1. No category

Journal of mathematics and computer science
14 (2015), 345-353
Best approximation by upward sets
Zeinab Soltani*, Hamid Reza Goudarzi**
Department of Mathematics, Faculty of Scinence, Yasouj University, Yasouj, Iran.
*[email protected]
**[email protected]
Article history:
Received November 2014
Accepted January 2015
Available online January 2015
Abstract
In this paper we prove some results on upward subsets of a Banach lattice with a strong unit. Also
we study the best approximation in by elements of upward sets, and we give the necessary and
sufficient conditions for any element of best approximation, by a closed subset of
Keywords: Banach lattice, Best approximation, Proximinal set, Upward set.
1. Introduction
The theory of best approximations is an important subject in Functional Analysis. It is a very
extensive field which has various applications ([2], [5], [7] and [10] ). Also the problems of best
approximation by elements of convex sets are welldeveloped and have many applications in different
areas of Mathematics ( [3], [4], [6], [8] and [12]). Downward and upward sets are not necessarily
convex and since, convexity is sometimes a very restrictive assumption, so we can use the so-called
downward and upward sets as good tools in the study of best approximation by closed and not
necessarily convex sets. Best approximation by downward sets and their properties have been studied
by several authors (for example see [1]).
Z. Soltani, H.R. Goudarzi / J. Math. Computer Sci. 14 (2015), 345-353
The approximation properties of upward sets play a crucial role in this paper. We study some aspects
of best approximation by elements of closed upward sets, in a Banach lattice with a strong unit. We
show that a closed upward set is proximinal, and we derive the necessary and su cient conditions for
the uniqueness of best approximation. As we reminded, an upward set is not necessarily convex. We
show that this set is abstract convex with respect to a certain set of elementary functions (for the
definition of abstract convexity, see [11]). This fact allows us to examine the separation properties of
upward sets and gives the necessary and sufficient conditions for best approximations.
2. Preliminaries
In this section, we introduce some basic definitions for the next sections. Let be a non empty subset
of a normed linear space . An element
is called a best approximation to
from if
for every
,
On the other hand, recall that (see e.g. [13]) a point
if
is called a best approximation to
Where
The set of all such elements
Thus
(called best approximations to
(x) = {
G:
x-
= d( x,G) }.
) is denoted by
(x).
(1)
Hence
defines a mapping from into the power set of , called metric projection onto ( other
names are nearest point mapping and proximity map). If
( contains at least one element, then
the subset is call a proximinal set. In other words, if (
, then is called a proximinal set.
Also if each element
has a unique best approximation in , then is called a Chebyshev
subset of .
It is well-known that
the boundary of .
(
is a closed and bounded subset of
. If
then.
(
is located in
Definition 2.1 ( [14]) A vector space which is ordered by a relation ≤ , is called a vector lattice if
any two elements
have a least upper bound denoted by
, a greatest lower
bound denoted by
and the following properties are satisfied:
(1)
implies that
(2
implies that
If a vector lattice
(3)
, for all
, for all
is equipped with a norm .
implies
and
.
for which
, for
,
643
Z. Soltani, H.R. Goudarzi / J. Math. Computer Sci. 14 (2015), 345-353
then (equipped with and
lattice is called a Banach lattice.
) is called a normed vector lattice. A complete normed vector
Example 2.2 Let be a set, equipped with a sigma-field . The space
of all bounded,
measurable real functions on is a vector lattice for the pointwise ordering
means
. The supremum norm, defined by
satisfies in 2.1(3)
and the space (
) is a normed vector lattice. The space
(
) is also complete by supremum
norm, in the sense that Cauchy sequences converge to bounded, measurable limit functions.
If
is a vector lattice, an element
R such that
.
is called a strong unit if for each
Using a strong unit 1, we can define a norm on
, there exists
by
(2)
Then
–
.
(3)
Also we have
(4)
(S, ,µ) denotes the lattice of all essentially bounded functions defined on the space with a
algebra of measurable sets, , and a measure µ. Also th lattice of all bounded functions defined on
is an example of Banach lattices with the strong units.
Recall that a subset
of an ordered set
is said to be upward if
If is a normed linear space and G is a subset of , we shall denote by
and
the
interior, the closure and the boundary of , respectively. If is a lattice and there exists the greatest
elements of , we shall denote it by
.
3. Upward sets and their approximation properties
Let be a Banach lattice with a strong unit 1. In this section we investigate the best approximations
in by elements of upward sets. In particular, we show that the greatest element, in the set of best
approximations, exists.
Proposition 3.1 Let
(1) If
, then
be an upward subset of
and
,
(2)
643
. Then the following statements are true:
Z. Soltani, H.R. Goudarzi / J. Math. Computer Sci. 14 (2015), 345-353
Proof: (1) Let
be given and
. suppose that
be an open neighborhood of (x + ε1). Then, by (3),
Since
is an upward set and
(2) Let
, it follows that
.
. Then there exists
such that the closed ball
Conversely, suppose that there exists
such that
, which completes the proof. □
saw
Corollary 3.2 Let G be a closed upward subset of
, for each
Lemma 3.3 Let
be a closed upward subset of
Proof: Let
that for each
Let
, and so
=
be arbitrary and
, there exists
and
Then
. In view of (3) ,
Then, as we
. Then
if and only if
is proximinal in .
