Asymptotic behavior of resolvents of equilibrium problems on complete geodesic spaces

Yasunori Kimura; Keisuke Shindo

doi:10.1515/dema-2022-0187

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Asymptotic behavior of resolvents of equilibrium problems on complete geodesic spaces

Yasunori Kimura and Keisuke Shindo

Published/Copyright: January 30, 2023

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From the journal Demonstratio Mathematica Volume 56 Issue 1

Abstract

In this article, we discuss equilibrium problems and their resolvents on complete geodesic spaces. In particular, we consider asymptotic behavior and continuity of resolvents with positive parameter in a complete geodesic space whose curvature is bounded above. Furthermore, we apply these results to resolvents of convex functions.

Keywords: asymptotic behavior; geodesic space; convex function; resolvent

MSC 2010: 52A41

1 Introduction

The geodesic spaces are a class of metric spaces with some convex structures and have been actively researched in recent years. A complete CAT(0) space is an example of such spaces and it is a generalization of a Hilbert space. A complete CAT(1) space is also a geodesic space and can be seen locally as a generalization of a complete CAT(0) space. Many researchers have been tried to apply theories of linear spaces such as Hilbert and Banach spaces to these geodesic spaces. The concept of resolvent is one of them. For example, we can define various types of resolvent maps for a convex function on complete CAT(0) spaces and complete CAT(1) spaces. In addition, it is defined for a bifunction of the equilibrium problems to be introduced later. The problem of asymptotic behavior of resolvents is one of the important subjects to know the property of resolvents. We describe famous results about asymptotic behavior of a resolvent for a convex function in a Hilbert space below.

Theorem 1.1

(See [1]) Let X be a Hilbert space, g : X → ] − ∞ , ∞ ] a proper lower semicontinuous convex function, and x ∈ X . Define x λ by

x λ = J λ g x = argmin y ∈ X { λ g ( y ) + ‖ y − x ‖ 2 }

for a positive number λ . If there exists { μ n } ⊂ ] 0 , ∞ [ such that μ n → ∞ and { d ( x μ n , x ) } is bounded, then argmin g ≠ ∅ and

lim λ → ∞ x λ = P argmin g x ,

where P K is a metric projection onto K for a nonempty closed convex subset K of X.

Theorem 1.2

(See [1]) Let X be a Hilbert space, g : X → ] − ∞ , ∞ ] a proper lower semicontinuous convex function, and x ∈ X . Define x λ by

x λ = J λ g x = argmin y ∈ X { λ g ( y ) + d ( x , y ) 2 }

for a positive number λ . Then

lim λ → 0 x λ = P dom g ¯ x .

Furthermore, the convergence of resolvents of maximal monotone operators has been considered in many papers. For example, see [2,3, 4,5,6, 7,8].

The equilibrium problem is an important problem which contains optimization problems, complementarity problems, fixed point problems, variational inequalities, and Nash equilibria. Let K be a set and f : K × K → R a bifunction. An equilibrium problem for a bifunction f : K × K → R is a problem of finding z ∈ K such that f ( z , y ) ≥ 0 for all y ∈ K . Resolvents of equilibrium problems are considered to understand this problem and to find their solutions. For example, for equilibrium problems in CAT(0) spaces, a resolvent is defined and studied in [9]. In CAT (1) spaces, it is studied in [10].

In this article, we study properties of resolvents of equilibrium problems with positive parameter λ . In Section 2, we introduce basic concepts. In Section 3, we consider asymptotic behavior and continuity of a resolvent on a complete CAT ( 0 ) space for λ . In Section 4, we study similar properties of resolvent on a complete CAT(1) space. In Section 5, we discuss the relationship between resolvent of a convex function and that of equilibrium problems.

2 Preliminaries

Let ( X , d ) be a metric space and x , y ∈ X . A mapping c x y : [ 0 , l ] → X is called a geodesic with endpoints x , y if it satisfies

c ( 0 ) = x , c ( l ) = y , and ∣ c ( u ) − c ( v ) ∣ = d ( u , v ) for all u , v ∈ [ 0 , l ] ,

where l is a distance between x and y . X is called a geodesic space if there exists a geodesic c x y for any two points x , y ∈ X . Furthermore, X is called a uniquely geodesic space if a geodesic exists uniquely for each two points. Let X be a uniquely geodesic space. Then the geodesic segment joining x and y is defined as the image of the geodesic c x y and denoted by [ x , y ] . That is, [ x , y ] = c x y ( [ 0 , l ] ) . Also, we define convex combination t x ⊕ ( 1 − t ) y for t ∈ [ 0 , 1 ] as follows:

t x ⊕ ( 1 − t ) y = c x y ( ( 1 − t ) l ) .

Let x 1 , x 2 , x 3 ∈ X . A geodesic triangle △ ( x 1 , x 2 , x 3 ) is defined by

△ ( x 1 , x 2 , x 3 ) = [ x 1 , x 2 ] ∪ [ x 2 , x 3 ] ∪ [ x 3 , x 1 ] .

