# Substochastic operators in symmetric spaces

Maciej Ciesielski and Grzegorz Lewicki

###### Abstract.

First, we solve a crucial problem under which conditions increasing uniform
$K$-monotonicity is equivalent to lower locally uniform $K$-monotonicity.
Next, we investigate properties of substochastic operators on
$L^{1}+L^{\infty}$ with applications. Namely, we show that a countable
infinite combination of substochastic operators is also substochastic. Using
$K$-monotonicity properties, we prove several theorems devoted to the
convergence of the sequence of substochastic operators in the norm of a
symmetric space $E$ under addition assumption on $E$. In our final discussion
we focus on compactness of admissible operators for arbitrary Banach couples.

2000 Mathematics Subjects Classification: 46E30, 46B20, 46B42.

Key Words and Phrases: Symmetric spaces, Banach couples, substochastic
operators, admissible operators, uniform $K$-monotonicity.

##  1\. Introduction

Substochastic operators have been researched and applied intensively in many
fields of mathematics since the early 20th century. Namely, the first
interesting result concerning a substochastic operator appeared in 1929 (see
[13, 19]), where Hardy, Littlewood and Pólya presented a connection between
existence of double stochastic matrix and a majorization one vector by another
with respect to the Hardy-Littlewood-Pólya relation $\prec$. Next, in 60-ties
Calderón and Ryff showed that for any two non-negative measurable functions
$f,g$ such that $f\prec{g}$ and $f\in{L^{1}+L^{\infty}}$ there exists a
substochastic operator $T$ such that $Tg=f$ a.e. (see [21, 3]). It is worth
mentioning that this result was a crucial key in the Calderón’s proof to
establish an equivalence criterion for the exact interpolation space between a
Banach couple $(L^{1},L^{\infty})$ [3]. It is necessary to recall some
investigations devoted to relationships between substochastic operators and
semigroups, among other used to explore properties of Markov processes or
Markov chains, which have a substantial history dating back also to the 1960s
(see [14, 15, 16]). The next motivation for this paper we find in [4, 7, 8,
11], where authors presented equivalent conditions for $K$-monotonicity
properties with applications in symmetric spaces. In the spirit of the
previous investigations we give a complete answer for the essential problem
about a relation between uniform $K$-monotonicity properties. Namely, we
provide complete criteria for equivalence of increasing uniform
$K$-monotonicity and lower local uniform $K$-monotonicity in symmetric spaces.
The second goal of this paper is to study a relationship between convergence
of substochastic operators and $K$-monotonicity properties in symmetric
spaces. First, we prove that a countable infinite combination of substochastic
operators is substochastic and discuss some basic examples of substochastic
operators in symmetric spaces. We start our investigation assuming the weakest
property lower locally uniform $K$-monotonicity in a symmetric space $E$.
Under additional assumption of the symmetric space $E$, we establish a norm
convergence of a sequence of specific substochastic operators given by a
countable infinite sum of the integral averaging on positive finite measurable
sets. Next, letting the stronger property decreasing uniform $K$-monotonicity
in a symmetric space $E$ we research norm and point convergence of the
sequence of substochastic operators given in the descending order with respect
to the Hardy-Littlewood-Pólya relation $\prec$. In view of the previous result
we also research norm convergence of $A^{n}$ a geometric sequence generated by
a fixed substochastic operator $A$ in a symmetric space $E$. We also focus our
attention on the convergence of a sequence of substochastic operator given in
the ascending order for the partial order $\leq$ in a symmetric space $E$,
equipped with uniformly $K$-monotone norm. Finally, we show compactness of the
admissible operator between Banach couples under additional assumptions of the
interpolation spaces.

##  2\. Preliminaries

Now we recall basic notation and definition. Let $\mathbb{R}$,
$\mathbb{R}^{+}$ and $\mathbb{N}$ be the sets of reals, nonnegative reals and
positive integers, respectively. In a Banach space
$(X,\left\|\cdot\right\|_{X})$ we denote the unit sphere and the closed unit
ball by $S(X)$ and $B(X))$, respectively. A function
$\phi:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ is called quasiconcave if
$\phi(t)$ is increasing and $\phi(t)/t$ is decreasing on $\mathbb{R}^{+}$ and
also $\phi(t)=0\Leftrightarrow{t=0}$. We denote as usual by $\mu$ the Lebesgue
measure on $I=[0,\alpha)$, where $\alpha=1$ or $\alpha=\infty$, and by $L^{0}$
the set of all (equivalence classes of) extended real valued Lebesgue
measurable functions on $I$. Let us use the notation $A^{c}=I\backslash A$ for
any measurable set $A$. By a Banach function space (or a Köthe space) we mean
a Banach lattice $(E,\|\cdot\|_{E})$ that is a sublattice of $L^{0}$ and
possesses the following conditions

  * (1)

If $x\in L^{0}$, $y\in E$ and $|x|\leq|y|$ a.e., then $x\in E$ and
$\|x\|_{E}\leq\|y\|_{E}$.

  * (2)

There exists a strictly positive $x\in E$.

To simplify the notation in the whole paper we use the symbol $E^{+}={\\{x\in
E:x\geq 0\\}}$. An element $x\in E$ is called a point of order continuity if
for any $(x_{n})\subset{}E^{+}$ such that $x_{n}\leq\left|x\right|$ and
$x_{n}\rightarrow 0$ a.e. we get $\left\|x_{n}\right\|_{E}\rightarrow 0.$ A
Banach function space $E$ is said to be order continuous (shortly
$E\in\left(OC\right)$) if any element $x\in{}E$ is a point of order continuity
(see [18]). We say that a Banach function space $E$ has the Fatou property if
for any $\left(x_{n}\right)\subset{}E^{+}$,
$\sup_{n\in\mathbb{N}}\|x_{n}\|_{E}<\infty$ and $x_{n}\uparrow x\in L^{0}$,
then we have $x\in E$ and $\|x_{n}\|_{E}\uparrow\|x\|_{E}$. Unless it is said
otherwise, in the whole article it is considered that $E$ has the Fatou
property. For any function $x\in L^{0}$ we denote the distribution function by

$d_{x}(\lambda)=\mu\left\\{s\in[0,\alpha):\left|x\left(s\right)\right|>\lambda\right\\},\qquad\lambda\geq
0.$

and also the decreasing rearrangement of $x$ by

$x^{\ast}\left(t\right)=\inf\left\\{\lambda>0:d_{x}\left(\lambda\right)\leq
t\right\\},\text{ \ \ }t\geq 0.$

The following notation is used
$x^{*}(\infty)=\lim_{t\rightarrow\infty}x^{*}(t)$ if $\alpha=\infty$ and
$x^{*}(\infty)=0$ if $\alpha=1$. For any function $x\in L^{0}$ we define the
maximal function of $x^{\ast}$ by

$x^{\ast\ast}(t)=\frac{1}{t}\int_{0}^{t}x^{\ast}(s)ds.$

For any element $x\in L^{0}$, it is well known that $x^{\ast}\leq
x^{\ast\ast},$ $x^{\ast\ast}$ is decreasing, continuous and subadditive. For
more details of $d_{x}$, $x^{\ast}$ and $x^{\ast\ast}$ the reader is referred
to see [1, 17]. Let us mention that two functions $x,y\in{L^{0}}$ are said to
be equimeasurable (shortly $x\sim y$) if $d_{x}=d_{y}$. A Banach function
space $(E,\|\cdot\|_{E})$ is called symmetric or rearrangement invariant (r.i.
for short) if whenever $x\in L^{0}$ and $y\in E$ such that $x\sim y,$ then
$x\in E$ and $\|x\|_{E}=\|y\|_{E}$. For a symmetric space $E$ we define
$\phi_{E}$ its fundamental function given by
$\phi_{E}(t)=\|\chi_{(0,t)}\|_{E}$ for any $t\in[0,\alpha)$ (see [1]). For any
two functions $x,y\in{}L^{1}+L^{\infty}$ the Hardy-Littlewood-Pólya relation
$\prec$ is given by

$x\prec y\Leftrightarrow x^{\ast\ast}(t)\leq y^{\ast\ast}(t)\text{ for all
}t>0.\text{ }$

A symmetric space $E$ is called $K$-monotone (shortly $E\in(KM)$) if for any
$x\in L^{1}+L^{\infty}$ and $y\in E$ with $x\prec y,$ then $x\in E$ and
$\|x\|_{E}\leq\|y\|_{E}.$ It is commonly known that a symmetric space $E$ is
$K$-monotone if and only if $E$ is exact interpolation space between $L^{1}$
and $L^{\infty}.$ It is worth mentioning that a symmetric space $E$ equipped
with an order continuous norm or with the Fatou property is $K$-monotone (see
[17]). Given $x\in{E}$ is called a point of lower $K$-monotonicity of $E$ for
short an $LKM$ point (resp. a point of upper $K$-monotonicity of $E$ shortly
an $UKM$ point) if for any $y\in{E}$, $y^{*}\neq{x^{*}}$ with $y\prec{x}$
(resp. $x\prec{y}$), then $\left\|y\right\|_{E}<\left\|x\right\|_{E}$ (resp.
$\left\|x\right\|_{E}<\left\|y\right\|_{E}$). Let us mention that given a
symmetric space $E$ is called strictly $K$-monotone (shortly $E\in(SKM)$) if
any element of $E$ is an $LKM$ point or equivalently any element of $E$ is a
$UKM$ point.

Given $x\in{E}$ is said to be a point of lower local uniform $K$-monotonicity
of $E$, shortly an $LLUKM$ point, if for any $(x_{n})\subset{E}$ with
$x_{n}\prec{x}$ for all $n\in\mathbb{N}$ and
$\left\|x_{n}\right\|_{E}\rightarrow\left\|x\right\|_{E}$, then
$\left\|x^{*}-x_{n}^{*}\right\|_{E}\rightarrow{0}$.

