# Measure theory and Perron-Frobenius operators for continued fractions

Jordan Bell
April 18, 2016

## 1 The continued fraction transformation

For $\xi\in\mathbb{R}$ let $[x]$ be the greatest integer $\leq\xi$, let $R(\xi)=\xi-[\xi]$, and let $\left\|\xi\right\|=\min(R(\xi),1-R(\xi))$, the distance from $\xi$ to a nearest integer. Let $I=[0,1]$ and define the continued fraction transformation $\tau:I\to I$ by

 $\tau(x)=\begin{cases}x^{-1}-[x^{-1}]&x\neq 0\\ 0&x=0.\end{cases}$

It is immediate that for $x\in I$, $x\in I\setminus\mathbb{Q}$ if and only if $\tau(x)\in I\setminus\mathbb{Q}$. For $x\in\mathbb{R}$, define $a_{0}(x)=[x]$, and for $n\geq 1$ define $a_{n}(x)\in\mathbb{Z}_{\geq 1}\cup\{\infty\}$ by

 $a_{n}(x)=\left[\frac{1}{\tau^{n-1}(x-a_{0}(x))}\right].$

For example, let $x=\frac{13}{71}$.

 $\tau(x)=\frac{71}{13}-\left[\frac{71}{13}\right]=\frac{71}{13}-5=\frac{6}{13}.$
 $\tau^{2}(x)=\frac{13}{6}-\left[\frac{13}{6}\right]=\frac{13}{6}-2=\frac{1}{6}.$
 $\tau^{3}(x)=\frac{6}{1}-\left[\frac{6}{1}\right]=0.$

Then $\tau^{n}(x)=0$ for $n\geq 3$. Thus, with $x=\frac{13}{71}$,

 $a_{0}(x)=0,\quad a_{1}(x)=\left[\frac{71}{13}\right]=5.$
 $a_{2}(x)=\left[\frac{1}{\tau(x)}\right]=\left[\frac{13}{6}\right]=2,\quad a_{3% }(x)=\left[\frac{1}{\tau^{2}(x)}\right]=\left[\frac{6}{1}\right]=6.$
 $a_{4}(x)=\left[\frac{1}{\tau^{3}(x)}\right]=\infty,\qquad a_{5}(x)=\infty,% \qquad\ldots.$

## 2 Convergents

For $x\in\Omega=I\setminus\mathbb{Q}$ write $a_{n}=a_{n}(x)$, and define

 $q_{-1}=0,\quad p_{-1}=1,\quad q_{0}=1,\quad p_{0}=0,$

and for $n\geq 1$,

 $q_{n}=a_{n}q_{n-1}+q_{n-2},\qquad p_{n}=a_{n}p_{n-1}+p_{n-2}.$

Thus

 $q_{1}=a_{1}q_{0}+q_{-1}=a_{1},\qquad p_{1}=a_{1}p_{0}+p_{-1}=1.$

One proves

 $p_{n}q_{n-1}-p_{n-1}q_{n}=(-1)^{n+1},\qquad n\geq 0.$

Also,11 1 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 9, Proposition 1.1.1.

 $x=\frac{p_{n}+\tau^{n}(x)p_{n-1}}{q_{n}+\tau^{n}(x)q_{n-1}},\qquad x\in\Omega,% \quad n\geq 0.$

From this,

 $x-\frac{p_{n}}{q_{n}}=\frac{(-1)^{n}\tau^{n}(x)}{q_{n}(q_{n}+\tau^{n}(x)q_{n-1% })}.$

Now,

 $a_{n+1}+\tau^{n+1}(x)=\left[\frac{1}{\tau^{n}(x)}\right]+\frac{1}{\tau^{n}(x)}% -\left[\frac{1}{\tau^{n}(x)}\right]=\frac{1}{\tau^{n}(x)},$

and using this,

 $\displaystyle\frac{\tau^{n}(x)}{q_{n}(q_{n}+\tau^{n}(x)q_{n-1})}$ $\displaystyle=\frac{1}{q_{n}(q_{n}\cdot(a_{n+1}+\tau^{n+1}(x))+q_{n-1})}$ $\displaystyle=\frac{1}{q_{n}(q_{n+1}+\tau^{n+1}(x)q_{n})}.$

Thus

 $\frac{1}{q_{n}(q_{n}+q_{n-1})}<\left|x-\frac{p_{n}}{q_{n}}\right|<\frac{1}{q_{% n}q_{n+1}}.$

For $n\geq 1$ let

 $r_{n}(x)=\frac{1}{\tau^{n-1}(x)}=a_{n}+\tau^{n}(x)$

and

 $s_{n}=\frac{q_{n-1}}{q_{n}},\qquad y_{n}=\frac{1}{s_{n}}$

and

 $\displaystyle u_{n}$ $\displaystyle=q_{n-1}^{-2}\left|x-\frac{p_{n-1}}{q_{n-1}}\right|^{-1}$ $\displaystyle=\frac{1}{q_{n-1}^{2}}\cdot\frac{q_{n-1}(q_{n-1}+\tau^{n-1}(x)q_{% n-2})}{\tau^{n-1}(x)}$ $\displaystyle=\frac{q_{n-1}+\tau^{n-1}(x)q_{n-2}}{\tau^{n-1}(x)q_{n-1}}$ $\displaystyle=\frac{q_{n-1}\cdot(a_{n}+\tau^{n}(x))+q_{n-2}}{q_{n-1}}$ $\displaystyle=a_{n}+\tau^{n}(x)+\frac{q_{n-2}}{q_{n-1}}.$

Let $s_{0}=0$. It is worth noting that

 $y_{1}\cdots y_{n}=\frac{q_{1}}{q_{0}}\cdots\frac{q_{n}}{q_{n-1}}=\frac{q_{n}}{% q_{0}}=q_{n}.$
 $\frac{1}{s_{n}}=\frac{q_{n}}{q_{n-1}}=a_{n}+\frac{q_{n-2}}{q_{n-1}}=a_{n}+s_{n% -1}.$
 $u_{n}=a_{n}+\tau^{n}(x)+\frac{q_{n-2}}{q_{n-1}}=r_{n}+s_{n-1}.$

## 3 Measure theory

Suppose that $(X,\mathscr{A})$ is a measurable space and $\mu,\nu$ are probability measures on $\mathscr{A}$. Let $\mathscr{D}=\{A\in\mathscr{A}:\mu(A)=\nu(A)\}$. First, $X\in\mathscr{D}$. Second, if $A,B\in\mathscr{D}$ and $A\subset B$ then

