Measure theory and Perron-Frobenius operators for continued fractions

1 The continued fraction transformation

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

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 $x\in ℝ$, define $a_0(x)=[x]$, and for $n\ge 1$ define $a_n(x)\in {\mathbb{Z}}_{\ge 1}\cup \{\mathrm{\infty }\}$ 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 $n\ge 3$. Thus, with $x=\frac{13}{71}$,

$$a_0(x)=0,a_1(x)=\left[\frac{71}{13}\right]=5.$$

$$a_2(x)=\left[\frac{1}{\tau (x)}\right]=\left[\frac{13}{6}\right]=2,a_3(x)=\left[\frac{1}{{\tau }^2(x)}\right]=\left[\frac{6}{1}\right]=6.$$

$$a_4(x)=\left[\frac{1}{{\tau }^3(x)}\right]=\mathrm{\infty },a_5(x)=\mathrm{\infty },\mathrm{\dots }.$$

2 Convergents

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

$$q_{-1}=0,p_{-1}=1,q_0=1,p_0=0,$$

and for $n\ge 1$,

$$q_n=a_n{q}_{n-1}+q_{n-2},p_n=a_n{p}_{n-1}+p_{n-2}.$$

Thus

$$q_1=a_1{q}_0+q_{-1}=a_1,p_1=a_1{p}_0+p_{-1}=1.$$

One proves

$$p_n{q}_{n-1}-p_{n-1}{q}_n={(-1)}^{n+1},n\ge 0.$$

Also,

$$x=\frac{p_n+{\tau }^n(x)p_{n-1}}{q_n+{\tau }^n(x)q_{n-1}},x\in \mathrm{\Omega },n\ge 0.$$

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

For $n\ge 1$ let

$$r_n(x)=\frac{1}{{\tau }^{n-1}(x)}=a_n+{\tau }^n(x)$$

and

$$s_n=\frac{q_{n-1}}{q_n},y_n=\frac{1}{s_n}$$

and

$u_n$ $=q_{n-1}^{-2}{\left|x-{\displaystyle \frac{p_{n-1}}{q_{n-1}}}\right|}^{-1}$

$={\displaystyle \frac{1}{q_{n-1}^2}}\cdot {\displaystyle \frac{q_{n-1}(q_{n-1}+{\tau }^{n-1}(x)q_{n-2})}{{\tau }^{n-1}(x)}}$

$={\displaystyle \frac{q_{n-1}+{\tau }^{n-1}(x)q_{n-2}}{{\tau }^{n-1}(x)q_{n-1}}}$

$={\displaystyle \frac{q_{n-1}\cdot (a_n+{\tau }^n(x))+q_{n-2}}{q_{n-1}}}$

$=a_n+{\tau }^n(x)+{\displaystyle \frac{q_{n-2}}{q_{n-1}}}.$

Let $s_0=0$. It is worth noting that

$$y_1\mathrm{\cdots }{y}_n=\frac{q_1}{q_0}\mathrm{\cdots }\frac{q_n}{q_{n-1}}=\frac{q_n}{q_0}=q_n.$$

$$\frac{1}{s_n}=\frac{q_n}{q_{n-1}}=a_n+\frac{q_{n-2}}{q_{n-1}}=a_n+s_{n-1}.$$

$$u_n=a_n+{\tau }^n(x)+\frac{q_{n-2}}{q_{n-1}}=r_n+s_{n-1}.$$

3 Measure theory

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

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

so $B\setminus A\in 𝒟$. Third, suppose that $A_n\in 𝒟$, $n\ge 1$, and $A_n\uparrow A$. Because $𝒜$ is a $\sigma$-algebra, $A\in 𝒜$, and then, setting $A_0=\mathrm{\varnothing }$,

$$\mu (A)=\mu \left(\bigcup _{n\ge 1}(A_n\setminus A_{n-1})\right)=\sum _{n\ge 1}(\mu (A_n)-\mu (A_{n-1})),$$

whence $\mu (A)=\nu (A)$. Therefore $𝒟$ is a Dynkin system. Dynkin’s theorem says that if $𝒟$ is a Dynkin system and $𝒞\subset 𝒟$ where $𝒞$ is a $\pi$-system (nonempty and closed under finite intersections), then $\sigma (𝒞)\subset 𝒟$.

