A note on singularity of a recently introduced family of Minkowski s question-mark functions

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1 Dynamical systems A note on singularity of a recently introduced family of Minkowski s question-mark functions Juan Fernández Sánchez a, Wolfgang Trutschnig b a Grupo de Investigación de Análisis Matemático, Universidad de Almería, La Cañada de San Urbano, Almería, Spain b Department for Mathematics, University Salzburg, Hellbrunnerstrasse 34, 5020 Salzburg, Austria. Tel: , Fax: Received *****; accepted after revision Presented by Abstract We point out a mistake in the proof of the main theorem in a recent article on a family of generalized Minkowski s question-mark functions, saying that each member of the family is a singular homeomorphism, and provide two alternative proofs, one based on ergodicity of the Gauss map G and the α-lüroth map L α, and another one focusing more on classical properties of continued fraction expansions. Résumé Note sur la singularité d une famille de fonctions Minkowski s question-mark récemment introduite. Nous mettons en évidence une erreur dans la démonstration du théroème principal dans un article récent traitant d une famille de fonctions Minkowski s question-mark généralisées, stipulant que chaque membre de la famille est un homéomorphisme singulier, et nous produisons deux preuves alternatives, l une basée sur l ergodicité de l application de Gauss G et de l application α-lüroth L α, l autre se focalisant plus sur des propriétés classiques des décompositions de fractions continues.. Introduction Based on continued fraction- and generalized Lüroth expansions a new family of Minkowski questionmark functions was recently introduced in []. When proving the main theorem of the paper (Theorem.3) the author correctly shows that each member? α of the family is a strictly increasing homeomorphism of the unit interval [0,] and then tackles proving that? α is singular (in the sense that? α has derivative zero λ-a.e). Unfortunately, the presented proof of singularity is incorrect, implying that the theorem is still formally unproven. Main objective of this note is to provide two correct proofs of singularity of? α for every partition α. The rest of this note is organized as follows: We first introduce some notation (essentially following [] with some slight modifications) in Section 2, point out what exactly went wrong in [] and then prove Theorem.3 in [] using two different methods in Section 3. addresses: juan.fernandez@ual.es (Juan Fernández Sánchez), wolfgang@trutschnig.net (Wolfgang Trutschnig). Preprint submitted to the Académie des sciences June 28, 207

2 2. Notation In the sequel, B([0,]) will denote the Borel σ-field on [0,] and λ will denote the Lebesgue measure on B([0,]). We will write N =,2,3,... and will refer to Σ := (N ) N as code-space. As usual, G : [0,] [0,] will denote the Gauss map, defined by G(0) = 0 and G(x) = x x for x (0,]. Set s i = i for every i N and s = 0. Then the intervals I = s,i = (s 2,s ],I 2 = (s 3,s 2 ],... form a partition γ G of [0,]. Coding orbits of G via γ G, the continued fraction expansion cf : [0,] Σ is defined by setting cf(x) = a = (a,a 2,a 3,...) Σ if and only if G i (x) I ai holds for every i N. It is well known that G is strongly mixing (hence ergodic) w.r.t. the absolutely continuous probability measure µ G with density 2 +x for x [0,] (see [3]). Inthesequelwewilllet α = J,J,J 2,J 3,...denotepartitionsoftheunit intervalinduced bystrictly decreasing sequences (t i ) i= converging to 0 =: t and fulfilling t =, i.e. we have J = t,j = (t 2,t ],J 2 = (t 3,t 2 ],... For each such partition α the α-lüroth map L α is defined by L α (0) = 0 as well as L α (x) = tj x t j t j+ for x (t j+,t j ] and j N. Coding orbits of L α via α, the α-lüroth expansion Lür α : [0,] Σ is defined by setting Lür α (x) = a = (a,a 2,a 3,...) Σ if and only if L i α (x) J a i holds for every i N. Additionally to being strongly mixing w.r.t. λ, the α-lüroth map is even (isomorphic to) a Bernoulli shift (see [2]). Moreover (again see [2]) the transformation Φ α : Σ [0,], defined by Φ α (a) = t a + m ( ) m+ t am (t aj t aj+), () m=2 with the convention t + = t, fulfills Φ α Lür α = id [0,]. j= 3. Two proofs of singularity of? α Based on cf : [0,] Σ and Φ α : Σ [0,], in [] the author introduces the (generalized) questionmark function? α by setting? α (x) = Φ α cf(x) for every x [0,] and then states the following theorem. Theorem 3. ([]) Given a partition α as above, the map? α : [0,] [0,] is an increasing singular homeomorphism fulfilling L α? α =? α G. Moreover, if t,t 2,... Q, then? α maps the set A 2 of all quadratic surds into Q. In [] a correct proof for the fact that? α is an increasing homeomorphism and for the assertion concerning A 2 is given. It is well known that singularity can be shown by establishing the existence of a Borel set B [0,] fulfilling λ( B) = 0 and λ(? α ( B)) =. Letting µ α denote the pull-back of λ via? α (or, equivalently, the push-forward of λ via? α ), defined by µ α (B) = λ(? α (B)) for every Borel set B B([0,]), the author then uses the identity dx =? α (x)dx (2)? α(b) toprovethatµ α andµ G aresingularwithrespecttoeachother.eq.(2),however,iseasilyseentobewrong and should be? dx = α(b) B dµ α = µ α (B) instead. In fact, considering, for instance, B = [0,], we get? dx = α([0,]) [0,] dx =, whereas the right-hand side of eq. (2) obviously fulfils [0,]? α is a homeomorphism of [0,] too. Since the rest of the proof in [] builds upon eq. (2), an alternative? α method is needed to show singularity of? α. 2 B (x)dx < since