For this triangle, a comparison triangle △ ¯ ( x ¯ 1 , x ¯ 2 , x ¯ 3 ) ⊂ E 2 is a triangle with the same side length as △ ( x 1 , x 2 , x 3 ) in the two-dimensional Euclidian space. Let p be a point on [ x i , x j ] ( i , j ∈ { 1 , 2 , 3 } , i ≠ j ) . For p , a comparison point p ¯ is a point on [ x ¯ i , x ¯ j ] with d ( p ¯ , x ¯ i ) E 2 = d ( p , x i ) . If for any x 1 , x 2 , x 3 ∈ X , p , q ∈ △ ( x 1 , x 2 , x 3 ) , and their comparison points p ¯ , q ¯ ∈ △ ¯ ( x ¯ 1 , x ¯ 2 , x ¯ 3 ) , it holds that

d ( p ¯ , q ¯ ) E 2 ≤ d ( p , q ) ,

then X is called a CAT(0) space. We know the following equivalence condition. A uniquely geodesic space X is a CAT(0) space if and only if it holds that

d ( t x ⊕ ( 1 − t ) y , z ) 2 ≤ t d ( x , z ) 2 + ( 1 − t ) d ( y , z ) 2 − t ( 1 − t ) d ( x , y ) 2

for all x , y , z ∈ X and t ∈ [ 0 , 1 ] . This inequality is called a parallelogram law in CAT(0) space.

Next, we define CAT(1) space. Let X be a uniquely geodesic space and x 1 , x 2 , x 3 ∈ X with d ( x 1 , x 2 ) + d ( x 2 , x 3 ) + d ( x 3 , x 1 ) < 2 π . We consider a geodesic triangle △ ( x 1 , x 2 , x 3 ) , its comparison triangle △ ¯ ( x ¯ 1 , x ¯ 2 , x ¯ 3 ) in the two-dimensional unit sphere S 2 , and a comparison point in a similar way. If for any x 1 , x 2 , x 3 ∈ X with d ( x 1 , x 2 ) + d ( x 2 , x 3 ) + d ( x 3 , x 1 ) < 2 π , p , q ∈ △ ( x 1 , x 2 , x 3 ) , and their comparison points p ¯ , q ¯ ∈ △ ¯ ( x ¯ 1 , x ¯ 2 , x ¯ 3 ) , it holds that

d ( p ¯ , q ¯ ) S 2 ≤ d ( p , q ) ,

then X is called a CAT(1) space. We also know that a uniquely geodesic space X is CAT ( 1 ) space if and only if for all x , y , z ∈ X with d ( x , y ) + d ( y , z ) + d ( z , x ) < 2 π and t ∈ [ 0 , 1 ] , it holds that

cos d ( t x ⊕ ( 1 − t ) y , z ) sin d ( x , y ) ≥ cos d ( x , z ) sin t d ( x , y ) + cos d ( y , z ) sin ( 1 − t ) d ( x , y ) ,

which is called a parallelogram law in CAT(1) space. From the above, considering the case that t = 1 / 2 , we have

cos d 1 2 x ⊕ 1 2 y , z cos d ( x , y ) 2 ≥ 1 2 cos d ( x , z ) + 1 2 cos d ( y , z )

if X is a CAT ( 1 ) space. This inequality is equivalent to

− log cos d 1 2 x ⊕ 1 2 y , z ≤ − 1 2 log cos d ( x , z ) − 1 2 log cos d ( y , z ) + log cos d ( x , y ) 2 .

A CAT(1) space X is said to be admissible if

d ( x , y ) < π 2

for all x , y ∈ X .

Let X be a uniquely geodesic space. A function g : X → R is said to be convex if

g ( t x ⊕ ( 1 − t ) y ) ≤ t g ( x ) + ( 1 − t ) g ( y )

for all x , y ∈ X and t ∈ [ 0 , 1 ] . Furthermore, g : X → R is said to be upper hemicontinuous if

limsup t → 0 g ( t x ⊕ ( 1 − t ) y ) ≤ g ( y )

for all x , y ∈ X . We say a subset C of X is convex if t x ⊕ ( 1 − t ) y ∈ C whenever x , y ∈ C . For a subset A of X , the convex hull co A of A is defined by

co A = ⋃ i = 1 ∞ A i ,

where A 1 = A and A i + 1 = { t x ⊕ ( 1 − t ) y ∣ x , y ∈ A i , t ∈ [ 0 , 1 ] } for i ∈ N . We say X has the convex hull finite property if for every finite subset A of X , continuous self-mapping T on co A ¯ has a fixed point. For the example of geodesic spaces satisfying this property, see [10]. Let X be a complete CAT(0) space or a complete admissible CAT(1) space, C a nonempty closed convex subset of X , and x ∈ X . Then, there exists a unique nearest point p ∈ C , which satisfies

d ( x , p ) ≤ d ( x , y )

for all y ∈ C . Therefore, for a closed convex subset C of X , we can define the metric projection P C onto C by

P C x = argmin y ∈ C d ( x , y )

for x ∈ X . The metric projection P C has the following properties:

in a complete CAT(0) space
d ( x , P C x ) 2 + d ( P C x , y ) 2 ≤ d ( x , y ) 2
for all x ∈ X and y ∈ C ;
in a complete admissible CAT(1) space,
cos d ( x , P C x ) cos d ( P C x , y ) ≤ cos d ( x , y )
for all x ∈ X and y ∈ C .

