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

Abstract

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.

1 Introduction

In 1985 Janusz Matkowski posed the problem asking if the equation

has a non-linear monotonic and continuous solution $\varphi :[0,1]\rightarrow {\mathbb {R}}$ (see [14]). During the 47th International Symposium on Functional Equations in 2009 Jacek Wesołowski asked whether the identity on [0, 1] is the only increasing and continuous solution $\varphi :[0,1]\rightarrow [0,1]$ of (e) satisfying

$$\begin{aligned} \varphi (0)=0\hbox { and }\varphi (1)=1. \end{aligned}$$

(1)

This question of Jacek Wesołowski 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 Wesołowski’s question (see [16]). A negative answer to Wesołowski’s question (as well as a positive answer to Matkowski’s problem) was obtained in [10] and reads as follows.

Theorem 1.1

• (i) The identity on [0, 1] is the only increasing and absolutely continuous solution $\varphi :[0,1]\rightarrow [0,1]$ of (e) satisfying (1).

• (ii) For every $p\in (0,1)$ the function $\varphi _p:[0,1]\rightarrow [0,1]$ given by

  $$\begin{aligned} \varphi _p\left( \sum _{n=1}^\infty \frac{x_n}{2^n}\right) =\sum _{n=1}^{\infty }x_n p^{n-\sum _{i=1}^{n-1}x_i}(1-p)^{\sum _{i=1}^{n-1}x_i}, \end{aligned}$$

  (2)

  where $x_n\in \{0,1\}$ for every $n\in {\mathbb {N}}$, is an increasing and continuous solution of (e) satisfying (1). Moreover, $\varphi _p$ is singular for every $p\ne \frac{1}{2}$.

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 $p\in (0,1)$ the symbol $\varphi _p$ will be reserved for the function $\varphi _p:[0,1]\rightarrow {\mathbb {R}}$ given by (2). Put

$$\begin{aligned} {\mathcal {W}}_0={{\,\mathrm{conv}\,}}(\{\varphi _p\,|\,p\in (0,1)\}), \end{aligned}$$

where ${{\,\mathrm{conv}\,}}(A)$ denotes the convex hull of a set A, and

$$\begin{aligned} {\mathcal {C}}=\{\varphi :[0,1]\rightarrow [0,1]\,|\,&\varphi \hbox { is a continuous and increasing solution}\\&\hbox { of } (\textsf {e}) \hbox { satisfying }(1)\}. \end{aligned}$$

From Theorem 1.1 we have ${\mathcal {W}}_0\subset {\mathcal {C}}$ and, so far, no element of the set ${\mathcal {C}}$ outside of the set ${\mathcal {W}}_0$ was known.

The main purpose of this paper is to show that

$$\begin{aligned} {\mathcal {C}}\setminus {\mathcal {W}}_0\ne \emptyset . \end{aligned}$$

To do this, we need a new approach for studying the set $\mathcal C$. The new approach is rooted in a paper by de Rham, published long before either the above problems was posed.

2 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.

Theorem 2.1

(see [4]; cf. [9]) For every $p\in (0,1)$ there exists exactly one bounded function $R_p:[0,1]\rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} R_p(\frac{x}{2})=p R_p(x)&{}\hbox { for every }x\in [0,1],\\ R_p(\frac{x+1}{2})=(1-p)R_p(x)+p &{}\hbox { for every }x\in [0,1]. \end{array}\right. } \end{aligned}$$

(3)

Moreover:

• (i) $R_p$ is continuous, strictly increasing and singular for every $p\ne \frac{1}{2}$.

• (ii) $R_\frac{1}{2}={{\,\mathrm{id}\,}}_{[0,1]}$.

• (iii) $R_p(0)=0$ and

  $$\begin{aligned} R_p\left( \sum _{n=0}^{\infty }2^{-\gamma _n}\right) =\sum _{n=0}^\infty \left( \frac{1-p}{p}\right) ^{n}p^{\gamma _n} \end{aligned}$$

  (4)

  for every strictly increasing sequence $(\gamma _n)_{n\in \mathbb N_0}$ of positive integer numbers.

From now on, for every $p\in (0,1)$ the symbol $R_p$ will be reserved for the function $R_p:[0,1]\rightarrow {\mathbb {R}}$ given by (4).

Fix $p\in (0,1)$. Adding the equations occurring in (3), we get

$$\begin{aligned} R_p\left( \frac{x}{2}\right) +R_p\left( \frac{x+1}{2}\right) =R_p(x)+p \end{aligned}$$

for every $x\in [0,1]$. Putting $x=1$ in both equations of (3), we obtain from the second one $R_p(1)=1$ and then from the first one $R_p(\frac{1}{2})=p$. Moreover, the representation (2) of $\varphi _p$ coincides with the representation (4) of $R_p$ (see [13, page 268]; see also [11, formula (6)]), and hence

$$\begin{aligned} R_p=\varphi _p \end{aligned}$$

(5)

for every $p\in (0,1)$. In consequence,

$$\begin{aligned} {\mathcal {W}}_0={{\,\mathrm{conv}\,}}(\{R_p\,|\,p\in (0,1)\}). \end{aligned}$$

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 Wesołowski problem yet. To find $\varphi \in \mathcal C\setminus {\mathcal {W}}_0$ 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 $\Phi :(0,1)\times [0,1]\rightarrow [0,1]$ by putting

$$\begin{aligned} \Phi (p,x)=R_p(x). \end{aligned}$$

From (5) we see that for all $p\in (0,1)$ and $x\in [0,1]$ we also have

$$\begin{aligned} \Phi (p,x)=\varphi _p(x). \end{aligned}$$

We will use both of the above formulas for $\Phi$, depending on the needs in the given situation.

