Positive coincidence points for a class of nonlinear operators and their applications to matrix equations

Lemma 2.1

Let x , y ∈ P ∘ . If x ≪ y , then for all ε > 0 , B ( y , ε ) ∩ [ x | y ] is a nonempty open set.

Proof

The element λ x + ( 1 − λ ) y belongs to the set B ( y , ε ) ∩ [ x | y ] , if λ ∈ ( 0 , 1 ) and λ ∥ x − y ∥ < ε .□

Definition 2.2

Let f , g : P ∘ → P ∘ be two operators. We say that:

1. f is strictly α -concave if there exists a constant α ∈ ( 0 , 1 ) such that f ( t x ) ≫ t α f x , or equivalently f ( t − 1 x ) ≪ t − α f x , for all t ∈ ( 0 , 1 ) and x ∈ P ∘ .

2. f is strictly ( − α ) -convex if there exists a constant α ∈ ( 0 , 1 ) such that f ( t x ) ≪ t − α f x , or equivalently f ( t − 1 x ) ≫ t α f x , for all t ∈ ( 0 , 1 ) and x ∈ P ∘ .

Definition 2.3

Let g : P ∘ → P ∘ be an operator. We say that g is strictly super-homogeneous if g ( t x ) ≪ t g x , or equivalently g ( t − 1 x ) ≫ t − 1 g x , for all t ∈ ( 0 , 1 ) and x ∈ P ∘ .

Definition 2.4

Let f : P ∘ → P ∘ be an operator. We say that:

1. f is strictly monotone (resp. antitone) if x ≪ y ⇒ f x ≪ f y (resp. f y ≪ f x ).

2. f is strictly reverse monotone (resp. antitone) if f x ≪ f y ⇒ x ≪ y (resp. y ≪ x ).

3. f is strictly inverse monotone if f is invertible and the inverse of f is strictly monotone.

Definition 2.5

Let f , g : P ∘ → P ∘ be two operators. We say that:

1. f is strictly g-monotone if g x ≪ g y ⇒ f x ≪ f y .

2. f is strictly g-antitone if g x ≪ g y ⇒ f y ≪ f x .

Remark 2.6

Every strictly inverse monotone (resp. antitone) operator is strictly reverse monotone (resp. antitone). The concept of strictly monotone (resp. antitone) is a particular case of the notion of strictly g-monotone (resp. strictly g-antitone), where g is the identity operator.

Lemma 2.7

Let f , g : P ∘ → P ∘ be two operators. Assume that g is invertible and its inverse is strictly monotone. Then, if f is strictly monotone (resp. strictly antitone), then f is strictly g-monotone (resp. strictly g-antitone).

Theorem 3.1

Let f , g : P ∘ → P ∘ be operators. Assume that g is surjective and strictly super-homogeneous and that f is strictly α -concave and strictly g-monotone. Then, C ( f , g ) is nonempty and g ( C ( f , g ) ) is a singleton. More precisely, if { ε n } is a decreasing real sequence convergent to zero, then there exist two sequences { u n } and { v n } in P ∘ such that for all n ≥ 1 , we have

then { g u n } and { g v n } converge to g ( C ( f , g ) ) . In addition, f and g have a unique common fixed point if and only if g ( C ( f , g ) ) ∈ C ( f , g ) . In particular, if C ( f , g ) ⊆ C ( f g , g f ) , then f and g have a unique common fixed point.

Theorem 3.2

Let f , g : P ∘ → P ∘ be operators. Assume that g is surjective and strictly super-homogeneous, and that f is strictly ( − α ) -convex and strictly g-antitone. Then, C ( f , g ) is nonempty and g ( C ( f , g ) ) is a singleton. More precisely, if { ε n } is a decreasing real sequence convergent to zero, then there exist two sequences { u n } and { v n } in P ∘ such that for all n ≥ 1 , we have

then { g u n } and { g v n } converge to g ( C ( f , g ) ) . In addition, f and g have a unique common fixed point if and only if g ( C ( f , g ) ) ∈ C ( f , g ) . In particular, if C ( f , g ) ⊆ C ( f g , g f ) , then f and g have a unique common fixed point.

Lemma 3.3

Let f , g : P ∘ → P ∘ be two operators such that g is surjective and f is strictly g-monotone. Assume that there exist x 0 and y 0 in P ∘ such that

Then, for all decreasing real sequences { ε n } converging to zero, there exist two sequences { u n } and { v n } in P ∘ such that (1) and (2) hold.

