A new approach with new solutions to the Matkowski and Wesołowski problem

Based on a result of de Rham, we give a family of functions solving the Matkowski and Wesołowski problem. This family consists of Hölder continuous functions, and it coincides with the whole family of solutions to the Matkowski and Wesołowski problem found earlier by a different method. Moreover, applying some results due to Hata and Yamaguti and due to Berg and Krüppel, we prove that there are functions solving the Matkowski and Wesołowski problem that are not Hölder continuous.

This question of Jacek Weso lowski was posed in connection with studying probability measures in the plane that are invariant under "winding" (see [15]). It is not difficult to check that Matkowski's problem is equivalent to Weso lowski's question (see [16]). A negative answer to Weso lowski's question (as well as a positive answer to Matkowski's problem) was obtained in [10] and reads as follows.  1] of (e) satisfying (1). (ii) For every p ∈ (0, 1) the function ϕ p : [0, 1] → [0, 1] given by where x n ∈ {0, 1} for every n ∈ N, is an increasing and continuous solution of (e) satisfying (1). Moreover, ϕ p is singular for every p = 1 2 . The proof of Theorem 1.1, and proofs of its generalizations obtained in [16,17] for a much more general functional equation than (e), are based deeply on self-similar measures.
Throughout this paper, for every p ∈ (0, 1) the symbol ϕ p will be reserved for the function ϕ p : [0, 1] → R given by (2). Put From Theorem 1.1 we have W 0 ⊂ C and, so far, no element of the set C outside of the set W 0 was known.
The main purpose of this paper is to show that To do this, we need a new approach for studying the set C. The new approach is rooted in a paper by de Rham, published long before either the above problems was posed.

Application of a result by de Rham
We begin this section by answering both problems based on the following well known result of de Rham. Moreover: for every strictly increasing sequence (γ n ) n∈N0 of positive integer numbers.
From now on, for every p ∈ (0, 1) the symbol R p will be reserved for the function R p : [0, 1] → R given by (4).
Fix p ∈ (0, 1). Adding the equations occurring in (3), we get 1]. Putting x = 1 in both equations of (3), we obtain from the second one R p (1) = 1 and then from the first one R p ( 1 2 ) = p. Moreover, the representation (2) of ϕ p coincides with the representation (4) of R p (see [13, page 268]; see also [11, formula (6)]), and hence for every p ∈ (0, 1). In consequence, Let us mention that the functions R p were investigated at the beginning of the 20th century in [3,6,18]; for more details see [9]. According to (5) we do not have a new solution to the Matkowski and Weso lowski problem yet. To find ϕ ∈ C \ W 0 we need two results. The first one is due to Hata and Yamaguti and the second one is due to Berg and Krüppel. Before we formulate these results, we define the function Φ : (0, 1) From (5) we see that for all p ∈ (0, 1) and x ∈ [0, 1] we also have We will use both of the above formulas for Φ, depending on the needs in the given situation. Theorem 2.2. (see [8, page 195]; see also [12, For every a ∈ (0, 1] we define the function φ a : [0, 1] → [0, 1] by putting The symbol φ a will be reserved for the just defined function. Note that (2) gives for all a ∈ (0, 1] and m ∈ N 0 . We are now in the position to prove the main result of this paper, which reads as follows.
If a ∈ ( 1 2 , 1], then a 0 |x − y| αp dp = If a ∈ (0, 1 2 ], then a similar calculation gives a 0 |x − y| αp dp = 1 This implies that φ a is continuous, and hence that φ a ∈ C. Now we want to prove that φ a / ∈ conv(W 0 ∪{φ b | b ∈ (0, 1) \{a}}). Suppose, towards a contradiction, that this is false. Then there are numbers N ∈ N, Evaluating the above equation at x = 1 2 m yields, jointly with (7) and (2), for every m ∈ N. Dividing both sides of the above equality by a m m+1 , and then passing with m to ∞ we obtain It remains to prove that φ a is not Hölder continuous. Suppose towards a contradiction that there are α, C > 0 such that |φ a (x) − φ a (y)| ≤ C|x − y| α for all x, y ∈ [0, 1]. Then making use of (8), we get for every m ∈ N, a contradiction.