Theorem 2.2

(see [8, page 195]; see also [12, Proposition 3.1]) The function $\Phi$ is differentiable with respect to the first variable.

Theorem 2.3

(see [2, Theorem 2.1]) The function $\Phi$ is Hölder continuous with respect to the second variable. More precisely, there is $C>0$ such that for every $p\in (0,1)$ we have

$$\begin{aligned} |\Phi (p,x)-\Phi (p,y)|\le C|x-y|^{\alpha _p} \end{aligned}$$

for all $x,y\in [0,1]$, where

$$\begin{aligned} \alpha _p=\min \left\{ -\frac{\log p}{\log 2},-\frac{\log (1-p)}{\log 2}\right\} . \end{aligned}$$

(6)

For every $a\in (0,1]$ we define the function $\phi _a:[0,1]\rightarrow [0,1]$ by putting

$$\begin{aligned} \phi _a(x)=\frac{1}{a}\int _0^a\Phi (p,x)\,dp. \end{aligned}$$

The symbol $\phi _a$ will be reserved for the just defined function. Note that (2) gives

$$\begin{aligned} \phi _a\left( \frac{1}{2^m}\right)&=\frac{1}{a}\int _0^a\varphi _p\left( \frac{1}{2^m}\right) \,dp =\frac{1}{a}\int _0^a p^m\,dp=\frac{a^m}{m+1}, \end{aligned}$$

(7)

$$\begin{aligned} \phi _a\left( 1-\frac{1}{2^m}\right)&=\frac{1}{a}\int _0^a\big (1-(1-p)^m\big )\,dp =1-\frac{1-(1-a)^{m+1}}{(m+1)a} \end{aligned}$$

(8)

for all $a\in (0,1]$ and $m\in {\mathbb {N}}_0$.

We are now in the position to prove the main result of this paper, which reads as follows.

Theorem 2.4

Assume $a\in (0,1]$. Then $\phi _a\in \mathcal C\setminus {{\,\mathrm{conv}\,}}(\mathcal W_0\cup \{\phi _b\,|\,b\in (0,1)\setminus \{a\}\})$. Moreover, $\phi _a$ is not Hölder continuous.

Proof

From Theorem 1.1 we see that $\phi _a$ is an increasing function satisfying (e) for every $x\in [0,1]$ with $\phi _a(0)=0$ and $\phi _a(1)=1$.

Fix $x,y\in [0,1]$. By Theorem 2.3 we have

$$\begin{aligned} |\phi _a(x)-\phi _a(y)|\le \frac{1}{a}\int _0^a|\Phi (p,x)-\Phi (p,y)|\,dp\le \frac{C}{a}\int _0^a|x-y|^{\alpha _p}\,dp. \end{aligned}$$

If $a\in (\frac{1}{2},1]$, then

$$\begin{aligned} \int _0^a|x-y|^{\alpha _p}\,dp&=\int _0^{\frac{1}{2}}|x-y|^{-\frac{\log (1-p)}{\log 2}}\,dp +\int _{\frac{1}{2}}^a|x-y|^{-\frac{\log p}{\log 2}}\,dp\\&=\int _0^{\frac{1}{2}} (1-p)^{-\frac{\log |x-y|}{\log 2}}\,dp +\int _{\frac{1}{2}}^a p^{-\frac{\log |x-y|}{\log 2}}\,dp\\&=\frac{1}{1-\frac{\log |x-y|}{\log 2}}\left[ 1+a^{1-\frac{\log |x-y|}{\log 2}} -2^{\frac{\log |x-y|}{\log 2}}\right] . \end{aligned}$$

If $a\in (0,\frac{1}{2}]$, then a similar calculation gives

$$\begin{aligned} \int _0^a|x-y|^{\alpha _p}\,dp=\frac{1}{1-\frac{\log |x-y|}{\log 2}}\left[ 1-a^{1-\frac{\log |x-y|}{\log 2}}\right] . \end{aligned}$$

This implies that $\phi _a$ is continuous, and hence that $\phi _a\in {\mathcal {C}}$.

Now we want to prove that $\phi _a\notin {{\,\mathrm{conv}\,}}(\mathcal W_0\cup \{\phi _b\,|\,b\in (0,1)\setminus \{a\}\})$. Suppose, towards a contradiction, that this is false. Then there are numbers $N\in {\mathbb {N}}$, $\alpha _1,\ldots ,\alpha _N,\beta _1,\ldots ,\beta _N\in [0,1]$, $p_1,\ldots ,p_N\in (0,1)$ and $b_1,\ldots ,b_N\in (0,1)\setminus \{a\}$ such that $\sum _{n=1}^N(\alpha _n+\beta _n)=1$ and

$$\begin{aligned} \phi _a=\sum _{n=1}^N(\alpha _n\varphi _{p_n}+\beta _n\phi _{b_n}). \end{aligned}$$

Evaluating the above equation at $x=\frac{1}{2^m}$ yields, jointly with (7) and (2),

$$\begin{aligned} \frac{a^m}{m+1}=\sum _{n=1}^N\left( \alpha _n(p_n)^m+\beta _n\frac{(b_n)^m}{m+1}\right) , \end{aligned}$$

for every $m\in {\mathbb {N}}$. Dividing both sides of the above equality by $\frac{a^m}{m+1}$, and then passing with m to $\infty$ we obtain

$$\begin{aligned} 1=\sum _{n=1}^N\left( \alpha _n\lim _{m\rightarrow \infty }(m+1)\left( \frac{p_n}{a}\right) ^m +\beta _n\lim _{m\rightarrow \infty }\left( \frac{b_n}{a}\right) ^m\right) \in \{0,\infty \}, \end{aligned}$$

a contradiction.