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

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

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

Theorem 1.

${\tau }_{*}\gamma =\gamma$.

Proof.

Let . For $x\in I$, if and only if if and only if if and only if . Then, as $0\notin {\tau }^{-1}((0,u])$,

$${\tau }^{-1}((0,u])=\bigcup _{i\ge 1}[\frac{1}{u+i},\frac{1}{i}).$$

We calculate

$\gamma ({\tau }^{-1}((0,u]))$ $={\displaystyle \sum _{i\ge 1}}\gamma \left([{\displaystyle \frac{1}{u+i}},{\displaystyle \frac{1}{i}})\right)$

$={\displaystyle \sum _{i\ge 1}}{\displaystyle {\int }_{[\frac{1}{u+i},\frac{1}{i})}}{\displaystyle \frac{1}{(1+x)\log 2}}𝑑\lambda (x)$

$={\displaystyle \frac{1}{\log 2}}{\displaystyle \sum _{i\ge 1}}\left(\log \left(1+{\displaystyle \frac{1}{i}}\right)-\log \left(1+{\displaystyle \frac{1}{u+i}}\right)\right).$

Using

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

this is

$\gamma ({\tau }^{-1}((0,u]))$ $={\displaystyle \frac{1}{\log 2}}{\displaystyle \sum _{i\ge 1}}\left(\log \left(1+{\displaystyle \frac{u}{i}}\right)-\log \left(1+{\displaystyle \frac{u}{i+1}}\right)\right)$

$={\displaystyle \frac{1}{\log 2}}{\displaystyle \sum _{i\ge 1}}{\displaystyle {\int }_{\frac{u}{i+1}}^{\frac{u}{i}}}{\displaystyle \frac{1}{1+x}}𝑑\lambda (x)$

$=\gamma ((0,u]).$

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

We remark that for a set $X$, $X^0$ is a singleton. For $i\in {\mathbb{Z}}_{\ge 1}^0$ let $I_0(i)=\mathrm{\Omega }$. For $n\ge 1$ and $i\in {\mathbb{Z}}_{\ge 1}^n$, let

$$I_n(i)=\{\omega \in \mathrm{\Omega }:a_k(x)=i_k,1\le k\le n\}.$$

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

$$[i_1,\mathrm{\dots },i_n]=\frac{1}{i_1+{\displaystyle \frac{1}{\mathrm{\cdots }+{\displaystyle \frac{1}{i_{n-1}+{\displaystyle \frac{1}{i_n}}}}}}}.$$

For $x\in I_n(i)$,

$$\frac{p_n(x)}{q_n(x)}=[i_1,\mathrm{\dots },i_n],\frac{p_{n-1}(x)}{q_{n-1}(x)}=[i_1,\mathrm{\dots },i_{n-1}].$$

The following is an expression for the sets $I_n(i)$.

Theorem 2.

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

and

Then

$$I_n(i)=\mathrm{\Omega }\cap (u_n(i),v_n(i)).$$

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

$\lambda (I_n(i))$ $=v_n(i)-u_n(i)$

$={\displaystyle \frac{p_n}{q_n}}-{\displaystyle \frac{p_n+p_{n-1}}{q_n+q_{n-1}}}$

$={\displaystyle \frac{p_n{q}_{n-1}-p_{n-1}{q}_n}{q_n(q_n+q_{n-1})}}$

$={\displaystyle \frac{{(-1)}^{n+1}}{q_n(q_n+q_{n-1})}}$

$={\displaystyle \frac{1}{q_n(q_n+q_{n-1})}},$

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.

Theorem 3.

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

Proof.