 $\mu(B\setminus A)=\mu(B)-\mu(A)=\nu(B)-\nu(A)=\nu(B\setminus A),$

so $B\setminus A\in\mathscr{D}$. Third, suppose that $A_{n}\in\mathscr{D}$, $n\geq 1$, and $A_{n}\uparrow A$. Because $\mathscr{A}$ is a $\sigma$-algebra, $A\in\mathscr{A}$, and then, setting $A_{0}=\emptyset$,

 $\mu(A)=\mu\left(\bigcup_{n\geq 1}(A_{n}\setminus A_{n-1})\right)=\sum_{n\geq 1% }(\mu(A_{n})-\mu(A_{n-1})),$

whence $\mu(A)=\nu(A)$. Therefore $\mathscr{D}$ is a Dynkin system. Dynkin’s theorem says that if $\mathscr{D}$ is a Dynkin system and $\mathscr{C}\subset\mathscr{D}$ where $\mathscr{C}$ is a $\pi$-system (nonempty and closed under finite intersections), then $\sigma(\mathscr{C})\subset\mathscr{D}$.22 2 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 136, Lemma 4.11.

Suppose now that $\sigma(\mathscr{C})=\mathscr{A}$, that $\mathscr{C}$ is closed under finite intersections, and that $\mu(A)=\nu(A)$ for all $A\in\mathscr{C}$. Then $\mathscr{C}\subset\mathscr{D}$, so by Dynkin’s theorem, $\mathscr{A}=\sigma(\mathscr{C})\subset\mathscr{D}$, hence $\mathscr{D}=\mathscr{A}$. That is, for any $A\in\mathscr{A}$, $\mu(A)=\nu(A)$, meaning $\mu=\nu$.

We shall apply the above with $(I,\mathscr{B}_{I})$, $I=[0,1]$. For

 $\mathscr{C}=\{(0,u]:0

it is a fact that $\sigma(\mathscr{C})=\mathscr{B}_{I}$. Therefore if $\mu$ and $\nu$ are probability measures on $\mathscr{B}_{I}$ such that $\mu((0,u])=\nu((0,u])$ for every $0, then $\mu=\nu$.

Let $\lambda$ be Lebesgue measure on $I=[0,1]$. Define

 $d\gamma(x)=\frac{1}{(1+x)\log 2}d\lambda(x),$

called the Gauss measure. If $\mu$ is a Borel probability measure on $I$, for measurable $T:I\to I$ and for $A\in\mathscr{B}_{I}$ let

 $T_{*}\mu(A)=\mu(T^{-1}(A)).$

$T_{*}\mu$, called the pushforward of $\mu$ by $T$, is itself a Borel probability measure on $I$. We prove that $\gamma$ is an invariant measure for $\tau$.33 3 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 17, Theorem 1.2.1; Manfred Einsiedler and Thomas Ward, Ergodic Theory with a view towards Number Theory, p. 77, Lemma 3.5.

###### Theorem 1.

$\tau_{*}\gamma=\gamma$.

###### Proof.

Let $0. For $x\in I$, $0<\tau(x)\leq u$ if and only if $0<\frac{1}{x}-\left[\frac{1}{x}\right]\leq u$ if and only if $\left[\frac{1}{x}\right]<\frac{1}{x}\leq u+\left[\frac{1}{x}\right]$ if and only if $\frac{1}{u+\left[\frac{1}{x}\right]}\leq x<\frac{1}{\left[\frac{1}{x}\right]}$. Then, as $0\not\in\tau^{-1}((0,u])$,

 $\tau^{-1}((0,u])=\bigcup_{i\geq 1}\left[\frac{1}{u+i},\frac{1}{i}\right).$

We calculate

 $\displaystyle\gamma(\tau^{-1}((0,u]))$ $\displaystyle=\sum_{i\geq 1}\gamma\left(\left[\frac{1}{u+i},\frac{1}{i}\right)\right)$ $\displaystyle=\sum_{i\geq 1}\int_{\left[\frac{1}{u+i},\frac{1}{i}\right)}\frac% {1}{(1+x)\log 2}d\lambda(x)$ $\displaystyle=\frac{1}{\log 2}\sum_{i\geq 1}\left(\log\left(1+\frac{1}{i}% \right)-\log\left(1+\frac{1}{u+i}\right)\right).$

Using

 $\frac{1+\frac{1}{i}}{1+\frac{1}{u+i}}=\frac{1+\frac{u}{i}}{1+\frac{u}{i+1}},$

this is

 $\displaystyle\gamma(\tau^{-1}((0,u]))$ $\displaystyle=\frac{1}{\log 2}\sum_{i\geq 1}\left(\log\left(1+\frac{u}{i}% \right)-\log\left(1+\frac{u}{i+1}\right)\right)$ $\displaystyle=\frac{1}{\log 2}\sum_{i\geq 1}\int_{\frac{u}{i+1}}^{\frac{u}{i}}% \frac{1}{1+x}d\lambda(x)$ $\displaystyle=\gamma((0,u]).$

Because $\gamma(\tau^{-1}((0,u]))=\gamma((0,u])$ for every $0, it follows that $\tau_{*}\gamma=\gamma$. ∎

We remark that for a set $X$, $X^{0}$ is a singleton. For $i\in\mathbb{Z}_{\geq 1}^{0}$ let $I_{0}(i)=\Omega$. For $n\geq 1$ and $i\in\mathbb{Z}_{\geq 1}^{n}$, let

 $I_{n}(i)=\{\omega\in\Omega:a_{k}(x)=i_{k},1\leq k\leq n\}.$

For $n\geq 1$ and for $i\in\mathbb{Z}_{\geq 1}^{n}$, define

 $[i_{1},\ldots,i_{n}]=\cfrac{1}{i_{1}+\cfrac{1}{\cdots+\cfrac{1}{i_{n-1}+\cfrac% {1}{i_{n}}}}}.$

For $x\in I_{n}(i)$,

 $\frac{p_{n}(x)}{q_{n}(x)}=[i_{1},\ldots,i_{n}],\qquad\frac{p_{n-1}(x)}{q_{n-1}% (x)}=[i_{1},\ldots,i_{n-1}].$

The following is an expression for the sets $I_{n}(i)$.44 4 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 18, Theorem 1.2.2.