Singularity of solutions
The following lemma is a crucial point in this section. we see that the class C is determined by two of its subclasses C a and C s , consisting of all absolutely continuous and all singular functions, respectively. From assertion (i) of Theorem 1.1 we know that the class C a consist of exactly one function, and it is the identity on [0, 1]. Hence there are α ∈ [0, 1] and ϕ sing ∈ C s such that which completes the proof. Now, we are in a position to prove that for every a ∈ (0, 1] the function φ a is singular. Next, we will have a closer look at the set where its derivative is zero and the set where its derivative is infinite.
For every a ∈ (0, 1] the function φ a is strictly increasing and singular.
Proof. The strict monotonicity of φ a is clear. It remains to prove that φ a is singular.
Assume first that a ∈ (0, 1 2 ]. Then, making use of (7), we obtain Therefore, φ a is singular, by Lemma 3.1. Assume now that a ∈ ( 1 2 , 1]. Since φ 1 2 is singular, what we have just proved, it follows that φ 1 for all p ∈ (0, 1) and y ∈ [0, 1]. This jointly with the monotonicity of φ a and the change of variable theorem gives The Matkowski-Wesolowski problem: new solutions 767 Before we formulate the next result, let us introduce some notions.
If x has two representations, we take the one with only finitely many x k non-vanishing. For all n ∈ N and p, a ∈ (0, 1) we put Moreover, for any increasing function f : assuming that the limits make sense. Our analysis of the sets where the derivative is 0 or ∞ is based on the idea of the proof of the following result, which is due to Kawamura.
2 . Now we can prove the following announced result.
Proof. We follow the idea of the proof of [11, Theorem 2.2] by Kawamura; note that our definition of I n differs from that of [11], because we only sum up to n − 1 instead of n as Kawamura does. Fix a ∈ (0, 1) and x = ∞ k=1 x k Let us first consider all the cases where φ a− (x) or φ a+ (x) vanishes. We begin with the cases where φ a+ (x) = 0. Observe that in each of these cases Theorem 3.3 yields R p+ (x) = 0 for almost all p ∈ (0, a). So if we could apply the Lebesgue dominated convergence theorem, we would get Therefore, we see that to conclude the proof, it is enough to verify that is bounded almost everywhere from above on (0, 1) × (0, 1 − x) by a constant. Let us start the required verification with the dyadic case of x, i.e. where there exists N ∈ N such that x k = 0 for every k > N. Fix As the polynomial Thus we have proved the first part of assertion (i). Now, we pass to the case where x is not dyadic and p D0(x) (1 − p) D1(x) < 1 2 for every p ≤ a. Let (n k ) k∈N be the subsequence of all natural numbers for which x n k = 0; i.e. n k is the address of the k-th 0 in the binary expansion of x. Clearly, lim k→∞ n k+1 n k = 1.
For every k ∈ N we have I n k = n k − k, and further for every m ∈ Z we also have Fix p ∈ (0, 1) (we will later restrict ourselves to p ≤ a) and k ∈ N, and let h be such that 2 −n k+1 ≤ h < 2 −n k . Then Let (n(l)) l∈N be the subsequence of all natural numbers greater than n k for which x n(l) = 1; i.e. n(l) is the address of the l-th 1 appearing after position n k in the binary expansion of x. As n(l) − n k − l ≥ 0 for every l ∈ N, we have Thus As a < D 0 (x), we conclude from (12) that there is k 0 ∈ N such that a ≤ k−1 n k for every k ∈ N with k ≥ k 0 . Assume now that k ≥ k 0 and that p ≤ a. As the polynomial p k−1 (1 − p) n k −k+1 is increasing on the interval [0, k−1 n k ] and p ≤ a, we have Applying (11) and (12) This implies that the sequence 2 n k+1 a k−1 (1 − a) n k −k+1 k∈N is bounded. In consequence Thus we have proved that φ a+ (x) = 0 in the case where x is not dyadic and p D0(x) (1 − p) D1(x) < 1 2 for every p ≤ a, which is the first step of the proof of assertion (iii).
We stay in the same case of assertion (iii), i.e. we assume that x is not dyadic and p D0(x) (1 − p) D1(x) < 1 2 for every p ≤ a. But now we want to prove that φ a− (x) = 0.
Observe first that (9) yields Since in the considered case Theorem 3.3 implies that R p+ (1 − x) = 0 for almost all p ∈ (1 − a, 1) as we will verify (in a moment) in (16), the idea is to follow the proof of φ a+ (x) = 0, replacing x by 1 − x with suitable changes, which we explain in details.
In the following n k refers to the address of the k-th 0 in the binary expansion of 1 − x instead of x. Using (12) for 1 − x, instead of x, we see that there exists k 0 ∈ N such that 1 − a ≥ k n k for every k ∈ N with k ≥ k 0 . Fix k ≥ k 0 and let h be such that 2 −n k+1 ≤ h < 2 −n k . We have shown before (see (13)) that As the polynomial p k−1 (1 − p) n k +1−k is decreasing on the interval [ k n k , 1] and p ≥ 1 − a, inequality (14) holds with a replaced by 1 − a, and hence also (15) holds with a replaced by 1 − a, which is what we need to apply (10) and conclude that φ a− (x) = 0.
It remains to prove all the cases where a one-sided derivative of φ a at the point x equals ∞. We prove these cases in more or less one go.
First of all, we show that in each of these cases for every N ∈ N we can find an open interval L N having 0 as its right or left endpoint (depending on the one-sided derivative of φ a at the point x we are interested in) and a Lebesgue where |P N | denotes the Lebesgue measure of the set P N .
In the case (i), we apply assertion (i) of Theorem 3.3 to find L N having 0 as right endpoint and P N ⊂ (0, 1 2 ); the same is true in the case (ii), and moreover, by assertion (ii) of Theorem 3.3, we can also find L N whose left endpoint is 0 and P N ⊂ ( 1 2 , 1). In the case (iii) we use assertion (iii) of Theorem 3.3 to find L N having 0 as right endpoint as well as to find L N having 0 as left endpoint and P N being the intersection of the interval (0, a) with a neighbourhood of this parameter p ≤ a for which p D0(x) (1 − p) D1(x) > 1 2 . If both sets L N and P N satisfy the above conditions, then Finally, in each of the considered cases we conclude that the one-sided derivative of φ a at the point x equals ∞.