Let X be a complete CAT(0) space or complete admissible CAT(1) space and K a nonempty closed convex subset of X . We consider an equilibrium problem for a bifunction f : K × K → R . We denote the set of solutions of an equilibrium problem for f by Equil f . That is,

Equil f = { z ∈ K ∣ inf y ∈ K f ( z , y ) ≥ 0 } .

In what follows, we suppose f satisfies the following four conditions:

f ( x , x ) = 0 for all x ∈ K ;
f ( x , y ) + f ( y , x ) ≤ 0 for all x , y ∈ K ;
f ( x , ⋅ ) : K → R is convex and lower semicontinuous for all x ∈ K ;
f ( ⋅ , x ) : K → R is upper hemicontinuous for all x ∈ K .

Minimization problem is an example of the equilibrium problems. Let g : K → R be a convex lower semicontinuous function. Define f : K × K → R by

f ( z , y ) = g ( y ) − g ( z )

for all y , z ∈ K . Then f satisfies (E1)–(E4) and the solutions of the equilibrium problem for f coincide with the solutions of the minimization problem for g , and hence

Equil f = argmin g .

Let X be a complete CAT(0) space having the convex hull finite property, K a nonempty closed convex subset of X , and f a bifunction K × K → R satisfying (E1)–(E4). A resolvent Q f of an equilibrium problem for f is defined by

Q f x = { z ∈ K ∣ inf y ∈ K ( f ( z , y ) + d ( y , x ) 2 − d ( z , x ) 2 ) ≥ 0 }

for x ∈ X . Then Q f x is a singleton; see [9]. Hence, we can consider Q f a single-valued mapping. We know that Q f is nonexpansive and the set of fixed points coincides with the solution set of the equilibrium problem. Therefore, approximation methods of a fixed point of nonexpansive mapping can be diverted to approximate a solution to the original problem; for instance, see [11,12,13, 14,15].

Let X be a complete admissible CAT(1) space having the convex hull finite property, K a nonempty closed convex subset of X , and f a bifunction K × K → R satisfying (E1)–(E4). A resolvent R f of equilibrium problems for f is defined by

R f x = { z ∈ K ∣ inf y ∈ K ( f ( z , y ) − log cos d ( y , x ) + log cos d ( z , x ) ) ≥ 0 }

for x ∈ X . Similarly, we know R f x is to be singleton (see [10]), and R f can be considered as a single-valued mapping.

3 Resolvent on CAT(0) space

In a complete CAT(0) space, we consider equilibrium problems for a bifunction and its resolvent. Let X be a complete CAT(0) space having the convex hull finite property and f a bifunction on K satisfying (E1)–(E4). For a positive number λ , we define a resolvent Q λ f as follows:

Q λ f x = { z ∈ K ∣ inf y ∈ K ( λ f ( z , y ) + ( d ( y , x ) 2 − d ( z , x ) 2 ) ) ≥ 0 }

for x ∈ X . We consider the asymptotic behavior and continuity of this resolvent J λ f x for λ . First we introduce useful lemmas to discuss the resolvent with parameter λ .

Lemma 3.1

Let X be a complete CAT(0) space having the convex hull finite property, K a nonempty closed convex subset of X, and f : K × K → R a bifunction satisfying (E1)–(E4). Let x ∈ X and λ , μ ∈ R with 0 ≤ λ ≤ μ . Then

d ( Q λ f x , x ) ≤ d ( Q μ f x , x ) , f ( Q μ f x , Q λ f x ) ≥ 0 , and f ( Q λ f x , Q μ f x ) ≤ 0 .

Proof

If λ = μ , it is obvious. Suppose λ < μ . Put x λ = Q λ f x and x μ = Q μ f x . From the definition of the resolvent, we have

(1) 0 ≤ f ( x λ , x μ ) + 1 λ { d ( x μ , x ) 2 − d ( x λ , x ) 2 } ,

(2) 0 ≤ f ( x μ , x λ ) + 1 μ { d ( x λ , x ) 2 − d ( x μ , x ) 2 } .

From these inequalities and (E2), we obtain

0 ≤ f ( x λ , x μ ) + f ( x μ , x λ ) + 1 λ − 1 μ { d ( x μ , x ) 2 − d ( x λ , x ) 2 } ≤ 0 + 1 λ − 1 μ { d ( x μ , x ) 2 − d ( x λ , x ) 2 } .

Since 1 / λ > 1 / μ , we obtain

d ( x λ , x ) ≤ d ( x μ , x ) .

From (1) and (E2), we have

0 ≤ f ( x μ , x λ ) + 1 μ { d ( x λ , x ) 2 − d ( x μ , x ) 2 } ≤ f ( x μ , x λ ) ≤ − f ( x λ , x μ ) .

It completes the proof.□

Lemma 3.2

Let X be a complete CAT(0) space having the convex hull finite property, K a nonempty closed convex subset of X, and f : K × K → R a bifunction satisfying (E1)–(E4). Let x ∈ X , λ , μ ∈ R with 0 < λ ≤ μ , and t ∈ [ 0 , 1 ] . Then

d ( Q λ f x , x ) ≤ d ( t Q λ f x + ( 1 − t ) Q μ f x , x ) ≤ d ( Q μ f x , x ) .