It remains to prove that $\phi _a$ is not Hölder continuous. Suppose towards a contradiction that there are $\alpha ,C>0$ such that $|\phi _a(x)-\phi _a(y)|\le C|x-y|^\alpha$ for all $x,y\in [0,1]$. Then making use of (8), we get

$$\begin{aligned} \frac{C}{2^{\alpha m}}\ge \left| \phi _a(1)-\phi _a\left( 1-\frac{1}{2^m}\right) \right| =\frac{1-(1-a)^{m+1}}{(m+1)a} \end{aligned}$$

for every $m\in {\mathbb {N}}$, a contradiction. $\square$

3 Singularity of solutions

The following lemma is a crucial point in this section.

Lemma 3.1

Assume that $\varphi \in {\mathcal {C}}$. If $\liminf _{x\rightarrow 0+}\frac{\varphi (x)}{x}=0$, then $\varphi$ is singular.

Proof

According to [16, Remark 2.2] both the absolutely continuous and the singular parts of $\varphi$ satisfy (e) for every $x\in [0,1]$. Comparing this with [16, Remark 2.1] we see that the class ${\mathcal {C}}$ is determined by two of its subclasses ${\mathcal {C}}_a$ and ${\mathcal {C}}_s$, consisting of all absolutely continuous and all singular functions, respectively. From assertion (i) of Theorem 1.1 we know that the class ${\mathcal {C}}_a$ consist of exactly one function, and it is the identity on [0, 1]. Hence there are $\alpha \in [0,1]$ and $\varphi _\text {sing}\in {\mathcal {C}}_s$ such that

$$\begin{aligned} \varphi (x)=\alpha x+(1-\alpha )\varphi _{\text {sing}}(x) \end{aligned}$$

for every $x\in [0,1]$. Therefore, $\varphi '(x)=\alpha$ for almost all $x\in [0,1]$. Since $\varphi _\text {sing}([0,1])\subset [0,1]$, we have

$$\begin{aligned} 0=\liminf _{x\rightarrow 0+}\frac{\varphi (x)}{x}=\alpha +(1-\alpha )\liminf _{x\rightarrow 0+} \frac{\varphi _{\text {sing}}(x)}{x}\ge \alpha , \end{aligned}$$

which completes the proof. $\square$

Now, we are in a position to prove that for every $a\in (0,1]$ the function $\phi _a$ 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.

Theorem 3.2

For every $a\in (0,1]$ the function $\phi _a$ is strictly increasing and singular.

Proof

The strict monotonicity of $\phi _a$ is clear. It remains to prove that $\phi _a$ is singular.

Assume first that $a\in (0,\frac{1}{2}]$. Then, making use of (7), we obtain

$$\begin{aligned} \liminf _{h\rightarrow 0+}\frac{\phi _a(h)}{h}\le \limsup _{n\rightarrow \infty } 2^n\phi _a\left( \frac{1}{2^n}\right) =\limsup _{n\rightarrow \infty }\frac{(2a)^n}{n+1}=0. \end{aligned}$$

Therefore, $\phi _a$ is singular, by Lemma 3.1.

Assume now that $a\in (\frac{1}{2},1]$. Since $\phi _{\frac{1}{2}}$ is singular, what we have just proved, it follows that $\phi _{\frac{1}{2}}'(x)=\phi _{\frac{1}{2}}'(1-x)=0$ for almost all $x\in [0,1]$. Fix $x\in [0,1]$ such that $\phi _{\frac{1}{2}}'(x)=\phi _{\frac{1}{2}}'(1-x)=0$. From [2, Proposition 2.3] we know that

$$\begin{aligned} \varphi _{p}(y)=1-\varphi _{1-p}(1-y) \end{aligned}$$

(9)

for all $p\in (0,1)$ and $y\in [0,1]$. This jointly with the monotonicity of $\phi _a$ and the change of variable theorem gives

$$\begin{aligned} 0&\le \frac{\phi _a(x+h)-\phi _a(x)}{h}=\frac{1}{a}\int _{0}^a\frac{\varphi _p(x+h)-\varphi _p(x)}{h}\,dp\\&=\frac{1}{2a}\frac{\phi _{\frac{1}{2}}(x+h)-\phi _{\frac{1}{2}}(x)}{h} +\frac{1}{a}\int _{\frac{1}{2}}^a\frac{\varphi _p(x+h)-\varphi _p(x)}{h}\,dp\\&=\frac{1}{2a}\frac{\phi _{\frac{1}{2}}(x+h)-\phi _{\frac{1}{2}}(x)}{h} +\frac{1}{a}\int _{\frac{1}{2}}^a\frac{\varphi _{1-p}(1-x)-\varphi _{1-p}(1-x-h)}{h}\,dp\\&=\frac{1}{2a}\frac{\phi _{\frac{1}{2}}(x+h)-\phi _{\frac{1}{2}}(x)}{h} +\frac{1}{a}\int _{1-a}^{\frac{1}{2}}\frac{\varphi _{p}(1-x) -\varphi _{p}(1-x-h)}{h}\,dp\\&\le \frac{1}{2a}\frac{\phi _{\frac{1}{2}}(x+h)-\phi _{\frac{1}{2}}(x)}{h} +\frac{1}{a}\int _{0}^{\frac{1}{2}}\frac{\varphi _{p}(1-x) -\varphi _{p}(1-x-h)}{h}\,dp\\&=\frac{1}{2a}\left( \frac{\phi _{\frac{1}{2}}(x+h)-\phi _{\frac{1}{2}}(x)}{h} +\frac{\phi _{\frac{1}{2}}(1-x)-\phi _{\frac{1}{2}}(1-x-h)}{h}\right) \end{aligned}$$

for every $h\in (0,1]$ with $x+h\in [0,1]$. Passing now with h to 0 we obtain $\phi _a'(x)=0$. $\square$

Before we formulate the next result, let us introduce some notions.

