A GAUSS-KUZMIN-LÉVY THEOREM FOR A CERTAIN CONTINUED FRACTION

We consider an interval map which is a generalization of the well-known Gauss transformation. In particular, we prove a result concerning the asymptotic behavior of the distribution functions of this map.


Introduction.
In 1800, Gauss studied the following problem.In modern notation, it reads as follows.Write x ∈ [0, 1) as a regular continued fraction ( Here, a k ∈ {1, 2,...}.Define the map (now known as the Gauss transformation or the continued fraction map) T G : [0, 1) → [0, 1) as follows: for x = 0, T G 0 := 0; for x = 0, we have Note that T G removes the first digit and the first level of x = [a 1 ,a 2 ,...] G .By T G , we generate a sequence of x k in the unit interval with Assuming that the "seed," x 0 , is a random real number uniformly distributed in the unit interval, define the distribution function In his notebook, Gauss remarked, after some numerical computations, that "they [F k ] come out so complicated that no hope appears to be left."(See Knuth [17, page 346]).Twelve years later, in a letter he wrote to Laplace, Gauss stated, without proof, that However, he was unable to describe the behavior of F k for a large but finite k.At the time, Gauss considered the study of F k a problem he could not resolve to his satisfaction.
A century later, a proof was finally provided by Kuzmin [18].In the same paper, he actually proved that, for large k, one has where r n (x) = O(q √ n ), and 0 < q < 1. Around the same time, Lévy [19], by using a different method that employs probabilistic notions, proved that Note that in the 60's, Szüsz [27] was able to prove the same result by using Kuzmin's approach.The asymptotic behavior of F k (x) was finally resolved in 1974 by Wirsing [30] and a complete solution to Gauss's problem was found a few years later by Babenko [1].For detailed introductions, see [6,8,10,11,12,13,14,15,16,17,23,25].For various extensions and generalizations, see [4,5,9,20,21,22,24,26,28,29].See also the monographs [15,25].Just as in the original Gauss-Kuzmin-Lévy Problem, the hard part of these generalizations often involves finding the explicit expressions of the distribution functions.Finally, we remark that the Gauss transformation has strong ties with chaos theory [2,3].
In this note, we consider generalization of the Gauss transformation and prove an analogous result.Write x ∈ [0, 1) as Here, a k are natural numbers (see the next section for details) and one should think of a k as the digits of x.Define the generalized Gauss transformation T : [0, 1) → [0, 1) as follows: for x = 0, T 0 := 0; for x = 0, we have Likewise, we can generate a sequence of x k in the unit interval with Assuming that the "seed," x 0 , is a random real number uniformly distributed in the unit interval, define the distribution function Note that G 0 (t) = t.Our main result is the following theorem.
In Section 2, we set up the necessary machinery, and in Section 3, we prove Theorem 1.1.

Preliminaries.
In this section, first, we show in Lemma 2.1 that x ∈ [0, 1) can be written in the form of (1.8).
Lemma 2.1.For all x ∈ [0, 1), there exist integers Proof.For any x ∈ [0, 1), we can find a natural number a 1 such that This implies, for some p ∈ [0, 1), ), we can repeat the same iteration and obtain Thus, the lemma is proven.
We remark that, from the above proof, it follows that the digits {a 1 (x), a 2 (x),...} are related by where Next, we want to prove the convergence of expansion of the type of (2.1).Define [a 1 ,a 2 ,...,a n ], the convergent of x, by truncating the expansion on the right-hand side of (2.1).We want to show (2.7) To this end, define integer-valued functions P n (x), Q n (x) by the following: ) (2.10) Note that the left-hand side of (2.9) is a compact notation of continued fractions of the type of (1.8) with k levels.By combining Lemma 2.1 and (2.9), we have, for x ∈ [0, 1), where t = T n x.Taking t = 0 in (2.11) gives By combining (2.10) and (2.12), we have where t = T n x.Note that this equation, which measures the difference between x and its convergent, is the key ingredient of the following estimate.
Proof.By using (2.13) and the fact that t −1 ≥ 1, we have We claim that (2.15) Indeed, (2.16) Note that, in obtaining the first inequality, we have used the fact that In obtaining the second inequality, we have used the fact that This proves the claim.By direct computation, we have s 1 = 2 −1−a 1 ≤ 2 −1 .This, with (2.15), shows that s n ≤ (1/2) n and so the lemma is proven.
A few remarks are as follows.First, it is clear that every irrational x ∈ [0, 1) has a unique expansion of the type of (2.1).
Second, we note that some particular cases of this type of continued fractions have been studied before.For example, by setting q = 1/2 and a k = k, the right-hand side of (1.8) gives the well-known continued fraction of Rogers and Ramanujan: (2.17 Another example is the beautiful result due to Davison [7].Let a k = F k , where F k is the kth Fibonacci number.Davison showed that where φ is the Golden Ratio and • denotes the floor function.Third, we give an example.In terms of the continued fraction of the type of (1.8), we have Here, in the first equality, we gave only the first three digits.In the second equality where we used the compact notation in (1.8), we gave the first ten digits.Next, we prove the following lemma.

.20)
Remark 2.5.As expected, the integral of ρ is the first term of the right-hand side of (1.13), that is,

Proof of Theorem 1.1.
Here, we follow [23, pages 152-155]; see also [10,17,25].First, by following the same trick that is used in the original Gauss-Kuzmin-Lévy problem, one can show that {G k (t)}, defined in (1.12), satisfy a Kuzmin-type equation Just as in the original Gauss-Kuzmin-Lévy problem, it is easier to work with the derivative of G k (t).To this end, we observe that since the derivative of G 0 (t) = t is bounded in the unit interval, we can show by induction that the derivative of G k (t) is also bounded in the unit interval.
This allows us to differentiate (3.1) term-by-term, obtaining Here, the prime denotes the derivative with respect to t. Next, we introduce f k (t) in such a way that In terms of f k (t), (3.2) can be written as where Here, ∆ m := γ m − γ 2m .Note that it follows from the definition of p m (t) that it is manifestly nonnegative for all t ∈ [0, 1) and for all natural numbers m.Also, we have used partial fraction decomposition in obtaining the second equality.These formulae fit into the overall strategy as follows.Introduce a function R k (t) such that Here, c is the constant in Theorem 1.1.Because G k (0) = 0 and G k (1) = 1, we have R k (0) = R k (1) = 0. To prove the theorem, we have to show that where 0 < q < 1.To achieve this goal, we proceed as follows.First, by comparing the derivatives of (3.3) and of (3.6), we obtain Next, by using (3.4), we can show that (the details will be given below) for 0 < q < 1.With (3.8), this implies R k = O(q k ).Finally, by an interpolation formula where 0 < ξ < 1, we arrive at (3.7), and Theorem 1.1 is proven.All we need to do is to prove (3.9), as promised.First, we note that from (3.4), we have This, with the application of the mean value theorem to the difference enables us to rewrite (3.11) as with where q := 1 + γ 2 This, with (3.20), implies f k (t) = O(q k ), that is, (3.9).This completes the proof of Theorem 1.1.