A symmetric space $E$ is called Lower locally uniformly $K$-monotone, shortly
$E\in(LLUKM)$, whenever any element $x\in{}E$ is a $LLUKM$ point.

We refer the reader for more information to see [4, 5, 6, 10]. In [11],
authors introduced some new notions which are a generalization of uniform
monotonicity properties in symmetric spaces. The generalization was obtained
by replacing a partial order $\leq$ by a weaker relation $\prec$, in
definition of monotonicity properties. It is worth mentioning that the new
properties characterizes a completely different geometric structure of
symmetric spaces than monotonicity properties. A symmetric space $E$ is said
to be uniformly $K$-monotone, shortly $E\in(UKM)$, if for any
$(x_{n}),(y_{n})\subset{E}$ such that $x_{n}\prec{y_{n}}$ for all
$n\in\mathbb{N}$ and
$\lim_{n\rightarrow\infty}\left\|x_{n}\right\|_{E}=\lim_{n\rightarrow\infty}\left\|y_{n}\right\|_{E}<\infty$
we have $\left\|x_{n}^{*}-y_{n}^{*}\right\|_{E}\rightarrow{0}$. A symmetric
space $E$ is called decreasing (resp. increasing) uniformly $K$-monotone,
shortly $E\in(DUKM)$ (resp. shortly $E\in(IUKM)$), if for any
$(x_{n}),(y_{n})\subset{E}$ such that $x_{n+1}\prec{}x_{n}\prec{y_{n}}$ for
all $n\in\mathbb{N}$ and
$\lim_{n\rightarrow\infty}\left\|x_{n}\right\|_{E}=\lim_{n\rightarrow\infty}\left\|y_{n}\right\|_{E}<\infty$
(resp. $x_{n}\prec{y_{n}}\prec{y_{n+1}}$ for every $n\in\mathbb{N}$ and
$\lim_{n\rightarrow\infty}\left\|x_{n}\right\|_{E}=\lim_{n\rightarrow\infty}\left\|y_{n}\right\|_{E}<\infty$),
we have $\left\|x_{n}^{*}-y_{n}^{*}\right\|_{E}\rightarrow{0}$.

An operator $T$ from a Banach function space $(X,\left\|\cdot\right\|_{X})$
into a Banach function space $(Y,\left\|\cdot\right\|_{Y})$ is called a
positive contraction if its norm is at most $1$ and it satisfies the property
that $T(x)\geq{0}$ whenever $x\geq{0}$. An admissible operator for a Banach
couple $(L^{1},L^{\infty})$ is said to be a substochastic operator whenever it
is positive contraction on $L^{1}$ and on $L^{\infty}$. In addition, on the
space $L^{1}+L^{\infty}$ we may employ an equivalent characterization of
substochastic operators, i.e. $T$ is substochastic if and only if
$T(x)\prec{x}$ for all $x\in L^{1}+L^{\infty}$ (see [1]).

Unless we say otherwise, we assume in the whole paper that two substochastic
operators $T$ and $S$ satisfies $T\leq S$ (resp. $T\prec S$) on
$L^{1}+L^{\infty}$, whenever for any $x\in{L^{1}+L^{\infty}}$ we have $Tx\leq
Sx$ (resp. $Tx\prec Sx$). For any quasiconcave function $\phi$ on $I$ the
Marcinkiewicz function space $M_{\phi}^{(*)}$ (resp. $M_{\phi}$) is a subspace
of $L^{0}$ such that

$\left\|x\right\|_{M_{\phi}^{(*)}}=\sup_{t>0}\\{x^{*}(t)\phi(t)\\}<\infty$
${\left(\textnormal{resp.
}\left\|x\right\|_{M_{\phi}}=\sup_{t>0}\\{x^{**}(t)\phi(t)\\}<\infty\right).}$

It is well known that $M_{\phi}\overset{1}{\hookrightarrow}M_{\phi}^{(*)}$ and
also the Marcinkiewicz space $M_{\phi}^{(*)}$ (resp. $M_{\phi}$) is a r.i.
quasi-Banach function space (r.i. Banach function space) with the fundamental
function $\phi$ on $I$. Let us recall the common fact that
$E\overset{1}{\hookrightarrow}M_{\phi}$ for any symmetric space $E$ with the
fundamental function $\phi$ (for more details see [1, 17]).

##  3\. Increasing uniform $K$-monotonicity

######  Theorem 3.1.

Let $E$ be a symmetric space and let $\phi_{E}$ be the fundamental function of
$E$. Then, $E$ is increasing uniformly $K$-monotone if and only if $E$ is
lower locally uniformly $K$-monotone and

$\lim_{t\rightarrow{0^{+}}}\frac{t}{\phi_{E}(t)}=0.$

###### Proof.

Necessity. Let $(x_{n}),(y_{n})\subset{E}$, $x_{n}\prec{y_{n}}\prec{y_{n+1}}$
and

(1)
$\lim_{n\rightarrow\infty}\left\|y_{n}\right\|_{E}=\lim_{n\rightarrow\infty}\left\|x_{n}\right\|_{E}=d,$

where $d\in(0,\infty)$. Clearly, by Proposition 5.9 in [1] we get $E$ is
embedded in the Marcinkiewicz space $M_{\phi_{E}}$ with the norm one, and so

$\phi_{E}(t)y_{n}^{*}(t)\leq\left\|y_{n}\right\|_{E}\leq{d}<\infty$

for all $t>0$, and $n\in\mathbb{N}$. Hence, by Helly’s Selection Theorem [22],
passing to subsequence if necessary we may assume that there is
$y_{0}^{*}=y_{0}\in{L^{0}}$ such that $y_{n}^{*}\rightarrow{y_{0}^{*}}$ a.e.
on $I$. Next, applying (1) and by assumption that $E$ has the Fatou property,
in view of Fatou’s Lemma [1] it follows that $y_{0}\in{E}$ and

(2)
$\left\|y_{0}\right\|_{E}\leq\liminf_{n\rightarrow\infty}\left\|y_{n}\right\|_{E}.$

Moreover, we have for any $t>0$ and $n\in\mathbb{N}$,

(3)
$x_{n}^{**}(t)\leq{y}_{n}^{**}(t)\leq{y_{n+1}^{**}(t)}\leq\frac{\left\|y_{n}\right\|_{E}}{\phi_{E}(t)}\leq\frac{d}{\phi_{E}(t)}.$

Consequently, for any $t>0$ we obtain

$\liminf_{n\rightarrow\infty}{y}_{n}^{**}(t)=\lim_{n\rightarrow\infty}{y}_{n}^{**}(t)\leq\frac{d}{\phi_{E}(t)}.$

Hence, since $y_{n}^{*}\rightarrow{y_{0}^{*}}$, by Fatou’s Lemma we conclude
that

(4) ${y_{0}^{**}(t)}\leq\lim_{n\rightarrow\infty}{y}_{n}^{**}(t)<\infty$

for any $t\in{I}$, $t>0$. Now, in view of Corollary 5.3 in [1], by assumption
$\lim_{t\rightarrow{0^{+}}}t/\phi_{E}(t)=0$ and by (3) it follows that for any
$\epsilon>0$ there exists $t_{\epsilon}>0$ such that for all
$t\in(0,t_{\epsilon}]$ and $n\in\mathbb{N}$,

(5)
$\frac{1}{d}\int_{0}^{t}y_{n}^{*}\leq\frac{t}{\phi_{E}(t)}<\frac{\epsilon}{4}.$

Furthermore, since $y_{n}^{*}\rightarrow{y_{0}^{*}}$ a.e. on $I$, by the
Lebesgue Dominated Convergence Theorem we observe that for any $0<s<t$, there
exists $N_{\epsilon}\in\mathbb{N}$ such that for all $n\geq{}N_{\epsilon}$,

$\left|\int_{s}^{t}{y_{n}^{*}}-\int_{s}^{t}{y_{0}^{*}}\right|<\frac{d\epsilon}{4}.$

Next, without loss of generality taking additionally $s<t_{\epsilon}$ for the
previous inequality and using (5) for any $n\geq{N_{\epsilon}}$ we get

$\displaystyle\frac{1}{d}\int_{0}^{t}y_{n}^{*}=\frac{1}{d}\int_{0}^{s}y_{n}^{*}+\frac{1}{d}\int_{s}^{t}y_{n}^{*}$
$\displaystyle\leq\frac{1}{d}\int_{0}^{t_{\epsilon}}y_{n}^{*}+\frac{1}{d}\left|\int_{s}^{t}y_{n}^{*}-\int_{s}^{t}y_{0}^{*}\right|+\frac{1}{d}\int_{s}^{t}y_{0}^{*}$
$\displaystyle<\frac{t_{\epsilon}}{\phi_{E}(t_{\epsilon})}+\frac{\epsilon}{4}+\frac{1}{d}\int_{s}^{t}y_{0}^{*}<\frac{\epsilon}{2}+\frac{1}{d}\int_{0}^{t}y_{0}^{*}.$

In consequence, applying (3) and (4) we conclude that for all $t\in{I}$ with
$t>0$,

(6) $\lim_{n\rightarrow\infty}y_{n}^{**}(t)={y_{0}}^{**}(t).$

Now, since $y_{n}\prec{y_{n+1}}$ for any $n\in\mathbb{N}$, we claim that
$y_{n}\prec{y_{0}}$ for all $n\in\mathbb{N}$. Indeed, if it is not true then
there exist $t_{0}>0$, $\epsilon_{0}>0$ and $(n_{k})\subset\mathbb{N}$ such
that for every $k\in\mathbb{N}$ we have

