# The Representations and Continuity of the Metric Projections on Two Classes of Half-Spaces in Banach Spaces

Authors: Zihou Zhang; Chunyan Liu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We show a necessary and sufficient condition for the existence of metric projection on a class of half-spaceKx0*,c={x∈X:x*(x)≤c} in Banach space. Two representations of metric projections PKx0*,c and PKx0,c are given, respectively, where Kx0,c stands for dual half-space of Kx0*,c in dual space X*. By these representations, a series of continuity results of the metric projections PKx0*,c and PKx0,c are given. We also provide the characterization that a metric projection is a linear bounded operator.


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## Body

## 1. Introduction

The metric projection in Banach space is an enduring question for study or discussion. It has been used in many areas of mathematics such as the theories of optimization and approximation, fixed point theory, nonlinear programming, and variational inequalities. On continuity of the metric projection, many mathematicians, for example, Nevesenko [1], Oshman [2], Wang [3], Fang and Wang [4], and Zhang and Shi [5] have done profound research. In practical application, giving the representations of metric projection is very necessary. Generally speaking, this is very difficult. In recent years, Wang and Yu [6] gave a representation of single-valued metric projection on a class of hyperplanes Hx0*,c={x∈X:x0*(x)=c} in reflexive, smooth, and strictly convex Banach space X. Song and Cao [7] gave a representation of metric projection on a class of half-space Kx0*,c in the reflexive, smooth, and strictly convex Banach space X. Wang [8] and Ni [9] extended the result of Wang and Yu to general Banach space, respectively. Wang [8] also discussed continuity of the metric projection on the hyperplane Hx0*,c in Banach space.In this paper, letX be a Banach space and let X* be the dual of X. Let S(X) and B(X) be the unit sphere and unit ball of X, respectively. Let x0*∈X*∖{θ},  let c∈ℝ, let Kx0*,c={x∈X:x0*(x)≤c}, let D-1(x0*)={x∈X:x0*(x)=∥x0*∥∥x∥=∥x0*∥2=∥x∥2}, let x0∈X∖{θ}, let Kx0,c={x*∈X*:x*(x0)≤c}, and let D(x0)={x*∈X*:x*(x0)=∥x*∥∥x0∥=∥x*∥2=∥x0∥2}. It is easily proved that D(αx)=αD(x), D-1(αx*)=αD-1(x*), and for all α∈ℝ. For M⊂X, the metric projection PM:X→2M is defined by PM(x)={y∈M:∥x-y∥=d(x,M)}, where d(x,M)=inf⁡{∥x-y∥:y∈M}. Obviously, PM is a set-valued mapping. If PM(x)≠∅ for each x∈X, then M is said to be proximinal. It is well known that PM is single-valued when X is strictly convex and M is proximal.Cabrera and Sadarangani [10] introduced the geometrical properties of Banach spaces as follows.A Banach spaceX is called nearly strictly convex (resp., weakly nearly strictly convex) whenever, for any x*∈S(X*), the set D-1(x*) is compact (resp., weakly compact). A Banach space X is called nearly smooth (resp., weakly nearly smooth) whenever, for any x∈S(X), the set D(x) is compact (resp., weakly compact).The metric projectionPM is said to be norm-norm (resp., norm-weakly) upper semicontinuous if, for all x in X and for all norm (resp., weakly) open set W⊃PM(x), there exists a norm neighborhood U of x such that PM(U)⊂W.In this paper, firstly, we established a necessary and sufficient condition for the existence of metric projection on a class of half-spaceKx0*,c in Banach space. Secondly, we give two representations of the metric projections PKx0*,c and PKx0,c by using a different method from the literatures [5–9]. Thirdly, by these representations, we prove that if X is weakly nearly strictly convex (resp., weakly nearly smooth), then metric projection PKx0*,c(resp., PKx0,c) is norm-weakly upper semicontinuous. Finally, the characterization of the metric projection PM from X to a subspace M to be a linear bounded operator is given. We extend the corresponding results in [5–9].


