 3.12 there exist η < 1 and K s < ∞ such that for every n ∈ N we have E(|W n | s ) ≤ K s η n . Observe now It immediately implies that E(|R| s ) < ∞, which in turn gives |R| < ∞ a.s.. Now we want to show that R is the unique solution of (1.1). It is enough to show that R * n , with arbitrary initial random variable for an arbitrary uniformly continuous function f defined on R d . Fix ε > 0, and choose δ > 0 such that since ε > 0 is arbitrary we have shown E(f (R * n )) − −−→ n→∞ E(f (R)), and Theorem 1.7 follows.

Application of Kesten's renewal theorem
In order to prove Theorem 1.8, as mentioned in the introduction, we will use Kesten's renewal theorem [16] which allows us to describe the desired tail asymptotic (1.9). Before we state Kesten's theorem we have to introduce necessary definitions and to prove a number of auxiliary results. They are contained in the three lemmas of Section 4.1 and they will be used later on to check that the assumptions of Kesten's renewal theorem are satisfied in our settings. The material presented in this section is adapted form [3], [5], [6] and [15]. 4.1. Some general results. At first we define the probability space Ω = G N . Bor(X) stands for the Borel σ-field of the space X. For any sequence ω = (a 1 , a 2 , . . .) ∈ Ω we write S n (ω) = a n · . . . · a 1 ∈ G, for n ∈ N and S 0 (ω) = Id ∈ G.
The cocycle property gives a very useful relation, i.e. for every m, n ∈ N, x ∈ S + and ω ∈ Ω we have q s m+n (x, ω) = q s n (x, S n (ω))q s m (S n (ω) · x, S m (θ n (ω))). (4.1) The Kolmogorov's consistency theorem guarantees the existence of the probability measure Q s x on Ω being the unique extension of measures q s k (x, g)µ * k (dg). Next we define the probability measure where π s is the unique Q s stationary measure on S + (see Theorem 2.3). By E s x we denote the expectation corresponding to Q s x . We extend the probability space Ω to a Ω = S + × Ω. Let a θ : a Ω → a Ω be the shift defined by a θ(x, ω) = (a 1 · x, θ(ω)), for every x ∈ S + and ω = (a 1 , a 2 , . . .) ∈ Ω.
We now define the probability measure a Q s on a Ω as follows In the same way, starting with µ * instead of µ, we define the measure Q s, * x , and E s, * x denotes its expectation. Moreover, the probabilities Q s, * and a Q s, * are defined similarly, i.e.
From now we will work with the measures Q s, * x , π s * , Q s, * and a Q s, * . Clearly, all the results stated below remain valid for the measures Q s x , π s , Q s and a Q s . We begin with following Proof. We can repeat the argument from Section 3 in [3].  Proof. Observe that (4.6) implies (4.5). Indeed, let and let Z c x be the complement of Z x . Then by Lemma 4.3 The proof of (4.6) is adapted from [15]. Conditions (1.3) yields the existence of n 0 ∈ N and 0 < τ < 1 such that where P * = µ ⊗N * . First of all we need to show that Q s, * x ({ω ∈ Ω : T (ω) < ∞}) = 1, for every x ∈ S + and (4.8) Notice that (4.8) immediately gives (4.9), since the event {T < ∞} does not depend on x ∈ S + and Q s, * = S + Q s, * x π s * (dx). Assume for a moment that (4.9) holds and prove (4.6). If x ∈ S + such that x > 0 then for any a ∈ G we have hence (4.6) holds with C = d −1/2 min 1≤i≤d x i > 0. Now fix an arbitrary x ∈ S + and let Ω 1 = {T < ∞} ⊆ Ω. By assumption, Q s, * (Ω 1 ) = 1. It is easy to see that S T (ω * )x > 0 for ω * ∈ Ω 1 . Thus for any n ≥ T and ω * ∈ Ω 1 we have It implies that |S n (ω * )x| ≥ C T,x (ω * ) S n (ω * ) holds with the constant C T,x (ω * ) > 0 independent of n ≥ T , for every ω * ∈ Ω 1 . Recall that G is the multiplicative semigroup of d × d invertible matrices with positive entries such that every row and every column contains a strictly positive element. Now take n ≤ T and notice that C n,x (ω * ) = |Sn(ω * )x| Sn(ω * ) > 0, for every ω * ∈ Ω 1 by the definition of G and x ∈ S + . Therefore, we take C(ω * ) = min{C 1,x (ω * ), . . . , C T,x (ω * )} > 0, and (4.6) follows.
We need only to prove (4.8). In this purpose we define the events We show that there exists γ ∈ [0, 1) such that for all l ∈ N Q s, * x ({T > ln 0 }) ≤ Q s, * x ({E jn0 does not occur for any 0 ≤ j < l}) ≤ γ l .