Proof

Put x λ = Q λ f x and x μ = Q μ f x . Since K is convex, we have t x λ ⊕ ( 1 − t ) x μ ∈ K for t ∈ [ 0 , 1 ] . From (E3), (E1), and Lemma 3.1, we have

0 ≤ λ f ( x λ , t x λ ⊕ ( 1 − t ) x μ ) + d ( t x λ ⊕ ( 1 − t ) x μ , x ) 2 − d ( x λ , x ) 2 ≤ λ t f ( x λ , x λ ) + λ ( 1 − t ) f ( x λ , x μ ) + d ( t x λ ⊕ ( 1 − t ) x μ , x ) 2 − d ( x λ , x ) 2 ≤ d ( t x λ ⊕ ( 1 − t ) x μ , x ) 2 − d ( x λ , x ) 2 .

Therefore, we obtain

d ( x λ , x ) ≤ d ( t x λ ⊕ ( 1 − t ) x μ , x ) ≤ t d ( x λ , x ) + ( 1 − t ) d ( x μ , x ) ≤ d ( x μ , x ) .

This is the desired result.□

Lemma 3.3

d ( Q λ f x , Q μ f x ) 2 ≤ d ( Q μ f x , x ) 2 − d ( Q λ f x , x ) 2 .

Proof

Put x λ = Q λ x and x μ = Q μ x . By Lemma 3.2 and the parallelogram law, we have

d ( x λ , x ) 2 ≤ d ( t x λ ⊕ ( 1 − t ) x μ , x ) 2 ≤ t d ( x λ , x ) 2 + ( 1 − t ) d ( x μ , x ) 2 − t ( 1 − t ) d ( x λ , x μ ) 2 ,

and hence

t ( 1 − t ) d ( x λ , x μ ) 2 ≤ ( 1 − t ) d ( x μ , x ) 2 − ( 1 − t ) d ( x λ , x ) 2 .

Dividing by 1 − t and letting t → 1 , we obtain

d ( x λ , x μ ) 2 ≤ d ( x μ , x ) 2 − d ( x λ , x ) 2 ,

which is the desired result.□

From these lemmas, we consider the asymptotic behavior of resolvent at λ → ∞ .

Theorem 3.4

x λ = Q λ f x = { z ∈ K ∣ inf y ∈ K ( λ f ( z , y ) + d ( y , x ) 2 − d ( z , x ) 2 ) ≥ 0 }

for a positive number λ . If there exists { μ n } ⊂ ] 0 , ∞ [ such that μ n → ∞ and { d ( x μ n , x ) } is bounded, then Equil f ≠ ∅ and

lim λ → ∞ x λ = P Equil f x ,

where P K is a metric projection onto K for a nonempty closed convex subset K of X .

Proof

Let { λ n } , { μ n } be positive sequences diverging to ∞ . Assume that { d ( x μ n , x ) } is bounded and { λ n } is increasing. Then { d ( x λ n , x ) } is bounded. In fact, suppose { d ( x λ n , x ) } is not bounded. Since { d ( x μ n , x ) } is bounded, there exists M > 0 such that

d ( x μ n , x ) ≤ M

for all n ∈ N . On the other hand, there exists a subsequence { λ n i } of { λ n } such that

d ( x λ n i , x ) > M

for all i ∈ N . Since { μ n } diverges to ∞ , we can find l ∈ N , such that λ n 1 ≤ μ l . From Lemma 3.2, we obtain

M < d ( x λ n 1 , x ) ≤ d ( x μ l , x ) ≤ M .

This is a contradiction. Put x n = x λ n for n ∈ N . Let n , m ∈ N satisfy n ≤ m . From Lemma 3.3, we have

d ( x n , x m ) 2 ≤ d ( x m , x ) 2 − d ( x n , x ) 2 .

By Lemma 3.2, { d ( x n , x ) } is bounded and increasing. Therefore, { x n } is a Cauchy sequence. Since { x n } ⊂ K and K is closed, there exists p ∈ K such that x n → p . By the definition of x n , we obtain

0 ≤ f ( x n , y ) + 1 λ n { d ( y , x ) 2 − d ( x n , x ) 2 }

for all y ∈ X . Since λ n → ∞ as n → ∞ , from (E2) and (E3), we have

0 ≤ limsup n → ∞ f ( x n , y ) ≤ limsup n → ∞ ( − f ( y , x n ) ) = − liminf n → ∞ f ( y , x n ) ≤ − f ( y , p ) .

That is, f ( y , p ) ≤ 0 for all y ∈ K . Let w ∈ K and t ∈ ] 0 , 1 [ . Since K is closed, f ( t w ⊕ ( 1 − t ) p , p ) ≤ 0 . From (E1) and (E2), we have

0 = f ( t w ⊕ ( 1 − t ) p , t w ⊕ ( 1 − t ) p ) ≤ t f ( t w ⊕ ( 1 − t ) p , w ) + ( 1 − t ) f ( t w ⊕ ( 1 − t ) p , p ) ≤ t f ( t w ⊕ ( 1 − t ) p , w ) .

Dividing by t and letting t → 0 , we have 0 ≤ f ( p , w ) for all w ∈ K from (E4). Therefore, we obtain

p ∈ Equil f ≠ ∅ .