3 We will now provide two proofs of singularity of? α for every partition α - the first one uses ergodicity of G and L α and explicitly constructs a Borel set B with the afore-mentioned properties, the second one is more elementary and directly derives the fact? α(x) = 0 for λ-a.e. x [0,] via properties of continued fraction expansions. Proof a: We distinguish two different types of partitions α: Type I: There exists k N such that t k t k+ (k+)2 2 k(k+2). Let the Borel sets Λ and Γ be defined by Λ= x [0,] : lim n ( k+, k ] G i (x) = (k +)2 2 k(k +2) holds for every k N n Γ= x [0,] : lim (tk+,t k ] L i α(x) = t k t k+ holds for every k N. (4) Ergodicity of G w.r.t. µ G and of L α w.r.t. λ (see [2,3]) and the fact that µ G is absolutely continuous with strictly positive density implies λ(λ) = λ(γ) =. Considering that? α (x) = Φ α cf(x) holds for every x (0,] and that for every x Λ we have cf(x) Lür α (Γ),? α (Λ) Γ c follows. Hence, choosing B = Λ c directly yields λ( B) = 0 as well as λ(? α ( B)) λ(γ) =, which completes the proof for all partitions α of Type I. Type II: For every k N we have t k t k+ = (k+)2 2 k(k+2). Taking into account t = we get that there is only one partition α of Type II, namely the one fulfilling t k = for every k N. Instead of considering asymptotic frequencies of single digits of the Lüroth- and continued-fraction expansions we now consider the asymptotic frequency of the block (, ) and set k+2 2 k+ Λ= x [0,] : lim n ( 2,]2 (Gi (x),g i+ (x)) = n Γ= x [0,] : lim ( 2,]2 (Li α (x),li+ ( 4 α (x)) = 3 2 (3) (5) ) 2. (6) According to Proposition 4..2 in [3] we have λ(λ) =. Moreover, using the fact that L α is (isomorphic to) a Bernoulli shift and that (t t 2 ) 2 = ( 4 ) 3 2 ( 4) 2 holds, λ(γ) = follows. Considering and proceeding as in the second part of the first case and setting B = Λ c completes the proof. Remark Instead of using Proposition 4..2 in [3] and the fact that L α is a Bernoulli shift in order to prove λ(λ) = λ(γ) = for α of Type II, we could as well consider the maps ε G,ε Lα : [0,] [0,] 2, defined by ε G (x) = (x,g(x)) and ε Lα (x) = (x,l α (x)), and directly work with ergodicity of G and L α. In fact, applying Birkhoff s ergodic theorem ([4]) to the indicator function f : x ( ε G(x) directly 2,]2 yields that for µ G -a.e. x [0,] (hence for λ-a.e. x [0,]) we get n lim n ( 2,]2 (Gi (x),g i+ (x))= lim f G i (x) = 2 [0,] Proceeding analogously with the function f : x ( 2,]2 ε L α (x) shows λ(γ) =. f(x) +x dx =