###### Theorem 2.

Let $n\geq 1$, $i\in\mathbb{Z}_{\geq 1}^{n}$, and define

 $u_{n}(i)=\begin{cases}\frac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}}&\textrm{n odd}\\ \frac{p_{n}}{q_{n}}&\textrm{n even}\end{cases}$

and

 $v_{n}(i)=\begin{cases}\frac{p_{n}}{q_{n}}&\textrm{n odd}\\ \frac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}}&\textrm{n even}.\end{cases}$

Then

 $I_{n}(i)=\Omega\cap(u_{n}(i),v_{n}(i)).$

From the above, if $n$ is odd and $i\in\mathbb{Z}_{\geq 1}$ then

 $\displaystyle\lambda(I_{n}(i))$ $\displaystyle=v_{n}(i)-u_{n}(i)$ $\displaystyle=\frac{p_{n}}{q_{n}}-\frac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}}$ $\displaystyle=\frac{p_{n}q_{n-1}-p_{n-1}q_{n}}{q_{n}(q_{n}+q_{n-1})}$ $\displaystyle=\frac{(-1)^{n+1}}{q_{n}(q_{n}+q_{n-1})}$ $\displaystyle=\frac{1}{q_{n}(q_{n}+q_{n-1})},$

and if $n$ is even then likewise

 $\lambda(I_{n}(i))=\frac{1}{q_{n}(q_{n}+q_{n-1})}.$

Kraaikamp and Iosifescu attribute the following to Torsten Brodén, in a 1900 paper.55 5 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 21, Corollary 1.2.6.

###### Theorem 3.

For $n\geq 1$, $i\in\mathbb{N}^{n}$, $x\in I$,

 $\lambda(\tau^{n}
###### Proof.

We have

 $\lambda(\tau^{n}

Using

 $\omega=\frac{p_{n}+\tau^{n}(\omega)p_{n-1}}{q_{n}+\tau^{n}(\omega)q_{n-1}},% \qquad\omega\in\Omega,\quad n\geq 0,$

if $n$ is odd then

 $\displaystyle(\tau^{n} $\displaystyle=\left\{\omega\in\Omega:\frac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}}<% \omega<\frac{p_{n}}{q_{n}},\tau^{n}(\omega) $\displaystyle=\left\{\omega\in\Omega:\frac{p_{n}+xp_{n-1}}{q_{n}+xq_{n-1}}<% \omega<\frac{p_{n}}{q_{n}}\right\}$

and if $n$ is even then

 $(\tau^{n}

Therefore if $n$ is odd,

 $\displaystyle\lambda((\tau^{n} $\displaystyle=\frac{p_{n}}{q_{n}}-\frac{p_{n}+xp_{n-1}}{q_{n}+xq_{n-1}}$ $\displaystyle=\frac{xp_{n}q_{n-1}-xp_{n-1}q_{n}}{q_{n}(q_{n}+xq_{n-1})}$ $\displaystyle=\frac{x}{q_{n}(q_{n}+xq_{n-1})}$

and likewise if $n$ is even then

 $\lambda((\tau^{n}

Therefore for $n\geq 1$,

 $\displaystyle\lambda(\tau^{n} $\displaystyle=\frac{x}{q_{n}(q_{n}+xq_{n-1})}\cdot q_{n}(q_{n}+q_{n-1})$ $\displaystyle=\frac{x(q_{n}+q_{n-1})}{q_{n}+xq_{n-1}}.$

Using $s_{n}+1=\frac{q_{n}+q_{n-1}}{q_{n}}$ and $s_{n}x+1=\frac{xq_{n-1}+q_{n}}{q_{n}}$,

 $\displaystyle\lambda(\tau^{n} $\displaystyle=\frac{xq_{n}(s_{n}+1)}{q_{n}(s_{n}x+1)}$ $\displaystyle=\frac{x(s_{n}+1)}{s_{n}x+1}.$

For $j\geq 1$ and $s\in I$ define

 $P_{j}(s)=\frac{s+1}{(s+j)(s+j+1)}.$

We now apply Theorem 3 to prove the following.66 6 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 22, Proposition 1.2.7.

###### Theorem 4.

For $j\geq 1$,

 $\lambda(a_{1}=j)=\frac{1}{j(j+1)}.$

For $n\geq 1$ and $i\in\mathbb{N}^{n}$,

 $\lambda(a_{n+1}=j|i)=P_{j}(s_{n}).$
###### Proof.

By Theorem 2,

 $\{\omega\in\Omega:a_{1}(\omega)=j\}=I_{1}(j)=\Omega\cap(u_{1}(j),v_{1}(j)).$

In this case, $q_{1}=j$, so $u_{1}(j)=\frac{p_{1}+p_{0}}{q_{1}+q_{0}}=\frac{1+0}{j+1}=\frac{1}{j+1}$ and $v_{1}(j)=\frac{p_{1}}{q_{1}}=\frac{1}{j}$, so

 $\{\omega\in\Omega:a_{1}(\omega)=j\}=\Omega\cap\left(\frac{1}{j+1},\frac{1}{j}% \right).$

Now,

 $a_{n+1}(\omega)=\left[\frac{1}{\tau^{n}(\omega)}\right]=a_{1}(\tau^{n}(\omega)).$

Thus

 $\{\omega\in\Omega:a_{n+1}(\omega)=j\}=\left\{\omega\in\Omega:\tau^{n}(\omega)% \in\left(\frac{1}{j+1},\frac{1}{j}\right)\right\}.$

Then using Theorem 3,

 $\displaystyle\lambda(a_{n+1}=j|i)$ $\displaystyle=\lambda\left(\tau^{n}<\frac{1}{j}\big{|}i\right)-\lambda\left(% \tau^{n}<\frac{1}{j+1}\big{|}i\right)$ $\displaystyle=\frac{\frac{1}{j}(s_{n}+1)}{s_{n}\frac{1}{j}+1}-\frac{\frac{1}{j% +1}(s_{n}+1)}{s_{n}\frac{1}{j+1}+1}$ $\displaystyle=\frac{s_{n}+1}{(s_{n}+1)(s_{n}+j+1)}.$

## 4 Perron-Frobenius operators

For a probability measure $\mu$ on $\mathscr{B}_{I}$ and for $f\in L^{1}(\mu)$ let $d\mu_{f}=fd\mu$. If $\tau_{*}\mu$ is absolutely continuous with respect to $\mu$, check that $\tau_{*}\mu_{f}$ is itself absolutely continuous with respect to $\mu$. Then applying the Radon-Nikodym theorem, let

 $P_{\mu}f=\frac{d(\tau_{*}\mu_{f})}{d\mu}.$

For $g\in L^{\infty}(\mu)$,

 $\int_{I}g\cdot P_{\mu}fd\mu=\int_{I}gd(\tau_{*}\mu_{f})=\int_{I}g\circ\tau d% \mu_{f}=\int_{I}(g\circ\tau)\cdot fd\mu.$

In particular, for $g=1_{A}$, $A\in\mathscr{B}_{I}$,

 $\int_{I}1_{A}\cdot P_{\mu}fd\mu=\int_{I}1_{\tau^{-1}(A)}\cdot fd\mu.$

For $g\in L^{\infty}(\mu)$,

 $\int_{I}g\cdot P_{\gamma}1d\gamma=\int_{I}g\circ\tau d\gamma=\int_{I}gd(\tau_{% *}\gamma),$

hence $P_{\gamma}1=1$ if and only if $\tau_{*}\gamma$.