. This implies
. Then, by (3), we have
such
+ r1. Then, we have
and hence
. Since G is upward and
it follows that
, for each ε > 0. Since G is closed, we have
G and then
( ). Thus the result follows. □
Remark 3.4 We proved that for each
with
. If
then
Proposition 3.5 Let
element
Proof: If
, the set
( ) contain the element
and
( ) = { }.
be a closed upward subset of and
( ) of the set
( ), namely
, then the result holds. Assume that
. Then there exists the greatest
, where
and
( ).
,
we get
for all
This implies that
is the greatest element of the closed ball
arbitrary. Then
, and so
. Therefore
greatest element of
( ). □
Corollary 3.6 Let
be a closed upward subset of
,
Corollary 3.7 Let
be a closed upward subset of
and
643
Now, let
and hence,
and
be arbitrary. Then
be
is the
( ). Then,
Z. Soltani, H.R. Goudarzi / J. Math. Computer Sci. 14 (2015), 345-353
Proof : Let
. If
Suppose that
. Thus, we have
then
then
Since, by proposition 3.5,
completes the proof. □
. Let
, it follows that
, and so
be arbitrary such that
Hence,
which
4. Characterization of best approximation by upward sets
In this section, we present the characterization of upward sets in terms of separation from outside
points. Throughout this section, X is a Banach lattice. Let
be a function defined
by
Since 1 is a strong unit, it follows that the set
} is non empty and bounded
from above ( by the number
). Clearly this set is closed. It follows from the aforesaid and
the definition of that the function enjoys the following properties.
For each
define the function
by
The function
is called topical if this function is increasing (
)) and
plus-homogeneous
for all
X and λ R). The definition of topical
function in finite dimensional case can be found in [11].
Lemma 4.1 The function
defined by (13) is topical.
Proof: we try to check the conditions. Let
Hence,
with
643
. Then
Z. Soltani, H.R. Goudarzi / J. Math. Computer Sci. 14 (2015), 345-353
Let
and
be arbitrary. Then
Proposition 4.2 The function
Proof: Let
is Lipschitz continuous.
be arbitrary. Since
, it follows that
In view of lemma 4.1, we have
and hence
Therefore,
is Lipschitz continuous. □
As a direct result of (14), we have:
Corollary 4.3 The function
Lemma 4.4 Let
(5). Then,
defined by (5) is continuous.
be a closed upward subset of
for all
and let
be the function defined by
Proof: Assume that there exists
such that
. Then
So there exists
such that
This means that
. Therefore
. Since is upward and
, it follows that
. So, by proposition 3.1 (2), we have
. This is a contradiction, which completes
the proof. □
Now, we give the characterization of upward sets in terms of separation from outside points. For
an easy reference we present the following theorem.
Theorem 4.5 Let be a subset of
statements are equivalent:
(1)
and
be the coupling function of (5). Then the following
is an upward set,
(2) For each
, we have
(3) For each
, there exists
such that
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Z. Soltani, H.R. Goudarzi / J. Math. Computer Sci. 14 (2015), 345-353
Proof:(1
(2). Suppose that is an upward set and that exists
,
that
. Then, by (7) we have
and so
upward set and
, it follows that
. This is a contradiction.
(2)
(3). Assume that (2) holds and
Now, let
is arbitrary. Then by hypothesis, we have
, using (9), for each
, we have :
(3)
1). Suppose that (3) holds and is not an upward set. Then there exists
\G with
. It follows, by hypothesis, that there exists
such that
Since
such
. Since G is an
and
is increasing, we have
This contradicts (15) . □
Theorem 4.6 Let
are equivalent:
be the function defined by (5). Then for a subset
(1
is a closed upward subset of
(2)
is upward, and for each
of
the following statements
the set
is closed in ,
(3) For each
, there exists
such that
(4) For each
there exists
such that
Proof: (1)
Since
Hence,
(2). Assume that
and
α
is a closed upward subset of
R. Then, we have
and
and let
is closed, it follows that
, and so
is a closed subset of .
(2) (3). Suppose that (2) holds and
is arbitrary. We claim that there exists
that
. Indeed, if
, for all
, then due to the closedness of H, we have
. This implies that
. This is a contradiction. Now, let
show that
, for all
. Assume that there exists
such that
653
such
. We
Z. Soltani, H.R. Goudarzi / J. Math. Computer Sci. 14 (2015), 345-353
Then by (7) we have
upward set and
, it follows that
contradiction. Hence,
, for all
, and so
and consequently
On the other hand, we have
( –
{
)
Since is an
. This is a
}
(3)
(4) is abvious.
(4)
(1). Suppose that (4) holds and that is not an upward set. Then there exists
with
. By hypothesis, there exists
such that
Since
and
is increasing, it followes that
This is a contradiction. Hence, is an upward set. Finally, assume that is not closed. Then there
exists a sequence
and
such that
, as
Since
, by hypothesis, there exists
such that
Thus, we have
By continuity of
, it follows that
which completes the proof. □
Lemma 4.7 Let
(5). Then
Proof: Since
. This is a contradiction,
be a closed upward subset of
and
, it follows, by lemma 4.4, that
Also, we have
653
Let
be defined by
Z. Soltani, H.R. Goudarzi / J. Math. Computer Sci. 14 (2015), 345-353
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