数理解析研究所講究録
1071 巻 1998 年 29-43

29

ALMOST EVERYWHERE CONVERGENCE THEOREMSFOR NONLINEAR OPERATORSSHOJI KAMIMURA

1. INTRODUCTION

There have been many important researches on the study of ergodic theorems
for nonlinear operators. However, most researches on this subject dealt only withquestions of mean ergodic theorems, while the problem of pointwize ergodic theorems
the study on this problem was started by Krengel
had been ignored. In theand Lin etc.
spaces are with respect to
Let $(E, \mu)$ be a -finite measure space and all thefor linear operators is
this measure space. The classical pointwise ergodic
the following theorem of Hopf:$1980’ \mathrm{s}$$L^{p}$$\sigma$$\mathrm{t}\mathrm{h}\mathrm{e}.\cdot \mathrm{o}\mathrm{r}\mathrm{e}\mathrm{m}$be a positive linear contraction on
Theorem 1.1. Let$T1=1$. Then for any $f\in L^{1}$ , the averages
that
$T$

$L^{1}$

with $\mu(E)<\infty$ such

$A_{n}f= \frac{1}{n+1}\sum_{i=0}^{n}T^{i}f$converge almos everywhere.
$t$is
In the nonlinear situation, instead of being a contraction we assume that
Krengelpreserving.
order
is
that
nonexpansive, and instead of positivity we assume
and Lin [3] obtained a result for this class of nonlinear operators:
$T$

$T$

Theorem 1.2. Let
creasing mapping on

nonexpansive andnorm debe an order preserving,$f\inL^{1}$, the averages
with $\mu(E)<\infty$ . Then for any$L^{1}$$T$$L^{\infty}$$L^{1}$(1)$A_{n}f-- \frac{1}{n+1}\sum_{i=0}T^{i}fn$converge weakly in$L^{1}$.

Further Krengel [2] gave an example that almost everywhere convergence of theaverages (1) fails. Namely, we can not expect the almost everywhere convergence fordefined by (1).
satisfy
In the linear case, the partial sums

$A_{n}f$$S_{n}f=\Sigma_{i=0^{T^{i}f}}^{n}$$S_{0}f=f$ ,$S_{n+1}f=f+\tau(s_{n}f)$$(n\geq 0)$

.

(2)

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SHOJI KAMIMURA

However in the nonlinear case, the partial sums
, of course, do not satisfy therecursive relations (2). Then Lin and Wittmann $[5, 7]$ adopted the definition (2) forand put$\{S_{n}f\}$$\{S_{n}f\}$(3)$A_{n}f= \frac{1}{n+1}S_{n}f$to get the almost everywhere convergence. Wittmann [7] obtained the following:
Theorem 1.3. Let be an order preserving, integral preserving, positively homogenonexpansive mapping on
neous andwith $\mu(E)<\infty$ . Thendefined by(3) converges almost everywhere for any $f\in L^{1}$ .
$T$$L^{1}$$L^{\infty}$$A_{n}f$

Recently Wittmann [6] improved his own result:
Theorem 1.4. Let be an order preserving andandnonexpansive mappingon . Thenconverges almost everywhere for any $f\in L^{1}$$T$$L^{1}$$L^{1}$$L^{\infty}$$A_{n}f$

In this paper, we prove another extention of Theorem 1.3 by assuming the existenceof a kind of strictly positive invariant functions. The method of the proof is due tothat of Wittmann [6]. Further we give some properties of the limit point of.
$A_{n}f$

2. PRELIMINARIES

Throughout this paper, $(E, \mu)$ is a -finite measure space and all the
spaces arewith respect to this measure space. And expressions involving measurable functions
or sets have to be understood in the almost everywhere sense.
Let be an operator on . The nonlinear ergodic average $A_{n}f(f\in L^{1})$ is definedby$L^{p}$$\sigma$$T$$L^{1}$$S_{0}f=f$ ,$S_{n+1}f=f+T(snf)$$(n\geq 0)$

and$A_{n}f= \frac{1}{n+1}S_{n}f$$(n\geq 0)$.

In the linear case,is equal to.
$f\leqg\Rightarrow
Tf\leq Tg(f, g\in L^{1})$ .
is said to be order preserving if
is said tobenonexpansive if $||Tf-Tg||p\leq||f-g||_{p}(f, g\in L^{1}\cap L^{P}, 1\leq p\leq\infty)$.
iscalled positively homogeneous if $T(\alpha f)=\alpha Tf(f\in L^{1}, \alpha\geq 0)$ .
is called integral
preserving if $\int Tfd\mu=\int fd\mu(f\in L^{1})$ . Note that an order preserving and integral
preserving mapping isnonexpansive (See Krengel and Lin [3]).
We denote by $F(T)$ the set of invariant functions of , i.e. $F(T)=\{f\in L^{1}$ :$Tf=f\}$ . We define $c(\tau)+^{\mathrm{b}G}\mathrm{y}(T)_{+}=\{k\in L^{1}$ : $k>0,$ $T(f+tk)=Tf+tk,$. Obviously if, then $G(T)_{+}\subset F(T)$ . We know from Lin andWittmann [5] that if $\mu(E)<$ oo and is order preserving, integral preserving andnonexpansive, then.
$A_{n}f$$\frac{1}{n+1}\Sigma_{i=0}^{n}T^{i}f$

$T$

$T$

$L^{p}$$T$$T$$L^{1}$$T$$\forall f\in$$L^{1},\forall t\in \mathrm{R}\}$$T\mathrm{O}=0$$T$$L^{\infty}$$G(T)_{+}\neq\emptyset$

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ALMOST EVERYWHERE CONVERGENCE THEOREMS

3. ALMOST EVERYWHERE CONVERGENCE THEOREMS FOR NONLINEAROPERATORS
For a sequence$\{f_{n}\}\subset L\mathrm{l}$, we define$f_{n}^{*}$$f_{n}^{*}= \sup_{0\leq i<n}f_{i}$by$(1\leq n\leq\infty)$.