We have

Using

$$\omega =\frac{p_n+{\tau }^n(\omega )p_{n-1}}{q_n+{\tau }^n(\omega )q_{n-1}},\omega \in \mathrm{\Omega },n\ge 0,$$

if $n$ is odd then

and if $n$ is even then

Therefore if $n$ is odd,

$={\displaystyle \frac{p_n}{q_n}}-{\displaystyle \frac{p_n+x{p}_{n-1}}{q_n+x{q}_{n-1}}}$

$={\displaystyle \frac{x{p}_n{q}_{n-1}-x{p}_{n-1}{q}_n}{q_n(q_n+x{q}_{n-1})}}$

$={\displaystyle \frac{x}{q_n(q_n+x{q}_{n-1})}}$

and likewise if $n$ is even then

Therefore for $n\ge 1$,

$={\displaystyle \frac{x}{q_n(q_n+x{q}_{n-1})}}\cdot q_n(q_n+q_{n-1})$

$={\displaystyle \frac{x(q_n+q_{n-1})}{q_n+x{q}_{n-1}}}.$

Using $s_n+1=\frac{q_n+q_{n-1}}{q_n}$ and $s_{n}x+1=\frac{x{q}_{n-1}+q_n}{q_n}$,

$={\displaystyle \frac{x{q}_n(s_n+1)}{q_n(s_{n}x+1)}}$

$={\displaystyle \frac{x(s_n+1)}{s_{n}x+1}}.$

∎

For $j\ge 1$ 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.

Theorem 4.

For $j\ge 1$,

$$\lambda (a_1=j)=\frac{1}{j(j+1)}.$$

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

$$\lambda (a_{n+1}=j|i)=P_j(s_n).$$

Proof.

By Theorem 2,

$$\{\omega \in \mathrm{\Omega }:a_1(\omega )=j\}=I_1(j)=\mathrm{\Omega }\cap (u_1(j),v_1(j)).$$

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

$$\{\omega \in \mathrm{\Omega }:a_1(\omega )=j\}=\mathrm{\Omega }\cap (\frac{1}{j+1},\frac{1}{j}).$$

Now,

$$a_{n+1}(\omega )=\left[\frac{1}{{\tau }^n(\omega )}\right]=a_1({\tau }^n(\omega )).$$

Thus

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

Then using Theorem 3,

$\lambda (a_{n+1}=j|i)$

$={\displaystyle \frac{\frac{1}{j}(s_n+1)}{s_n\frac{1}{j}+1}}-{\displaystyle \frac{\frac{1}{j+1}(s_n+1)}{s_n\frac{1}{j+1}+1}}$

$={\displaystyle \frac{s_n+1}{(s_n+1)(s_n+j+1)}}.$

∎

4 Perron-Frobenius operators

For a probability measure $\mu$ on ${ℬ}_I$ 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 $g\in L^{\mathrm{\infty }}(\mu )$,

$${\int }_{I}g\cdot P_{\mu }f𝑑\mu ={\int }_{I}gd({\tau }_{*}{\mu }_f)={\int }_{I}g\circ \tau 𝑑{\mu }_f={\int }_I(g\circ \tau )\cdot f𝑑\mu .$$

In particular, for $g=1_A$, $A\in {ℬ}_I$,

$${\int }_I{1}_A\cdot P_{\mu }f𝑑\mu ={\int }_I{1}_{{\tau }^{-1}(A)}\cdot f𝑑\mu .$$

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

$${\int }_{I}g\cdot P_{\gamma }1𝑑\gamma ={\int }_{I}g\circ \tau 𝑑\gamma ={\int }_{I}gd({\tau }_{*}\gamma ),$$

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 $Uf(x)$.

Theorem 5.

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

$$Uf(x)=\sum _{i\ge 1}{P}_i(x)f\left(\frac{1}{x+i}\right).$$

Proof.

Let $I_i=(\frac{1}{i+1},\frac{1}{i}]$ and let ${\tau }_i$ be the restriction of $\tau :I\to I$ to $I_i$. For $u\in I_i$, , 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 $A\in {ℬ}_I$, if $0\notin A$ then

$${\tau }^{-1}(A)={\tau }^{-1}\left(\bigcup _{i\ge 1}(A\cap I_i)\right)=\bigcup _{i\ge 1}{\tau }^{-1}(A\cap I_i),$$

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

$${\int }_{{\tau }^{-1}(A)}f𝑑\gamma =\sum _{i\ge 1}{\int }_{{\tau }^{-1}(A\cap I_i)}f𝑑\gamma =\sum _{i\ge 1}{\int }_{{\tau }_i^{-1}(A)}f𝑑\gamma .$$