$y_{0}^{**}(t_{0})<y_{n_{k}}^{**}(t_{0})-\epsilon_{0}.$

On the other hand, in view of (6) there exists $N_{0}\in\mathbb{N}$ such that
for each $n\geq{N_{0}}$,

$y_{n}^{**}(t_{0})<y_{0}^{**}(t_{0})+\epsilon_{0}/2,$

which yields a contradiction and proves the claim. Hence, by assumption
$x_{n}\prec{y_{n}}\prec{y_{n+1}}$ for all $n\in\mathbb{N}$ we get
$x_{n}\prec{y_{0}}$ for any $n\in\mathbb{N}$. Therefore, by symmetry of $E$ we
have $\left\|x_{n}\right\|_{E}\leq\left\|y_{0}\right\|_{E}$, and so in view of
the facts (1) and (2) we observe that

(7)
$d=\lim_{n\rightarrow\infty}\left\|x_{n}\right\|_{E}=\lim_{n\rightarrow\infty}\left\|y_{n}\right\|_{E}=\left\|y_{0}\right\|_{E}.$

In consequence, sine $x_{n}\prec{y_{0}}$ and $y_{n}\prec{y_{0}}$ for any
$n\in\mathbb{N}$, by assumption that $E$ is lower locally uniformly
$K$-monotone it follows that

$\lim_{n\rightarrow\infty}\left\|x_{n}^{*}-y_{0}^{*}\right\|_{E}=\lim_{n\rightarrow\infty}\left\|y_{n}^{*}-y_{0}^{*}\right\|_{E}=0.$

Finally, using the triangle inequality of the norm in $E$ we obtain

$\lim_{n\rightarrow\infty}\left\|y_{n}^{*}-x_{n}^{*}\right\|_{E}=0.$

Sufficiency. Immediately, by Remark 4.1 in [11] it is enough to prove that
increasing uniform $K$-monotonicity implies

$\lim_{t\rightarrow{0^{+}}}\frac{t}{\phi_{E}(t)}=0.$

By Corollary 5.3 in [1], we suppose for a contrary that
$\lim_{t\rightarrow{0^{+}}}{\phi_{E}(t)/t}=c>0$. Define for any
$n\in\mathbb{N}$,

$x_{n}=n\chi_{[0,\frac{1}{n})}\qquad\textnormal{and}\qquad
y_{n}=2n\chi_{[0,\frac{1}{2n})}.$

Then, it is easy to see that $x_{n}\prec{y_{n}}\prec{y_{n+1}}$ for all
$n\in\mathbb{N}$ and

$\lim_{n\rightarrow\infty}\left\|x_{n}\right\|_{E}=\lim_{n\rightarrow\infty}\frac{\phi_{E}(1/n)}{1/n}=c\qquad\textnormal{and}\qquad\lim_{n\rightarrow\infty}\left\|y_{n}\right\|_{E}=\lim_{n\rightarrow\infty}\frac{\phi_{E}(1/2n)}{1/2n}=c.$

In consequence, by assumption that $E$ is increasing uniformly $K$-monotone it
follows that

$\lim_{n\rightarrow\infty}\left\|y_{n}^{*}-x_{n}^{*}\right\|_{E}=0.$

On the other hand, by symmetry of $E$ we have

$\displaystyle\lim_{n\rightarrow\infty}\left\|y_{n}^{*}-x_{n}^{*}\right\|_{E}$
$\displaystyle=\lim_{n\rightarrow\infty}\left\|2n\chi_{[0,\frac{1}{2n})}-n\chi_{[0,\frac{1}{n})}\right\|_{E}=\lim_{n\rightarrow\infty}\left\|n\chi_{[0,\frac{1}{2n})}-n\chi_{[\frac{1}{2n},\frac{1}{n})}\right\|_{E}$
$\displaystyle=\lim_{n\rightarrow\infty}\frac{\phi_{E}(1/n)}{1/n}=c>0,$

which contradicts and completes the proof. ∎

##  4\. Substochastic operators

This section is devoted to investigation of convergence of the substochastic
operators in symmetric spaces. First, we prove that any countable sum of
substochastic operators is substochastic.

######  Remark 4.1.

Let us recall some basic facts about substochastic operators, which will use
in our further investigation. Letting $\mathbb{S}$ is a family of all
substochastic operators on $L^{1}+L^{\infty}$ it is easy to show that
$\mathbb{S}$ is convex and closed under the composition. Indeed, by
subadditivity of the Hardy operator (see Theorem 3.4 in [1]) taking
$A,B\in\mathbb{S}$, for any $\theta\in[0,1]$ and $x\in L^{1}+L^{\infty}$ we
have

$\displaystyle\theta A(x)+(1-\theta)B(x)\prec x\quad\textnormal{and}\quad
A\circ B(x)\prec B(x)\prec x.$

######  Proposition 4.2.

Let $(T_{j})$ be a sequence of substochastic operators on a Banach pair
$(L^{1},L^{\infty})$ and let $(E_{j})_{j\in\mathbb{J}}$ be a countable
collection of pairwise disjoint subsets of $I$, each with finite positive
measure. Assume that $E=I\setminus{\bigcup_{j\in\mathbb{J}}}E_{j}$ and
$f\in{L^{0}}$ be locally integrable function on every $E_{j}$. Then the
operator $\mathcal{A}:L^{0}\rightarrow{L^{0}}$ given by

$\mathcal{A}(f)=f\chi_{E}+\sum_{j\in J}T_{j}(f\chi_{E_{j}})\chi_{E_{j}}$

holds $\mathcal{A}(f)\prec{f}$.

###### Proof.

Assume first $\mathcal{A}(f)=f\chi_{E}+T_{1}(f\chi_{E_{1}})\chi_{E_{1}}$ for
$0<\mu(E_{1})<\infty$ and $E=I\setminus{E_{1}}$. Then, taking $F\subset{I}$,
$t=\mu(F)\in{I}\setminus\\{0\\}$, $t_{0}=\mu(F\cap E_{1})$ we observe that

$\displaystyle\int_{F}|\mathcal{A}f|d\mu$
$\displaystyle=\int_{F\cap{E}}|\mathcal{A}f|d\mu+\int_{F\cap
E_{1}}|\mathcal{A}f|d\mu$ $\displaystyle=\int_{F\cap{E}}|f|d\mu+\int_{F\cap
E_{1}}|T_{1}(f\chi_{E_{1}})|d\mu.$

Next, by assumption that $T_{j}$ is a substochastic operator on
$L^{1}+L^{\infty}$ and by Lemma 2.5 in[1] we have

$\displaystyle\int_{F\cap E_{1}}|T_{1}(f\chi_{E_{1}})|d\mu$
$\displaystyle\leq\mu(F\cap E_{1})(T_{1}(f\chi_{E_{1}}))^{**}(\mu(F\cap
E_{1}))d\mu$ $\displaystyle\leq\mu(F\cap E_{1})(f\chi_{E_{1}})^{**}(\mu(F\cap
E_{1}))d\mu$ $\displaystyle=\int_{0}^{\mu(F\cap
E_{1})}(f\chi_{E_{1}})^{*}(s)ds.$

Since $t_{0}=\mu(F\cap E_{1})\leq\mu(E_{1})$, by Proposition 3.3(b) in [1]
there exists a subset $B\subset{E_{1}}$ such that $\mu(B)=t_{0}$ and

$\int_{0}^{\mu(F\cap
E_{1})}(f\chi_{E_{1}})^{*}(s)ds=\int_{B}|f\chi_{E_{1}}|d\mu=\int_{B}|f|d\mu.$

In consequence, we get

$\displaystyle\int_{F}|\mathcal{A}f|d\mu$
$\displaystyle\leq\int_{F\cap{E}}|f|d\mu+\int_{0}^{\mu(F\cap
E_{1})}(f\chi_{E_{1}})^{*}(s)ds$
$\displaystyle\leq\int_{F\cap{E}}|f|d\mu+\int_{B}|f|d\mu=\int_{B\cup
F\cap{E}}|f|d\mu$

Next, since $B\cap(F\cap E)=\emptyset$ and $\mu(F\cap E\cup B)=\mu(E\cup
F)+\mu(B)=t$, by the Hardy-Littlewood inequality (Theorem 2.2 in [1]) it
follows that

$\int_{F}|\mathcal{A}f|d\mu\leq\int_{0}^{t}f^{*}(s)ds.$

In consequence, by Proposition 3.3 [1] taking supremum over all sets $F$ of a
measure $t$ we conclude that for all $t>0$,

$\int_{0}^{t}(\mathcal{A}f)^{*}(s)ds\leq\int_{0}^{t}f^{*}(s)ds.$

Now, we consider the case when $J=\\{1,2,3,\dots,N\\}$ is a finite index set.
Define

$I_{n}=I\setminus\bigcup_{m=1}^{n}E_{m}\quad\textnormal{and}\quad\mathcal{A}_{n}f=f\chi_{E}+\sum_{m=1}^{n}T_{m}(f\chi_{E_{m}})\chi_{E_{m}}$

for all $n\in{J}$. We easily observe that

$\mathcal{A}_{n}f\prec\mathcal{A}_{n-1}f\prec\mathcal{A}_{n-2}f\prec\dots\prec\mathcal{A}_{1}f\prec
f.$

Now, assume that $J=\\{1,2,\dots\\}$ is countable infinite. We claim that
$|\mathcal{A}f|\leq\mathcal{A}|f|$. Indeed, by assumption that $T_{j}$ is a
positive contraction on a pair $(L^{1},L^{\infty})$ and since
$-|f|\leq{f}\leq|f|$ it follows that
$-\mathcal{A}|f|\leq{\mathcal{A}f}\leq\mathcal{A}|f|$. Now, it is sufficient
to prove that $\mathcal{A}f\prec{f}$ for all nonnegative $f$. Consider for all
$n\in\mathbb{N}$,