Our first aim is to prove the nonlinear version of the maximal ergodic theorem.
The maximal ergodic theorem plays important role in the proof of pointwize ergodic
theorems. Wittmann [6] proved the following theorem and lemma.
, and letTheorem 3.1. Let be an order preserving mapping on . And let(both functions may attainbe two measurable functions with value inon a set of positive measure) such thatthe value$L^{1}$$T$$h,$ $h’$$K\in \mathrm{R}$$(-\infty, \infty]$$\infty$$\int(Tg-h’)_{+}d\mu+\leq\int(g-h)+^{d}\mu+I\{’<\infty$

Further, let$\{f_{n}\}$be a sequence in$f_{0}\leq h$,$L^{1}$$(g\in L^{1})$(4).

and let $f\in L^{1}$ such that$(n\geq 1)$$1_{\{f_{n}>h\}}f_{n}\leq 1_{\{f_{n}>h\}}(f+Tf_{n-1})$.

(5)

Then $1_{\{f_{n}^{*}>h\}}(f+h-h’)_{-}\dot{i}S$ integrable (since $\{h=\infty\}\cap\{f_{n}^{*}>h\}=\emptyset$ , this is welldefined) and$\int_{\{f_{n}^{*}h\}}>d(f+h-h’)\mu\geq-K$

If$1_{\{f_{\infty}^{*}>h\}}(f+h’-h)+or1_{\{f_{\infty}^{*}>h\}}(f+h’-h)_{-}\dot{i}S$$n=\infty$.

$(1\leq n<\infty)$(6)

integrable, then (6) holds also fornonexpansive $mapp\dot{i}ng$ on
Lemma 3.2. Let
be anmeasurable functions such that$T$.

$L^{1}$$L^{1}$. Further, let $h,$

$h’$

be

$f\leq h\Rightarrow Tf\leq h’$

for any$f\in L^{1}$Then$\int(Tf-h’)+d\mu\leq\int(f-h)_{+}d\mu<\infty$$(f\in L^{1})$.

We prepare one more lemma.
Lemma 3.3. Let be an order preserving and, and letnonempty. Let
that
that
$T$

$G(T)_{+}\dot{i}S$$K\in \mathrm{R}$$L^{1}$$h,$ $h’$nonexpansive mapping on
such
be two measurable functions such$\int(\tau_{g}-h^{;})+^{d\mu}\leq\int(g-h)_{+}d\mu+I\{’<\infty$$L^{1}$$(g\in L^{1})$.

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SHOJI KAMIMURA

Then$\int(Tf-(h’+\alpha k))+^{d\mu}\leq$$\int(f-(h+\alpha k))_{+^{dK<}}\mu+\infty$

Proof. We fix$h_{-}\in L^{1}$

fact,$(f\in L^{1}, k\in G(T)+,$ $\alpha\geq 0)$.

and $f\in L^{1}$ . Since $\int(g-h)_{+}d\mu<\infty$ , we obtainby putting $g=0$ . Setting $f_{\alpha}= \sup(f-\alpha k, -h-)$ , we can show
. In$\alpha\geq 0,$$k\in G(T)_{+}$$f_{\alpha}\in L^{1}$$\int|f_{\alpha}|d\mu=\int|\sup(f-\alpha k, -h_{-})|d\mu$$= \int_{\{f\geq\alpha k\}}(f-\alpha k)d\mu-\int_{\{f<\alpha k\}}\sup(f-\alpha k, -h_{-})d\mu$$\leq\int_{\{f\geq\alpha k\}}(f-\alpha k)d\mu+\int_{\{f<\alpha k\}}h_{-}d\mu<\infty$.

Further we obtain$(f_{\alpha}-h)+=( \sup(f-\alpha k, -h-)-h)_{+}$$=( \sup(f-(h+\alpha k), -h_{-}-h))+$$=(f-(h+\alpha k))+$and$\sup(f,$ $-h_{-)}-f \alpha=\sup(f, -h-)-\sup(f-\alpha k, -h_{-})\leq\alpha k$.

Since is order preserving and $k\in G(T)_{+},$ $T \sup(f, -h_{-)}\leq T(f_{\alpha}+\alpha k)=Tf\alpha+\alpha k$ .
This implies $Tf_{\alpha}+ \alpha k\geq T\sup(f, -h-)\geq Tf$ . The assertion follows from$T$$\int(Tf-(h’+\alpha k))_{+^{d\mu}}\leq\int(Tf_{\alpha}+\alpha k-(h’+\alpha k))_{+^{d\mu}}$$= \int(Tf_{\alpha}-h’)_{+}d\mu$$\leq\int(f_{\alpha}-h)+d\mu+I4^{r}$$= \int(f-(h+\alpha k))_{+^{dI}}\mu+4’<\infty$.

$\square$Now we can prove the nonlinear maximal ergodic theorem.
Theorem 3.4. Let
be an order preserving mapping onsuch that $G(T)_{+}$ isnonempty. Let
and letbe two measurable functions with values
such that $1_{\{h<\infty\}(}h’-h$ )integrable and$T$$L^{1}$$h,$ $h’$$K\in \mathrm{R}$$(-\infty, \infty]$$+\dot{i}S$$\int(Tg-h^{;})+^{d}\mu\leq\int(g-h)+d\mu+K<\infty$$(g\in L^{1})$

.

(7)

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ALMOST EVERYWHERE CONVERGENCE THEOREMS

F..urther, le.

$t\{f_{n}\}|$

.

be a

$sequenC|e,r\dot{iF.}nL^{1}$

and $f\in L^{1}$ such that’

$1_{\{f_{n}>0\}}fn\leq 1_{\{f_{n}\}}^{\cdot}.>0(f+‘\tau fn-1^{\cdot})$$(n\geq 1)$.