We shall be especially interested in

 $U=P_{\gamma},$

where $\gamma$ is the Gauss measure on $I$. We establish almost everywhere an expression for $Uf(x)$.77 7 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 59, Proposition 2.1.2.

###### Theorem 5.

For $f\in L^{1}(\gamma)$, for $\gamma$-almost all $x\in I$,

 $Uf(x)=\sum_{i\geq 1}P_{i}(x)f\left(\frac{1}{x+i}\right).$
###### Proof.

Let $I_{i}=\left(\frac{1}{i+1},\frac{1}{i}\right]$ and let $\tau_{i}$ be the restriction of $\tau:I\to I$ to $I_{i}$. For $u\in I_{i}$, $i\leq\frac{1}{u}, hence $\tau_{i}(u)=\tau(u)=\frac{1}{u}-i$, i.e. $u=\frac{1}{\tau_{i}(u)+i}$, i.e. $\tau_{i}^{-1}(x)=\frac{1}{x+i}$.

For $A\in\mathscr{B}_{I}$, if $0\not\in A$ then

 $\tau^{-1}(A)=\tau^{-1}\left(\bigcup_{i\geq 1}(A\cap I_{i})\right)=\bigcup_{i% \geq 1}\tau^{-1}(A\cap I_{i}),$

and the sets $\tau^{-1}(A\cap I_{i})$ are pairwise disjoint, hence

 $\int_{\tau^{-1}(A)}fd\gamma=\sum_{i\geq 1}\int_{\tau^{-1}(A\cap I_{i})}fd% \gamma=\sum_{i\geq 1}\int_{\tau_{i}^{-1}(A)}fd\gamma.$

Applying the change of variables formula, as $\frac{d}{dx}\tau_{i}^{-1}(x)=-(x+i)^{-2}$,

 $\displaystyle\int_{\tau_{i}^{-1}(A)}fd\gamma$ $\displaystyle=\frac{1}{\log 2}\int_{\tau_{i}^{-1}(A)}\frac{f(u)}{u+1}d\lambda(u)$ $\displaystyle=\frac{1}{\log 2}\int_{A}\frac{f\circ\tau_{i}^{-1}(x)}{\tau_{i}^{% -1}(x)+1}\cdot(x+i)^{-2}d\lambda(x)$ $\displaystyle=\frac{1}{\log 2}\int_{A}f\left(\frac{1}{x+i}\right)\cdot\frac{1}% {(x+i+1)(x+i)}d\lambda(x)$ $\displaystyle=\frac{1}{\log 2}\int_{A}f\left(\frac{1}{x+i}\right)\cdot P_{i}(x% )\cdot\frac{1}{x+1}d\lambda(x)$ $\displaystyle=\int_{A}f\left(\frac{1}{x+i}\right)\cdot P_{i}(x)d\gamma(x).$

Therefore

 $\displaystyle\int_{\tau^{-1}(A)}fd\gamma$ $\displaystyle=\sum_{i\geq 1}\int_{A}f\left(\frac{1}{x+i}\right)\cdot P_{i}(x)d% \gamma(x)$ $\displaystyle=\int_{A}\sum_{i\geq 1}f\left(\frac{1}{x+i}\right)\cdot P_{i}(x)d% \gamma(x).$

Then

 $\int_{A}P_{\gamma}fd\gamma=\int_{A}\sum_{i\geq 1}f\left(\frac{1}{x+i}\right)% \cdot P_{i}(x)d\gamma(x).$

Because this is true for any $A\in\mathscr{B}_{I}$ with $0\not\in A$, it follows that for $\gamma$-almost all $x\in I$,

 $P_{\gamma}f(x)=\sum_{i\geq 1}f\left(\frac{1}{x+i}\right)\cdot P_{i}(x).$

The following gives an expression for $P_{\mu}f(x)$ under some hypotheses.88 8 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 60, Proposition 2.1.3.

###### Theorem 6.

Let $\mu$ be a probability measure on $\mathscr{B}_{I}$ that is absolutely continuous with respect to $\lambda$ and suppose that $d\mu=hd\lambda$ with $h(x)>0$ for $\mu$-almost all $x\in I$. Let $f\in L^{1}(\mu)$ and define $g(x)=(x+1)h(x)f(x)$. For $\mu$-almost all $x\in I$,

 $P_{\mu}f(x)=\frac{1}{h(x)}\sum_{i\geq 1}\frac{h((x+i)^{-1})}{(x+i)^{2}}f\left(% \frac{1}{x+i}\right)=\frac{Ug(x)}{(x+1)h(x)}.$

For $n\geq 1$, for $\mu$-almost all $x\in I$,

 $P_{\mu}^{n}f(x)=\frac{U^{n}g(x)}{(x+1)h(x)}.$

We prove an expression for $\mu(\tau^{-n}(A))$.99 9 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 61, Proposition 2.1.5.

###### Theorem 7.

Let $\mu$ be a probability measure on $\mathscr{B}_{I}$ that is absolutely continuous with respect to $\lambda$. Let $h=\frac{d\mu}{d\lambda}$ and let $f(x)=(x+1)h(x)$. For $A\in\mathscr{B}_{I}$ and $n\geq 1$,

 $\mu(\tau^{-n}(A))=\int_{A}\frac{U^{n}f(x)}{x+1}d\lambda(x).$
###### Proof.