Assume that $x=\sum _{k=1}^{\infty }\frac{x_k}{2^k}$, where $x_k\in \{0,1\}$ for every $k\in {\mathbb {N}}$. If x has two representations, we take the one with only finitely many $x_k$ non-vanishing. For all $n\in {\mathbb {N}}$ and $p,a\in (0,1)$ we put

$$\begin{aligned} I_n=\sum _{k=1}^{n-1}x_k,\quad D_1(x)=\lim _{n \rightarrow \infty }\frac{I_n}{n},\quad D_0(x)=1-D_1(x). \end{aligned}$$

Moreover, for any increasing function $f:[0,1]\rightarrow [0,1]$ we put

$$\begin{aligned} f_{+}'(x)=\lim _{h\rightarrow 0+}\frac{f(x+h)-f(x)}{h}\quad \hbox {and}\quad f_{-}'(x)=\lim _{h\rightarrow 0-}\frac{f(x+h)-f(x)}{h}, \end{aligned}$$

assuming that the limits make sense.

Our analysis of the sets where the derivative is 0 or $\infty$ is based on the idea of the proof of the following result, which is due to Kawamura.

Theorem 3.3

(see [11, Theorem 2.2]) Assume that $x\in [0,1]$.

• (i) If x is dyadic and $p\in (0,\frac{1}{2})$, then $R'_{p+}(x)=0$ and $R'_{p-}(x)=\infty$.

• (ii) If x is dyadic and $p\in (\frac{1}{2},1)$, then $R'_{p-}(x)=0$ and $R'_{p+}(x)=\infty$.

• (iii) If x is not dyadic and $0<D_1(x)< 1$, then

  $$\begin{aligned} R_p'(x)={\left\{ \begin{array}{ll} 0, &{} \text{ if } p^{D_0(x)}(1-p)^{D_1(x)}<\frac{1}{2},\\ \infty , &{} \text{ if } p^{D_0(x)}(1-p)^{D_1(x)}>\frac{1}{2}. \end{array}\right. } \end{aligned}$$

Now we can prove the following announced result.

Theorem 3.4

Assume that $x\in [0,1]$.

• (i) If x is dyadic and $a\in (0,\frac{1}{2})$, then $\phi '_{a+}(x)=0$ and $\phi '_{a-}(x)=\infty$.

• (ii) If x is dyadic and $a\in (\frac{1}{2},1)$, then $\phi '_{a}(x)=\infty$.

• (iii) If x is not dyadic with $D_0(x)\in (0,1)$ and $a\in (0,D_0(x))$, then

  $$\begin{aligned} \phi _a'(x)={\left\{ \begin{array}{ll} 0, &{} \text{ if } p^{D_0(x)}(1-p)^{D_1(x)}<\frac{1}{2}\text { for all }p\le a,\\ \infty , &{} \text{ if } p^{D_0(x)}(1-p)^{D_1(x)}>\frac{1}{2}\text { for some }p\le a.\end{array}\right. } \end{aligned}$$

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 $a\in (0,1)$ and $x=\sum _{k=1}^\infty \frac{x_k}{2^k}$, where $x_k\in \{0,1\}$ for every $k\in {\mathbb {N}}$.

Let us first consider all the cases where $\phi _{a-}'(x)$ or $\phi _{a+}'(x)$ vanishes.

We begin with the cases where $\phi _{a+}'(x)=0$. Observe that in each of these cases Theorem 3.3 yields $R'_{p+}(x)=0$ for almost all $p\in (0,a)$. So if we could apply the Lebesgue dominated convergence theorem, we would get

$$\begin{aligned} 0\le \phi _{a+}'(x)=\lim _{h\rightarrow 0+}\frac{1}{a}\int _0^a\frac{\Phi (p,x+h)-\Phi (p,x)}{h}\,dp =\frac{1}{a}\int _0^a R_{p+}'(x)\,dp=0. \end{aligned}$$

(10)

Therefore, we see that to conclude the proof, it is enough to verify that

$$\begin{aligned} (p,h)\mapsto \frac{\varphi _p(x+h)-\varphi _p(x)}{h} \end{aligned}$$

is bounded almost everywhere from above on $(0,1)\times (0,1-x)$ by a constant.

Let us start the required verification with the dyadic case of x, i.e. where there exists $N\in {\mathbb {N}}$ such that $x_k=0$ for every $k>N$.

Fix $k\in {\mathbb {N}}$ and $h\in {\mathbb {R}}$ such that $k>N$, $\frac{k-I_{N+1}}{k}>a$ and $2^{-(k+1)}\le h<2^{-k}$. Then

$$\begin{aligned} \frac{\varphi _p\left( x+h\right) -\varphi _p(x)}{h}&\le 2^{k+1}\left( \varphi _p\left( x+2^{-k}\right) -\varphi _p(x)\right) =2^{k+1}p^{k-I_{N+1}}(1-p)^{I_{N+1}}. \end{aligned}$$

As the polynomial $p^{k-I_{N+1}}(1-p)^{I_{N+1}}$ is increasing on the interval $[0,\frac{k-I_{N+1}}{k}]$ and $p\le a$, we have

$$\begin{aligned} \frac{\varphi _p\left( x+h\right) -\varphi _p(x)}{h}\le 2^{k+1}a^{k-I_{N+1}}(1-a)^{I_{N+1}}. \end{aligned}$$

Since $\lim _{k\rightarrow \infty }2^{k+1}a^{k-I_{N+1}}(1-a)^{I_{N+1}} =\lim _{k\rightarrow \infty }2\cdot (2a)^k\left( \frac{1-a}{a}\right) ^{I_{N+1}}=0$, we see that the sequence $\big (2^{k+1}a^{k-I_{N+1}}(1-a)^{I_{N+1}}\big )_{k\in {\mathbb {N}}}$ is bounded. In consequence

$$\begin{aligned} \frac{\Phi (p,x+h)-\Phi (p,x)}{h}&\le \sup \{2^{k+1}a^{k-I_{N+1}}(1-a)^{I_{N+1}}\,|\,k\in \mathbb N\}<\infty . \end{aligned}$$

Thus we have proved the first part of assertion (i).