(8)

Then we have(9)$\int_{\{f_{\infty}^{*}=\infty\}\backslash }\{h=\infty\})(f+h’-hd\mu\geq-K.$

Proof. We assume $k\in G(T)_{+}$ . As observed in the last proof, we have

. Thus.
}$h_{-}\in L^{1}$such that, there exists
for a given}$\tilde{h}=\sup(h, -\alpha_{0^{k})}+(f_{0}-\alpha_{0}k)_{+}$ ,we have $\tilde{h}=\sup(h, -\alpha_{0}k)+(f_{0}-$Hence, setting$\alpha_{0}k)_{+}=(-h-\alpha_{0}k)_{+}+(f_{0}-\alpha_{0}k)_{+}+h\geq h$ and$\alpha_{0}>0$$\epsilon>0$$\int_{\{h_{-}>0}\alpha k$$h-d \mu+\int_{\{f_{0>\alpha 0}}k$$f\mathrm{o}\leq\epsilon$$\int|\tilde{h}-h|d\mu=\int(-h-\alpha_{0}k)_{+^{d}}\mu+\int(f_{0}-\alpha_{o}k)_{+^{d\mu}}$$= \int_{\{\}}-h-\alpha_{0}k>0-(h-\alpha_{0}k)_{+^{d}}\mu+\int_{\{f_{0}\}}-\alpha 0k>0f_{0^{-}}(\alpha 0k)_{+}d\mu$$\leq\int_{\{-h_{>}\alpha_{0}k\}}h-d\mu+\int_{\{f_{0}\}}>\alpha_{0}k.f\mathrm{o}^{d}\mu$.

$\leq\int_{\{h_{-}>0}\alpha k\}h_{-d}\mu+\int_{\{f_{0>\alpha 0k}\}}f0d\mu\leq\epsilon$

Together with (7), this implies$\int(Tg-h’)+^{d}\mu\leq\int(g-h)_{+}d\mu+I${’$\leq\int(g-\tilde{h})_{+}d\mu+\int(\tilde{h}-h)_{+^{d}}\mu+K$.
$\leq\int(g-\tilde{h})_{+^{d}}\mu+K+\epsilon<\infty$$(g\in L^{1})$.

Combining this with Lemma 3.3, we obtain$\int(Tg-(h’+\alpha k))_{+}d\mu$$(g\in L^{1}, k\in G(\tau)_{+},$ $\alpha>0)$$\leq\int(g-(\tilde{h}+\alpha k))_{+}d\mu+K+\epsilon<\infty$.

From the definition of $h-f_{0}=* \sup(h, -\alpha_{0)}k+(f_{0^{-\alpha}}\mathrm{o})+-f0\geq-\alpha_{0}+(-\alpha_{0}k)\vee$$(-f_{0})\geq-2\alpha_{0}k$ . This implies$\tilde{h},$$f_{0}\leq\tilde{h}+\alpha k$$(\alpha\geq 2\alpha_{0)}$.

Further we also have $\tilde{h}+2\alpha_{0}k=\sup(h, -\alpha_{0}k)+(f\mathrm{o}-\alpha_{0}k)++2\alpha_{0}k=\sup(h+$. Together with (8),. So we obtain
this implies$2\alpha_{0},$$\alpha_{0}k)+(f_{0}-\alpha_{0}k)_{+}\geq 0$$\tilde{h}+\alpha k\geq 0(\alpha\geq 2\alpha_{0})$$1_{\{f_{n}\tilde{h}\alpha k}>+\}\leq 1_{\{f_{n}\geq\tilde{h}\alpha}+k\}(f+Tf_{n-1})$$(n\geq 1, \alpha\geq 2\alpha_{0})$.

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SHOJI KAMIMURA

Thus the assumption of Theorem 3.1 are satisfied for any, if we replace$h’+\alpha k$byby. Hence we obtain, from Theorem 3.1 andandthat
is integrable and that$\alpha\geq 2\alpha_{0}$$\tilde{h}+\alpha k,$$I<^{r}$$I\{’+\epsilon$$h,$ $h’$$h\leq\tilde{h}$,$1_{\{f_{\infty}^{*}>\tilde{h}}(+\alpha k\}f+h’-h)$$\int_{\{f_{\infty}^{*}>\tilde{h}+\alpha k\}}(f+h’-h)d\mu\geq\int_{\{f_{\infty}^{*}+}>\overline{h}\alpha k\}-(f+(h’+\alpha k)(\tilde{h}+\alpha k))d\mu$$\geq-I\{^{r}-\epsilon$

Since$\infty$$\{f_{\infty}^{*}>\tilde{h}+\alpha k\}\downarrow\{f_{\infty}^{*}=\infty\}\backslash \{h=\infty\}$$(\alpha\geq 2\alpha_{0})$as$\alpha$.

tends to$\infty$in the above inequalities to obtain$\int_{\{f_{\infty}^{*}=}\infty\}\backslash \{h=\infty\}-(f+h’h)d\mu\geq-I\{’-\epsilon$

Since$\epsilon>0$

is arbitrary, the assertion (9) follows.

, we may let$\alpha$tend to.

$\square$Using the above theorem, we obtain the following lemma which is crucial for our
main result.
Lemma 3.5. Let be an order preserving andnonempty. Then for any $f\in L^{1}$
that
$T$

$G(T)_{+}\dot{i}S$$L^{1}$

non.. expansive mapping on$\mu(\{\lim_{narrow}\sup_{\infty}A_{n}f>0\}\cap\{\lim_{narrow}\inf A_{n}f\infty<0\})=0$

Proof. We assume $k\in G(T)_{+}$ . It suffices to show that

where

$A_{\epsilon}= \{\lim_{narrow}\sup_{\infty}A_{n}f>\epsilon k\}$\‘and$L^{1}$such.

$\mu(A_{\epsilon}\cap\tilde{A}_{\epsilon})=0$

for any$\tilde{A}_{\epsilon}=\{\lim_{narrow}\inf A_{n}f\infty<-\epsilon k\}$$\epsilon>0$,.

. We define a mapping on
by setting $\tilde{T}f=-T(-f)(f\in L^{1})$ . Let
We fixbe defined as
but with instead of . Putting $f_{n}=(S_{n}f-(n+1)\epsilon k)_{+}$ and$\tilde{f}_{n}=(\tilde{S}_{n}(-f)-(n+1)\epsilon k)_{+}=(S_{n}f+(n+1)\epsilon k)_{-}$ ,we have$\tilde{T}$$\epsilon>0$$\tilde{S}_{n}$$S_{n}$$L^{1}$$T$$\tilde{T}$$A_{\epsilon}\subset A:=\{f_{\infty}^{*}=\infty\}$and$\tilde{A}_{\epsilon}\subset\tilde{A}:=\{\tilde{f}_{\infty}^{*}=\infty\}$.