4 Proof b: Let Λ denote the set of all points x (0,) at which? α is differentiable and define Λ according to eq. (3). For every k N define two new Borel sets Λ k,λk 3 by Λ k = x [0,] : lim n n Λ k 3 = x [0,] : lim ( 2,] ( k+, k ] (G i (x),g i+ (x)) = 2 ( 4, 3 ] ( k+, k ] (G i (x),g i+ (x)) = 2 k+ + k+2 + k k+ + k+ 3k+4 + k 3k+ Following the same reasoning as in Remark (or using Proposition 4..2 in [3]) we get λ(λ k Λ k 3) =, implyingthat Ω := Λ Λ k= (Λk Λk 3 ) fulfillsλ(ω) =.Fixx Ω andseta := cf(x) Σ. Additionally, for every n 3 let x n,y n Q [0,] be defined by x n = [a,a 2,...,a n ] := a + a a n, y n = [a,a 2,...,a n +] := a + a a n+. (7) Then we have lim n x n = lim n y n = x as well as x (x n,y n ) for n even and x (y n,x n ) for n odd. Setting x n = pn q n with p n,q n relatively prime and using the recurrence relations (.2) in [2] we get y n = (an+)pn +pn 2 (a n+)q n +q n 2 = pn+pn q n q n, which implies x n y n = ( ) n+ q n(q n+q n ). On the other hand, considering? α (x n ) = Φ α ((a,a 2,...,a n,,,...)) and? α (y n ) = Φ α ((a,a 2,...,a n +,,,...)), using eq. () yields n δ n :=? α(x n )? α (y n ) x n y n = ( )n+ n j= (t a j t aj+) ( ) n+ q n(q n+q n ) = q n (q n +q n ) j= (t aj t aj+) 0 (8) Since by construction? α is differentiable at x, obviously lim n δ n =? α(x) 0 holds. Suppose now that? α (x) > 0. Then lim n δn δ n = follows, from which, again using the recurrence relations (.2) in [2], we get = lim n δ n δ n = lim t an+) = lim t an+) q n (q n +q n ) q n (q n +q n 2 ) = lim t an+) qn 2 + qn qn 2 q n + qn 2 q n ( an + qn 2 q n ) 2 +(an + qn 2 q n ) + qn 2 q n. (9) Fix an arbitrary k N. Letting let (n j ) j N denote the subsequence of all indices with a nj = k, eq. (9) simplifies to ( q nj 2) 2 q nj 2 k + q nj +(k + q ) nj = (t k t k+ ) lim j + qn j 2 q nj By constructionofωwehavethat (a nj,a nj ) = (,k) is fulfilled infinitely often and that the sameis true for (a nj,a nj ) = (3,k). Using the same notation as in eq. (7), accordingto [2], qn j 2 q nj = [a n j,...,a 2,a ] holds, from which we conclude that qn j 2 q lies infinitely often in ( nj 2,] and infinitely often in ( 4, 3 ]. This contradicts eq. (0), implying that we can not have? α (x) > 0 if x Ω. Since x Ω was arbitrary, we have shown that? α = 0 holds λ-a.e., which completes the proof. 4 (0)

5 References [] A. Arroyo, Generalised Lüroth expansions and a family of Minkowski s question-mark functions, C. R. Acad.Sci.Paris, Ser.I 353 (205) [2] K. Dajani, C. C. Kraaikamp, Ergodic Theory of Numbers, Carus Mathematical Monographs 29, The Mathematical Association of America, [3] M. Iosifescu, C. Kraaikamp, Metrical Theory of Continued Fractions, Springer Science+Business Media Dordrecht, [4] P. Walters, An Introduction to Ergodic Theory, Springer New York,