Proof

Let u 0 = x 0 and v 0 = y 0 . By hypothesis, using the surjectivity of g and Lemma 2.1, there exist u 1 , v 1 ∈ P ∘ such that g u 1 ∈ B ( f u 0 , ε 0 ) ∩ [ g u 0 | f u 0 ] and g v 1 ∈ B ( f v 0 , ε 0 ) ∩ [ f v 0 | g v 0 ] , then the elements u 0 , u 1 , v 0 , v 1 satisfy g u 0 ≪ g u 1 ≪ f u 0 ≪ f v 0 ≪ g v 1 ≪ g v 0 . As f is strictly g-monotone, we obtain g u 0 ≪ g u 1 ≪ f u 0 ≪ f u 1 ≪ f v 1 ≪ f v 0 ≪ g v 1 ≪ g v 0 . Using again the surjectivity of g there exist u 2 , v 2 such that g u 2 ∈ B ( f u 1 , ε 1 ) ∩ [ f u 0 | f u 1 ] and g v 2 ∈ B ( f v 1 , ε 1 ) ∩ [ f v 1 | f v 0 ] . By construction, the elements u 0 , u 1 , u 2 , v 0 , v 1 , v 2 satisfy (1) and (2) for n = 1 . For the induction step, assume that until some positive integer n and for all k ≤ n , there exist { u k } , { v k } in P ∘ satisfying (1) and (2). Hence, it follows that

As f is strictly g-monotone, we get

Combining (6) and (7), we obtain

As before, from the surjectivity of g, there exist u n + 1 , v n + 1 ∈ P ∘ such that g u n + 1 ∈ B ( f u n , ε n ) ∩ [ f u n − 1 | f u n ] and g v n + 1 ∈ B ( f v n , ε n ) ∩ [ f v n | f v n − 1 ] . Consequently, (1) and (2) hold for all n.□

Lemma 3.4

Let f , g : P ∘ → P ∘ be two operators such that g is surjective and f is g-antitone. Assume that there exist x 0 and y 0 in P ∘ such that

Then, for all decreasing real sequences { ε n } convergent to zero, there exist two sequences { u n } , { v n } in P ∘ such that (3) and (4) hold.

Proof

Let u 0 = x 0 and v 0 = y 0 . By hypothesis, using the surjectivity of g and Lemma 2.1, there exist u 1 , v 1 ∈ P ∘ such that g u 1 ∈ B ( f v 0 , ε 0 ) ∩ [ g u 0 | f v 0 ] and g v 1 ∈ B ( f u 0 , ε 0 ) ∩ [ f u 0 | g v 0 ] , then we have u 0 , u 1 , v 0 , v 1 satisfy g u 0 ≪ g u 1 ≪ f v 0 ≪ f u 0 ≪ g v 1 ≪ g v 0 . As f is strictly g-antitone, we obtain g u 0 ≪ g u 1 ≪ f v 0 ≪ f v 1 ≪ f u 1 ≪ f u 0 ≪ g v 1 ≪ g v 0 . Using again the surjectivity of g there exist u 2 , v 2 such that g u 2 ∈ B ( f v 1 , ε 1 ) ∩ [ f v 0 | f v 1 ] and g v 2 ∈ B ( f u 1 , ε 1 ) ∩ [ f u 1 | f u 0 ] . By construction, the elements u 0 , u 1 , u 2 , v 0 , v 1 , v 2 satisfy (3) and (4) for n = 1 . For the induction step, assume that until some positive integer n and for all k ≤ n − 1 , there exist { u k } , { v k } in P ∘ satisfying (3) and (4). Hence, it follows that

As f is strictly g-antitone, we get

Combining (10) and (11), we obtain

As before, from the surjectivity of g, there exist u n + 1 , v n + 1 ∈ P ∘ such that g u n + 1 ∈ B ( f v n , ε n ) ∩ [ f v n − 1 | f v n ] and g v n + 1 ∈ B ( f u n , ε n ) ∩ [ f u n | f u n − 1 ] . Consequently, (3) and (4) hold for all n.□

Lemma 3.5

Let f , g : P ∘ → P ∘ be two operators such that g is surjective and strictly super-homogeneous and f is strictly α -concave and strictly g-monotone (resp. strictly ( − α ) -convex and strictly g-antitone). Then, there exist x 0 , y 0 ∈ P ∘ satisfying (5) (resp. (9)).