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

${\displaystyle {\int }_{{\tau }_i^{-1}(A)}}f𝑑\gamma$ $={\displaystyle \frac{1}{\log 2}}{\displaystyle {\int }_{{\tau }_i^{-1}(A)}}{\displaystyle \frac{f(u)}{u+1}}𝑑\lambda (u)$

$={\displaystyle \frac{1}{\log 2}}{\displaystyle {\int }_A}{\displaystyle \frac{f\circ {\tau }_i^{-1}(x)}{{\tau }_i^{-1}(x)+1}}\cdot {(x+i)}^{-2}𝑑\lambda (x)$

$={\displaystyle \frac{1}{\log 2}}{\displaystyle {\int }_A}f\left({\displaystyle \frac{1}{x+i}}\right)\cdot {\displaystyle \frac{1}{(x+i+1)(x+i)}}𝑑\lambda (x)$

$={\displaystyle \frac{1}{\log 2}}{\displaystyle {\int }_A}f\left({\displaystyle \frac{1}{x+i}}\right)\cdot P_i(x)\cdot {\displaystyle \frac{1}{x+1}}𝑑\lambda (x)$

$={\displaystyle {\int }_A}f\left({\displaystyle \frac{1}{x+i}}\right)\cdot P_i(x)𝑑\gamma (x).$

Therefore

${\displaystyle {\int }_{{\tau }^{-1}(A)}}f𝑑\gamma$ $={\displaystyle \sum _{i\ge 1}}{\displaystyle {\int }_A}f\left({\displaystyle \frac{1}{x+i}}\right)\cdot P_i(x)𝑑\gamma (x)$

$={\displaystyle {\int }_A}{\displaystyle \sum _{i\ge 1}}f\left({\displaystyle \frac{1}{x+i}}\right)\cdot P_i(x)d\gamma (x).$

Then

$${\int }_A{P}_{\gamma }f𝑑\gamma ={\int }_A\sum _{i\ge 1}f\left(\frac{1}{x+i}\right)\cdot P_i(x)d\gamma (x).$$

Because this is true for any $A\in {ℬ}_I$ with $0\notin A$, it follows that for $\gamma$-almost all $x\in I$,

$$P_{\gamma }f(x)=\sum _{i\ge 1}f\left(\frac{1}{x+i}\right)\cdot P_i(x).$$

∎

The following gives an expression for $P_{\mu }f(x)$ under some hypotheses.

Theorem 6.

Let $\mu$ be a probability measure on ${ℬ}_I$ 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 $x\in I$,

$$P_{\mu }f(x)=\frac{1}{h(x)}\sum _{i\ge 1}\frac{h({(x+i)}^{-1})}{{(x+i)}^2}f\left(\frac{1}{x+i}\right)=\frac{Ug(x)}{(x+1)h(x)}.$$

For $n\ge 1$, 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 $\mu ({\tau }^{-n}(A))$.

Theorem 7.

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

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

Proof.

For $n=0$,

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

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

$\mu ({\tau }^{-n-1}(A))$ $=\mu ({\tau }^{-n}({\tau }^{-1}(A)))$

$={\displaystyle {\int }_{{\tau }^{-1}(A)}}{\displaystyle \frac{U^{n}f(x)}{x+1}}𝑑\lambda (x)$

$=\log 2\cdot {\displaystyle {\int }_{{\tau }^{-1}(A)}}{U}^{n}f(x)𝑑\gamma (x)$

$=\log 2\cdot {\displaystyle {\int }_A}{U}^{n+1}f(x)𝑑\gamma (x)$

$=\log 2\cdot {\displaystyle {\int }_A}{\displaystyle \frac{U^{n+1}f(x)}{x+1}}𝑑\lambda (x).$

∎

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

${\displaystyle {\int }_A}{P}_{\lambda }f𝑑\lambda$ $={\displaystyle {\int }_{{\tau }^{-1}(A)}}{\displaystyle \frac{1}{x+1}}𝑑\lambda (x)$