$f_{n}=f\chi_{E}+\sum_{m=1}^{n}f\chi_{E_{m}}.$

It is easy to see that $0\leq{f_{n}}\uparrow{f}$ a.e. and since $T_{j}$ for
each $j\in{J}$ is a positive contraction we provide

$\displaystyle\mathcal{A}f_{n}$
$\displaystyle=\mathcal{A}(f\chi_{E})+\mathcal{A}\left(\sum_{m=1}^{n}f\chi_{E_{m}}\right)$
$\displaystyle=f\chi_{E}+\sum_{m=1}^{n}T_{m}\left(f\chi_{E_{m}}\right)\chi_{E_{m}}\rightarrow
f\chi_{E}+\sum_{m\in J}T_{m}\left(f\chi_{E_{m}}\right)\chi_{E_{m}}$

a.e. on $I$. Moreover, we have

$\mathcal{A}f_{n}=\mathcal{A}_{n}f_{n}=\mathcal{A}_{n}f\prec{f}$

for all $n\in\mathbb{N}$, whence by Proposition 3.2 [1] this yields
$\mathcal{A}f\prec{f}$. ∎

######  Remark 4.3.

Let us observe that the above theorem is a more general version of the well-
known Proposition 3.7 in [1] shown for integer operators. Namely, in
Proposition 4.2 replacing the operator $T_{j}$ by the integral operator

$\frac{1}{\mu(E_{j})}\int_{E_{j}}{x(t)}d\mu(t),$

given for all integrable $x$ on each $E_{j}$, we constitute the conclusion
immediately.

######  Remark 4.4.

Now, we recall some examples of substochastic operators. First, we observe
that for any measure preserving transformation $\sigma:I\rightarrow{I}$ there
exists a substochastic operator
$T_{\sigma}:L^{1}+L^{\infty}\rightarrow{L^{1}+L^{\infty}}$ given by
$T_{\sigma}(x)=x\circ\sigma$ for any $x\in{L^{1}+L^{\infty}}$. Namely,
$T_{\sigma}(x)\prec x$ for any $x\in{L^{1}+L^{\infty}}$. Next, taking
$T=\\{e^{i\theta}:-\pi\leq\theta\leq\pi\\}$, by Lemma 6.1 in [1], we notice
that the convolution given for any $f\in{L^{1}(T)}$ by

$(f*g)(e^{i\theta})=\int_{-\pi}^{\pi}f(e^{i(\theta-\phi)})g(e^{i\phi})d\phi$

where $g\in L^{1}(T)$ with $\left\|g\right\|_{L^{1}}=1$ is a substochastic
operator in $L^{1}(T)$ i.e. for any $f\in{L^{1}(T)}$ we have
$(f*g)\prec\left\|g\right\|_{L^{1}}f=f$. Furthermore, by Theorem 3.8 in [1] we
see that the Hardy-Littlewood maximal operator defined for any locally
integrable function $f$ on $\mathbb{R}^{n}$ by

$Mf(t)=\sup_{\Omega\in t}\frac{1}{\mu(\Omega)}\int_{\Omega}|f(s)|ds$

constitutes the substochastic operator $T(f)=cM(f)\prec{f}$ for any $f\in
L^{1}(\mathbb{R}^{n})+L^{\infty}(\mathbb{R}^{n})$, where constant $c$ depends
on $n\in\mathbb{N}$. Finally, ti is worth mentioning that any permutation
$\gamma:{\mathbb{N}\rightarrow\mathbb{N}}$ generates a substochastic operator
$P_{\gamma}(x)=x_{\gamma(n)}$ for all $x\in\ell^{1}+\ell^{\infty}$.

Now, we discuss a relation between lower local uniform $K$-monotonicity and
convergence of the sequence of substochastic operators in symmetric spaces.

######  Theorem 4.5.

Let $E$ be a symmetric space on $I$. Assume that $(\mathcal{S}_{n})$ is a
sequence of operators given by

$\mathcal{S}_{n}(x)=x\chi_{\Omega_{n}}+\sum_{j\in
J_{n}}\left(\frac{1}{\mu(E_{j}^{(n)})}\int_{E_{j}^{(n)}}xd\mu\right)\chi_{E_{j}^{(n)}}$

for any $x\in{E}$, where for any $n\in\mathbb{N}$,
$\mathcal{A}_{n}=\\{E_{j}^{(n)}\\}_{j\in J_{n}}$ is a countable collection of
pairwise disjoint measurable subsets of positive and finite measure on $I$
such that each set of $\mathcal{A}_{n}$ is the union of a finite subcollection
of $\mathcal{A}_{n+1}$ and

$\Omega_{n}=I\setminus\bigcup_{j\in
J_{n}}E_{j}^{(n)},\qquad\lim_{n\rightarrow\infty}\sup_{j\in
J_{n}}\mu(E_{j}^{(n)})=0.$

If $E$ is lower locally uniformly $K$-monotone, then for any $x\in{E}$ we have

$\left\|\mathcal{S}_{n}(x)-x\right\|_{E}\rightarrow{0}\qquad\textnormal{as}\qquad{n\rightarrow\infty}.$

Additionally, there exists a sequence of finite rank operators $(T_{n})$ from
$E$ into $E$ such that

$\left\|(\mathcal{S}_{n}-T_{n})(x)\right\|\rightarrow{0}\qquad\textnormal{as}\qquad{n\rightarrow\infty}$

for any $x\in{E}$.

###### Proof.

First, we prove that for any $x\in{E}$,

(8)
$\left\|\mathcal{S}_{n}(x)^{*}-x^{*}\right\|_{E}\rightarrow{0}\qquad\textnormal{as}\qquad{n\rightarrow\infty}.$

Let $|J_{n}|$ means the cardinality of the set $J_{n}$ for any
$n\in\mathbb{N}$. Then, by definition of a sequence $(\mathcal{A}_{n})$ we are
able to find a sequence $(\sigma_{n})\subset\mathbb{N}$ such that
$\sigma_{n}\leq|J_{n}|$ and for every $n\in\mathbb{N}$ we have

$\bigcup_{j=1}^{\sigma_{n}}E_{j}^{(n)}\subset\bigcup_{j=1}^{\sigma_{n+1}}E_{j}^{(n+1)}.$

We define the finite rank operator $T_{n}:E\rightarrow{E}$ that corresponds to
$\mathcal{S}_{n}$ given by

$T_{n}(x)=x\chi_{\Theta_{n}}+\sum_{j=1}^{\sigma_{n}}\left(\frac{1}{\mu(E_{j}^{(n)})}\int_{E_{j}^{(n)}}xd\mu\right)\chi_{E_{j}^{(n)}}$

for any $x\in{E}$, where $\Theta_{n}=\Omega_{n}\cap[0,n]$. It is easy to see
that $\left|T_{n}(x)\right|\leq\left|\mathcal{S}_{n}(x)\right|$ for any
$n\in\mathbb{N}$ and $x\in{E}$. Furthermore, by Proposition 3.7 [1] it follows
that

(9) $\mathcal{S}_{n}(x)\prec{x}\qquad\textnormal{for all
}n\in\mathbb{N}\textnormal{ and }x\in{E}.$

Hence, since $E$ is symmetric we conclude

(10)
$\left\|T_{n}(x)\right\|\leq\left\|\mathcal{S}_{n}(x)\right\|_{E}\leq\left\|x\right\|_{E}$

for any $n\in\mathbb{N}$ and $x\in{E}$. Next, since $x\in{E}$ is locally
integrable on $I$, in view of Lebesgue’s differentiation theorem (see Theorem
3.4 [1]) and by assumption that

(11) $\lim_{n\rightarrow\infty}\sup_{j\in J_{n}}\mu(E_{j}^{(n)})=0$

we get that for a.e. on $I$,

(12)
$\mathcal{S}_{n}(x)\rightarrow{x}\qquad\textnormal{and}\qquad{T}_{n}(x)\rightarrow{x}\qquad\textnormal{as}\qquad
n\rightarrow\infty.$

Moreover, since $E$ is lower locally uniformly $K$-monotone, by Proposition
3.4 [8] this yields that $E$ is order continuous. Now, for any fixed $x\in{E}$
we can find a sequence of bounded functions $(x_{n})\subset{E}$ that
corresponds to $(T_{n})$ such that $|x_{n}|\leq\min\\{|x|,{n}\\}$ and
$T_{n}(x_{n})=x_{n}$ for every $n\in\mathbb{N}$ and $x_{n}\rightarrow{x}$ a.e.
on $I$ as $n\rightarrow\infty$. Namely, taking $(y_{n})$ a sequence of
truncations of the element $x$ we may construct $(x_{n})$ by

$x_{n}=y_{n}\chi_{\Theta_{n}}+\sum_{j=1}^{\sigma_{n}}c_{j}^{(n)}\chi_{E_{j}^{(n)}}$

where a sequence $(c_{j}^{(n)})\subset\mathbb{R}$ is chosen in such way that
$|x_{n}|\leq|y_{n}|\leq|x|$ on $I$ and it guarantees that $x_{n}$ converges to
$x$ a.e. on $I.$ Additionally, by definition of $x_{n}$ on the set
$\bigcup_{j=1}^{\sigma_{n}}E_{j}^{(n)}$ it is easy to observe that
$T_{n}(x_{n})=x_{n}$ and $|x_{n}|\leq{|T_{n}(x)|}$ a.e. on $I$ for all
$n\in\mathbb{N}$. Therefore, by order continuity and symmetry of $E$ we have

$\left\|x_{n}\right\|_{E}\leq\left\|T_{n}(x)\right\|_{E}\quad\textnormal{for
all
}n\in\mathbb{N},\qquad\left\|x-x_{n}\right\|_{E}\rightarrow{0}\quad\textnormal{as}\quad
n\rightarrow\infty.$