Continuity of the function Φ
Before we construct more new solutions to the Matkowski and Weso lowski problem, let us look at the function Φ. By Theorem 2.1 this function is continuous with respect to the second variable, and from Theorem 2.2 we see that it is also continuous with respect to the first variable. Thus the question reads: Is it continuous? To answer this question we need a result due to Grushka. Theorem 4.1. (see [7]) Let N ∈ N and let a n , b n ∈ R with a n < b n for every n ∈ {1, . . . , N +1}. If f : (a 1 , b 1 )×· · ·×(a N , b N )×(a N +1 , b N +1 ) → R is continuous with respect to each variable separately and monotonous with respect to each of the first N variables separately, then f is continuous. Fix p 0 ∈ (0, 1) and ε > 0. Next choose n ∈ N such that 0 < (p 0 + 1 2 n ) n ≤ ε. Fix (p, x) ∈ (0, 1)×[0, 1] and assume that (p−p 0 ) 2 +x 2 ≤ 1 4 n . Hence p ≤ p 0 + 1 AEM and x ≤ 1 2 n . Then, making use of Theorem 2.1, we obtain which proves that Φ is continuous at the point (p 0 , 0).
Observe that the function Φ cannot be extended continuously to the closed unit square [0, 1] 2 ; indeed and the continuity breaks at the points (0, 1) and (1, 0).

More new solutions
We want to construct more new solutions to the Matkowski and Weso lowski problem different from those presented in Theorem 2.4. For this purpose we fix a Borel probability measure μ on (0, 1). Next, given a Borel measurable function g : (0, 1) → (0, 1), we define a function ψ g : [0, 1] → [0, 1] by putting note that the function is well defined, because by Theorem 2.2 the map p → Φ(g(p), x) is Borel measurable and bounded for every x ∈ [0, 1]. We keep the symbol ψ g for the above defined function to the end of this paper. Given a Borel measurable function g : (0, 1) → (0, 1), we define the nonincreasing rearrangement g * : (0, 1) → [0, 1] of g by putting It is clear that the function g * is non-increasing and right-side continuous.
In view of the above observation let us assume, to the end of this paper, that μ is the one-dimensional Lebesgue measure. Theorem 5.1. Assume that g : (0, 1) → (0, 1) is a Borel measurable function. Then ψ g ∈ C, and it is strictly increasing.
Proof. From Theorem 1.1 we see that ψ g is a strictly increasing function satisfying (e) for every x ∈ [0, 1] with ψ g (0) = 0 and ψ g (1) = 1. To prove that ψ g is continuous we fix x ∈ [0, 1] and a sequence (x n ) n∈N of elements of [0, 1] converging to x. Applying the Lebesgue dominated convergence theorem jointly with Theorem 4.2 we obtain, noting that the integrand is bounded, Therefore, ψ g ∈ C.
Theorem 5.1 produces a very large class of new solutions to the Matkowski and Weso lowski problem. Put and for every s ∈ (0, ∞) we define the function g s : (0, 1) → (0, 1) by putting g s (p) = p s ; the symbol g s will be reserved just for this function. Obviously, for all s ∈ (0, ∞) and m ∈ N.
We evaluate the above expression at the point 1 2 m by applying (17), (2) and (7): α n (p n ) m + β n (a n ) m m + 1 AEM for every m ∈ N. Multiplying both sides by (m + 1)(sm + 1) N n=1 (r n m + 1) we find that there are polynomials P 1 , P 2 and P 3 in m such that P 1 (m) = N n=1 α n P 2 (m)(p n ) m + β n P 3 (m)(a n ) m (18) for every m ∈ N. Let us assume for a contradiction that the leading term of P 1 is α K m K for some K ≥ 1. Then dividing both sides by α K m K and letting m tend to ∞ we have that 1 = 0, which is impossible. Hence P 1 is constant.
Taking the limit as m tends to ∞ on the right-hand side of (18) gives P 1 = 0.
Again, the symbol sin λ will be reserved for the just defined function. One can check that for all λ ∈ (0, 1) and m ∈ N.