For all z ∈ Equil f , we have

0 ≤ λ n f ( x n , z ) + d ( z , x ) 2 − d ( x n , x ) 2 ≤ − λ n f ( z , x n ) + d ( z , x ) 2 − d ( x n , x ) 2 ≤ d ( z , x ) 2 − d ( x n , x ) 2 ,

and hence d ( x n , x ) ≤ d ( z , x ) . Letting n → ∞ , we obtain

d ( p , x ) ≤ liminf n → ∞ d ( x n , x ) ≤ d ( z , x )

for all z ∈ Equil f . This implies p = P Equil f x . From the above, we know that x n → P Equil f x for every increasing sequence { λ n } diverging to infinity. Hence, we obtain

lim λ → ∞ x λ = P Equil f x .

This is the desired result.□

Furthermore, we discuss the asymptotic behavior of resolvent when λ → 0 .

Theorem 3.5

x λ = Q λ f x = { z ∈ K ∣ inf y ∈ K ( λ f ( z , y ) + d ( y , x ) 2 − d ( z , x ) 2 ) ≥ 0 }

for a positive number λ . Then

lim λ → 0 x λ = P K x .

Proof

Let { λ n } be a decreasing positive sequence converging to 0 and put x n = x λ n for n ∈ N . By Lemma 3.3,

d ( x n , x m ) 2 ≤ ∣ d ( x m , x ) 2 − d ( x n , x ) 2 ∣

for n , m ∈ N . On the other hand, from Lemma 3.2, { d ( x n , x ) } is decreasing and bounded below. Therefore, { x n } is a Cauchy sequence, and hence { x n } converges to some p ∈ K . By the definition of the resolvent, we have

0 ≤ λ n f ( x n , y ) + d ( y , x ) 2 − d ( x n , x ) 2 ≤ − λ n f ( y , x n ) + d ( y , x ) 2 − d ( x n , x ) 2 .

From the lower semicontinuity of f for the second argument, letting n → ∞ , we obtain 0 ≤ d ( y , x ) 2 − d ( p , x ) 2 , and hence

d ( p , x ) ≤ d ( y , x )

for all y ∈ X . This implies p = P K x . Hence, for every decreasing sequence { λ n } convergent to 0, { x n } converges to P K x . From this fact, we conclude that

lim λ → 0 x λ = P K x ,

which is the desired result.□

Next, we consider the continuity of the function λ ↦ Q λ f x .

Theorem 3.6

x λ = Q λ f x = { z ∈ K ∣ inf y ∈ K ( λ f ( z , y ) + d ( y , x ) 2 − d ( z , x ) 2 ) ≥ 0 }

for a positive number λ . Then

lim λ → λ 0 x λ = x λ 0 ,

where λ 0 is a positive number.

Proof

Let { λ n } be an increasing or decreasing positive sequence converging to λ 0 and put x n = x λ n for n ∈ N . We know by Lemma 3.3,

d ( x n , x m ) 2 ≤ ∣ d ( x m , x ) 2 − d ( x n , x ) 2 ∣

for n , m ∈ N and { d ( x n , x ) } is convergent. Hence, x n is a Cauchy sequence and we can find x 0 ∈ K such that x n → x 0 . From the definition of x n , we obtain

0 ≤ λ n f ( x n , y ) + d ( y , x ) 2 − d ( x n , x ) 2 ≤ − λ n f ( y , x n ) + d ( y , x ) 2 − d ( x n , x ) 2

for all y ∈ K . By the lower semicontinuity of f ( y , ⋅ ) , we have

limsup n → ∞ ( − f ( y , x n ) ) = − liminf n → ∞ f ( y , x n ) ≤ − f ( y , x 0 ) ,

and hence

0 ≤ limsup n → ∞ { − λ n f ( x n , y ) + d ( y , x ) 2 − d ( x n , x ) 2 } ≤ λ 0 limsup n → ∞ ( − f ( y , x n ) ) + d ( y , x ) 2 − d ( x 0 , x ) 2 ≤ − λ 0 f ( y , x 0 ) + d ( y , x ) 2 − d ( x 0 , x ) 2 ,

for all y ∈ K . Let w ∈ K and t ∈ ] 0 , 1 [ . From (E1) and (E2), we have

0 = λ 0 f ( t w ⊕ ( 1 − t ) x 0 , t w ⊕ ( 1 − t ) x 0 ) ≤ λ 0 t f ( t w ⊕ ( 1 − t ) x 0 , w ) + λ 0 ( 1 − t ) f ( t w ⊕ ( 1 − t ) x 0 , x 0 ) ≤ λ 0 t f ( t w ⊕ ( 1 − t ) x 0 , w ) + ( 1 − t ) { d ( t w ⊕ ( 1 − t ) x 0 , x ) 2 − d ( x 0 , x ) 2 } ≤ λ 0 t f ( t w ⊕ ( 1 − t ) x 0 , w ) + ( 1 − t ) { t d ( w , x ) 2 + ( 1 − t ) d ( x 0 , x ) 2 − d ( x 0 , x ) 2 } = λ 0 t f ( t w ⊕ ( 1 − t ) x 0 , w ) + t ( 1 − t ) d ( w , x ) 2 − t ( 1 − t ) d ( x 0 , x ) 2 .

Dividing by t and letting t → 0 , we obtain

0 ≤ λ 0 f ( x 0 , w ) + d ( w , x ) 2 − d ( x 0 , x ) 2

for all w ∈ K . This implies x 0 = x λ 0 , and hence we conclude that

lim λ → λ 0 x λ = x λ 0 .