Proof

Let h ∈ P ∘ . Choose t 0 ∈ ( 0 , 1 ) sufficiently small such that

Assume that g is strictly super-homogeneous. For x 0 = t 0 h and y 0 = t 0 − 1 h , we deduce that

which implies that

Since f is strictly α -concave, it follows from (13) that

Now, using (14) combined with the fact that f is strictly g-monotone, we obtain (5). Assume now that f is strictly g-antitone and strictly ( − α ) -convex. Then, it follows from (13) that

Now, from (14) and the fact that f is strictly g-antitone, we obtain (9).□

Lemma 3.6

Let f , g : P ∘ → P ∘ be two operators. Assume that g is strictly super-homogeneous and f is strictly α -concave and strictly g-monotone. Let { u n } , { v n } be sequences satisfying (1) and (2) and t n ≔ sup { 0 < t ≤ 1 : g u n ≥ t g v n } , for all n ≥ 0 . Then, lim n → ∞ t n = 1 .

Proof

Observe first that from definition of t n and (2) it follows that g u n + 1 ≥ g u n ≥ t n g v n ≥ t n g v n + 1 , so we have t n ≤ t n + 1 for all n, which implies 0 < t 0 ≤ t 1 ≤ ⋯ ≤ t n ≤ ⋯ ≤ 1 . Then, there is t ∗ ∈ ( 0 , 1 ] such that lim n → ∞ t n = t ∗ . We claim that t ∗ = 1 . Assume by contradiction that t ∗ < 1 .

Observe first that from the super-homogeneity of g, we have

Since f is strictly α -concave and strictly g-monotone, it follows from (15) that

By definition of t n , it follows that

which yields a contradiction when n tends to infinity. Consequently, we have lim n → ∞ t n = 1 .□

Lemma 3.7

Let f , g : P ∘ → P ∘ be two operators. Assume that g is strictly super-homogeneous and f is strictly ( − α ) -convex and g-antitone. Let { u n } , { v n } be sequences satisfying (3) and (4) and t n ≔ sup { 0 < t ≤ 1 : g u n ≥ t g v n } , for all n ≥ 0 . Then, lim n → ∞ t n = 1 .

Proof

Observe first that from definition of t n , we have g u n ≥ t n g v n for all n ≥ 0 . In particular, from (4) follows that g u n + 1 ≥ g u n ≥ t n g v n ≥ t n g v n + 1 , so we have t n ≤ t n + 1 for all n, which implies 0 < t 0 ≤ t 1 ≤ ⋯ ≤ t n ≤ ⋯ ≤ 1 . Then, there is t ∗ ∈ [ t n , 1 ] for all n ≥ 0 such that lim n → ∞ t n = t ∗ . We claim that t ∗ = 1 . Assume by contradiction that t ∗ < 1 .

Observe first that from the super-homogeneity of g, we have

Since f is strictly ( − α ) -convex and g-antitone, it follows from (16) that

By definition of t n + 2 , it follows that

which yields a contradiction when n tends to infinity. Consequently, we have lim n → ∞ t n = 1 .□

Lemma 3.8

Under the hypotheses of Lemma 3.6 (resp. Lemma 3.7). If { u n } , { v n } be sequences satisfying (2) (resp. (4)), then the sequences { g u n } and { g v n } converge to g v ∗ , where v ∗ ∈ C ( f , g ) .

Proof

For any positive integers n and k, it follows from (2) (resp. (4)) and Lemma 3.6 (resp. Lemma 3.7) that