Hence in view of Remark 4.2, on the one hand, by the Birkhoff ergodic theorem (applied to a Q s, * and a θ) we obtain a Q s, * (x, ω) ∈ a Ω : lim On the other hand by the Kingman subadditive ergodic theorem (applied to Q s, * and θ) we have Q s, * ω ∈ Ω : lim n→∞ 1 n log S n (ω) = α s = 1.
A function g : S + × R → R is called direct Riemann integrable (dRi), if it is Bor(S + ) × Bor(R) measurable and for every fixed x ∈ S + and 0 < L < ∞ the function t → g(x, t) is Riemann integrable on [−L, L], and satisfies For the reader's convenience we formulate Kesten's renewal theorem [16]. There exists π s * ∈ M 1 (S + ) such that π s * Q s * = π s * and for every open set U ⊆ S + with π s * (U ) > 0, Q s, * x (X n ∈ U for some n ∈ N) = 1 for every x ∈ S + . • Condition I.2 and for all x ∈ S + , • Condition I.3 There exists a sequence {ζ i } ⊂ R such that the group generated by ζ i is dense in R and such that for each ζ i and λ > 0 there exists y = y(ζ i , λ) ∈ S + with the following property: for each ε > 0, there exists an A ∈ Bor(S + ) with π s * (A) > 0 and m 1 , m 2 ∈ N, τ ∈ R such that for any • Condition I.4 For each fixed x ∈ S + , ε > 0 there exists r 0 = r 0 (x, ε) > 0 such that for all real valued functions f measurable with respect to Bor (S + × R) N and for all y ∈ S + with |x − y| < r 0 one has: If a function g : S + × R → R is jointly continuous and (dRi), then for every for α(s) defined in (4.19).
In the next four subsections we indicate how the material developed in Section 2 and 4.1, under the hypotheses of Theorem 1.8, may be used to check the assumptions of Theorem 4.18. From now we will work with the measures Q χ, *
Therefore, by the Borel-Cantelli lemma we obtain that for every x ∈ S + Q χ, * x ω ∈ Ω : lim inf This shows that α(χ) > 0 Q χ, *  Proof. We give only a sketch of the proof, for more details we refer to [3]. First of all we prove that C k = S + , for some sufficiently large k ∈ N, (C k was defined in (4.17)). Then obviously (4.23) implies (4.16). There is a finite number N 1 of points such that Then Q χ, * (Ω ′ ) = 1, by Lemma 4.4 and 4.13. Take any y ∈ S + , then there exists 1 ≤ i ≤ N 1 such that y ∈ B(x i , 2). This implies the existence of m 0 ∈ N such that Q χ, * ω ∈ Ω : log |S n (ω)y| n > α(χ)/2, for all n ≥ m 0 with the constant c > 0 defined in Lemma 4.3. Taking any 1/k ≤ min{α(s)/2, 1/m 0 } Lemma 4.22 follows.

Proof of the main Theorem
In this section we give a detailed proof of Theorem 1.8. For that we consider the following smooth version of P({ R, u > t}) be the space of all bounded measurable functions on S + × R. Define a linear operator Θ : B(S + × R) → B(S + × R) given by the formula Observe that for every n ∈ N Θ n f (u, t) = E χ, * u (f (X n , t − V n )) . First we express G(u, t) as a potential of a function g(u, t) that turns out later on to be direct Riemann integrable. Proof. First of all we show G 0 (u, t) = ΘG(u, t). Indeed, e χ (u) |A * u| χ = ΘG(u, t).

Now we have
where S n = A n · . . . · A 1 . By the continuity of I µ ∋ s → κ(s) (see Theorem 2.3) we can find p < χ, such that κ(p) = 1−ε N , for some ε > 0, then This implies that Θ n G(u, t) = N n e t e χ (u) Now it is easy to see that for any n ∈ N we have G(u, t) = g(u, t) + Θg(u, t) + Θ 2 g(u, t) + . . . + Θ n−1 g(u, t) + Θ n G(u, t), and (5.5) follows. This completes the proof of Lemma 5.4.