This completes the proof.□

4 Resolvent on CAT(1) space

In a complete admissible CAT(1) space, we consider a resolvent with parameter λ and its convergence. Let X be a complete admissible CAT(1) space having the convex hull finite property and f a bifunction on K satisfying (E1)–(E4). For a positive number λ , we define a resolvent R λ f as follows:

R λ f x = { z ∈ K ∣ inf y ∈ K ( λ f ( z , y ) − log cos d ( y , x ) + log cos d ( z , x ) ) ≥ 0 }

for x ∈ X . We describe some lemmas and show the result for CAT(1) spaces.

Lemma 4.1

Let X be a complete admissible CAT(1) space satisfying the convex hull finite property, K a nonempty closed convex subset of X, and f : K × K → R a bifunction satisfying (E1)–(E4). Let x ∈ X and λ , μ ∈ R with 0 < λ ≤ μ . Then

d ( Q λ f x , x ) ≤ d ( Q μ f x , x ) , f ( Q μ f x , Q λ f x ) ≥ 0 , and f ( Q λ f x , Q μ f x ) ≤ 0 .

Proof

Put x λ = R λ f x and x μ = R μ f x . Then, we have

(3) 0 ≤ f ( x λ , x μ ) + 1 λ { − log cos d ( x μ , x ) + log cos d ( x λ , x ) }

and

(4) 0 ≤ f ( x μ , x λ ) + 1 μ { − log cos d ( x λ , x ) + log cos d ( x μ , x ) } .

From these inequalities and (E2), we obtain

0 ≤ f ( x λ , x μ ) + f ( x μ , x λ ) + 1 λ − 1 μ { − log cos d ( x μ , x ) + log cos d ( x λ , x ) } ≤ 0 + 1 λ − 1 μ { − log cos d ( x μ , x ) + log cos d ( x λ , x ) } .

Assume λ < μ . From the inequality above, we obtain

− log cos d ( x λ , x ) ≤ − log cos d ( x μ , x ) ,

which is equivalent to d ( x λ , x ) ≤ d ( x μ , x ) . Therefore, we obtain

0 ≤ f ( x μ , x λ ) + 1 μ { − log cos d ( x λ , x ) + log cos d ( x μ , x ) 2 } ≤ f ( x μ , x λ ) ≤ − f ( x λ , x μ ) .

It completes the proof.□

Lemma 4.2

d ( R λ x , x ) ≤ d ( t R λ x + ( 1 − t ) R μ x , x ) ≤ d ( R μ x , x ) .

Proof

Put x λ = R λ x and x μ = R μ x . From Lemma 4.1, we have

0 ≤ λ f ( x λ , t x λ ⊕ ( 1 − t ) x μ ) − log cos d ( t x λ ⊕ ( 1 − t ) x μ , x ) + log cos d ( x λ , x ) ≤ λ t f ( x λ , x λ ) + λ ( 1 − t ) f ( x λ , x μ ) − log cos d ( t x λ ⊕ ( 1 − t ) x μ , x ) + log cos d ( x λ , x ) ≤ − log cos d ( t x λ ⊕ ( 1 − t ) x μ , x ) + log cos d ( x λ , x ) ,

and hence − log cos d ( x λ , x ) ≤ − log cos d ( t x λ ⊕ ( 1 − t ) x μ , x ) . From the convexity of the function − log cos d ( ⋅ , x ) , we have

− log cos d ( x λ , x ) ≤ − log cos d ( t x λ ⊕ ( 1 − t ) x μ , x ) ≤ − t log cos d ( x λ , x ) − ( 1 − t ) log cos d ( x μ , x ) ≤ − log cos d ( x μ , x ) .

This implies

d ( x λ , x ) ≤ d ( t x λ ⊕ ( 1 − t ) x μ , x ) ≤ d ( x μ , x ) ,

which is the desired result.□

Lemma 4.3

− 2 log cos d ( R λ f x , R μ f x ) 2 ≤ − log cos d ( R μ f x , x ) + log cos d ( R λ f x , x ) .

Proof

Put x λ = R λ f x and x μ = R μ f x . By Lemma 4.2 and the parallelogram law at t = 1 / 2 , we have

− log cos d ( x λ , x ) ≤ − log cos d 1 2 x λ ⊕ 1 2 x μ , x ≤ − 1 2 log cos d ( x λ , x ) − 1 2 log cos d ( x μ , x ) + log cos d ( x λ , x μ ) 2 ,

and hence

− log cos d ( x λ , x μ ) 2 ≤ − 1 2 log cos d ( x μ , x ) + 1 2 log cos d ( x λ , x ) ,

which is the desired result.□

Theorem 4.4

x λ = R λ f x = { z ∈ K ∣ inf y ∈ K ( λ f ( z , y ) − log cos d ( y , x ) + log cos d ( z , x ) ) ≥ 0 }

for a positive number λ . If there exists { μ n } ⊂ ] 0 , ∞ [ such that μ n → ∞ and sup d ( x μ n , x ) < π / 2 , then Equil f ≠ ∅ and

lim λ → ∞ x λ = P Equil f x ,

where P K is a metric projection onto a nonempty closed convex subset K of X.