In fact, if $x\not\in A$ , there exists
such that $f_{n}(x)\leq M$ $(0\leq n<\infty)$ .
Then we have $S_{n}f(x)\leq(n+1)\epsilon k(x)+M$ and $A_{n}(x) \leq\epsilon k(x)+\frac{M}{n+1}$ . This implies$\lim\sup_{narrow\varphi}A_{n}f(x)\leq\epsilon k(x)$ , and hence. So we obtain. Similarly we. Thus the proof is complete if we can show that $\mu(A\cap\tilde{A})=0$ .
can show
Next we will show
$M\in \mathrm{R}$$x\not\in A_{\epsilon}$$A_{\epsilon}\subset A$$A_{\epsilon}\subset\tilde{A}$

1 $\{f_{n}>0\}\leq 1_{\{f_{n}>0\}(}(f-\epsilon k)+\tau f_{n-1})$$(n\geq 1)$

and
1

Since

$T$

$\{\overline{f}_{n}>0\}\leq 1_{\{\overline{f}_{n}>0\}}((-f-\epsilon k)+\tilde{\tau}\tilde{f}n-1)$$(n\geq 1)$.

is order preserving, $k\in G(T)_{+}$ and $(S_{n-1}f)+-f_{n-}1=(s_{n-1}f)+-(S_{n-1}f-$, we have $T((sn-1f)_{+})\leq Tf_{n-1}+n\epsilon k$ . This implies $Tf_{n-1}+n\epsilon k\geq$$n\epsilon k)_{+}\leq n\epsilon k$

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ALMOST EVERYWHERE CONVERGENCE THEOREMS

and therefore $(f-\epsilon k)+\tau f_{n-}1\geq f+T(Sn-1f)-(n+1)\epsilon k=$. Hence we obtain$T((S_{n-1}f)+)\geq T(S_{n-1}f)$$S_{n}f-(n+1)\epsilon k$$1_{\{f_{n}>0}\}((f-\epsilon k)+Tfn-1)\geq 1_{\{f_{n}>0}\}(s_{n}f-(n+1)\epsilon k)$$=1_{\{f_{n}\}}(>0Snf-(n+1)\epsilon k)_{+}$$=1_{\{f_{n}0\}}>f_{n}$Similarly we can obtain
Putting$(n\geq 1)$$1_{\{\overline{f}_{n}>0\}}\leq 1_{\{\overline{f}_{n}0}(>\}(-f-\epsilon k)+\tilde{T}\tilde{f}_{n-1})$$h_{\alpha,\beta}=-\alpha k1\overline{A}\backslash A+\beta k1_{A\backslash \overline{A}}$

and

.

.

$\tilde{h}_{\alpha,\beta}=-\alpha k1_{A\backslash \tilde{A}}+\beta k1_{\overline{A}\backslash A}$$(0\leq\alpha<\infty, 0\leq\beta<\infty)$,

we define$h_{\alpha}= \lim_{\infty\betaarrow}h_{\alpha,\beta}$

Further, since$h_{\alpha,\beta},\tilde{h}_{\alpha,\beta}$

belong to$h_{\alpha}’= \sup_{n\in \mathrm{N}}Th_{\alpha},n$

Since
that

$L^{1}$

and$\tilde{h}_{\alpha}=\lim_{\betaarrow\infty}\tilde{h}_{\alpha,\beta}$.

, we can defineand$\tilde{h}_{\alpha}’=\sup_{n\in \mathrm{N}}\tilde{T}\tilde{h}_{\alpha},n$$(\alpha>0)$.

, we get from Lemma 3.2$f \leq h_{\alpha}(f\in L^{1})\Rightarrow Tf\leq Th_{\alpha}=\sup_{n\in \mathrm{N}\alpha,n}Th--h_{\alpha}’$$\int(Tg-h’\alpha)_{+}d\mu\leq\int(g-h_{\alpha})_{+}d\mu$$(g\in L^{1}, \alpha\geq 0)$

.

(10)

$(g\in L^{1}, \alpha\geq 0)$

.

(11)

Similarly we can get$\int(\tilde{\tau}_{g-}\tilde{h}_{\alpha}’)+^{d\mu}\leq\int(g-\tilde{h}_{\alpha})+d\mu$

To see the principal idea, we postpone the proof of the following statements until
later,$1_{\{h_{\alpha}<\}}h^{;}\infty\alpha$is integrable for any$\alpha\geq 0$

,

(12)

$1_{\{\tilde{h}_{\alpha}<\}}\tilde{h}’\infty\alpha$is integrable for any$\alpha\geq 0$

,

(13)

$- \infty<\inf_{\alpha>0}\int_{A\cap\overline{A}}h_{\alpha\mu}’d\leq\sup\alpha>0\int_{A\cap\overline{A}}h\prime d\mu\alpha<\infty$

,

(14)

and
.

$\lim_{\alphaarrow\infty\beta}\lim_{arrow\infty}\int_{A\cap\tilde{A}}Th_{\alpha},\beta d\mu=\lim_{\betaarrow\infty}\lim_{arrow\alpha\infty}\int_{A\cap\overline{A}}Th_{\alpha},\beta d\mu$

(15)

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SHOJI KAMIMURA

Because of (10) and (12), we can apply Theorem 3.4 to. Then we obtain
instead of$\{f_{n}\}$with $K=0$ and$h_{\alpha},$$h_{\alpha}’$$h,$ $h’$$0 \leq\int\{f_{\infty}*\}=\infty\backslash \{h_{\alpha_{k))\mu}}=\infty\},-=\int_{A\mathrm{n}}\overline{A}((f-\epsilon+h_{\alpha}((f-d\epsilon k)+h’\alpha h\alpha)d\mu$

(16)

since

$h_{\alpha}=0$

on$A\cap\tilde{A}$. Analogously, we can show that.

$0 \leq\int_{A\cap\overline{A}}((-f-\epsilon k)+h_{\alpha}’)d\mu$Since$Th_{\alpha,\beta}=T(\beta k1_{\overline{A}\backslash A\backslash }-A\alpha k1)\overline{A}=T(-\tilde{h}\alpha,\beta)=-\tilde{T}\tilde{h}_{\alpha,\beta},$(17)(15) implies$\lim_{\alphaarrow\infty}\int_{A\cap\tilde{A}}h_{\alpha}\prime d\mu=\alphaarrow\lim_{\infty\betaarrow}\lim_{\infty}\int_{A\cap\tilde{A}}Th_{\alpha},\beta d\mu$$= \lim_{\betaarrow\infty}\lim_{\alphaarrow\infty}\int_{A\cap\tilde{A}}Th_{\alpha},\beta d\mu$$= \lim_{\betaarrow\infty}\lim_{aarrow\infty}\int_{A\cap\tilde{A}^{-}}\tilde{\tau}\tilde{h}\beta,\alpha d\mu$

(18)

.