$=\log 2\cdot {\displaystyle {\int }_{{\tau }^{-1}(A)}}𝑑\gamma$

$=\log 2\cdot {\displaystyle {\int }_A}𝑑\gamma$

$={\displaystyle {\int }_A}f𝑑\lambda .$

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

$${\mathrm{\Pi }}_{1}f(x)=\frac{1}{(x+1)\log 2}{\int }_{I}f𝑑\lambda .$$

Define

$$T_0=P_{\lambda }-{\mathrm{\Pi }}_1.$$

For $n\ge 1$, ${\mathrm{\Pi }}_1^n={\mathrm{\Pi }}_1$. For $f\in L^1(\lambda )$,

$$P_{\lambda }{\mathrm{\Pi }}_{1}f=\frac{1}{\log 2}{\int }_{I}f𝑑\lambda \cdot P_{\lambda }\frac{1}{x+1}=\frac{1}{\log 2}{\int }_{I}f𝑑\lambda \cdot \frac{1}{x+1}={\mathrm{\Pi }}_{1}f(x)$$

and

$${\mathrm{\Pi }}_1{P}_{\lambda }f=\frac{1}{(x+1)\log 2}{\int }_I{P}_{\lambda }f𝑑\lambda =\frac{1}{(x+1)\log 2}{\int }_{I}f𝑑\lambda ={\mathrm{\Pi }}_{1}f(x),$$

hence

$$P_{\lambda }{\mathrm{\Pi }}_1={\mathrm{\Pi }}_1={\mathrm{\Pi }}_1{P}_{\lambda }.$$

Moreover,

$$T_0{\mathrm{\Pi }}_1=(P_{\lambda }-{\mathrm{\Pi }}_1){\mathrm{\Pi }}_1=P_{\lambda }{\mathrm{\Pi }}_1-{\mathrm{\Pi }}_1^2=0$$

and

$${\mathrm{\Pi }}_1{T}_0={\mathrm{\Pi }}_1(P_{\lambda }-{\mathrm{\Pi }}_1)={\mathrm{\Pi }}_1{P}_{\lambda }-{\mathrm{\Pi }}_1^2=0.$$

Because $P_{\lambda }={\mathrm{\Pi }}_1+T_0$, using ${\mathrm{\Pi }}_1^2={\mathrm{\Pi }}_1$, $T_0{\mathrm{\Pi }}_1=0$, and ${\mathrm{\Pi }}_1{T}_0=0$, we have

$$P_{\lambda }^n={\mathrm{\Pi }}_1+T_0^n,n\ge 1.$$

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

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

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

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

Thus

$U^{n}g$ $=h{P}_{\lambda }^{n}f$

$=h{\mathrm{\Pi }}_{1}f+h{T}_0^{n}f$

$={\displaystyle \frac{1}{\log 2}}{\displaystyle {\int }_I}f𝑑\lambda +h{T}_0^{n}f$

$={\displaystyle {\int }_I}g𝑑\gamma +h{T}_0^n(g/h).$

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

$$U{I}_{\gamma }f={\int }_{I}f𝑑\gamma \cdot U1={\int }_{I}f𝑑\gamma \cdot 1=I_{\gamma }f,$$

meaning $U{I}_{\gamma }=I_{\gamma }$.

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

$$Tg=h\cdot T_0(g/h),$$

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

$T^{2}g$ $=T(h\cdot T_0(g/h))$

$=h\cdot T_0\left({\displaystyle \frac{h\cdot T_0(g/h)}{h}}\right)$

$=h\cdot T_0^2(g/h).$

For $n\ge 1$,

$$T^{n}g=h\cdot T_0^n(g/h).$$

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

$$U^{n}g=I_{\gamma }g+h{T}_0^n(g/h)=I_{\gamma }g+T^{n}g,$$

meaning

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

It is a fact that $T^n$ converges to $0$ in the strong operator topology on $ℒ(L^1(\gamma ))$, 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. ${\parallel T^{n}f\parallel }_{L^1}\to 0$. 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 $f:I\to ℂ$ and write ${\parallel f\parallel }_{\mathrm{\infty }}={sup}_{x\in I}|f(x)|$. For $f\in B(I)$, define for $x\in I$,

$$Uf(x)=\sum _{i\ge 1}{P}_i(x)f\left(\frac{1}{x+i}\right)=\sum _{i\ge 1}\frac{x+1}{(x+i)(x+i+1)}f\left(\frac{1}{x+i}\right).$$