Now, we pass to the case where x is not dyadic and $p^{D_0(x)}(1-p)^{D_1(x)}<\frac{1}{2}$ for every $p\le a$.

Let $(n_k)_{k\in {\mathbb {N}}}$ 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,

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{n_{k+1}}{n_k}=1. \end{aligned}$$

(11)

For every $k\in {\mathbb {N}}$ we have $I_{n_k}=n_k-k$, and further for every $m\in \mathbb {Z}$ we also have

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{k-m}{n_k}=D_0(x). \end{aligned}$$

(12)

Fix $p\in (0,1)$ (we will later restrict ourselves to $p\le a$) and $k\in \mathbb {N}$, and let h be such that $2^{-n_{k+1}}\le h<2^{-n_k}$. Then

$$\begin{aligned} \frac{\varphi _p\left( x+h\right) -\varphi _p(x)}{h}&\le 2^{n_{k+1}}\left( \varphi _p\left( x+2^{-n_k}\right) -\varphi _p(x)\right) \\&=2^{n_{k+1}}\bigg [p^{k}(1-p)^{n_k-k}\\&\quad +\frac{1-2p}{p}\sum _{n=n_k+1}^{\infty }x_np^{n-I_n}(1-p)^{I_n}\bigg ]. \end{aligned}$$

Let $(n(l))_{l\in {\mathbb {N}}}$ 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 $n(l)-n_k-l\ge 0$ for every $l\in {\mathbb {N}}$, we have

$$\begin{aligned} \sum _{n=n_k+1}^{\infty }x_np^{n-I_n}(1-p)^{I_n}&=p^{k}(1-p)^{n_k-k}\sum _{l=1}^{\infty }p^{n(l)-n_k-l+1}(1-p)^{l-1}\\&\le p^{k+1}(1-p)^{n_k-k}\sum _{l=1}^{\infty }(1-p)^{l-1}=p^{k}(1-p)^{n_k-k}. \end{aligned}$$

Thus

$$\begin{aligned} \begin{aligned} \frac{\varphi _p\left( x+h\right) -\varphi _p(x)}{h}&\le 2^{n_{k+1}}p^{k}(1-p)^{n_k-k}\left( 1+\frac{1-2p}{p}\right) \\&= 2^{n_{k+1}}p^{k}(1-p)^{n_k-k}\left( \frac{1-p}{p}\right) \\&=2^{n_{k+1}}p^{k-1}(1-p)^{n_k-k+1}. \end{aligned} \end{aligned}$$

(13)

As $a<D_0(x)$, we conclude from (12) that there is $k_0\in {\mathbb {N}}$ such that $a\le \frac{k-1}{n_k}$ for every $k\in {\mathbb {N}}$ with $k\ge k_0$. Assume now that $k\ge k_0$ and that $p\le a$. As the polynomial $p^{k-1}(1-p)^{n_k-k+1}$ is increasing on the interval $[0,\frac{k-1}{n_k}]$ and $p\le a$, we have

$$\begin{aligned} \frac{\varphi _p\left( x+h\right) -\varphi _p(x)}{h}\le 2^{n_{k+1}}a^{k-1}(1-a)^{n_k-k+1}. \end{aligned}$$

(14)

Applying (11) and (12), we obtain

$$\begin{aligned} \lim _{k\rightarrow \infty }2^{\frac{n_{k+1}}{n_k}}a^{\frac{k-1}{n_{k}}}(1-a)^{1-\frac{k-1}{n_k}}=2a^{D_0(x)}(1-a)^{D_1(x)}<1, \end{aligned}$$

and hence

$$\begin{aligned} \lim _{k\rightarrow \infty }2^{n_{k+1}}a^{k-1}(1-a)^{n_k-k+1}= \lim _{k\rightarrow \infty }\Big (2^{\frac{n_{k+1}}{n_k}}a^{\frac{k-1}{n_{k}}}(1-a)^{1-\frac{k-1}{n_k}}\Big )^{n_k}=0. \end{aligned}$$

This implies that the sequence $\big (2^{n_{k+1}}a^{k-1}(1-a)^{n_k-k+1}\big )_{k\in {\mathbb {N}}}$ is bounded. In consequence

$$\begin{aligned} \frac{\Phi (p,x+h)-\Phi (p,x)}{h}\le \sup \{2^{n_{k+1}}a^{k-1}(1-a)^{n_k-k+1}\,|\,k\in {\mathbb {N}}\}<\infty . \end{aligned}$$

(15)

Thus we have proved that $\phi _{a+}'(x)=0$ in the case where x is not dyadic and $p^{D_0(x)}(1-p)^{D_1(x)}<\frac{1}{2}$ for every $p\le a$, 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 $p^{D_0(x)}(1-p)^{D_1(x)}<\frac{1}{2}$ for every $p\le a$. But now we want to prove that $\phi _{a-}'(x)=0$.