$=- \lim_{\betaarrow\infty}\int_{A\cap\overline{A}}\tilde{h}_{\beta}’d\mu$

Adding (16) and (17), we obtain$0 \leq\int_{A\cap\tilde{A}}((f-\epsilon k)+h’a)d\mu+\int_{A\cap\tilde{A}}((-f-\epsilon k)+\tilde{h}_{\alpha}’)d\mu$$= \int_{A\cap\overline{A}}(-2\epsilon k+h’\tilde{h}_{\alpha}+;)\alpha d\mu$Letting$\alpha$tend to$\infty$.

and using (14) and (18), we obtain$0 \leq\int_{A\cap\overline{A}}(-2\epsilon k)d\mu$

and therefore $\mu(A\cap\tilde{A})=0$ . Thus the proof is completed if we can show (12) $-(15)$ .
At first, we will prove (15). Because of Lemma 3.2, we can apply Theorem 3.4 towithinstead ofand $K=0$ . Then we obtain$\{f_{n}\}$$h_{\alpha,0},$$Th_{\alpha,0}$$h,$ $h’$$0 \leq\int_{\{f\infty}*\}(f-\epsilon k+\tau_{d\mu}h\alpha,0-h_{\alpha},0=\int_{A}(f-\epsilon k+\tau\infty^{=}h_{\alpha},0),)d\mu$

because$h_{\alpha,0}=0$

on

$A$

. It follows that$- \infty<\int_{A}(\epsilon k-f)d\mu\leq\int_{A}Th_{\alpha},0d\mu$$(\forall\alpha\geq 0)$

.

(19)

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ALMOST EVERYWHERE CONVERGENCE THEOREMS

Since

and is order preserving, we have $Th_{\alpha,0}-\tau 0\leq 0$ . Together with.
, this implies$T$$h_{\alpha,0}\leq 0$$Th_{\alpha,0}\leq h_{\alpha}’$$\int_{A}(Th\alpha,0-\tau 0)d\mu\leq\int_{A\cap\overline{A}}(\tau h\alpha,0^{-T0})d\mu\leq\int_{A\cap\tilde{A}}(h_{\alpha}’-\tau \mathrm{o})d\mu$Therefore
Because of (19), the$\int_{A}Th_{\alpha},0d\mu\leq\int_{A\cap\overline{A}}h_{\alpha}’d\mu+\int_{A}T\mathrm{O}d\mu-\int_{A\cap\tilde{A}}T\mathrm{O}d\mu\leq\int_{A\cap\overline{A}}h^{;}d\mu\alpha+||T0||_{1}$$”-\infty$

half” of (14) follows from$(\forall\alpha\geq 0)$$- \infty<\int_{A}(\epsilon k-f)d\mu\leq\int_{A}Th_{\alpha,0}d\mu\leq\int_{A\cap\overline{A}}h_{\alpha}’d\mu+||\tau \mathrm{o}||_{1}$Replacing$T$by$\tilde{T}$and$h_{\beta,0},$$Th\beta,0$by$\tilde{h}_{\beta,0},\tilde{T}\tilde{h}\beta,0$, we obtain analogously$- \infty<\int_{\overline{A}}(\epsilon k+f)d\mu\leq\int_{\tilde{A}}\tilde{T}\tilde{h}_{\beta,0\mu}d$Since$\tilde{T}\tilde{h}_{\beta,0}=-Th_{0,\beta}$.

$(\forall\beta\geq 0)$.

and $Th_{0,\beta}-\tau 0\geq 0$ , this implies$\int_{A\cap\tilde{A}}\tau h_{0,\beta}d\mu-||T0||_{1}\leq\int_{A}(-\tilde{T}\tilde{h}_{\beta,0})d\mu$$(\forall\beta\geq 0)$$\leq\int_{\tilde{A}}(-f-\epsilon k)d\mu<\infty$

and therefore the$‘(+\infty$

(20)

,

half” of (14) follows from$\sup_{\alpha\geq 0}\int_{A\cap\tilde{A}}h_{a}’d\mu=\int_{A\cap\overline{A}}h_{0}’d\mu=\sup\beta\geq 0\int_{A\cap\tilde{A}}Th0,\beta d\mu<\infty$.

Our next aim is to show that
$C_{1}-- \sup_{0\alpha,,\beta\geq}\int_{A\backslash \overline{A}}(h_{\alpha,\beta}-Th\alpha,\beta)d\mu<\infty$

To prove this, we put
instead of$K=0$ and(21).

and apply Theorem 3.4 to. Then we obtain$g_{\alpha,\beta}=-\alpha k1_{\overline{A}\backslash A}+\beta k1_{A}$$g_{\alpha,\beta},$$Tg_{\alpha,\beta}$$h,$ $h’$$\{f_{n}\}$with$\int_{\{f_{\infty}^{*}\}}=\infty((f-\epsilon k)+Tg_{\alpha,\beta}-g\alpha,\beta)\geq 0$

and therefore$\sup_{\alpha,\beta\geq 0}\int_{A}(g\alpha,\beta-^{\tau}g\alpha,\beta)d\mu\leq\int_{A}(f-\epsilon k)d\mu<\infty$

Since$g_{\alpha,\beta}-g_{\alpha},0=\beta k1_{A}\leq\beta k,$$\beta k\leq T\mathrm{O}+\beta k$. Therefore$g_{\alpha},0\leq 0$

and $k\in G(T)_{+}$ , we have$g_{\alpha,\beta}-\tau_{g_{\alpha},\beta}=\beta k-Tg\alpha,\beta\geq-T\mathrm{O}$$\int_{A}(g_{\alpha,\beta}-\tau g\alpha,\beta)_{-d}\mu\leq\int_{A}|T0|d\mu\leq||T0||_{1}$

(22)

.

on

$A$

$Tg_{\alpha,\beta}\leq Tg_{\alpha,0}+$. It follows that

. Together with (22), this implies$C_{2}-- \sup_{0\alpha,,\beta\geq}\int_{A}(g_{\alpha,\beta}-Tg\alpha,\beta)_{+}d\mu$$\leq\int_{A}(f-\epsilon k)d\mu+||T0||_{1}<\infty$

.