For $n=0$,

 $\mu(A)=\int_{A}d\mu=\int_{A}hd\lambda=\int_{A}\frac{f(x)}{x+1}d\lambda(x)=\int% _{A}\frac{U^{0}f(x)}{x+1}d\lambda(x).$

Suppose by hypothesis that the claim is true for some $n\geq 0$. Then

 $\displaystyle\mu(\tau^{-n-1}(A))$ $\displaystyle=\mu(\tau^{-n}(\tau^{-1}(A)))$ $\displaystyle=\int_{\tau^{-1}(A)}\frac{U^{n}f(x)}{x+1}d\lambda(x)$ $\displaystyle=\log 2\cdot\int_{\tau^{-1}(A)}U^{n}f(x)d\gamma(x)$ $\displaystyle=\log 2\cdot\int_{A}U^{n+1}f(x)d\gamma(x)$ $\displaystyle=\log 2\cdot\int_{A}\frac{U^{n+1}f(x)}{x+1}d\lambda(x).$

For $f(x)=\frac{1}{x+1}$ and $A\in\mathscr{B}_{I}$,

 $\displaystyle\int_{A}P_{\lambda}fd\lambda$ $\displaystyle=\int_{\tau^{-1}(A)}\frac{1}{x+1}d\lambda(x)$ $\displaystyle=\log 2\cdot\int_{\tau^{-1}(A)}d\gamma$ $\displaystyle=\log 2\cdot\int_{A}d\gamma$ $\displaystyle=\int_{A}fd\lambda.$

Because this is true for all Borel sets $A$,

 $P_{\lambda}\frac{1}{x+1}=\frac{1}{x+1}.$

For $f\in L^{1}(\lambda)$ and $x\in I$, let

 $\Pi_{1}f(x)=\frac{1}{(x+1)\log 2}\int_{I}fd\lambda.$

Define

 $T_{0}=P_{\lambda}-\Pi_{1}.$

For $n\geq 1$, $\Pi_{1}^{n}=\Pi_{1}$. For $f\in L^{1}(\lambda)$,

 $P_{\lambda}\Pi_{1}f=\frac{1}{\log 2}\int_{I}fd\lambda\cdot P_{\lambda}\frac{1}% {x+1}=\frac{1}{\log 2}\int_{I}fd\lambda\cdot\frac{1}{x+1}=\Pi_{1}f(x)$

and

 $\Pi_{1}P_{\lambda}f=\frac{1}{(x+1)\log 2}\int_{I}P_{\lambda}fd\lambda=\frac{1}% {(x+1)\log 2}\int_{I}fd\lambda=\Pi_{1}f(x),$

hence

 $P_{\lambda}\Pi_{1}=\Pi_{1}=\Pi_{1}P_{\lambda}.$

Moreover,

 $T_{0}\Pi_{1}=(P_{\lambda}-\Pi_{1})\Pi_{1}=P_{\lambda}\Pi_{1}-\Pi_{1}^{2}=0$

and

 $\Pi_{1}T_{0}=\Pi_{1}(P_{\lambda}-\Pi_{1})=\Pi_{1}P_{\lambda}-\Pi_{1}^{2}=0.$

Because $P_{\lambda}=\Pi_{1}+T_{0}$, using $\Pi_{1}^{2}=\Pi_{1}$, $T_{0}\Pi_{1}=0$, and $\Pi_{1}T_{0}=0$, we have

 $P_{\lambda}^{n}=\Pi_{1}+T_{0}^{n},\qquad n\geq 1.$

Theorem 6 tells us that for $f\in L^{1}(\lambda)$, for $\lambda$-almost all $x\in I$,

 $P_{\lambda}f(x)=\sum_{i\geq 1}\frac{1}{(x+i)^{2}}f\left(\frac{1}{x+i}\right).$

With $h(x)=x+1$ and $g=hf$, for $n\geq 1$, for $\lambda$-almost all $x\in I$,

 $P_{\lambda}^{n}f(x)=\frac{U^{n}g(x)}{x+1}.$

Thus

 $\displaystyle U^{n}g$ $\displaystyle=hP_{\lambda}^{n}f$ $\displaystyle=h\Pi_{1}f+hT_{0}^{n}f$ $\displaystyle=\frac{1}{\log 2}\int_{I}fd\lambda+hT_{0}^{n}f$ $\displaystyle=\int_{I}gd\gamma+hT_{0}^{n}(g/h).$

Define $I_{\gamma}:L^{1}(\gamma)\to L^{1}(\gamma)$ by

 $I_{\gamma}f=1\cdot\int_{I}fd\gamma.$

We have

 $I_{\gamma}Uf=\int_{I}P_{\gamma}fd\gamma=\int_{I}fd\gamma=I_{\gamma}f,$

meaning $I_{\gamma}U=I_{\gamma}$. Furthermore, because $\tau_{*}\gamma=\gamma$ we have $P_{\gamma}1=1$, so

 $UI_{\gamma}f=\int_{I}fd\gamma\cdot U1=\int_{I}fd\gamma\cdot 1=I_{\gamma}f,$

meaning $UI_{\gamma}=I_{\gamma}$.

Let $h(x)=x+1$. $h,\frac{1}{h}\in L^{\infty}(\gamma)$. Now define $T:L^{1}(\gamma)\to L^{1}(\gamma)$ by

 $Tg=h\cdot T_{0}(g/h),$

which makes sense because $\frac{1}{h}\in L^{\infty}(\gamma)$. Then

 $\displaystyle T^{2}g$ $\displaystyle=T(h\cdot T_{0}(g/h))$ $\displaystyle=h\cdot T_{0}\left(\frac{h\cdot T_{0}(g/h)}{h}\right)$ $\displaystyle=h\cdot T_{0}^{2}(g/h).$

For $n\geq 1$,

 $T^{n}g=h\cdot T_{0}^{n}(g/h).$

Recapitulating the above, for $n\geq 1$ and $g\in L^{1}(\gamma)$,

 $U^{n}g=I_{\gamma}g+hT_{0}^{n}(g/h)=I_{\gamma}g+T^{n}g,$

meaning

 $U^{n}=I_{\gamma}+T^{n},\qquad n\geq 1.$

It is a fact that $T^{n}$ converges to $0$ in the strong operator topology on $\mathscr{L}(L^{1}(\gamma))$, the bounded linear operators $L^{1}(\gamma)\to L^{1}(\gamma)$, that is, for each $f\in L^{1}(\gamma)$, $T^{n}f\to 0$ in $L^{1}(\gamma)$, i.e. $\left\|T^{n}f\right\|_{L^{1}}\to 0$.1010 10 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 63, Proposition 2.1.7. Then $U^{n}\to I_{\gamma}$ in the strong operator topology: for $f\in L^{1}(\gamma)$,

 $\int_{I}\left|U^{n}f(x)-\int_{I}fd\gamma\right|d\lambda\to 0.$

Iosifescu and Kraaikamp state that has not been determined whether for $\gamma$-almost all $x\in I$, $U_{n}f(x)\to I_{\gamma}f$.