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

$$\sum _{1\le i\le m}\frac{x+1}{(x+i)(x+i+1)}=\frac{m}{m+x+1},$$

hence

$$U1(x)=\sum _{i\ge 1}\frac{x+1}{(x+i)(x+i+1)}=1.$$

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

$$|Uf(x)|\le {\parallel f\parallel }_{\mathrm{\infty }}\cdot U1(x),$$

hence

$${\parallel U\parallel }_{B(I)\to B(I)}=1.$$

Say that $f:I\to ℝ$ is increasing if $x\le y$ implies $f(x)\le f(y)$. An increasing function $f:I\to ℝ$ belongs to $B(I)$. We prove that if $f$ is increasing then $Uf$ is decreasing.

Theorem 8.

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

Proof.

Take and let

$$S_1=\sum _{i\ge 1}{P}_i(y)\left(f\left(\frac{1}{y+i}\right)-f\left(\frac{1}{x+i}\right)\right)$$

and

$$S_2=\sum _{i\ge 1}(P_i(y)-P_i(x))f\left(\frac{1}{x+i}\right).$$

Then

$Uf(y)-Uf(x)$ $={\displaystyle \sum _{i\ge 1}}\left(P_i(y)f\left({\displaystyle \frac{1}{y+i}}\right)-P_i(x)f\left({\displaystyle \frac{1}{x+i}}\right)\right)$

$=S_1+S_2.$

Because $f$ is increasing, $S_1\le 0$. Using ${\sum }_{i\ge 1}{P}_i(u)=1$ for any $u\in I$,

$$\sum _{i\ge 1}(P_i(y)-P_i(x))f\left(\frac{1}{x+1}\right)=0,$$

and therefore

$S_2$ $={\displaystyle \sum _{i\ge 1}}\left(f\left({\displaystyle \frac{1}{x+i}}\right)-f\left({\displaystyle \frac{1}{x+1}}\right)\right)(P_i(y)-P_i(x))$

$=\left(f\left({\displaystyle \frac{1}{x+2}}\right)-f\left({\displaystyle \frac{1}{x+1}}\right)\right)(P_2(y)-P_2(x))$

$+{\displaystyle \sum _{i\ge 3}}\left(f\left({\displaystyle \frac{1}{x+i}}\right)-f\left({\displaystyle \frac{1}{x+1}}\right)\right)(P_i(y)-P_i(x)).$

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

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

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 $i^2-i-1\ge 0$ and $i^2-i-4\le 0$. 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 $P_1^{\prime }(u)\le 0$ for all $u\in I$ and for $i\ge 3$, $P_i^{\prime }(u)\ge 0$ for all $u\in I$. For $i=2$, check that if $0\le u\le \sqrt{2}-1$ then $P_2^{\prime }(u)\ge 0$ and if $\sqrt{2}-1\le u\le 1$ then $P_2^{\prime }(u)\le 0$. Then

$S_2$ $\le \left(f\left({\displaystyle \frac{1}{x+2}}\right)-f\left({\displaystyle \frac{1}{x+1}}\right)\right)(P_2(y)-P_2(x))$

$+{\displaystyle \sum _{i\ge 3}}\left(f\left({\displaystyle \frac{1}{x+2}}\right)-f\left({\displaystyle \frac{1}{x+1}}\right)\right)(P_i(y)-P_i(x))$

$=\left(f\left({\displaystyle \frac{1}{x+2}}\right)-f\left({\displaystyle \frac{1}{x+1}}\right)\right)(P_2(y)-P_2(x))$

$+\left(f\left({\displaystyle \frac{1}{x+2}}\right)-f\left({\displaystyle \frac{1}{x+1}}\right)\right)(-P_1(y)-P_2(y)-(-P_1(x)-P_2(x)))$