(23)

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SHOJI KAMIMURA

From (20) and$Th_{\alpha,\beta}\leq Th_{0,\beta}$, we get$C_{3}= \sup_{\alpha,\beta\geq 0}\int_{A\cap\tilde{A}}Th_{\alpha},\beta d\mu<\infty$

Since

$T$

is order preserving,
impliesand isnonexpansive, we have$h_{\alpha,\beta}\leq g_{\alpha,\beta}$$\beta\int_{A\cap\tilde{A}}kd\mu$

$T$

.

(24)

$Th_{\alpha,\beta}\leq Tg_{\alpha,\beta}$. Since$||g_{\alpha,\beta}-h\alpha,\beta||=$$L^{1}$$\int(Tg_{\alpha,\beta}-Th_{\alpha,\beta})d\mu=||\tau_{g_{\alpha,\beta}-}Th_{\alpha,\beta}||_{1}\leq\beta\int_{A\cap\tilde{A}}kd\mu$

.

(25)

On the other hand, by (23) and (24),$\int_{A\cap\tilde{A}}(Tg\alpha,\beta-\tau h_{\alpha},\beta)d\mu$$= \int_{A\cap\tilde{A}}g_{\alpha,\beta}d\mu-\int_{A\cap\tilde{A}}Th\alpha,\beta d\mu-\int_{A\cap\tilde{A}}(g\alpha,\beta-Tg\alpha,\beta)d\mu$$\geq\beta\int_{A\cap\tilde{A}}kd\mu-C3-C_{2}$.

Together with (25), this implies$\int_{A\backslash \overline{A}}(Tg_{\alpha},\beta-Th_{\alpha,\beta})d\mu\leq\int_{E\backslash (A\cap\tilde{A})}(Tg\alpha,\beta-Th_{\alpha},\beta)d\mu$$= \int_{E}(Tg_{\alpha,\beta}-Th_{\alpha,\beta})d\mu-\int_{A\cap\tilde{A}}(\tau_{g_{\alpha}},\beta-\tau h_{\alpha},\beta)d\mu$$\leq\beta\int_{A\cap\tilde{A}}kd\mu-\beta\int_{A\cap\tilde{A}}kd\mu+c_{2}+C_{3}$$=C_{2}+C_{3}.-$Since$h_{\alpha,\beta}=g_{\alpha,\beta}=\beta k$on$A\backslash \tilde{A}$

(26)

.

, we obtain from (23) and (26) that$\int_{A\backslash \tilde{A}}(h_{\alpha},\beta-Th_{\alpha},\beta)d\mu=\int_{A\backslash \overline{A}}(g_{\alpha,\beta}-\tau h_{\alpha},\beta)d\mu$$\leq\int_{A\backslash \overline{A}}(g_{\alpha},\beta-\tau_{g)d\mu+}\alpha,\beta\int_{A\backslash \tilde{A}}(Tg_{\alpha,\beta}-Th_{\alpha},\beta)d\mu$$\leq C_{2}+(c_{2}+C\mathrm{s})$$(\forall\alpha, \beta>0)$.

So we obtain (21).
Our next aim is to show$\alpha_{1}\leq\alpha_{2},$If$T$$\beta_{1}\leq\beta_{2}\Rightarrow 1_{A\backslash \overline{A}}(h_{\alpha_{1}},\beta_{1}-Th_{\alpha_{1},\beta 2})\leq 1_{A\backslash \tilde{A}}(h_{\alpha_{2}},\beta_{2}-Th_{\alpha_{2},\beta 2})$, then. Therefore we haveis order preserving. On the other hand,$\alpha_{1}\leq\alpha_{2}$$h_{\alpha_{1},\beta}\geq h_{\alpha_{2},\beta}$.

$Th_{\alpha_{1},\beta}\geq Th_{\alpha_{2},\beta}(\alpha_{1}\leq\alpha_{2})$$h_{\alpha_{1},\beta}=h_{\alpha_{2},\beta}$

on

$A$

, and hence$\alpha_{1}\leq\alpha_{2}\Rightarrow 1_{A\backslash \tilde{A}}(h_{\alpha,\beta}-\tau h\alpha_{1},\beta)1\leq 1_{A\backslash \tilde{A}}(h_{\alpha_{2},\beta}-\tau h\alpha 2,\beta)$.

(27), since(28)

39
ALMOST EVERYWHERE CONVERGENCE THEOREMS

If$\beta_{1}\leq\beta_{2}$, then. Therefore we obtainsince $k\in G(T)_{+}$ . On the other hand,
and therefore$h_{\alpha,\beta_{2}}-h_{\alpha,\beta_{1}}=(\beta_{2}-\beta 1)k1_{A\backslash \overline{A}}\leq(\beta_{2^{-}}\beta_{1})k$$Th_{\alpha,\beta_{2}}\leq Th_{\alpha,\beta_{1}}+(\beta_{2}-\beta_{1})k(\beta_{1}\leq\beta_{2})$$h_{a,\beta_{2}}-h_{\alpha,\beta_{1}}=(\beta_{2^{-}}\beta_{1})k$on$A\backslash \tilde{A}$$\beta_{1}\leq\beta_{2}\Rightarrow 1_{A\backslash \overline{A}}(h\alpha,\beta_{1}-Th_{\alpha,\beta 1})\leq 1_{A\backslash \overline{A}}(h_{\alpha,\beta 2}-\tau h\alpha,\beta_{2})$

(29)

.

Now (27) follows from (28) and (29).
Analogously to (21) and (27), we can show$\sup_{\alpha,\beta\geq 0}\int_{\tilde{A}\backslash A}(\tilde{h}_{\alpha,\beta}-^{\tilde{\tau}\tilde{h}}\alpha,\beta)d\mu<\infty$

and$\alpha_{1}\leq\alpha_{2},$

Since

.