Lemmas 5.8 and 5.16 below imply that g(u, t) is direct Riemann integrable. Lemmas 5.7, 5.12 and 5.14 contain some necessary technicalities. Proof. At the beginning, we assume that the law η of B is nonsingular, i.e. η s < 1. Let ν be the law of R and µ be the law of A ∈ G. Let * be the classical convolution on R d . Moreover, we define ξ = µ * G ν, where µ * G ν(D) = G R d 1 D (ax)ν(dx)µ(da) and D ∈ Bor(R d ). Obviously ξ defines a probability measure on R d which coincide with the distribution of AR. Notice that ν = ξ * N * η, since R D = N i=1 A i R i + B, and observe that by the Lebesgue decomposition we obtain ν a + ν s = ν = (ξ a + ξ s ) * N * (η a + η s ) = N n=0 N n ξ * n a * ξ * (N −n) and by its uniqueness ν s = (ξ * N s * η s ) s . This gives ν s ≤ ξ s N η s . Again by the Lebesgue decomposition and its uniqueness we have ξ = µ * G ν = µ * G ν a +µ * G ν s , hence ξ s = (µ * G ν) s ≤ µ * G ν s ≤ ν s . Now combining ν s ≤ ξ s N η s and ξ s ≤ ν s we get ν s ≤ ν s N η s , if ν s > 0, then 1 ≤ ν s N −1 η s ≤ η s < 1. This contradiction shows that ν s = 0 hence ν is absolutely continuous with respect to the Lebesgue measure, which in turn implies that P({ R, u = r}) = 0, for every (u, r) ∈ S + × R + ∪ {0}.
If the law η of B is singular, i.e. η s = 1, then for fixed (u, r) ∈ S + × R + ∪ {0}, we have P({ R, u = r}) = 0, since P({ B, u = r}) = 0. Now we prove that S d−1 ∋ u → P({ R, u > r}) is continuous. Take any (u n ) n∈N ⊆ S + such that lim n→∞ u n = u ∈ S + and consider The same arguments work for u → P({ AR, u > r}), since A ∈ G is independent of R.
Lemma 5.8. Under the assumptions of Theorem 1.8, there exists 0 < β 1 < 1 such that for every β ∈ [0, β 1 ), there is a finite constant C β > 0, such that for every (u, t) ∈ S + × R we have and In the proof we extend the approach developed in [12].
Proof. Let β 1 ∈ (0, min{1, χ/2}) and take any 0 ≤ β < β 1 . Then for every t > 0 Let us define F (y) = P({ AR, u > y}), and γ = χ + β − β 1 , and notice N P , and for some c > 0 Clearly, 1 < χ+β γ , and β 1 < χ/2 implies γ = χ + β − β 1 ≥ χ/2 + β/2, hence χ+β γ < 2. Then We have shown that I 1 ≤ C β e −βt , for every β ∈ [0, β 1 ) with the constant C β > 0 which does not depend on u ∈ S + . A straightforward applications of Fubini theorem yields In order to show the continuity of S + × R ∋ (u, t) → g 1 (u, t) it is enough to prove the continuity of is continuous. Observe that then arguing in a similar way as above with β = 0, and using Lebesgue dominated convergence theorem we obtain the continuity of (5.11) and the lemma follows. Now we are going to prove inequality (5.13) and (5.15), that will provide necessary estimates for Lemma 5.16. The first one was proved in [12] and was sufficient in the one dimensional case discussed there. The second one is more subtle and allows us to deal with our situation.
Proof. As mentioned before the proof is contained in [12].
Then f is µ integrable on X and lim n→∞ X f n dµ = X f dµ.
We will apply Theorem 5.20 with f n (r) = r χ−1 P({ R, u n > r}) − P max Clearly, |f n | ≤ h n for every n ∈ N, and by the previous part of the lemma (h n ) n∈N and h are all integrable. Lemma 5.7 guarantees that lim n→∞ f n (r) = 0 and lim n→∞ h n (r) = h(r). In order to show that lim n→∞  As a consequence of Lemma 5.8 and Lemma 5.16 the function S + × R ∋ (u, t) → g(u, t) is jointly continuous. Moreover, it is possible to find β > 0 and a positive constant C β < ∞ such that |g(u, t)| ≤ C β e −β|t| , for every (u, t) ∈ S + × R, since |g(u, t)| ≤ g 1 (u, t) + g 2 (u, t), for g 1 (u, t) and g 2 (u, t) defined in Lemma 5.8 and Lemma 5.16 respectively. This shows that g(u, t) satisfies condition (4.23). By the Kesten's renewal theorem 4.18 we obtain lim t→∞ G(u, t) = lim t→∞ E χ, * x ∞ n=0 g(X n , t − V n ) = 1 α(χ) S + R g(y, x)dx π χ * (dy) = C χ .
In other words we have proved that for every u ∈ S + lim t→∞ G(u, t) = lim t→∞ 1 e t e χ (u) e t 0 r χ P({ R, u > r})dr = C χ ≥ 0.
It remains to prove that C χ > 0 for every χ ≥ 1. In this purpose notice that We need only to show that We will show that there exists c χ > 0 such that for every x ∈ R d + . Observe that S + ∋ x → S + x, u χ π χ * (du) is continuous and nonzero for every x ∈ S + , since suppπ χ * is not contained in any proper subspace of S + (see Section (2)). This allows us to conclude that x → S + x, u χ π χ * (du) attains its minimum c χ > 0 on S + , and in fact this proves (5.23).