$=\left(f\left({\displaystyle \frac{1}{x+2}}\right)-f\left({\displaystyle \frac{1}{x+1}}\right)\right)(P_1(x)-P_1(y))$

$\le 0.$

We have shown that $S_1\le 0$ and $S_2\le 0$, so

$$Uf(y)-Uf(x)=S_1+S_2\le 0,$$

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

For $J=[a,b]\subset I$, a partition of $J$ is a sequence $P=(t_0,\mathrm{\dots },t_n)$ such that . For $f:I\to ℝ$ define

$$V(f,P)=\sum _{1\le i\le n}|f(t_i)-f(t_{i-1})|.$$

Define

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

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 , and denote by $BV(I)$ the set of functions $f:I\to ℝ$ with bounded variation. It is a fact that with the norm

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

$BV(I)$ 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.

Lemma 9.

If $f:I\to ℝ$ is increasing then

$$V_I(Uf)\le \frac{1}{2}{V}_{I}f.$$

Proof.

Because $Uf$ is decreasing,

$$V_I(Uf)=Uf(0)-Uf(1)=\sum _{i\ge 1}\left(P_i(0)f\left(\frac{1}{i}\right)-P_i(1)f\left(\frac{1}{1+i}\right)\right).$$

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

$V_I(Uf)$ $={\displaystyle \sum _{i\ge 1}}\left(P_i(0)f\left({\displaystyle \frac{1}{i}}\right)-P_i(1)f\left({\displaystyle \frac{1}{1+i}}\right)\right)$

$={\displaystyle \sum _{i\ge 1}}\left(P_i(0)f\left({\displaystyle \frac{1}{i}}\right)-P_{i+1}(0)f\left({\displaystyle \frac{1}{1+i}}\right)\right)$

$-{\displaystyle \sum _{i\ge 1}}{P}_{i+1}(0)f\left({\displaystyle \frac{1}{1+i}}\right)$

$=P_1(0)f(1)-{\displaystyle \sum _{i\ge 1}}{P}_{i+1}(0)f\left({\displaystyle \frac{1}{1+i}}\right)$

$={\displaystyle \frac{1}{2}}f(1)-{\displaystyle \sum _{i\ge 1}}{P}_{i+1}(0)f\left({\displaystyle \frac{1}{1+i}}\right).$

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

$$V_I(Uf)\le \frac{1}{2}f(1)-f(0)\sum _{i\ge 1}{P}_{i+1}(0)=\frac{1}{2}f(1)-\frac{1}{2}f(0),$$

using ${\sum }_{i\ge 1}{P}_i(0)=1$ and $P_1(0)=\frac{1}{2}$. As $f$ is increasing this means

$$V_I(Uf)\le \frac{1}{2}(f(1)-f(0))=\frac{1}{2}{V}_{I}f.$$

∎

Theorem 10.

If $f\in BV(I)$ then

$$V_I(Uf)\le \frac{1}{2}{V}_{I}f.$$

Proof.

Let

$$p_f(x)=\frac{v_f(x)+f(x)-f(0)}{2},n_f(x)=\frac{v_f(x)-f(x)+f(0)}{2},$$

the positive variation of $f$ and the negative variation of $f$. It is a fact that $0\le p_f\le v_f$, $0\le n_f\le v_f$, and $p_f$ and $n_f$ are increasing. Using this,

$V_I(Uf)$ $=V_I(U{p}_f+U{n}_f)$

$\le {\displaystyle \frac{1}{2}}{V}_I{p}_f+{\displaystyle \frac{1}{2}}{V}_I{n}_f$

$={\displaystyle \frac{1}{2}}(p_f(1)-p_f(0))+{\displaystyle \frac{1}{2}}(n_f(1)-n_f(0))$

$={\displaystyle \frac{1}{2}}(v_f(1)-v_f(0))$

$={\displaystyle \frac{1}{2}}{V}_{I}f.$

∎

For $f:I\to ℂ$, let

$$s(f)=\underset{x,y\in I,x\ne y}{sup}\frac{|f(x)-f(y)|}{|x-y|}.$$

We denote by $\mathrm{Lip}(I)$ the set of $f:I\to ℂ$ such that .