Hence, by (10) we conclude

$\lim_{n\rightarrow\infty}\left\|T_{n}(x)\right\|_{E}=\lim_{n\rightarrow\infty}\left\|\mathcal{S}_{n}(x)\right\|_{E}=\left\|x\right\|_{E}.$

Thus, since $E$ is lower locally uniformly $K$-monotone, in view of (9) we
show (8). Now, we prove that $\mathcal{S}_{n}(x)$ converges to $x$ globally in
measure on $I$. Since $E$ is order continuous, by Lemma 2.5 [9] it follows
that $\lim_{t\rightarrow{a}}x^{*}(t)=x^{*}(a)=0$. So, for any $\epsilon>0$
there exists a subset $\Omega\subset{I}$ such that $|x(t)|<\epsilon$ for any
$t\in\Omega$. It is easy to notice that
$\mu(\Omega^{c})=d_{x}(\epsilon)<\infty$ and $\mathcal{S}_{n}(x)=x$ on
$\Omega_{n}$ for all $n\in\mathbb{N}$, whence

$\displaystyle\mu\left\\{s\in{I}:\left|\mathcal{S}_{n}(x)(s)-x(s)\right|\geq\epsilon\right\\}=$
$\displaystyle\mu\left\\{s\in{\Omega^{c}}\cap\Omega_{n}^{c}:\left|\mathcal{S}_{n}(x)(s)-x(s)\right|\geq\epsilon\right\\}$
$\displaystyle+\mu\left\\{s\in{\Omega}\cap\Omega_{n}^{c}:\left|\mathcal{S}_{n}(x)(s)-x(s)\right|\geq\epsilon\right\\}$

for every $n\in\mathbb{N}$. Next, in view of (12) we observe that

(13)
$\mu\left\\{s\in{\Omega^{c}}\cap\Omega_{n}^{c}:\left|\mathcal{S}_{n}(x)(s)-x(s)\right|\geq\epsilon\right\\}\rightarrow{0}\quad\textnormal{as}\quad{n\rightarrow\infty}.$

We suppose temporarily for simplicity of our investigation that $x$ is a
simple function with finite measure support. Then, without loss of generality
we may consider that

$x=\sum_{j=1}^{m}a_{i}\chi_{B_{j}}$

where $\\{B_{1},\dots,B_{m}\\}\subset{I}$ are pairwise disjoint subsets of
finite positive measure and $a_{j}\in\mathbb{R}$ for any
$j\in\\{1,\dots,m\\}$. Thus, by Proposition 8 and Proposition 15 in [20] for
any $j\in\\{1,\dots,m\\}$ there exists a finite collection of open intervals
$\\{U_{i}\\}_{i=1}^{m_{j}}\subset{I}$ such that

$B_{j}\subset\bigcup_{i=1}^{m_{j}}U_{i}\qquad\textnormal{and}\qquad\mu\left(\bigcup_{i=1}^{m_{j}}U_{i}\setminus{B_{j}}\right)<\frac{\epsilon}{m}$

for any $j\in\\{1,\dots,m\\}$. Thus, there exists $0<\gamma<\infty$ such that
$\operatorname{supp}(x)\subset[0,\gamma]$. Hence, by (12) we get
$\mathcal{S}_{n}(x)$ converges to $x$ in measure on $[0,2\gamma]$, which
implies that

$\mu\left\\{s\in{\Omega}\cap\Omega_{n}^{c}:\left|\mathcal{S}_{n}(x)(s)-x(s)\right|\geq\epsilon\right\\}\rightarrow{0}\qquad\textnormal{as}\qquad{n\rightarrow\infty.}$

In consequence, by (13) we obtain

$\mu\left\\{s\in{I}:\left|\mathcal{S}_{n}(x)(s)-x(s)\right|\geq\epsilon\right\\}\rightarrow{0}\qquad\textnormal{as}\qquad{n\rightarrow\infty,}$

which proves that $\mathcal{S}_{n}(x)$ converges to $x$ globally in measure on
$I$ in case when $x$ is a simple function with finite measure support.
Therefore, since $E$ is order continuous, by (8) and by Proposition 2.4 in
[12] it follows that

(14)
$\left\|\mathcal{S}_{n}(x)-x\right\|_{E}\rightarrow{0}\qquad\textnormal{as}\qquad{n\rightarrow\infty}$

for any simple function $x$ with finite measure support. Finally, taking
arbitrary $x$ in $E$ we may find $(y_{n})$ a sequence of simple functions with
finite measure support such that $0\leq|y_{n}|\leq|x|$ a.e. on $I$ for any
$n\in\mathbb{N}$ and $y_{n}\rightarrow{x}$ a.e. on $I$ as
$n\rightarrow\infty$. So, by order continuity of $E$ there is
$M_{\epsilon}\in\mathbb{N}$ such that for all $m\geq{M_{\epsilon}}$ we have
$\left\|x-y_{m}\right\|_{E}<\epsilon/3$. Next, by (10) and by definition of
$\mathcal{S}_{n}$ it is easy to see that

$\left\|\mathcal{S}_{n}(x)-\mathcal{S}_{n}(y_{m})\right\|_{E}=\left\|\mathcal{S}_{n}(x-y_{m})\right\|_{E}\leq\left\|x-y_{m}\right\|_{E}$

for any $n,m\in\mathbb{N}$. Therefore, choosing $m\geq{M_{\epsilon}}$ we
observe that

$\displaystyle\left\|\mathcal{S}_{n}(x)-x\right\|_{E}$
$\displaystyle\leq\left\|\mathcal{S}_{n}(x)-\mathcal{S}_{n}(y_{m})\right\|_{E}+\left\|\mathcal{S}_{n}(y_{m})-y_{m}\right\|_{E}+\left\|y_{m}-x\right\|_{E}$
$\displaystyle=\left\|\mathcal{S}_{n}(x-y_{m})\right\|_{E}+\left\|\mathcal{S}_{n}(y_{m})-y_{m}\right\|_{E}+\left\|y_{m}-x\right\|_{E}$
$\displaystyle\leq{2}\left\|x-y_{m}\right\|_{E}+\left\|\mathcal{S}_{n}(y_{m})-y_{m}\right\|_{E}$
$\displaystyle\leq\frac{2\epsilon}{3}+\left\|\mathcal{S}_{n}(y_{m})-y_{m}\right\|_{E}.$

Hence, by (14) there exists $N_{\epsilon}\in\mathbb{N}$ such that for any
$n\geq{N_{\epsilon}}$ we have

$\displaystyle\left\|\mathcal{S}_{n}(x)-x\right\|_{E}\leq\frac{2\epsilon}{3}+\left\|\mathcal{S}_{n}(y_{m})-y_{m}\right\|_{E}<\epsilon,$

which shows that $\mathcal{S}_{n}(x)$ converges to $x$ in $E$. Now, proceeding
analogously for a sequence $(T_{n})$ as in the case of a sequence
$(\mathcal{S}_{n})$ we provide that $T_{n}$ converges to $x$ in $E$, which
yields that for any $x\in{E}$,

$\left\|(T_{n}-\mathcal{S}_{n})(x)\right\|\rightarrow{0}\qquad\textnormal{as}\qquad{n\rightarrow\infty}.$

∎

In the next theorem we investigate criteria which guarantee that the
generalization of Lebesgue’s differentiation theorem holds in symmetric
spaces.

######  Theorem 4.6.

Let $E$ be a symmetric space on $I$. Assume that $(\mathcal{H}_{n})$ is a
sequence of operators given by

$\mathcal{H}_{n}(x)=\sum_{j\in
J_{n}}\left(\frac{1}{\mu(E_{j}^{(n)})}\int_{E_{j}^{(n)}}xd\mu\right)\chi_{E_{j}^{(n)}}$

for any $x\in{E}$, where for any $n\in\mathbb{N}$,
$\mathcal{A}_{n}=\\{E_{j}^{(n)}\\}_{j\in J_{n}}$ is a countable collection of
pairwise disjoint measurable subsets of positive and finite measure on $I$
such that each set of $\mathcal{A}_{n}$ is the union of a finite subcollection
of $\mathcal{A}_{n+1}$ and

$\Omega_{n}=I\setminus\bigcup_{j\in
J_{n}}E_{j}^{(n)},\qquad\lim_{n\rightarrow\infty}\mu\left(\Omega_{n}\right)=0,\qquad\lim_{n\rightarrow\infty}\sup_{j\in
J_{n}}\mu(E_{j}^{(n)})=0.$

If $E$ is lower locally uniformly $K$-monotone, then for any $x\in{E}$ we have

$\left\|\mathcal{H}_{n}(x)-x\right\|_{E}\rightarrow{0}\qquad\textnormal{as}\qquad{n\rightarrow\infty}.$

Additionally, there exists a sequence of finite rank operators $(T_{n})$ from
$E$ into $E$ such that

$\left\|(\mathcal{H}_{n}-T_{n})(x)\right\|\rightarrow{0}\qquad\textnormal{as}\qquad{n\rightarrow\infty}.$

###### Proof.

Immediately, since $E$ is lower locally uniformly $K$-monotone, applying
Proposition 3.4 [8] we obtain $E$ is order continuous. Therefore, since
$\mu(\Omega_{n})\rightarrow{0}$ as $n\rightarrow\infty$, we conclude that

$\left\|x\chi_{\Omega_{n}}\right\|_{E}\rightarrow{0}\qquad\textnormal{as
}\qquad n\rightarrow\infty.$

Next, by assumption that $\Omega_{n}=I\setminus\bigcup_{j\in
J_{n}}E_{j}^{(n)}$ and by definition of $\mathcal{H}_{n}$ for any
$n\in\mathbb{N}$ it follows that