$\beta_{1}\leq\beta_{2}\Rightarrow 1_{\tilde{A}\backslash A}(\tilde{h}_{\alpha_{1}},\beta_{1}-\tilde{T}\tilde{h}_{\alpha_{1},\beta 1})\leq 1_{\tilde{A}\backslash A}(\tilde{h}_{\alpha_{2}},\beta_{2}-\tilde{T}\tilde{h}_{\alpha_{2},\beta 2})$$\tilde{h}_{\alpha,\beta}=-h_{\beta,\alpha}$and$\tilde{T}\tilde{h}_{\alpha,\beta}=-Th_{\beta,a}$, we obtain(30)$C_{4}= \sup_{\beta\alpha,\geq 0}\int_{\tilde{A}\backslash A}(\tau h_{\alpha,\beta}-h\alpha,\beta)d\mu<\infty$

and.

$\alpha_{1}\leq\alpha_{2},$ $\beta_{1}\leq\beta_{2}\Rightarrow 1_{\overline{A}\backslash A}(h_{\alpha_{1}},\beta_{1}-Th_{a_{1},\beta_{1}})\leq 1_{\tilde{A}\backslash A}(h_{\alpha_{2}},\beta_{2}-Th_{\alpha_{2},\beta 2})$

Let $\delta>0$ be given. By (21) and (27) (resp. (30), and (31)) there exists
such that$\int_{A\backslash \tilde{A}}(h_{a},\beta-\tau h\alpha,\beta)d\mu\geq C_{1}-\delta$$(\alpha, \beta\geq\alpha_{\delta})$$\int_{\overline{A}\backslash A}(Th_{\alpha,\beta}-h_{\alpha},\beta)d\mu\geq C_{4}-\delta$$(\alpha, \beta\geq\alpha_{\delta})$(31)$\alpha_{\delta}\geq 0$

(32)

and

Let

$\alpha_{1},$$\alpha_{2},$$\beta_{1},$be given. Then we haveand therefore$\beta_{2}\geq\alpha_{\delta}$$| \beta_{1}-\beta_{2}|\int_{A}\backslash \overline{A}\mu kd$

(33)

.

$||h_{\alpha_{1},\beta_{1}}-ha_{2,\beta}2|| \leq|\alpha_{1}-\alpha_{2}|\int_{\overline{A}}\backslash Akd\mu+$$\int|Th_{\alpha_{1},\beta_{1}}-Th_{\alpha_{2}},\beta_{2}|d\mu\leq|\alpha_{1}-\alpha 2|\int_{\tilde{A}\backslash A}kd\mu+|\beta_{1}-\beta_{2}|\int_{A\backslash \tilde{A}}kd\mu$

.

(34)

40
SHOJI KAMIMURA

Using (32), we obtain$\int_{A\backslash \tilde{A}}(Th_{\alpha_{1},\beta 1}-Th_{\alpha_{2\beta 2}},)d\mu$$= \int_{A\backslash \tilde{A}}(h\alpha_{1},\beta_{1}-h_{\alpha_{2},\beta_{2}})d\mu-\int_{A\backslash \tilde{A}}(h\alpha_{1},\beta_{1}-Th_{\alpha_{1},\beta 1})d\mu$$+ \int_{A\backslash \overline{A}}(ha_{2},\beta_{2^{-}}\tau h_{a2},\beta 2)d\mu$$\geq(\beta_{2}-\beta 1)\int_{A\backslash \overline{A}}kd\mu-C_{1}+(c1-\delta)$$\geq(\beta_{2}-\beta_{1})\int_{A\backslash \overline{A}}kd\mu-\delta$

Interchanging$\alpha_{1},$$\beta_{1}$with$\alpha_{2},$$\beta_{2}$

(35)

.

, we also get
.

(36)

.

(37)

$\int_{A\backslash \tilde{A}}(\tau ha_{2},\beta_{2}-Th_{\alpha_{1},\beta 1})d\mu\geq(\beta_{1}-\beta_{2})\int_{A\backslash \overline{A}}kd\mu-\delta$

From (35) and (36), we obtain$\int_{A\backslash \tilde{A}}|Th_{a_{1},\beta 1}-Th_{\alpha_{2},\beta_{2}}|d\mu\geq|\beta_{1}-\beta_{2}|\int_{A\backslash \overline{A}}kd\mu-\delta$

Analogously, using (33) instead of (32), we can show that

.

(38)

$\int_{\overline{A}\backslash A}|Th_{\alpha_{1},\beta_{1}}-Th_{\alpha_{2},\beta 2}|d\mu\geq|\alpha_{1}-\alpha_{2}|\int_{\overline{A}\backslash A}kd\mu-\delta$

Combining (37) and (38) with (34), we obtain$\int|Th\alpha_{1},\beta 1-Th_{\alpha_{2},\beta}|2d\mu$$\leq 2\delta+\int_{A\backslash \tilde{A}}|Th_{\alpha_{1},\beta_{1}}-Th_{\alpha_{2},\beta_{2}}|d\mu+\int_{\tilde{A}\backslash A}|Th_{\alpha_{1},\beta_{1}}-Th_{a_{2},\beta_{2}}|d\mu$

and therefore$\int_{A\cap\overline{A}}|Th_{a_{1},\beta_{1}}-Th_{\alpha_{2},\beta_{2}}|d\mu\leq 2\delta$

Since$\delta>0$

is arbitrarily, wc have (15).

$(\alpha_{2}, \alpha_{2}, \beta_{1}, \beta_{2}\geq\alpha_{\delta})$.

41
ALMOST EVERYWHERE CONVERGENCE THEOREMS

Using (21), we obtain$\int_{E\backslash (A\backslash \overline{A}})|\tau h0,\beta|d\mu\leq||Th_{0},\beta||_{1}-\int_{A\backslash \tilde{A}}Th_{0,\beta}d\mu$$\leq||T0||_{1}+||Th0,\beta-\tau \mathrm{o}||_{1}-IA\backslash \tilde{A}Th_{0,\beta}d\mu$$\leq||T0||1+||h_{0},\beta||_{1}-\int_{A\backslash \overline{A}}Th0,\beta d\mu$$=||T0||1+ \int_{A\backslash \overline{A}}(h_{0,\beta}-\tau h_{0,\beta})d\mu$$\leq||T0||1+c_{1}$Therefore we obtain$Th_{0,\beta} \leq h_{\alpha}’\leq\sup_{\beta\geq 0\beta}\tau h_{0},$,$\int_{\{\}}h_{\alpha}<\infty d|h_{\alpha}’|\mu\leq||Th_{0},\beta||1+\sup_{\beta\geq 0}\int_{E\backslash (A\backslash )}\overline{A}||\tau h_{0},\beta|d\mu$$\leq||Th_{0},\beta||_{1}+||\tau \mathrm{o}||1+c_{1}$Thus we have (12). The proof of (13) is analogous.