Observe first that (9) yields

$$\begin{aligned} \begin{aligned} \phi _{a-}'(x)&=\lim _{h\rightarrow 0+}\frac{1}{a}\int _0^a\frac{\varphi _p(x)-\varphi _p(x-h)}{h}\,dp\\&=\frac{1}{a}\lim _{h\rightarrow 0+}\int _0^a\frac{\varphi _{1-p}(1-x+h)-\varphi _{1-p}(1-x)}{h}\,dp\\&=\frac{1}{a}\lim _{h\rightarrow 0+}\int _{1-a}^1\frac{\varphi _p(1-x+h)-\varphi _p(1-x)}{h}\,dp\\&=\frac{1}{a}\lim _{h\rightarrow 0+}\int _{1-a}^1\frac{\Phi (p,1-x+h)-\Phi (p,1-x)}{h}\,dp. \end{aligned} \end{aligned}$$

Since in the considered case Theorem 3.3 implies that $R'_{p+}(1-x)=0$ for almost all $p\in (1-a,1)$ as we will verify (in a moment) in (16), the idea is to follow the proof of $\phi _{a+}(x)=0$, replacing x by $1-x$ with suitable changes, which we explain in details.

Fix $1-a<p<1$. Since $D_0(1-x)=D_1(x)$, we have

$$\begin{aligned} p^{D_0(1-x)}(1-p)^{D_1(1-x)}=p^{D_1(x)}(1-p)^{D_0(x)}<\frac{1}{2}. \end{aligned}$$

(16)

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 $k_0\in \mathbb N$ such that $1-a\ge \frac{k}{n_k}$ for every $k\in {\mathbb {N}}$ with $k\ge k_0$. Fix $k\ge k_0$ and let h be such that $2^{-n_{k+1}}\le h<2^{-n_k}$. We have shown before (see (13)) that

$$\begin{aligned} \frac{\varphi _p\left( 1-x+h\right) -\varphi _p(1-x)}{h} \le 2^{n_{k+1}}p^{k-1}(1-p)^{n_k+1-k}. \end{aligned}$$

As the polynomial $p^{k-1}(1-p)^{n_k+1-k}$ is decreasing on the interval $[\frac{k}{n_k},1]$ and $p\ge 1-a$, 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 $\phi _{a-}'(x)=0$.

It remains to prove all the cases where a one-sided derivative of $\phi _a$ at the point x equals $\infty$. We prove these cases in more or less one go.

First of all, we show that in each of these cases for every $N\in {\mathbb {N}}$ we can find an open interval $L_N$ having 0 as its right or left endpoint (depending on the one-sided derivative of $\phi _a$ at the point x we are interested in) and a Lebesgue measurable set $P_N\subset (0,a)$ such that

$$\begin{aligned} \frac{\varphi _p(x+h)-\varphi _p(x)}{h}\ge N\quad \hbox {for all }p\in P_N\hbox { and }h\in L_N \end{aligned}$$

and

$$\begin{aligned} \lim _{N\rightarrow \infty } N|P_N|=\infty , \end{aligned}$$

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 $P_N\subset (0,\frac{1}{2})$; 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 $P_N\subset (\frac{1}{2},1)$. 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 $p\le a$ for which $p^{D_0(x)}(1-p)^{D_1(x)}>\frac{1}{2}$.

If both sets $L_N$ and $P_N$ satisfy the above conditions, then

$$\begin{aligned} \frac{\phi _a(x+h)-\phi _a(x)}{h} \ge \frac{1}{a}\int _{P_N} \frac{\varphi _p(x+h)-\varphi _p(x)}{h}\, dp \ge \frac{1}{a}N|P_N| \end{aligned}$$

for all $h\in L_N$ and $N\in {\mathbb {N}}$.

Finally, in each of the considered cases we conclude that the one-sided derivative of $\phi _a$ at the point x equals $\infty$. $\square$

4 Continuity of the function $\Phi$

Before we construct more new solutions to the Matkowski and Wesołowski problem, let us look at the function $\Phi$. 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 $N\in {\mathbb {N}}$ and let $a_n,b_n\in {\mathbb {R}}$ with $a_n<b_n$ for every $n\in \{1,\ldots ,N+1\}$. If $f:(a_1,b_1)\times \cdots \times (a_N,b_N)\times (a_{N+1},b_{N+1})\rightarrow \mathbb R$ is continuous with respect to each variable separately and monotonous with respect to each of the first N variables separately, then f is continuous.

Theorem 4.2

The function $\Phi$ is continuous.

Proof

According to Theorems 2.1,  2.2 and 4.1 it suffices to prove that $\Phi$ is continuous at every point of the set $(0,1)\times \{0,1\}$.

Fix $p_0\in (0,1)$ and $\varepsilon >0$. Next choose $n\in {\mathbb {N}}$ such that $0<(p_0+\frac{1}{2^n})^n\le \varepsilon$. Fix $(p,x)\in (0,1)\times [0,1]$ and assume that $(p-p_0)^2+x^2\le \frac{1}{4^{n}}$. Hence $p\le p_0+\frac{1}{2^n}$ and $x\le \frac{1}{2^n}$. Then, making use of Theorem 2.1, we obtain

$$\begin{aligned} |\Phi (p,x)-\Phi (p_0,0)|=R_p(x)\le R_p\left( \frac{1}{2^n}\right) =p^n\le \left( p_0+\frac{1}{2^n}\right) ^n\le \varepsilon , \end{aligned}$$

which proves that $\Phi$ is continuous at the point $(p_0,0)$.