Thus, we deduce from the normality of P that

where c N > 0 is the constant of normality of P. Hence, by using Lemma 3.6 (resp. Lemma 3.7), we deduce that { g u n } and { g v n } are Cauchy sequences. Therefore, the sequences { g u n } and { g v n } converge to some w 1 , w 2 ∈ E , respectively. As the sequences { g u n } and { g v n } are included in the closed order interval [ g u 0 , g v 0 ] , we deduce that their limits are in [ g u 0 , g v 0 ] ⊂ P ∘ . Therefore, there exist u ∗ , v ∗ ∈ P ∘ such that w 1 = g u ∗ and w 2 = g v ∗ . Now, since g u ∗ , g v ∗ ∈ [ g u n , g v n ] and g u n ≥ t n g v n for all n ≥ 0 , it follows that 0 E ≤ g v ∗ − g u ∗ ≤ g v n − g u n ≤ ( 1 − t n ) g v 0 . So, by the normality of P, we get ∥ g v ∗ − g u ∗ ∥ ≤ c N ( 1 − t n ) ∥ g v 0 ∥ , and then, g v ∗ = g u ∗ . Hence, for all n ≥ 0 we have g u n ≪ g v ∗ ≪ g v n , then from the fact that f is strictly g-monotone (resp. f is strictly g-antitone), we obtain

Consequently, as n tends to infinity, we obtain g v ∗ ≤ f v ∗ ≤ g v ∗ . Therefore, g v ∗ = f v ∗ which proves that v ∗ ∈ C ( f , g ) .□

Lemma 3.9

Let f , g : P ∘ → P ∘ be two operators such that g is surjective and strictly super-homogeneous, and f is strictly α -concave and strictly g-monotone (resp. strictly ( − α ) -convex and strictly g-antitone). Then, g ( C ( f , g ) ) is a singleton.

Proof

From Lemma 3.5, there exist x 0 , y 0 ∈ P ∘ satisfying (5) (resp. (9)). Then by Lemma 3.3 (resp. Lemma 3.4), there exist two sequences { u n } and { v n } satisfying (1) and (2) (resp. (3) and (4)). Then using Lemma 3.8, we see that C ( f , g ) is nonempty. Let x ∗ , y ∗ ∈ C ( f , g ) and define t ∗ ≔ sup { 0 < t ≤ 1 : g y ∗ ∈ [ t g x ∗ , t − 1 g x ∗ ] } . Observe first that we have g y ∗ ∈ [ t ∗ g x ∗ , t ∗ − 1 g x ∗ ] with t ∗ ∈ ( 0 , 1 ] . Next, we claim that t ∗ = 1 . To this purpose assume that 0 < t ∗ < 1 . Since g is strictly super-homogeneous, then we get

So, if we assume that f is strictly α -concave and strictly g-monotone, we have

However, if we assume that f is strictly ( − α ) -convex and strictly g-antitone, we obtain

Consequently, from the definition of t ∗ , we obtain t ∗ α ≤ t ∗ where α ∈ ( 0 , 1 ) , which is a contradiction. Therefore, t ∗ = 1 and so g x ∗ = g y ∗ , which implies g ( C ( f , g ) ) is a singleton.□

Lemma 3.10

Under the hypotheses of Lemma 3.9, f and g have a unique common fixed point z in P ∘ if and only if z = g ( C ( f , g ) ) ∈ C ( f , g ) . In particular, if C ( f , g ) ⊆ C ( f g , g f ) , then f and g have a unique common fixed point.

Proof

If f and g have a unique common fixed point z in P ∘ , then we have f ( z ) = g ( z ) = z which obviously means that the unique point z = g ( C ( f , g ) ) ∈ C ( f , g ) . Conversely, if g ( C ( f , g ) ) ∈ C ( f , g ) , then f z = g z = g ( C ( f , g ) ) = z . If z 1 , z 2 are two common fixed points, then z 1 = g z 1 , z 2 = g z 2 ∈ g ( C ( f , g ) ) , which is a singleton. In particular, if C ( f , g ) ⊆ C ( f g , g f ) , for z ∗ ∈ C ( f , g ) , we have g z ∗ = f z ∗ . It follows from C ( f , g ) ⊆ C ( f g , g f ) that g 2 z ∗ = g f z ∗ = f g z ∗ . Thus, g z ∗ ∈ C ( f , g ) .□

Proof of Theorem 3.1

(resp. Theorem 3.2). At first, observe that C ( f , g ) is nonempty follows from Lemma 3.8 and g ( C ( f , g ) ) is a singleton follows from Lemma 3.9. More precisely, we deduce, by lemmas and { v n } in P ∘ such that (1) and (2) (resp. (3)and (4)) and 3.5, that there exist two sequences { u n } and { v n } satisfying (1) and (2) (resp. (3) and (4)). Moreover, by lemmas 3.8 and 3.9, the sequences { g ( u n ) } and { g ( v n ) } converge to g ( C ( f , g ) ) . Finally, the uniqueness result of the common fixed point follows from Lemma 3.10.□

Theorem 4.1

[18, Theorem 3.1] Let X and Y be strictly positive operators on ℋ . If X ≪ Y , then f ( X ) ≪ f ( Y ) for any nonconstant operator monotone function f on ( 0 , ∞ ) .