$\left\|\mathcal{H}_{n}(x)-x\right\|_{E}=\left\|(\mathcal{H}_{n}(x)-x)\chi_{\Omega_{n}^{c}}+x\chi_{\Omega_{n}}\right\|_{E}\leq\left\|\mathcal{H}_{n}(x)-x\chi_{\Omega_{n}^{c}}\right\|_{E}+\left\|x\chi_{\Omega_{n}}\right\|_{E}$

for all $n\in\mathbb{N}$. Hence, by Theorem 4.5 we have

$\left\|\mathcal{H}_{n}(x)-x\right\|_{E}\rightarrow{0}\qquad\textnormal{as}\qquad{n\rightarrow\infty}.$

We can easily observe that there exists a sequence of finite rank operators
$(T_{n})$, that is given analogously as in the previous theorem and
corresponds to a sequence $(\mathcal{H}_{n})$, such that for any
$n\in\mathbb{N}$ we

$\left\|\mathcal{H}_{n}(x)-T_{n}(x)\right\|_{E}\leq\left\|\mathcal{H}_{n}(x)+x\chi_{\Omega_{n}}-T_{n}(x)\right\|_{E}+\left\|x\chi_{\Omega_{n}}\right\|_{E}.$

In consequence, by definition of $\mathcal{H}_{n}$ for any $n\in\mathbb{N}$
and by Theorem 4.5 it follows that
$\left\|\mathcal{H}_{n}(x)-T_{n}(x)\right\|_{E}\rightarrow{0}$ as
$n\rightarrow\infty$ and ends the proof. ∎

In the next theorems we investigate a convergence of admissible operators in
symmetric spaces.

######  Theorem 4.7.

Let $E$ be a symmetric space and $(T_{n})$ be a sequence of admissible
operators on $L^{1}+L^{\infty}$ such that $T_{n+1}\prec{T_{n}}$ for every
$n\in\mathbb{N}$. If $E$ is decreasing uniformly $K$-monotone then for any
$x\in{E}$ there exists $y\in{E}$ such that

$\left\|(T_{n}x)^{*}-y^{*}\right\|_{E}\rightarrow{0}\quad\textnormal{ and
}\quad(T_{n}x)^{*}\rightarrow{y^{*}}\quad\textnormal{globally in measure as
}n\rightarrow\infty.$

###### Proof.

Let $x\in{E}$. Since $T_{n}$ is admissible and $T_{n+1}\prec{T_{n}}$ for every
$n\in\mathbb{N}$ there exists a constant $C$ such that

$T_{n+1}x\prec{T_{n}x}\prec{T_{1}x}\prec{Cx}.$

for all $n\in\mathbb{N}$. Hence, by symmetry of $E$ we get for any
$n\in\mathbb{N}$,

$\left\|T_{n+1}x\right\|_{E}\leq\left\|T_{n}x\right\|_{E}\leq\left\|Cx\right\|_{E}.$

Thus, since any monotone and bounded real sequence is convergent we obtain
that

(15) $\lim_{n\rightarrow\infty}\left\|T_{n}x\right\|_{E}=d<\infty.$

Next, taking two arbitrary sequences $(k_{n}),(j_{n})\subset\mathbb{N}$ such
that $k_{n}\leq j_{n}$ for each $n\in\mathbb{N}$ and denoting
$x_{n}=T_{k_{n}}x$ and $y_{n}=T_{j_{n}}x$ for any $n\in\mathbb{N}$ we can
easily see that $x_{n+1}\prec x_{n}\prec{y_{n}}$ for any $n\in\mathbb{N}$.
Therefore, by assumption that $E$ is decreasing uniformly $K$-monotone and by
the fact (15) it follows that

$\left\|(T_{j_{n}}x)^{*}-(T_{k_{n}}x)^{*}\right\|_{E}=\left\|y_{n}^{*}-x_{n}^{*}\right\|_{E}\rightarrow{0}\quad\textnormal{as}\quad
n\rightarrow\infty$

So, since $(T_{n}x)^{*}$ is Cauchy in $E$, by completeness of $E$ there exists
$y\in{E}$ such that

$\left\|(T_{n}x)^{*}-y\right\|_{E}\rightarrow{0}\quad\textnormal{as}\quad
n\rightarrow\infty.$

Next, by Lorentz and Shimogaki Theorem (see Theorem 7.4 in [1]) we conclude
that $y=y^{*}$ a.e. on $I$. Finally, by Proposition 4.11 in [11], since $E$ is
a symmetric Banach function space, applying double extract sequence theorem we
get that

${(T_{n}x)^{*}}\rightarrow{y^{*}}\quad\textnormal{ globally in measure
as}\quad n\rightarrow\infty.$

∎

######  Theorem 4.8.

Let $E$ be a symmetric space and $A$ be a substochastic operator on
$L^{1}+L^{\infty}$. If $E$ is decreasing uniformly $K$-monotone then for any
$x\in{E^{+}}$ there exists a sequence of substochastic operators
$(B_{n})_{n\geq 0}$ such that for every $n\in\mathbb{N}$,

$B_{n}\circ{A^{n}}x=B_{0}x\textnormal{
a.e.}\quad\textnormal{and}\quad\left\|(B_{0}x)^{*}-(A^{n}x)^{*}\right\|_{E}\rightarrow{0}\quad\textnormal{
as }\quad n\rightarrow\infty.$

Additionally, if $\left\|A^{n}x\right\|_{E}\rightarrow\left\|x\right\|_{E}$
then

$B_{n}\circ{A^{n}}x=x\textnormal{
a.e.}\quad\textnormal{and}\quad\left\|({A^{n}}x)^{*}-x^{*}\right\|_{E}\rightarrow{0}\quad\textnormal{
as }\quad n\rightarrow\infty.$

###### Proof.

First, by Proposition 4.11 in [11] and by Theorem 4.7 we conclude that

(16) $\left\|(A^{n}x)^{*}-y^{*}\right\|_{E}\rightarrow{0}\quad\textnormal{ and
}\quad(A^{n}x)^{*}\rightarrow{y^{*}}\quad\textnormal{a.e. as
}n\rightarrow\infty.$

We claim that $y\prec{x}$. Indeed, taking an arbitrary $t\in I$ and assuming
that $\phi_{E}$ is the fundamental function of the space $E$, by Proposition
5.9 in [1] and by subadditivity of the Hardy operator (see Theorem 3.4 in [1])
we get

$|(A^{n}x)^{**}(t)-y^{**}(t)|\phi_{E}(t)\leq((A^{n}x)^{*}-y^{*})^{**}(t)\phi_{E}(t)\leq\left\|(A^{n}x)^{*}-y^{*}\right\|_{E}.$

In consequence, $(A^{n}x)^{**}(t)\rightarrow{y^{**}(t)}$ as
$n\rightarrow\infty$ for any $t\in{I}$. Hence, since $A^{n}x\prec{x}$ for
every $n\in\mathbb{N}$ we provide the claim. Thus, since $x\in{E}^{+}$, by
Calderón and Ryff Theorem (see Theorem 2.10 in [1]) there exists a
substochastic operator $B_{0}$ on $L^{1}+L^{\infty}$ such that $B_{0}x=|y|$
a.e. Furthermore, since $A^{n+1}x\prec{A^{n}x}$ for any $n\in\mathbb{N}$, it
follows that

$(B_{0}x)^{**}(t)=y^{**}(t)=\lim_{n\rightarrow\infty}(A^{n}x)^{**}(t)\leq(A^{n}x)^{**}(t)$

for any $t\in{I}$ and $n\in\mathbb{N}$. Applying again Calderón and Ryff
Theorem we employ a sequence of the substochastic operators $(B_{n})_{n\geq
1}$ such that

$B_{n}\circ A^{n}x=B_{0}x=|y|$

a.e. on $I$ and for every $n\in\mathbb{N}$. Hence, applying (16) we prove the
first part of our theorem. Now, assume that
$\left\|A^{n}x\right\|_{E}\rightarrow\left\|x\right\|_{E}$. Then, since
$y\prec{x}$, by assumption that $E$ is decreasing uniformly $K$-monotone and
by Remark 4.1 in [11] it follows that $x=|y|$ a.e. on $I$ and completes the
proof. ∎

The immediate consequence of the previous theorem is the following result.

######  Corollary 4.9.

Let $E$ be a symmetric space with the fundamental function $\phi$ such that
$\phi(\infty)=\infty$ and $A$ be a substochastic operator on
$L^{1}+L^{\infty}$. If $E$ is decreasing uniformly $K$-monotone and if for any
$x\in{E^{+}}$ we have $Ax\leq{x}$ and
$\left\|A^{n}x\right\|_{E}\rightarrow\left\|x\right\|_{E}$ as
$n\rightarrow\infty$ then $\left\|A^{n}x-x\right\|_{E}\rightarrow{0}$ as
$n\rightarrow\infty$.

###### Proof.

Directly, by Theorem 4.8 we obtain

(17) $\left\|x^{*}-(A^{n}x)^{*}\right\|_{E}\rightarrow{0}\quad\textnormal{ as
}\quad n\rightarrow\infty.$

Next, since $Ax\leq{x}$ for any $x\in{E^{+}}$ it follows that

$A^{n}x\leq\frac{A^{n}x+x}{2}\leq{x}$

for every $n\in\mathbb{N}$. Hence, by symmetry of $E$ and by assumption that
$\left\|A^{n}x\right\|_{E}\rightarrow\left\|x\right\|_{E}$ as
$n\rightarrow\infty$ we have

$\frac{1}{2}\left\|{A^{n}}x+x\right\|_{E}\rightarrow\left\|x\right\|_{E}\quad\textnormal{
as }\quad n\rightarrow\infty.$

In consequence, since $\frac{1}{2}(A^{n}x+x)\prec{x}$ for all
$n\in\mathbb{N}$, in view of the assumption that $E$ is decreasing uniformly
$K$-monotone we conclude that

$\left\|\frac{1}{2}({A^{n}}x+x)^{*}-x^{*}\right\|_{E}\rightarrow{0}\quad\textnormal{
as }\quad n\rightarrow\infty.$

Therefore, by (17) and by Lemma 2.2 in [12] this yields that
$A^{n}x\rightarrow{x}$ globally in measure. Furthermore, since
$\phi(\infty)=\infty$, by Proposition 4.11 in [11] we provide $E$ is order
continuous. Finally, by (17) and by Proposition 2.4 in [12] we get end of the
proof. ∎

Now we focus our attention on conditions that guarantee the convergence of
substochastic operators in symmetric spaces.