Noting that $R_p(x)=1-R_{1-p}(1-x)$ for every $x\in [0,1]$, as stated for example in [1, Proposition 2.3], we can easily see that $\Phi$ is continuous at every point of the set $(0,1)\times \{1\}$. $\square$

Observe that the function $\Phi$ cannot be extended continuously to the closed unit square $[0,1]^2$; indeed

$$\begin{aligned} \begin{aligned} \lim _{p\rightarrow 0}\Phi (p,x) ={\left\{ \begin{array}{ll} 0,&{}\hbox {if } x\in [{0}, {1})\\ 1,&{}\hbox {if } x=1 \end{array}\right. } \quad \text{ and } {\quad } \lim _{p\rightarrow 1}\Phi (p,x) ={\left\{ \begin{array}{ll} 0,&{}\hbox {if } x=0\\ 1,&{}\hbox {if } x\in (0,1] \end{array}\right. }, \end{aligned} \end{aligned}$$

and the continuity breaks at the points (0, 1) and (1, 0).

5 More new solutions

We want to construct more new solutions to the Matkowski and Wesołowski problem different from those presented in Theorem 2.4. For this purpose we fix a Borel probability measure $\mu$ on (0, 1). Next, given a Borel measurable function $g:(0,1)\rightarrow (0,1)$, we define a function $\psi _g:[0,1]\rightarrow [0,1]$ by putting

$$\begin{aligned} \psi _g(x)=\int _{(0,1)}\Phi (g(p),x)\,d\mu (p); \end{aligned}$$

note that the function is well defined, because by Theorem 2.2 the map $p\mapsto \Phi (g(p),x)$ is Borel measurable and bounded for every $x\in [0,1]$. We keep the symbol $\psi _g$ for the above defined function to the end of this paper.

Given a Borel measurable function $g:(0,1)\rightarrow (0,1)$, we define the non-increasing rearrangement $g^*:(0,1)\rightarrow [0,1]$ of g by putting

$$\begin{aligned} g^*(t)=\inf \{\lambda \ge 0\,|\,\mu (\{p\in (0,1)\,|\,g(p)>\lambda \})\le t\}. \end{aligned}$$

It is clear that the function $g^*$ is non-increasing and right-side continuous. From [1, Proposition 2.3] and Theorem 2.2 we deduce that the function $\Phi$ is strictly increasing and continuous with respect to the first variable. Therefore, according to [5, Corollary 3.2.8] we have

$$\begin{aligned} \psi _g(x)=\int _{(0,1)}\Phi (g(p),x)\,d\mu (p)=\int _{(0,1)}\Phi (g^*(p),x)\, dp \end{aligned}$$

for every $x\in [0,1]$.

In view of the above observation let us assume, to the end of this paper, that $\mu$ is the one-dimensional Lebesgue measure.

Theorem 5.1

Assume that $g:(0,1)\rightarrow (0,1)$ is a Borel measurable function. Then $\psi _g\in {\mathcal {C}}$, and it is strictly increasing.

Proof

From Theorem 1.1 we see that $\psi _g$ is a strictly increasing function satisfying $(\mathsf{e})$ for every $x\in [0,1]$ with $\psi _g(0)=0$ and $\psi _g(1)=1$. To prove that $\psi _g$ is continuous we fix $x\in [0,1]$ and a sequence $(x_n)_{n\in \mathbb N}$ 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,

$$\begin{aligned} \lim _{n\rightarrow \infty }\psi _g(x_n)=\int _{(0,1)}\lim _{n\rightarrow \infty }\Phi (g(p),x_n)\,dp =\psi _g(x). \end{aligned}$$

Therefore, $\psi _g\in {\mathcal {C}}$. $\square$

Theorem 5.1 produces a very large class of new solutions to the Matkowski and Wesołowski problem.

Put

$$\begin{aligned} {\mathcal {W}}_1={{\,\mathrm{conv}\,}}(\{\phi _a\,|\,a\in (0,1)\}) \end{aligned}$$

and for every $s\in (0,\infty )$ we define the function $g_s:(0,1)\rightarrow (0,1)$ by putting

$$\begin{aligned} g_s(p)=p^s; \end{aligned}$$

the symbol $g_s$ will be reserved just for this function. Obviously,

$$\begin{aligned} \psi _{g_s}\left( \frac{1}{2^m}\right) =\frac{1}{sm+1} \end{aligned}$$

(17)

for all $s\in (0,\infty )$ and $m\in {\mathbb {N}}$.

Corollary 5.2

Assume that $\mu$ is the one-dimensional Lebesgue measure. Then for every $s\in (0,\infty )$ we have $\psi _{g_s}\in \mathcal C\setminus {{\,\mathrm{conv}\,}}({\mathcal {W}}_0\cup \mathcal W_1\cup \{\psi _{g_r}\,|\,r\in (0,\infty )\setminus \{s\}\})$.

Proof

According to Theorem 5.1 we only need to show that $\psi _{g_s}\notin {{\,\mathrm{conv}\,}}({\mathcal {W}}_0\cup \mathcal W_1\cup \{\psi _{g_r}\,|\,r\in (0,\infty )\setminus \{s\}\})$ for every $s\in (0,\infty )$.