Let $B(I)$ be the set of bounded Borel measurable functions $f:I\to\mathbb{C}$ and write $\left\|f\right\|_{\infty}=\sup_{x\in I}|f(x)|$. For $f\in B(I)$, define for $x\in I$,

 $Uf(x)=\sum_{i\geq 1}P_{i}(x)f\left(\frac{1}{x+i}\right)=\sum_{i\geq 1}\frac{x+% 1}{(x+i)(x+i+1)}f\left(\frac{1}{x+i}\right).$

$1\in B(I)$, and for $x\in I$,

 $\sum_{1\leq i\leq m}\frac{x+1}{(x+i)(x+i+1)}=\frac{m}{m+x+1},$

hence

 $U1(x)=\sum_{i\geq 1}\frac{x+1}{(x+i)(x+i+1)}=1.$

For $f\in B(I)$ and $x\in I$,

 $|Uf(x)|\leq\left\|f\right\|_{\infty}\cdot U1(x),$

hence

 $\left\|U\right\|_{B(I)\to B(I)}=1.$

Say that $f:I\to\mathbb{R}$ is increasing if $x\leq y$ implies $f(x)\leq f(y)$. An increasing function $f:I\to\mathbb{R}$ belongs to $B(I)$. We prove that if $f$ is increasing then $Uf$ is decreasing.1111 11 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 65, Proposition 2.1.11.

###### Theorem 8.

If $f:I\to\mathbb{R}$ is increasing then $Uf$ is decreasing.

###### Proof.

Take $x and let

 $S_{1}=\sum_{i\geq 1}P_{i}(y)\left(f\left(\frac{1}{y+i}\right)-f\left(\frac{1}{% x+i}\right)\right)$

and

 $S_{2}=\sum_{i\geq 1}(P_{i}(y)-P_{i}(x))f\left(\frac{1}{x+i}\right).$

Then

 $\displaystyle Uf(y)-Uf(x)$ $\displaystyle=\sum_{i\geq 1}\left(P_{i}(y)f\left(\frac{1}{y+i}\right)-P_{i}(x)% f\left(\frac{1}{x+i}\right)\right)$ $\displaystyle=S_{1}+S_{2}.$

Because $f$ is increasing, $S_{1}\leq 0$. Using $\sum_{i\geq 1}P_{i}(u)=1$ for any $u\in I$,

 $\sum_{i\geq 1}(P_{i}(y)-P_{i}(x))f\left(\frac{1}{x+1}\right)=0,$

and therefore

 $\displaystyle S_{2}$ $\displaystyle=\sum_{i\geq 1}\left(f\left(\frac{1}{x+i}\right)-f\left(\frac{1}{% x+1}\right)\right)(P_{i}(y)-P_{i}(x))$ $\displaystyle=\left(f\left(\frac{1}{x+2}\right)-f\left(\frac{1}{x+1}\right)% \right)(P_{2}(y)-P_{2}(x))$ $\displaystyle+\sum_{i\geq 3}\left(f\left(\frac{1}{x+i}\right)-f\left(\frac{1}{% x+1}\right)\right)(P_{i}(y)-P_{i}(x)).$

For $i\geq 2$, using that $f$ is increasing,

 $f\left(\frac{1}{x+i}\right)-f\left(\frac{1}{x+1}\right)\leq f\left(\frac{1}{x+% 2}\right)-f\left(\frac{1}{x+1}\right)\leq 0.$

We calculate

 $P_{i}^{\prime}(u)=-\frac{-i^{2}+i+(u+1)^{2}}{(u+i)^{2}(u+i+1)^{2}}.$

The roots of the above rational function are $u=-\sqrt{(i-1)i}-1,\sqrt{(i-1)i}-1$. Thus, $P_{i}^{\prime}(u)=0$ if and only if $u=\sqrt{(i-1)i}-1$. But $\sqrt{(i-1)i}-1\in I$ if and only if $i^{2}-i-1\geq 0$ and $i^{2}-i-4\leq 0$. This is possible if and only if $i=2$. And

 $P_{i}^{\prime}(0)=\frac{i^{2}-i-1}{i^{2}(i+1)^{2}},$

so $P_{1}^{\prime}(u)\leq 0$ for all $u\in I$ and for $i\geq 3$, $P_{i}^{\prime}(u)\geq 0$ for all $u\in I$. For $i=2$, check that if $0\leq u\leq\sqrt{2}-1$ then $P_{2}^{\prime}(u)\geq 0$ and if $\sqrt{2}-1\leq u\leq 1$ then $P_{2}^{\prime}(u)\leq 0$. Then

 $\displaystyle S_{2}$ $\displaystyle\leq\left(f\left(\frac{1}{x+2}\right)-f\left(\frac{1}{x+1}\right)% \right)(P_{2}(y)-P_{2}(x))$ $\displaystyle+\sum_{i\geq 3}\left(f\left(\frac{1}{x+2}\right)-f\left(\frac{1}{% x+1}\right)\right)(P_{i}(y)-P_{i}(x))$ $\displaystyle=\left(f\left(\frac{1}{x+2}\right)-f\left(\frac{1}{x+1}\right)% \right)(P_{2}(y)-P_{2}(x))$ $\displaystyle+\left(f\left(\frac{1}{x+2}\right)-f\left(\frac{1}{x+1}\right)% \right)(-P_{1}(y)-P_{2}(y)-(-P_{1}(x)-P_{2}(x)))$ $\displaystyle=\left(f\left(\frac{1}{x+2}\right)-f\left(\frac{1}{x+1}\right)% \right)(P_{1}(x)-P_{1}(y))$ $\displaystyle\leq 0.$

We have shown that $S_{1}\leq 0$ and $S_{2}\leq 0$, so

 $Uf(y)-Uf(x)=S_{1}+S_{2}\leq 0,$

which means that $Uf:I\to\mathbb{R}$ is decreasing. ∎

For $J=[a,b]\subset I$, a partition of $J$ is a sequence $P=(t_{0},\ldots,t_{n})$ such that $a=t_{0}<\cdots. For $f:I\to\mathbb{R}$ define

 $V(f,P)=\sum_{1\leq i\leq n}|f(t_{i})-f(t_{i-1})|.$

Define

 $V_{J}f=\sup\{V(f,P):\textrm{P is a partition of J}\}.$

Let $v_{f}(x)=V_{[0,x]}f$, the variation of $f$. $v_{f}(1)=V_{[0,1]}f$. We say that $f$ has bounded variation if $v_{f}(1)<\infty$, and denote by $BV(I)$ the set of functions $f:I\to\mathbb{R}$ with bounded variation. It is a fact that with the norm

 $\left\|f\right\|_{BV}=|f(0)|+V_{I}f,$

$BV(I)$ is a Banach algebra.