######  Theorem 4.10.

Let $X,Y$ be symmetric spaces and let $T_{n},T\in B(X,Y)$ be substochastic
operators such that $T_{n}\leq{T_{n+1}}\leq{T}$ for all $n\in\mathbb{N}$. If
$Y$ is uniformly $K$-monotone, then

$\left\|T_{n}(x)-T(x)\right\|_{Y}\rightarrow{0}\quad\textnormal{as}\quad
n\rightarrow\infty,$

for some $x\in{X^{+}}$ such that
$\left\|x\right\|_{X}=\left\|T(x)\right\|_{Y}=1$.

###### Proof.

Let $\epsilon>0$. Then, there exists $x_{1}\in X$ such that
$\left\|x_{1}\right\|_{X}=1$, $x_{1}\geq{0}$ and

$\left|\left\|T_{1}(x_{1})\right\|_{Y}-\left\|T_{1}\right\|\right|<\frac{\epsilon}{2}.$

Moreover, without loss of generality passing to subsequence and relabeling we
may assume that for all $n\in\mathbb{N}$,

$\left|\left\|T_{n}\right\|-1\right|<\frac{\epsilon}{2},$

whence we get

(18) $\left|\left\|T_{1}(x_{1})\right\|_{Y}-1\right|<{\epsilon}.$

Next, since $T_{n}\leq{T_{n+1}}\leq{T}$ for all $n\in\mathbb{N}$, it follows
that

$T_{1}(x_{1})\leq{T_{n}(x_{1})}\leq{T_{n+1}(x_{1})}\leq{T(x_{1})},$

for any $n\in\mathbb{N}$. By symmetry of $Y$, we conclude

(19)
$\left\|T_{1}(x_{1})\right\|_{Y}\leq\left\|T_{n}(x_{1})\right\|_{Y}\leq\left\|T_{n+1}(x_{1})\right\|_{Y}\leq\left\|T(x_{1})\right\|_{Y},$

for any $n\in\mathbb{N}$. Hence, in view of (18) we easily get

$1-\epsilon<\left\|T_{1}(x_{1})\right\|_{Y}\leq\left\|T_{n}(x_{1})\right\|_{Y}\leq\left\|T(x_{1})\right\|_{Y}\leq{1},$

for all $n\in\mathbb{N}$. So, since $\epsilon$ is arbitrary chosen we obtain

(20)
$\left\|T_{n}(x_{1})\right\|_{Y}\rightarrow\left\|T(x_{1})\right\|_{Y}=1\quad\textnormal{as}\quad
n\rightarrow\infty.$

Now, applying analogous techniques as in the proof of Theorem 3.1 and passing
to subsequence and also relabelling if necessary we may show that there exists
$S(x_{1})\in{Y}$ such that

(21) ${T_{n}(x_{1})}^{*}\rightarrow{S(x_{1})}^{*}\quad\textnormal{a.e. on
$I$,}\qquad{T_{n}(x_{1})}^{**}(t)\rightarrow{S(x_{1})}^{**}(t)\quad\textnormal{for
all }t\in{I}$

as $n\rightarrow\infty$ and

(22)
$\left\|T_{n}(x_{1})\right\|_{Y}\rightarrow\left\|S(x_{1})\right\|_{Y}\quad\textnormal{as}\quad
n\rightarrow\infty\qquad\textnormal{and}\qquad T_{n}(x_{1})\prec{S(x_{1})}$

for each $n\in\mathbb{N}$. Furthermore, since $T_{n}\leq{T}$ for all
$n\in\mathbb{N}$ on the positive cone of $X$ we have
$T_{n}(x_{1})^{*}\leq{T(x_{1})}^{*}$. Hence, by (20), (21) and (22) this
yields $S(x_{1})^{*}\leq{T(x_{1})^{*}}$ a.e. on $I$ and
$\left\|S(x_{1})\right\|_{y}=1$. Consequently, since $Y$ is uniformly
$K$-monotone by Remark 4.1 in [11] it follows that $Y$ is strictly
$K$-monotone and so $T(x_{1})^{*}=S(x_{1})^{*}$ a.e. on $I$. Moreover, it is
clearly obvious that $T_{n}(x_{1})\prec T_{n+1}(x_{1})\prec T(x_{1})$ for any
$n\in\mathbb{N}$ and in view of (20) and by assumption that $Y$ is uniformly
$K$-monotone we get

(23)
$\left\|T_{n}(x_{1})^{*}-T(x_{1})^{*}\right\|_{Y}\rightarrow{0}\quad\textnormal{as}\quad
n\rightarrow\infty.$

Now, we shall prove that

(24)
$\left\|\frac{(T_{n}(x_{1})+T(x_{1}))}{2}^{*}-T(x_{1})^{*}\right\|_{Y}\rightarrow{0}\quad\textnormal{as}\quad
n\rightarrow\infty.$

First, we observe that for all $n\in\mathbb{N}$,

$T_{n}(x_{1})\leq\frac{T_{n}(x_{1})+T(x_{1})}{2}\leq T(x_{1}).$

Hence, by symmetry of $Y$ this implies

$\left\|T_{n}(x_{1})\right\|_{Y}\leq\frac{\left\|T_{n}(x_{1})+T(x_{1})\right\|_{Y}}{2}\leq\left\|T(x_{1})\right\|_{Y}=1,$

for all $n\in\mathbb{N}$, which provides in view of (22) that

$\frac{\left\|T_{n}(x_{1})+T(x_{1})\right\|_{Y}}{2}\rightarrow{1}\quad\textnormal{as}\quad
n\rightarrow\infty.$

Moreover, for any $n\in\mathbb{N}$ we have $T_{n}(x_{1})+T(x_{1})\leq
2T(x_{1})$. In consequence, proceeding analogously as in the proof of Theorem
3.1 we may see that

$\frac{T_{n}(x_{1})+T(x_{1})}{2}\prec
T(x_{1})\quad\textnormal{and}\quad\frac{(T_{n}(x_{1})+T(x_{1}))}{2}^{*}\rightarrow
T(x_{1})^{*}\quad\textnormal{a.e. on }I.$

Thus, by assumption that $Y$ is uniformly $K$-monotone we prove (24). Next,
according to Lemma 2.2 [12] it follows that $T_{n}(x_{1})$ converges to
$T(x_{1})$ globally in measure on $I$. Consequently, in view of Remark 4.1 in
[11] and by assumption that $Y$ is uniformly $K$-monotone we get $Y$ is order
continuous. Finally, by (23) and by Proposition 2.4 [12] we obtain that

$\left\|T_{n}(x_{1})-T(x_{1})\right\|_{Y}\rightarrow{0}\quad\textnormal{as}\quad
n\rightarrow\infty.$

∎

######  Theorem 4.11.

Let $X,Y$ be symmetric spaces and let $T_{n},T\in{B(X,Y)}$ be substochastic
operators such that for any $x\in{X^{+}}$ we have
$\left\|T_{n}(x)\right\|_{Y}\rightarrow\left\|T(x)\right\|_{Y}$ and
$T_{n}\leq{T_{n+1}}\leq{T}$ for all $n\in\mathbb{N}$. If $Y$ is uniformly
$K$-monotone, then for any $x\in{X}$ we have

$\left\|T_{n}(x)-T(x)\right\|_{Y}\rightarrow{0}\quad\textnormal{as}\quad
n\rightarrow\infty.$

###### Proof.

Since the convergence is satisfied for $x=0$ we may assume without loss of
generality that $x\neq{0}$. Proceeding analogously as in the proof of Theorem
4.10 and assuming that
$\left\|T_{n}(x)\right\|_{Y}\rightarrow\left\|T(x)\right\|_{Y}=1$ as
$n\rightarrow\infty$ we obtain that for any $x\in X^{+}\setminus\\{0\\}$,

$\left\|T_{n}(x)-T(x)\right\|_{Y}\rightarrow{0}.$

Since for any substochastic operator $S:X\rightarrow Y$ we have
$S(x)=S(x^{+}-x^{-})=S(x^{+})-S(x^{-})$ where $x\in{X}$, we easily get for any
$x\in{X}$,

$0\leq\left\|T_{n}(x)-T(x)\right\|_{Y}\leq\left\|T_{n}(x^{+})-T(x^{+})\right\|_{Y}+\left\|T_{n}(x^{-})-T(x^{-})\right\|_{Y}\rightarrow{0}$

as $n\rightarrow\infty$, which completes the proof. ∎

######  Theorem 4.12.

Let $\overline{X}=(X_{0},X_{1})$, $\overline{Y}=(Y_{0},Y_{1})$ be two Banach
couples and let $X,Y$ be interpolation spaces with respect to the couples
$\overline{X}$ and $\overline{Y}$, where $X,Y$ are of exponent
$\theta\in[0,1]$. Assume that $Y_{1}$ is lower locally uniformly $K$-monotone
and $T:\overline{X}\rightarrow\overline{Y}$ is an admissible operator such
that $T:{X_{1}}\rightarrow{Y_{1}}$ compactly. Then $T:{X}\rightarrow{Y}$ is
compact and there exists a sequence of admissible operators
$T_{n}:\overline{X}\rightarrow\overline{Y}$ such that