## 2. The Representations of the Metric Projection on Two Classes of Half-Spaces in Banach Spaces

Lemma 1.
LetX be a Banach space and let x0*∈X*∖{θ}; then d(x,Kx0*,c)=|x0*(x)-c|/∥x0*∥ for all x∈X∖Kx0*,c.Proof.
Firstly, suppose that∥x0*∥=1. Let x∈X/Kx0*,c. For any y∈Kx0*,c, since
(1)x0*(x-y)≥x0*(x)-c>0,
we deduce that
(2)d(x,Kx0*,c)≥|x0*(x)-c|.
On the other hand, for anyε>0(ε<1/4), there exists uε in S(X) such that 1-ε<x0*(uε)≤1. Set yε=x-(1+2ε)(x0*(x)-c)uε. Then
(3)x0*(yε)=x0*(x)-(1+2ε)(x0*(x)-c)x0*(uε)<x0*(x)-(1+2ε)(x0*(x)-c)(1-ε)=x0*(x)-(1+ε-2ε2)(x0*(x)-c)≤x0*(x)-(x0*(x)-c)=c.
Consequently, yε∈Kx0*,c and
(4)∥x-yε∥=(1+2ε)|x0*(x)-c|.
It follows that
(5)d(x,Kx0*,c)≤(1+2ε)|x0*(x)-c|.
By arbitrariness of ε, we deduce that
(6)d(x,Kx0*,c)≤|x0*(x)-c|.
This means that
(7)d(x,Kx0*,c)=|x0*(x)-c|.
Secondly, for x*∈X*∖θ and ∥x*∥≠1, since
(8)Kx0*,c={x∈X:x0*(x)≤c}={x∈X:x0*∥x0*∥(x)≤c∥x0*∥},
from (7), we may obtain that
(9)d(x,Kx0*,c)=|x0*(x)-c|∥x0*∥,∀x∈X∖Kx0*,c.Remark 2.
For givenx0*∈X*∖{θ} and  c∈ℝ, by Lemma 1, we have that
(10)d(x,Kx0*,c)=d(x,Hx0*,c),
for any x∈X∖Kx0*,c.Theorem 3.
LetX be a Banach space, let x0*∈X*∖θ, and let c∈ℝ; then PKx0*,c(x)≠∅ for any x∈X if and only if D-1(x0*)≠∅.Proof.
On necessity: takex∈X∖Kx0*,c; then there exists a y∈PKx0*,c(x). Set u=(∥x0*∥2/(x0*(x)-c))(x-y); by Lemma 1, we have that
(11)∥u∥=∥x0*∥2|x0*(x)-c|∥x-y∥=∥x0*∥2|x0*(x)-c||x0*(x)-c|∥x0*∥=∥x0*∥.
Hence, x0*(u)⩽∥x0*∥∥u∥=∥x0*∥2.
On the other hand,(12)x0*(u)=∥x0*∥2x0*(x)-c(x0*(x)-x0*(y))⩾∥x0*∥2x0*(x)-c(x0*(x)-c)=∥x0*∥2.
This shows that x0*(u)=∥x0*∥2=∥u∥2, that is, u∈D-1(x0*) and D-1(x0*)≠∅.
On sufficiency: takex∈S(X) such that x0*(x)=∥x0*∥∥x∥=∥x0*∥2=∥x∥2. We discuss that in two cases.
Case  1. If x∈Kx0*,c, then x∈PKx0*,c(x).
Case  2. If x∉Kx0*,c, since
(13)x0*(x-x0*(x)-c∥x0*∥2x0)=x0*(x)-(x0*(x)-c)=c;then we have that x-((x0*(x)-c)/∥x0*∥2)x0∈Kx0*,c. By Lemma 1,
(14)∥x-(x-x0*(x)-c∥x0*∥2x0)∥=x0*(x)-c∥x0*∥=d(x,Kx0*).
It follows that x-((x0*(x)-c)/∥x0*∥2)x0∈PKx0*,c(x).Theorem 4.
LetX be a Banach space, let x0*∈X*∖{θ}, let x0* attain its norm on S(X), and let c∈ℝ. Then
(15)PKx0*,c(x)=x-max⁡{0,x0*(x)-c∥x0*∥2}D-1(x0*).Proof.
Takex∈X. We discuss that in two cases.
Case  1. If x∈Kx0*,c, then PKx0*,c(x)={x}.
Case  2. If x∉Kx0*,c, we arbitrarily take x0∈D-1(x0*). Let y=x-((x0*(x)-c)/∥x0*∥2)x0. Similar to the proof of Theorem 3, we may obtain that y∈PKx0*,c(x). Therefore,
(16)x-x0*(x)-c∥x0*∥2D-1(x0*)⊂PKx0*,c(x).
On the other hand, we arbitrarily takey∈PKx0*,c(x). Let u=(∥x0*∥2/(x0*(x)-c))(x-y); similar to the proof of Theorem 3, we may obtain that u∈D-1(x0*). Therefore,
(17)y=x-x0*(x)-c∥x0*∥2u∈x-x0*(x)-c∥x0*∥2D-1(x0*),
that is,
(18)PKx0*,c(x)⊂x-x0*(x)-c∥x0*∥2D-1(x0*).
By Case 1 and Case 2, we have(19)PKx0*,c(x)=x-max⁡{0,x0*(x)-c∥x0*∥2}D-1(x0*),
for any x∈X.By the similar proof to that in Lemma1, we can obtain the following result.Lemma 5.
LetX be a Banach space, let x0∈X∖{θ}, and let  c∈ℝ. Then
(20)d(x*,Kx0,c)=|x*(x0)-c|∥x0∥,
for any x*∈X*∖Kx0,c.By a similar proof to that in Theorem4, we can also prove the following result according to Lemma 5.Theorem 6.
LetX be a Banach space, let x0∈X∖{θ}, and let c∈ℝ. Then
(21)PKx0,c(x*)=x*-max⁡{0,x*(x0)-c∥x0∥2}D(x0),
for any x*∈X*.