If $f$ is increasing then $V_{I}f=f(1)-f(0)$. We will use the following to prove the theorem coming after it.1212 12 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 66, Proposition 2.1.12.

###### Lemma 9.

If $f:I\to\mathbb{R}$ is increasing then

 $V_{I}(Uf)\leq\frac{1}{2}V_{I}f.$
###### Proof.

Because $Uf$ is decreasing,

 $V_{I}(Uf)=Uf(0)-Uf(1)=\sum_{i\geq 1}\left(P_{i}(0)f\left(\frac{1}{i}\right)-P_% {i}(1)f\left(\frac{1}{1+i}\right)\right).$

As $P_{i}(u)=\frac{u+1}{(u+i)(u+i+1)}$,

 $P_{i}(1)=\frac{2}{(i+1)(i+2)}=2P_{i+1}(0),$

hence

 $\displaystyle V_{I}(Uf)$ $\displaystyle=\sum_{i\geq 1}\left(P_{i}(0)f\left(\frac{1}{i}\right)-P_{i}(1)f% \left(\frac{1}{1+i}\right)\right)$ $\displaystyle=\sum_{i\geq 1}\left(P_{i}(0)f\left(\frac{1}{i}\right)-P_{i+1}(0)% f\left(\frac{1}{1+i}\right)\right)$ $\displaystyle-\sum_{i\geq 1}P_{i+1}(0)f\left(\frac{1}{1+i}\right)$ $\displaystyle=P_{1}(0)f(1)-\sum_{i\geq 1}P_{i+1}(0)f\left(\frac{1}{1+i}\right)$ $\displaystyle=\frac{1}{2}f(1)-\sum_{i\geq 1}P_{i+1}(0)f\left(\frac{1}{1+i}% \right).$

Because $f\left(\frac{1}{1+i}\right)\geq f(0)$ we have $-f\left(\frac{1}{1+i}\right)\leq-f(0)$, hence

 $V_{I}(Uf)\leq\frac{1}{2}f(1)-f(0)\sum_{i\geq 1}P_{i+1}(0)=\frac{1}{2}f(1)-% \frac{1}{2}f(0),$

using $\sum_{i\geq 1}P_{i}(0)=1$ and $P_{1}(0)=\frac{1}{2}$. As $f$ is increasing this means

 $V_{I}(Uf)\leq\frac{1}{2}(f(1)-f(0))=\frac{1}{2}V_{I}f.$

###### Theorem 10.

If $f\in BV(I)$ then

 $V_{I}(Uf)\leq\frac{1}{2}V_{I}f.$
###### Proof.

Let

 $p_{f}(x)=\frac{v_{f}(x)+f(x)-f(0)}{2},\qquad n_{f}(x)=\frac{v_{f}(x)-f(x)+f(0)% }{2},$

the positive variation of $f$ and the negative variation of $f$. It is a fact that $0\leq p_{f}\leq v_{f}$, $0\leq n_{f}\leq v_{f}$, and $p_{f}$ and $n_{f}$ are increasing. Using this,

 $\displaystyle V_{I}(Uf)$ $\displaystyle=V_{I}(Up_{f}+Un_{f})$ $\displaystyle\leq\frac{1}{2}V_{I}p_{f}+\frac{1}{2}V_{I}n_{f}$ $\displaystyle=\frac{1}{2}(p_{f}(1)-p_{f}(0))+\frac{1}{2}(n_{f}(1)-n_{f}(0))$ $\displaystyle=\frac{1}{2}(v_{f}(1)-v_{f}(0))$ $\displaystyle=\frac{1}{2}V_{I}f.$

For $f:I\to\mathbb{C}$, let

 $s(f)=\sup_{x,y\in I,x\neq y}\frac{|f(x)-f(y)|}{|x-y|}.$

We denote by $\mathrm{Lip}(I)$ the set of $f:I\to\mathbb{C}$ such that $s(f)<\infty$.1313 13 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 67, Proposition 2.1.14.

###### Theorem 11.

For $f\in\mathrm{Lip}(I)$,

 $s(Uf)\leq(2\zeta(3)-\zeta(2))s(f).$
###### Proof.

Suppose $x,y\in I$, $x>y$. We calculate

 $\begin{split}&\displaystyle\frac{Uf(y)-Uf(x)}{y-x}\\ \displaystyle=&\displaystyle\frac{1}{y-x}\sum_{i\geq 1}\left(P_{i}(y)f\left(% \frac{1}{y+i}\right)-P_{i}(y)f\left(\frac{1}{x+i}\right)\right)\\ &\displaystyle+\frac{1}{y-x}\sum_{i\geq 1}\left(P_{i}(y)f\left(\frac{1}{x+i}% \right)-P_{i}(x)f\left(\frac{1}{x+i}\right)\right)\\ \displaystyle=&\displaystyle\sum_{i\geq 1}P_{i}(y)\cdot\frac{f\left(\frac{1}{y% +i}\right)-f\left(\frac{1}{x+i}\right)}{y-x}\\ &\displaystyle+\sum_{i\geq 1}\frac{P_{i}(y)-P_{i}(x)}{y-x}f\left(\frac{1}{x+i}% \right).\end{split}$

Calculating further,

 $\displaystyle\frac{Uf(y)-Uf(x)}{y-x}$ $\displaystyle=-\sum_{i\geq 1}P_{i}(y)\cdot\frac{f\left(\frac{1}{y+i}\right)-f% \left(\frac{1}{x+i}\right)}{\frac{1}{y+i}-\frac{1}{x+i}}\cdot\frac{1}{(x+i)(y+% i)}$ $\displaystyle+\sum_{i\geq 1}\frac{P_{i}(y)-P_{i}(x)}{y-x}f\left(\frac{1}{x+i}% \right).$

Now,

 $P_{i}(u)=\frac{u+1}{(u+i)(u+i+1)}=\frac{i}{u+i+1}-\frac{i-1}{u+i},$

whence

 $P_{i}(y)-P_{i}(x)=\frac{(x-y)i}{(x+i+1)(y+i+1)}+\frac{(y-x)(i-1)}{(x+i)(y+i)},$

therefore

 $\begin{split}&\displaystyle\sum_{i\geq 1}\frac{P_{i}(y)-P_{i}(x)}{y-x}f\left(% \frac{1}{x+i}\right)\\ \displaystyle=&\displaystyle\sum_{i\geq 1}\left(\frac{i-1}{(x+i)(y+i)}-\frac{i% }{(x+i+1)(y+i+1)}\right)f\left(\frac{1}{x+i}\right).\end{split}$