Theorem 11.

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

$$s(Uf)\le (2\zeta (3)-\zeta (2))s(f).$$

Proof.

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

Calculating further,

${\displaystyle \frac{Uf(y)-Uf(x)}{y-x}}$ $=-{\displaystyle \sum _{i\ge 1}}{P}_i(y)\cdot {\displaystyle \frac{f\left(\frac{1}{y+i}\right)-f\left(\frac{1}{x+i}\right)}{\frac{1}{y+i}-\frac{1}{x+i}}}\cdot {\displaystyle \frac{1}{(x+i)(y+i)}}$

$+{\displaystyle \sum _{i\ge 1}}{\displaystyle \frac{P_i(y)-P_i(x)}{y-x}}f\left({\displaystyle \frac{1}{x+i}}\right).$

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

Summation by parts tells us

$$\sum _{i\ge 1}{f}_i(g_{i+1}-g_i)=-f_1{g}_1-\sum _{i\ge 1}{g}_{i+1}(f_{i+1}-f_i),$$

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

Recapitulating the above,

Then

$\left|{\displaystyle \frac{Uf(y)-Uf(x)}{y-x}}\right|$ $\le s(f){\displaystyle \sum _{i\ge 1}}{P}_i(y){\displaystyle \frac{1}{(x+i)(y+i)}}$

$+s(f){\displaystyle \sum _{i\ge 1}}{\displaystyle \frac{i}{(x+i){(x+i+1)}^2(y+i+1)}}.$

Then, using that $x>y$,

$$\left|\frac{Uf(y)-Uf(x)}{y-x}\right|\le s(f)\sum _{i\ge 1}\left(P_i(y)\frac{1}{{(y+i)}^2}+\frac{i}{(y+i){(y+i+1)}^3}\right).$$

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

$$\sum _{i\ge 1}\frac{i}{(y+i){(y+i+1)}^3}\le \sum _{i\ge 1}\frac{1}{{(i+1)}^3}=-1+\zeta (3).$$

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

$$\sum _{i\ge 1}{P}_i(y)\frac{1}{{(y+i)}^2}=Uh(y).$$

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

$$\sum _{i\ge 1}{P}_i(y)\frac{1}{{(y+i)}^2}=Uh(y)\le Uh(0)=\sum _{i\ge 1}{P}_i(0)\frac{1}{i^2}=\sum _{i\ge 1}\frac{1}{i^3(i+1)}.$$

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

$$\sum _{i\ge 1}\frac{1}{i^3(i+1)}=\zeta (3)-\zeta (2)+1.$$

Therefore

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

∎

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

$$Uf(x)=\sum _{i\ge 1}{P}_i(x)\frac{1}{x+i}.$$

We remind ourselves that

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

Then

${(Uf)}^{\prime }(x)$ $={\displaystyle \sum _{i\ge 1}}\left(P_i^{\prime }(x){\displaystyle \frac{1}{x+i}}-P_i(x){\displaystyle \frac{1}{{(x+i)}^2}}\right)$

$={\displaystyle \sum _{i\ge 1}}\left({\displaystyle \frac{i^2-i-{(x+1)}^2}{{(x+i)}^3{(x+i+1)}^2}}-{\displaystyle \frac{x+1}{{(x+i)}^3(x+i+1)}}\right)$

$={\displaystyle \sum _{i\ge 1}}{\displaystyle \frac{i^2-i-{(x+1)}^2-(x+1)(x+i+1)}{{(x+i)}^3{(x+i+1)}^2}}$

$={\displaystyle \sum _{i\ge 1}}{\displaystyle \frac{-2x^2-ix-4x+i^2-2i-2}{{(x+i)}^3{(x+i+1)}^2}}.$

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

$${(Uf)}^{\prime }(0)=\sum _{i\ge 1}\frac{i^2-2i-2}{i^3{(i+1)}^2}=-2\zeta (3)+\zeta (2).$$

Therefore for $f(x)=x$,

$$s(f)={\parallel {(Uf)}^{\prime }\parallel }_{\mathrm{\infty }}=2\zeta (3)-\zeta (2),$$

which shows that the above theorem is sharp.