.

$\square$

The following theorem is our main result.
Theorem 3.6. $LetT$ be an order preserving andnonexpansive mapping on such
$G(T)_{+}\dot{i}S$
nonempty. Thenconverges a. . to an element
thatof forany $f\in L^{1}$ . Further $\dot{i}fF(T)$ is nonempty, then we have
. In paticular if, then we also have $\lim_{narrow\infty}||A_{n}f-f^{*}||_{1}=0$ . And further
is positively$h_{omog}eneous_{f}$ then $f^{*}\in F(T)$ .
$L^{1}$$A_{n}f$$L^{1}$$f^{*}$$e$$L^{1}$$f^{*}\in L^{1}$$\mathrm{O}\in F(T)$Proof. Let$\dot{i}fT$$f\in L^{1}$

and $k\in G(T)_{+}$ be given. We assume$\mu(\{\lim_{narrow}\inf_{\infty}A_{n}f<\lim_{narrow}\sup_{\infty}A_{n}f\})>0$

Since $k>0$ , there exists
$\mu$

(

$\alpha\in \mathrm{R}$$\{\lim\inf_{\infty narrow}$

An

$f$

.

such that$< \alpha k\}\cap\{\lim_{narrow}\sup_{\infty}A_{n}f>\alpha k\}$) $>0$ .

Since $k\in G(T)_{+}$ , we have $A_{n}(f-\alpha k)=A_{n}$ f–ak and therefore$\mu(\{\lim_{narrow}\inf A_{n}(f-ak\infty)<0\}\cap\{\lim_{narrow}\sup_{\infty}A(nf-ak)>0\})>0$.

But this contradicts Lemma 3.5. So.
converges
We assume $l\in F(T)$ . And we set.
Since
is order preserving, we have $(-f_{-})^{*}\leq f^{*}\leq(f_{+})^{*}$ . Using Fatou’s lemma,
nonexpansive and $A_{n}l=l$ , we obtainsinceis$A_{n}f$$\mathrm{a}.\mathrm{e}.$$(f_{+})^{*}= \lim_{narrow\infty}A_{n}(f_{+}),$$A_{n}$$A_{n}$$L^{1}$$\int(f^{*})_{+}d\mu\leq\int(f_{+})^{*}d\mu$$(f_{-})^{*}= \lim_{narrow\infty}A_{n}(f_{-})$

42
SHOJI KAMIMURA$\leq\int\lim_{narrow}\inf_{\infty}A_{n}(f+)d\mu$$\leq\lim_{narrow}\inf_{\infty}\int A(nf+)d\mu$$\leq\lim_{narrow}\inf_{\infty}||A_{n}(f+)-l||_{1}+||l||_{1}$$\leq||f_{+}-l||_{1}+||l||_{1}<\infty$

and$\int(f^{*})_{-d}\mu\leq-\int(-f_{-})d\mu$$\leq-\int\lim_{narrow}\sup_{\infty}A_{n}(-f_{-})d\mu$$\leq-\lim\sup\int A_{n}(-f-)d\mu$$= \lim_{narrow}\inf_{\infty}^{n}\int(-arrow\infty A_{n}(-f-))d\mu$$= \lim_{narrow}\inf_{\infty}||A(n-f-)||1$$\leq\lim_{narrow}\inf_{\infty}||A_{n}(-f-)-l||_{1}+||l||_{1}$.

$\leq||f_{-}+l||1+||l||_{1}<\infty$

Therefore$f^{*}\mathrm{b}\mathrm{e}1_{\mathrm{o}\mathrm{n}}\mathrm{g}\mathrm{s}$to$L^{1}$. Next we will show that if$\lim_{narrow\infty}||A_{n}f-f*||_{1}=0$$\mathrm{O}\in F(T)$, then.

We set $f_{m}=$ ( A $mk$ ) $(-mk)(m\in \mathrm{N})$ and
A n $m=f_{m}^{*}$ . Then we obtainby Lebesgue’s convergence theorem, because $A_{n}0=0$ and$k\in G(T)_{+}$ implie $|A_{n}f_{m}|\leq mk(n\in \mathrm{N})$ . Further we have$f$$\lim_{narrow\infty}$$f$$\lim_{narrow\infty}||A_{n}f_{m}-f_{m}^{*}||_{1}=0$$||A_{n}f-f^{*}||_{1}\leq||A_{n}f-A_{n}fm||_{1}+||A_{n}f_{m}-f_{m}^{*}||_{1}+||f_{m}^{*}-f^{*}||1$$\leq 2||f-f_{m}||_{1}+||A_{n}fm-f*m||_{1}$ .

Letting$n$tend to$\infty$, we obtain$\lim_{narrow\infty}||Afn-f*1|_{1}\leq 2||f-fm||_{1}$.

Therefore, since $\lim_{marrow\infty}||f-f_{m}||1=0$ , we get$\lim_{marrow\infty}||A_{n}f-f^{*}||_{1}=0$

Moreover if$T$.

is positively homogeneous, we have$T(A_{n}f)= \frac{n+2}{n+1}A_{n}+1f-\frac{1}{n+1}f$

Hence we obtain $Tf^{*}=f^{*}$ by letting$n$tend to$\infty$.

$\square$.

43
ALMOST EVERYWHERE CONVERGENCE THEOREMSREFERENCES(1985).
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170-192.
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on , Israel J.Math., 71 (1990), 181-191.
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$L^{1}$$L^{1}$

DEPARTMENT OF MATHEMATICAL AND COMPUTING SCIENCES, TOKYO INSTITUTE OF TECHNOLOGY, OHOKAYAMA, MEGURO-KU, TOKYO 152-8552, JAPAN
$E$-mail address: kamimura@is.titech.ac.jp