Lemma 4.2

If X ≪ Y , then A ∗ X A ≪ A ∗ Y A for every invertible operator A.

Proof

The result follows from the fact that for all nonzero vector u, if A is invertible, then

Therefore,

□

Lemma 4.3

Let X and Y be strictly positive operators on ℋ . If X ≪ Y , then Y − 1 ≪ X − 1 .

Proof

At first, note that if X is strictly positive operator, then 〈 X v , v 〉 > 0 for all v ≠ 0 , which means that X is invertible. We show next that T ↦ T − 1 is a strictly antitone on commuting positive operators. Let 0 ≪ X ≪ Y such that X and Y commute, then X 1 / 2 , Y 1/2 , X and Y also commute. Since, for all nonzero vector v, we have

Then, for all nonzero vector v and w = Y − 1 / 2 X − 1 / 2 v , we have

Now, for non-necessarily commuting operators X and Y which satisfy 0 ≪ X ≪ Y , we have, by Lemma 4.2, Y − 1 / 2 X Y − 1 / 2 ≪ I . As Y − 1 / 2 X Y − 1 / 2 and I commute, then I ≪ Y 1 / 2 X − 1 Y 1 / 2 . Using again Lemma 4.2, we obtain Y − 1 ≪ X − 1 .□

Corollary 4.4

Let X and Y be strictly positive operators on ℋ . If X ≪ Y , then g ( Y ) ≪ g ( X ) for any nonconstant operator antitone function g on ( 0 , ∞ ) .

Proof

The function f = 1 / g is operator monotone on P ∘ . Then, the result is a direct consequence of Theorem 4.1 and Lemma 4.3.□

Theorem 4.5

Let F , G : P ∘ → P ∘ be nonconstant operators. Assume that

1. F is strictly α -concave (resp. strictly ( − α ) -convex) and strictly monotone (resp. strictly antitone) operator.

2. G is surjective, strictly super-homogeneous and strictly inverse monotone operator.

Then, the equation

has a solution.

Proof

Since from (i) and (ii), F is a G-strictly monotone (resp. antitone) operator, then all hypotheses of Theorem 3.1 are fulfilled.□

Lemma 4.6

Let X , Y be two commuting strictly positive operators such that X ≪ Y , and g : ( 0 , ∞ ) → ( 0 , ∞ ) be a strictly monotone function. Then, g ( X ) ≪ g ( Y ) .

Proof

The commutativity hypothesis implies that there exists a unitary matrix Q such that X = Q ∗ D X Q and Y = Q ∗ D Y Q , where D X = diag ( λ 1 , … , λ n ) and D Y = diag ( μ 1 , … , μ n ) are diagonal matrices. If X ≪ Y , then using Lemma 4.2, we deduce that λ i < μ i for all i, and as g is monotone it follows that g ( λ i ) < g ( μ i ) for all i. Therefore, g ( D X ) ≪ g ( D Y ) , which implies that g ( X ) ≪ g ( Y ) .□

Lemma 4.7

Let X, Y be two commuting operators and f a real function. Then, the operators f ( X ) and f ( Y ) commute.

Proof

The proof is straightforward, hence omitted.□

Proposition 4.8

If g , g ′ : ( 0 , ∞ ) → ( 0 , ∞ ) are two surjective strictly super-homogeneous inverse monotone operators, then so is g ∘ g ′ .

Proof

Since g and g ′ are surjective and inverse monotone, then we deduce that g ∘ g ′ is surjective and inverse monotone operator. It remains to prove that g ∘ g ′ is strictly super-homogeneous. By hypothesis, g ′ is strictly super-homogeneous and by Lemma 4.7, g ′ ( t X ) and g ′ ( X ) commute. Then using Lemma 4.6, we deduce that ( g ∘ g ′ ) ( t X ) = g ( g ′ ( t X ) ) ≪ g ( t g ′ ( X ) ) ≪ t ( g ∘ g ′ ) ( X ) , for all t ∈ ( 0 , 1 ) .□