Summation by parts tells us

 $\sum_{i\geq 1}f_{i}(g_{i+1}-g_{i})=-f_{1}g_{1}-\sum_{i\geq 1}g_{i+1}(f_{i+1}-f% _{i}),$

and here this yields, for $g_{i}=\frac{i-1}{(x+i)(y+i)}$ and $f_{i}=f\left(\frac{1}{x+i}\right)$,

 $\begin{split}&\displaystyle\sum_{i\geq 1}\left(\frac{i-1}{(x+i)(y+i)}-\frac{i}% {(x+i+1)(y+i+1)}\right)f\left(\frac{1}{x+i}\right)\\ \displaystyle=&\displaystyle\sum_{i\geq 1}g_{i+1}(f_{i+1}-f_{i})\\ \displaystyle=&\displaystyle\sum_{i\geq 1}\frac{i}{(x+i+1)(y+i+1)}\left(f\left% (\frac{1}{x+i+1}\right)-f\left(\frac{1}{x+i}\right)\right)\\ \displaystyle=&\displaystyle\sum_{i\geq 1}\frac{i}{(x+i+1)(y+i+1)}\cdot\frac{f% \left(\frac{1}{x+i+1}\right)-f\left(\frac{1}{x+i}\right)}{\frac{1}{x+i+1}-% \frac{1}{x+i}}\cdot\frac{-1}{(x+i)(x+i+1)}.\end{split}$

Recapitulating the above,

 $\begin{split}&\displaystyle\frac{Uf(y)-Uf(x)}{y-x}\\ \displaystyle=&\displaystyle-\sum_{i\geq 1}P_{i}(y)\cdot\frac{f\left(\frac{1}{% y+i}\right)-f\left(\frac{1}{x+i}\right)}{\frac{1}{y+i}-\frac{1}{x+i}}\cdot% \frac{1}{(x+i)(y+i)}\\ &\displaystyle-\sum_{i\geq 1}\frac{i}{(x+i)(x+i+1)^{2}(y+i+1)}\cdot\frac{f% \left(\frac{1}{x+i+1}\right)-f\left(\frac{1}{x+i}\right)}{\frac{1}{x+i+1}-% \frac{1}{x+i}}.\end{split}$

Then

 $\displaystyle\left|\frac{Uf(y)-Uf(x)}{y-x}\right|$ $\displaystyle\leq s(f)\sum_{i\geq 1}P_{i}(y)\frac{1}{(x+i)(y+i)}$ $\displaystyle+s(f)\sum_{i\geq 1}\frac{i}{(x+i)(x+i+1)^{2}(y+i+1)}.$

Then, using that $x>y$,

 $\left|\frac{Uf(y)-Uf(x)}{y-x}\right|\leq s(f)\sum_{i\geq 1}\left(P_{i}(y)\frac% {1}{(y+i)^{2}}+\frac{i}{(y+i)(y+i+1)^{3}}\right).$

Because $y\in I=[0,1]$, $y\geq 0$ so

 $\sum_{i\geq 1}\frac{i}{(y+i)(y+i+1)^{3}}\leq\sum_{i\geq 1}\frac{1}{(i+1)^{3}}=% -1+\zeta(3).$

Let $h(u)=u^{2}$, with which

 $\sum_{i\geq 1}P_{i}(y)\frac{1}{(y+i)^{2}}=Uh(y).$

$h:I\to\mathbb{R}$ is increasing, so $Uh$ is decreasing. Because $P_{i}(0)=\frac{1}{i(i+1)}$,

 $\sum_{i\geq 1}P_{i}(y)\frac{1}{(y+i)^{2}}=Uh(y)\leq Uh(0)=\sum_{i\geq 1}P_{i}(% 0)\frac{1}{i^{2}}=\sum_{i\geq 1}\frac{1}{i^{3}(i+1)}.$

Doing partial fractions,

 $\frac{1}{i^{3}(i+1)}=\frac{1}{i^{3}}-\frac{1}{i^{2}}+\frac{1}{i}-\frac{1}{1+i},$

so

 $\sum_{i\geq 1}\frac{1}{i^{3}(i+1)}=\zeta(3)-\zeta(2)+1.$

Therefore

 $\left|\frac{Uf(y)-Uf(x)}{y-x}\right|\leq s(f)\left(\zeta(3)-\zeta(2)+1-1+\zeta% (3)\right)=s(f)(2\zeta(3)-\zeta(2)).$

For example, let $f(x)=x$, for which $s(f)=1$. Now,

 $Uf(x)=\sum_{i\geq 1}P_{i}(x)\frac{1}{x+i}.$

We remind ourselves that

 $P_{i}(x)=\frac{x+1}{(x+i)(x+i+1)},\quad P_{i}^{\prime}(x)=\frac{i^{2}-i-(x+1)^% {2}}{(x+i)^{2}(x+i+1)^{2}}.$

Then

 $\displaystyle(Uf)^{\prime}(x)$ $\displaystyle=\sum_{i\geq 1}\left(P_{i}^{\prime}(x)\frac{1}{x+i}-P_{i}(x)\frac% {1}{(x+i)^{2}}\right)$ $\displaystyle=\sum_{i\geq 1}\left(\frac{i^{2}-i-(x+1)^{2}}{(x+i)^{3}(x+i+1)^{2% }}-\frac{x+1}{(x+i)^{3}(x+i+1)}\right)$ $\displaystyle=\sum_{i\geq 1}\frac{i^{2}-i-(x+1)^{2}-(x+1)(x+i+1)}{(x+i)^{3}(x+% i+1)^{2}}$ $\displaystyle=\sum_{i\geq 1}\frac{-2x^{2}-ix-4x+i^{2}-2i-2}{(x+i)^{3}(x+i+1)^{% 2}}.$

Check that $x\mapsto(Uf)^{\prime}(x)$ is increasing and negative. Then $\left\|(Uf)^{\prime}\right\|\leq|(Uf)^{\prime}(0)|$, with

 $(Uf)^{\prime}(0)=\sum_{i\geq 1}\frac{i^{2}-2i-2}{i^{3}(i+1)^{2}}=-2\zeta(3)+% \zeta(2).$

Therefore for $f(x)=x$,

 $s(f)=\left\|(Uf)^{\prime}\right\|_{\infty}=2\zeta(3)-\zeta(2),$

which shows that the above theorem is sharp.