## 3. Continuity of the Metric Projection on the Two Classes of Half-Spaces in Banach Spaces

Theorem 7.
Letx0*∈X∖{θ},  let x0* attain its norm on S(X), and let c∈ℝ. If X is weakly nearly strictly convex, then the metric projection PKx0*,c is norm-weakly upper semicontinuous.Proof.
Letx,  xn∈X, and let xn→x as n→∞. Our proof will be divided into two cases.
Case  1. Suppose that {xn}⊂Kx0*,c. Since Kx0*,c is a closed set, x∈Kx0*,c. Clearly, PKx0*,c(xn)=xn→x=PKx0*,c(x).
Case  2. Suppose that {xn}⊄Kx0*,c.
If there are an infinite number ofn for which xn∈Kx0*,c, then we can choose a subsequence {xnk}⊂{xn} with {xnk}⊂Kx0*,c. Therefore, PKx0*,c(xnk)=xnk→x=PKx0*,c(x) as k→∞.
If there are an infinite number ofn for which xn∉Kx0*,c, without loss of generality, we may assume that {xn}⊂X∖Kx0*,c. Taking yn∈PKx0*,c(xn), by Theorem 4, we have
(22)PKx0*,c(xn)=xn-x0*(xn)-c∥x0∥2D-1(x0*).
We assume thatyn=xn-((x0*(xn)-c)/∥x0*∥2)zn, where zn∈D-1(x0*). Since X is weakly nearly strictly convex, we know that {zn} has a weakly convergent subsequence {znk} with znk→wz as k→∞. Consequently,
(23)ynk=xnk-x0*(xnk)-c∥x0*∥znk⟶wx-x0*(x)-c∥x0*∥z.
Notingx0*(z)=lim⁡kx0*(znk)=lim⁡k∥x0*∥2=∥znk∥2 and ∥z∥⩽lim⁡k¯∥znk∥,  we know that x0*(z)⩾∥x0*∥·∥z∥. Therefore,
(24)x0*(z)=∥x0*∥·∥z∥=∥x0*∥2=∥z∥2,
where z∈D-1(x0*). This shows that ynk→wx-((x0*(x)-c)/∥x0*∥)z∈PKx0*,c(x).
Now, we will show thatPKx0*,c is norm-weakly upper semicontinuous at x. Otherwise, there exist a weakly open set W0⊃PKx0*,c(x) and a sequence {xm} with xm→x as m→∞, but PKx0*,c(xm)⊄W0 for all m. Taking ym∈PKx0*,c(xm)∖W0,  m=1,2,…, similar to previous arguments, we can observe the fact that there exists a subsequence {ymk} of {ym} such that ymk→wy as k→∞ and y∈PKx0*,c(x). This means that there exists ymk∈W0 for some k large enough, which is a contradiction.Similar to the proof of Theorem8, we may prove the following theorem.Theorem 8.
LetX be a Banach space. (1)
Letx0*∈X*∖{θ},  let x0* attain its norm on S(X), and let c∈ℝ. If X is nearly strictly convex, then the metric projection PKx0*,c is norm-norm upper semicontinuous.(2)
Letx0∈X∖{θ}  and let c∈ℝ. If X is weakly nearly smooth, then the metric projection PKx0,c is norm-weakly upper semicontinuous.(3)