Proof

Suppose that there exists a sequence { μ n } which satisfies the assumption above. In a similar way to the proof of Theorem 3.4, we have

sup n ∈ N d ( Q λ n x , x ) < π 2

for all positive increasing sequence { λ n } . Put x n = x λ n for n ∈ N . Let n , m ∈ N with n ≤ m . From Lemma 4.3, we have

− 2 log cos d ( x n , x m ) 2 ≤ − log cos d ( x m , x ) + log cos d ( x n , x ) .

By Lemma 4.2, { − log cos d ( x n , x ) } is convergent and thus { x n } is Cauchy. From the completeness of K , there exists p ∈ K such that x n → p . By the definition of x n , we obtain

0 ≤ f ( x n , y ) + 1 λ n { − log cos d ( y , x ) + log cos d ( x n , x ) } ≤ − f ( y , x n ) + 1 λ n { − log cos d ( y , x ) + log cos d ( x n , x ) }

for all y ∈ X . From (E3), we obtain

0 ≤ limsup n → ∞ f ( x n , y ) ≤ − f ( y , p ) .

This implies

f ( y , p ) ≤ 0

for all y ∈ K . Let w ∈ K and t ∈ ] 0 , 1 [ . Put w t = t w ⊕ ( 1 − t ) p . Then we obtain f ( w t , p ) ≤ 0 . From (E1) and (E2), we have

0 = f ( w t , w t ) ≤ t f ( w t , w ) + ( 1 − t ) f ( w t , p ) ≤ t f ( w t , w ) .

Dividing by t and letting t → 0 , we have

0 ≤ f ( p , w )

for all w ∈ K from (E4). Therefore, p is a solution of the equilibrium problem for f . For all z ∈ Equil f , we have

0 ≤ λ n f ( x n , z ) − log cos d ( z , x ) + log cos d ( x n , x ) ≤ − λ n f ( z , x n ) − log cos d ( z , x ) + log cos d ( x n , x ) ≤ − log cos d ( z , x ) + log cos d ( x n , x ) ,

and hence d ( x n , x ) ≤ d ( z , x ) . From the lower semicontinuity of d ( ⋅ , x ) , we obtain

d ( p , x ) ≤ liminf n → ∞ d ( x n , x ) ≤ d ( z , x )

for all z ∈ Equil f and this means p = P Equil f x . We have x n → P Equil f x for every increasing sequence { λ n } diverging to infinity, and hence we obtain

lim λ → ∞ x λ = P Equil f x .

This is the desired result.□

Theorem 4.5

x λ = R λ f x = { z ∈ K ∣ inf y ∈ K ( λ f ( z , y ) − log cos d ( y , x ) + log cos d ( z , x ) ) ≥ 0 }

for a positive number λ . Then

lim λ → 0 x λ = P K x .

Proof

Let { λ n } be a decreasing positive sequence converging to 0 and put x n = x λ n for n ∈ N . By Lemma 4.3,

− 2 log cos d ( x n , x m ) 2 ≤ ∣ − log cos d ( x m , x ) + log cos d ( x n , x ) ∣ ,

for n , m ∈ N . On the other hand, from Lemma 4.2, { − log cos d ( x n , x ) } is decreasing and bounded below. Therefore, { x n } converges to p ∈ K since K is closed. By the definition of the resolvent, we have

0 ≤ λ n f ( x n , y ) − log cos d ( y , x ) + log cos d ( x n , x ) ≤ − λ n f ( y , x n ) − log cos d ( y , x ) + log cos d ( x n , x ) .

From the lower semicontinuity of f for the second argument, letting n → ∞ , we obtain 0 ≤ − log cos d ( y , x ) + log cos d ( x n , x ) , and hence

d ( p , x ) ≤ d ( y , x )

for all y ∈ X . This implies p = P K x . Therefore, for every decreasing sequence { λ n } convergent to 0, { x n } converges to P K x . Thus, we conclude that

lim λ → 0 x λ = P K x ,

which is the desired result.□

Theorem 4.6

x λ = R λ f x = { z ∈ K ∣ inf y ∈ K ( λ f ( z , y ) − log cos d ( y , x ) + log cos d ( z , x ) ) ≥ 0 }

for a positive number λ . Then

lim λ → λ 0 x λ = x λ 0 ,

where λ 0 is a positive number.

Proof

Let { λ n } be an increasing or decreasing positive sequence converging to λ 0 and put x n = x λ n for n ∈ N . By Lemma 4.3,

− 2 log cos d ( x n , x m ) ≤ ∣ − log cos d ( x m , x ) + log cos d ( x n , x ) ∣

for n , m ∈ N and { − log cos d ( x n , x ) } is convergent. Hence, we can find x 0 ∈ K such that x n → x 0 . From the definition of x n , we obtain

0 ≤ λ n f ( x n , y ) + d ( y , x ) 2 − d ( x n , x ) 2 ≤ − λ n f ( y , x n ) + d ( y , x ) 2 − d ( x n , x ) 2 .