$\left\|T-T_{n}\right\|_{B(X,Y)}\rightarrow{0}\quad\textnormal{as}\quad{n}\rightarrow\infty.$

###### Proof.

Since every finite rank operator is compact and the norm-limit of compact
operators is also a compact operator, it is enough to prove that
$T:\overline{X}\rightarrow\overline{Y}$ an arbitrary admissible operator is a
limit of the finite rank admissible operators. Define analogously as in
Theorem 4.5 the operator $F_{n}$ by

$F_{n}(x)=x\chi_{\Omega_{n}}+\sum_{j\in
J_{n}}\left(\frac{1}{\mu\left(E_{j}^{(n)}\right)}\int_{E_{j}^{(n)}}xd\mu\right)\chi_{E_{j}^{(n)}}$

for any $n\in\mathbb{N}$ and $x\in{Y_{0}+Y_{1}}$. Let
$T:{X_{0}+X_{1}}\rightarrow{Y_{0}+Y_{1}}$ be admissible operator and let
$(X,Y)$ be an intermediate couple for $(X_{0},X_{1})$ and $(Y_{0},Y_{1})$.
Denote $T_{n}=F_{n}\circ{T}$ for any $n\in\mathbb{N}$. Clearly, $T_{n}$ is a
finite rank operator and

(25)
$\left\|T-T_{n}\right\|_{B(X_{i},Y_{i})}=\sup_{\left\|x\right\|_{X_{i}}\leq{1}}\left\|T(x)-F_{n}\circ{T}(x)\right\|_{Y_{i}}$

for any $n\in\mathbb{N}$. Since $V=\\{T(z):\left\|z\right\|_{X_{1}}\leq{1}\\}$
has a compact closure in $Y_{1}$, for each $v\in V$ there exists $(z_{n})\in
X$, $\left\|z_{n}\right\|_{X_{1}}\leq{1}$ such that $T(z_{n})\rightarrow{v}$
as $n\rightarrow\infty$ in $Y_{1}$. Therefore, we have

$\sup_{\left\|x\right\|_{X_{1}}\leq{1}}\left\|T(x)-F_{n}\circ{T}(x)\right\|_{Y_{1}}=\sup_{v\in{V}}\left\|v-F_{n}(v)\right\|_{Y_{1}}$

for any $n\in\mathbb{N}$. Next, by Theorem 4.5, since $Y_{1}$ is lower locally
uniformly $K$-monotone, it follows that
$\left\|v-F_{n}(v)\right\|_{Y_{1}}\rightarrow{0}$ as $n\rightarrow\infty$.
Hence, since $V$ has a compact closure in $Y_{1}$, by (25) we get

(26)
$\left\|T-T_{n}\right\|_{B(X_{1},Y_{1})}=\sup_{v\in{V}}\left\|v-F_{n}(v)\right\|_{Y_{1}}\rightarrow{0}\quad\textnormal{as}\quad
n\rightarrow\infty.$

By assumption that $(X,Y)$ are the interpolation spaces of exponent
$\theta\in[0,1]$, we conclude

$\displaystyle\left\|T-T_{n}\right\|_{B(X,Y)}$
$\displaystyle\leq\left\|T-T_{n}\right\|_{B(X_{0},Y_{0})}^{1-\theta}\left\|T-T_{n}\right\|_{B(X_{1},Y_{1})}^{\theta}$
$\displaystyle\leq\left(\left\|T\right\|+\left\|T_{n}\right\|\right)^{1-\theta}\left\|T-T_{n}\right\|_{B(X_{1},Y_{1})}^{\theta}$
$\displaystyle\leq\left(\left\|T\right\|+\left\|F_{n}\right\|\left\|T\right\|\right)^{1-\theta}\left\|T-T_{n}\right\|_{B(X_{1},Y_{1})}^{\theta}.$

Since $F_{n}(x_{i})\prec{x_{i}}$ for all $n\in\mathbb{N}$ and
$x_{i}\in{X_{i}}$, this yields that
$\left\|F_{n}\right\|_{B(X_{i},Y_{i})}\leq{1}$, whence

$\displaystyle\left\|T-T_{n}\right\|_{B(X,Y)}$
$\displaystyle\leq\left(2\left\|T\right\|_{B(X_{0},Y_{0})}\right)^{1-\theta}\left\|T-T_{n}\right\|_{B(X_{1},Y_{1})}^{\theta}$
$\displaystyle\leq\left(2\left\|T\right\|_{B(X_{0},Y_{0})}\right)^{1-\theta}\left(2\left\|T\right\|_{B(X_{1},Y_{1})}\right)^{\theta}$
$\displaystyle=2\left\|T\right\|_{B(X_{0},Y_{0})}^{1-\theta}\left\|T\right\|_{B(X_{1},Y_{1})}^{\theta}.$

Moreover, by (26) we obtain

$\left\|T-T_{n}\right\|_{B(X,Y)}\leq\left(2\left\|T\right\|_{B(X_{0},Y_{0})}\right)^{1-\theta}\left\|T-T_{n}\right\|_{B(X_{1},Y_{1})}^{\theta}\rightarrow{0}\quad\textnormal{as}\quad
n\rightarrow\infty.$

To show the second part of the theorem we define $F_{n}$ as follows

$F_{n}(x)=\sum_{j=1}^{\sigma_{n}}\left(\frac{1}{\mu\left(E_{j}^{(n)}\right)}\int_{E_{j}^{(n)}}xd\mu\right)\chi_{E_{j}^{(n)}}$

for any $n\in\mathbb{N}$ and $x\in{X_{0}+X_{1}}$. Then, we proceed as
previously to get

$\left\|T-T_{n}\right\|_{B(X,Y)}\rightarrow{0}\quad\textnormal{as}\quad
n\rightarrow\infty.$

Furthermore, $T_{n}=F_{n}\circ{T}$ is admissible with respect to the couples
$(X_{0},X_{1})$ and $(Y_{0},Y_{1})$, i.e. for any $x\in{X_{i}}$ and
$n\in\mathbb{N}$,

$\displaystyle\left\|T_{n}(x)\right\|_{Y_{i}}=\left\|F_{n}\circ
T(x)\right\|_{Y_{i}}\leq\left\|F_{n}\circ
T\right\|_{B(X,Y)}\left\|x\right\|_{X_{i}}\leq\left\|T\right\|_{B(X,Y)}\left\|x\right\|_{X_{i}},$

and consequently

$\displaystyle\left\|T_{n}\right\|_{\mathcal{A}}=\max_{i=0,1}\left\|F_{n}\circ
T\right\|_{B(X_{i},Y_{i})}\leq\left\|T\right\|_{\mathcal{A}}.$

∎

######  Remark 4.13.

Now, taking an arbitrary Banach couple $(A_{0},A_{1})$ we recall a
construction

$K_{\theta,q}(\overline{A})=\\{a\in
A_{0}+A_{1}:\Phi_{\theta,q}(K(t,a))<\infty\\}$

for $\theta\in[0,1]$ and $q\in[1,\infty]$, where

$\Phi_{\theta,q}(f(t))=\left(\int_{0}^{\infty}(t^{-\theta}f(t))^{q}\frac{dt}{t}\right)^{1/q}\quad\textnormal{and}\quad
K(t,a)=\inf_{a=a_{0}+a_{1}}\\{\left\|a_{0}\right\|_{A_{0}}+t\left\|a_{1}\right\|_{A_{1}}\\}.$

It is well known that $K_{\theta,q}$ is an exact interpolation functor of
exponent $\theta$, see [2].

######  Theorem 4.14.

Let $\overline{X}=(X_{0},X_{1})$, $\overline{Y}=(Y_{0},Y_{1})$ be two Banach
couples such that $X_{i},Y_{i}\hookrightarrow{L^{1}+L^{\infty}}$ for each
$i=0,1$. If $Y_{1}$ is lower locally uniformly $K$-monotone and
$T:\overline{X}\rightarrow\overline{Y}$ is an admissible operator such that
$T:{X_{1}}\rightarrow{Y_{1}}$ compactly, then
$T:K_{\theta,q}(\overline{X})\rightarrow{K_{\theta,q}(\overline{Y})}$ is
compact.

###### Proof.

Let $\theta\in[0,1]$ and $g\in[1,\infty)$. Then, by Theorem 3.1.2 [2] it
follows that $K_{\theta,q}(\overline{X})$ and $K_{\theta,q}(\overline{Y})$ are
exact interpolation functors of exponent $\theta$ on the category
$\mathcal{N}$. Hence, by Theorem 4.12, since $Y_{1}$ is lower locally
uniformly $K$-monotone this yields that any admissible operator
$T:K_{\theta,q}(\overline{X})\rightarrow{K}_{\theta,q}(\overline{Y})$ is
compact. ∎

### Acknowledgement

The first author (Maciej Ciesielski) is supported by Poznan University of
Technology, Poland, Grant No. 0213/SBAD/0118.

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$\begin{array}[]{l}\textnormal{\small Maciej CIESIELSKI}\\\ \textnormal{\small
Institute of Mathematics}\\\ \textnormal{\small Pozna\'{n} University of
Technology}\\\ \textnormal{\small Piotrowo 3A, 60-965 Pozna\'{n}, Poland}\\\
\textnormal{\small email: maciej.ciesielski@put.poznan.pl;}\\\ \textnormal{
}\end{array}$

$\begin{array}[]{l}\textnormal{\small Grzegorz LEWICKI}\\\ \textnormal{\small
Department of Mathematics and Computer Science}\\\ \textnormal{\small
Jagiellonian University}\\\ \textnormal{\small 30-348 Krak\'{o}w, \L
ojasiewicza 6, Poland}\\\ \textnormal{\small email:
grzegorz.lewicki@im.uj.edu.pl;}\end{array}$