Fix $s\in (0,\infty )$ and suppose towards a contradiction that there are $N\in {\mathbb {N}}$, $\alpha _1,\ldots ,\alpha _N,\beta _1,\ldots ,\beta _N, \gamma _1,\ldots ,\gamma _N\in [0,1]$, $p_1,\ldots ,p_N,a_1,\ldots ,a_N\in (0,1)$ and $r_1,\ldots ,r_N\in (0,\infty )\setminus \{s\}$ such that $\sum _{n=1}^N(\alpha _n+\beta _n+\gamma _n)=1$ and

$$\begin{aligned} \psi _{g_s}=\sum _{n=1}^N(\alpha _n\varphi _{p_n}+\beta _n\phi _{a_n} +\gamma _n\psi _{g_{r_n}}). \end{aligned}$$

We evaluate the above expression at the point $\frac{1}{2^m}$ by applying (17), (2) and (7):

$$\begin{aligned} \frac{1}{sm+1}-\sum _{n=1}^N\gamma _n\frac{1}{r_n m+1} =\sum _{n=1}^N\left( \alpha _n(p_n)^m+\beta _n\frac{(a_n)^m}{m+1}\right) \end{aligned}$$

for every $m\in {\mathbb {N}}$. Multiplying both sides by $(m+1)(sm+1)\prod _{n=1}^{N}(r_nm+1)$ we find that there are polynomials $P_1$, $P_2$ and $P_3$ in m such that

$$\begin{aligned} P_1(m)=\sum _{n=1}^{N}\big (\alpha _n P_2(m)(p_n)^m+\beta _nP_3(m)(a_n)^m\big ) \end{aligned}$$

(18)

for every $m\in \mathbb {N}$. Let us assume for a contradiction that the leading term of $P_1$ is $\alpha _K m^K$ for some $K\ge 1$. Then dividing both sides by $\alpha _K m^K$ and letting m tend to $\infty$ we have that $1=0$, which is impossible. Hence $P_1$ is constant. Taking the limit as m tends to $\infty$ on the right-hand side of (18) gives $P_1=0$. Since $P_1(m)=(m+1)\big (\prod _{n=1}^{N}(r_nm+1)+(sm+1)Q(m)\big )$ with a polynomial Q in m, it follows that

$$\begin{aligned} \prod _{n=1}^{N}(r_nm+1)+(sm+1)Q(m)=0 \end{aligned}$$

(19)

for every $m\in \mathbb {R}$. Finally, putting $m=-\frac{1}{s}$ in (19) we conclude that $s=r_n$ for some $n\in \{1,\ldots ,N\}$, a contradiction. $\square$

Put

$$\begin{aligned} {\mathcal {W}}_2={{\,\mathrm{conv}\,}}(\{\psi _{g_s}\,|\,s\in (0,\infty )\}) \end{aligned}$$

and for every $\lambda \in (0,1)$ define the function $\sin _{\lambda }:(0,1)\rightarrow (0,1)$ by putting

$$\begin{aligned} \sin _{\lambda }(p)=\lambda \sin \pi p. \end{aligned}$$

Again, the symbol $\sin _{\lambda }$ will be reserved for the just defined function. One can check that

$$\begin{aligned} \psi _{\sin _{\lambda }}\left( \frac{1}{4^{m}}\right) =\left( \frac{\lambda }{2}\right) ^{2m}\frac{(2m)!}{(m!)^2} \end{aligned}$$

(20)

for all $\lambda \in (0,1)$ and $m\in {\mathbb {N}}$.

Corollary 5.3

Assume that $\mu$ is the one-dimensional Lebesgue measure. Then for every $\lambda \in (0,1)$ we have $\psi _{\sin _{\lambda }}\in \mathcal C\setminus {{\,\mathrm{conv}\,}}({\mathcal {W}}_0\cup {\mathcal {W}}_1\cup \mathcal W_2)$.

Proof

According to Theorem 5.1 we only need to show that $\psi _{\sin _\lambda }\notin {{\,\mathrm{conv}\,}}({\mathcal {W}}_0\cup \mathcal W_1\cup {\mathcal {W}}_2)$.

Fix $\lambda \in (0,\infty )$ and suppose towards a contradiction that there are $N\in {\mathbb {N}}$, $\alpha _1,\ldots ,\alpha _N,\beta _1,\ldots ,\beta _N, \gamma _1,\ldots ,\gamma _N\in [0,1]$, $p_1,\ldots ,p_N,a_1,\ldots ,a_N\in (0,1)$ and $s_1,\ldots ,s_N\in (0,\infty )$ such that $\sum _{n=1}^N(\alpha _n+\beta _n+\gamma _n)=1$ and

$$\begin{aligned} \psi _{\sin _\lambda }=\sum _{n=1}^N(\alpha _n\varphi _{p_n}+\beta _n\phi _{a_n} +\gamma _n\psi _{g_{s_n}}). \end{aligned}$$

Applying (20), (2), (7), and (17), we obtain

$$\begin{aligned} \left( \frac{\lambda }{2}\right) ^{2m}\frac{(2m)!}{(m!)^2}= \sum _{n=1}^N\left( \alpha _n(p_n)^{2m}+\beta _n\frac{(a_n)^{2m}}{2m+1} +\gamma _n\frac{1}{2s_n m+1}\right) \end{aligned}$$

for every $m\in {\mathbb {N}}$. Dividing both sides of the above equality by $\big (\frac{\lambda }{2}\big )^{2m}\frac{(2m)!}{(m!)^2}$, and then passing with m to $\infty$ we obtain, by for example using (20),

$$\begin{aligned} 1&=\lim _{m\rightarrow \infty }\sum _{n=1}^N\Bigg (\alpha _n\frac{(m!)^2}{(2m)!}\left( \frac{2p_n}{\lambda }\right) ^{2m}+\beta _n\frac{(m!)^2}{(2m+1)!}\left( \frac{2a_n}{\lambda }\right) ^{2m}\\&\quad +\gamma _n\frac{(m!)^2}{(2m)!(2s_n m+1)}\left( \frac{2}{\lambda }\right) ^{2m}\Bigg )\in \{0,\infty \}, \end{aligned}$$

a contradiction. $\square$