By the lower semicontinuity of f ( y , ⋅ ) , we have

0 ≤ limsup n → ∞ { − λ n f ( x n , y ) − log cos d ( y , x ) + log cos d ( x n , x ) } ≤ λ 0 limsup n → ∞ ( − f ( y , x n ) ) − log cos d ( y , x ) + log cos d ( x 0 , x ) ≤ − λ 0 f ( y , x 0 ) − log cos d ( y , x ) + log cos d ( x 0 , x ) .

for all y ∈ K . Let w ∈ K and t ∈ ] 0 , 1 [ . Put w t = t w ⊕ ( 1 − t ) x 0 . From (E1) and (E2), we have

0 = λ 0 f ( w t , w t ) ≤ λ 0 t f ( w t , w ) + λ 0 ( 1 − t ) f ( w t , x 0 ) ≤ λ 0 t f ( w t , w ) + ( 1 − t ) { − log cos d ( w t , x ) + log cos d ( x 0 , x ) } ≤ λ 0 t f ( w t , w ) + ( 1 − t ) { − t log cos d ( w , x ) − ( 1 − t ) log cos d ( x 0 , x ) + log cos d ( x 0 , x ) } = λ 0 t f ( w t , w ) − t ( 1 − t ) log cos d ( w , x ) + t ( 1 − t ) log cos d ( x 0 , x ) .

Dividing by t and letting t → 0 , we obtain

0 ≤ λ 0 f ( x 0 , w ) − log cos d ( w , x ) + log cos d ( x 0 , x )

for all w ∈ K . This implies x 0 = x λ 0 , and hence we conclude that

lim λ → λ 0 x λ = x λ 0 .

This completes the proof.□

5 Applications

We discuss applications of our results to a convergence of resolvent for a convex function in minimization problems.

Let X be a complete CAT(0) space satisfying the convex hull finite property and g : X → ] − ∞ , ∞ ] a lower semicontinuous convex function. The effective domain dom g of g is defined by

dom g = { x ∈ X ∣ g ( x ) ∈ R } .

Furthermore, we define a resolvent J g for g by

J g x = argmin y ∈ X { g ( y ) + d ( y , x ) 2 } ,

for all x ∈ X . In what follows, we suppose dom g is closed. We assume f : dom g × dom g → R satisfies

f ( z , y ) = g ( y ) − g ( z )

for all y , z ∈ K . Then J g = R f . In fact, if z = J g x , it holds that

g ( z ) + d ( z , x ) 2 ≤ g ( y ) + d ( y , x ) 2

for all y ∈ X , and hence

0 ≤ g ( y ) − g ( z ) + d ( y , x 2 ) − d ( z , x ) 2 .

Therefore, z = J g x implies z = R f x , and vice versa.

We remark that it is not easy to calculate the exact value of the resolvent operator, even if a convex function is explicitly given, because it is defined by a solution to another convex minimization problem. However, this problem is much easier to solve than the original minimization problem since the corresponding convex function satisfies more vital conditions, such as coercivity and strict convexity. If the given convex function meets additional requirements, such as differentiability, then the Newton method and other approximation techniques will work well.

From the argument of Section 3, we can derive the following theorems.

Theorem 5.1

(See [16]) Let X be a complete CAT(0) space having the convex hull finite property, g : X → ] − ∞ , ∞ ] a lower semicontinuous convex function, and x ∈ X . Suppose dom f is closed. Define x λ by

x λ = J λ g x = argmin y ∈ X { λ g ( y ) + d ( x , y ) 2 }

for a positive number λ . If there exists { μ n } ⊂ ] 0 , ∞ [ such that μ n → ∞ and { d ( J μ n g x , x ) } is bounded, then Equil f ≠ ∅ and

lim λ → ∞ x λ = P argmin f x .

Theorem 5.2

x λ = J λ g x = argmin y ∈ X { λ g ( y ) + d ( x , y ) 2 }

for a positive number λ . Then

lim λ → 0 x λ = P dom f x .

Theorem 5.3

(See [16]) Let X be a complete CAT(0) space having the convex hull finite property, g : X → ] − ∞ , ∞ ] a lower semicontinuous convex function, and x ∈ X . Suppose dom f is closed. Define x λ by

x λ = J λ g x = argmin y ∈ X { λ g ( y ) + d ( x , y ) 2 }

for a positive number λ . Then

lim λ → λ 0 x λ = x λ 0 ,

where λ 0 is a positive number.

Strictly speaking, in the minimization problems, we need not assume the convex hull finite property to show the convergence of the resolvent for convex functions (see [16]).

We can consider the convergence of resolvent of a convex function on a complete admissible CAT(1) space. Let X be a complete admissible CAT(1) space satisfying the convex hull finite property, g : X → ] − ∞ , ∞ ] a lower semicontinuous convex function, and x ∈ X . We define L λ g x by

L λ g x = argmin x ∈ X { λ g ( y ) − log cos d ( y , x ) }

for x ∈ X . We can obtain similar results to the theorems of CAT ( 0 ) spaces in this section. The problem of asymptotic behavior of convex functions on a complete CAT(1) space is discussed in [17].

Funding information: This work was partially supported by JSPS KAKENHI Grant Number JP21K03316.
Author contributions: The authors have contributed to this work equally. All authors read and approved the final manuscript.
Conflict of interest: The authors declare no conflict of interest.
Data availability statement: Not applicable.

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Received: 2022-07-01

Revised: 2022-11-18

Accepted: 2022-11-24

Published Online: 2023-01-30

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/dema-2022-0187

Keywords for this article

asymptotic behavior; geodesic space; convex function; resolvent

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