Letx0∈X∖{θ}  and let c∈ℝ. If X is nearly smooth, then the metric projection PKx0,c is norm-norm upper semicontinuous.Lemma 9 (see [11]).
LetM be a proximal subspace. Then for any x∈X, one has the decomposition
(25)x=x1+x2,x1∈PM(x),x2∈D-1(M⊥),
where M⊥={x*∈X*:x*(x)=0,∀x∈M} and
(26)D-1(M⊥)={x∈X:D(x)∩M⊥≠∅}.
If M is a Chebyshev subspace, the decomposition is unique, and
(27)x=PM(x)+x2,x2∈D-1(M⊥).Lemma 10.
LetX be a strictly convex Banach space and let M be a proximal subspace. Then, for any x∈X, one has
(28)PM(x+y)=PM(x)+y,y∈M.Proof.
Lety∈M, for any z∈M, we have that w=z-y∈M. Consider
(29)∥PM(x)+y-(x+y)∥=∥PM(x)-x∥≤∥w-x∥=∥(w+y)-(x+y)∥=∥z-(x+y)∥.
By the definition of PM, we obtain PM(x)+y∈PM(x+y). Since X is strictly convex, we know that PM is single-valued, and hence we have PM(x+y)=PM(x)+y.Similar to the proof Theorem2.1(1) in [6], we can prove the following result by Lemmas 9 and 10.Lemma 11.
LetX be a strictly convex Banach space and let M be a proximal subspace. P is single-valued operator from X into M, and PM is a metric projection from X into M. Then P=PM if and only if the following conditions are satisfied:(1)
P-1(θ)=D-1(M⊥);(2)
P(x+y)=P(x)+y,forally∈M.Theorem 12.
LetX be a strictly convex Banach space and let M be a proximal subspace. Then the metric projection PM is a linear bounded operator if and only if D-1(M⊥) is a linear subspace.Proof.
On necessity: letPM be a linear operator. Since X is strictly convex and M is proximal, then PM is single valued. By Lemma 11(1), for any x,y∈D-1(M⊥)=PM-1(θ),  α,  β∈ℝ, then
(30)PM(αx+βy)=αPM(x)+βPM(y)=0,
and hence αx+βy∈PM-1(θ)=D-1(M⊥). This shows that D-1(M⊥) is a linear subspace.
On sufficiency: letD-1(M⊥) be a linear subspace and let PM be a metric projection; since X is strictly convex, by Lemma 11(1), PM-1(θ) is also a linear subspace. For any x,y∈X,x-PM(x),y-PM(y)∈{x-PM(x):x∈X}, we have that
(31)(x+y)-(PM(x)+PM(y))=(x-PM(x))+(y-PM(y))∈{z-PM(z):z∈X}=PM-1(θ).
By Lemma 11(2), we have that
(32)0=PM((x+y)-(PM(x)+PM(y)))=PM(x+y)-(PM(x)+PM(y)).
It follows that PM(x+y)=PM(x)+PM(y). Note that PM is homogeneous; we obtain that PM is a linear operator. In addition, for any x∈X, since θ∈M, we have that
(33)∥PM(x)∥=∥PM(x)-x+x∥≤∥PM(x)-x∥+∥x∥≤∥θ-x∥+∥x∥=2∥x∥.
This shows that PM is a bounded operator.



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