A common fixed point theorem for a commuting family of nonexpansive mappings one of which is multivalued

Abstract

Bruck [Pac. J. Math. 53, 59-71 1974 Theorem 1] proved that for a nonempty closed convex subset E of a Banach space X, if E is weakly compact or bounded and separable and suppose that E has both (FPP) and (CFPP), then for any commuting family S of nonexpansive self-mappings of E, the set F(S) of common fixed points of S is a nonempty nonexpansive retract of E. In this paper, we extend the above result when one of its elements in S is multivalued. The result extends previously known results (on common fixed points of a pair of single valued and multivalued commuting mappings) to infinite number of mappings and to a wider class of spaces.

2000 Mathematics Subject Classification: 47H09; 47H10

1 Introduction

For a pair (t, T) of nonexpansive mappings t : E → E and T : E → 2 ^X defined on a bounded closed and convex subset E of a convex metric space or a Banach space X, we are interested in finding a common fixed point of t and T. In [1], Dhompongsa et al. obtained a result for the CAT(0) space setting:

Theorem 1.1. [[1], Theorem 4.1] Let E be a nonempty bounded closed and convex subset of a complete CAT(0) space X, and let t : E → E and T : E → 2 ^Xbe nonexpansive mappings with T(x) a nonempty compact convex subset of X. Assume that for some p ∈ Fix(t),

$$\alpha p\oplus \left(1-\alpha \right)Txisconvexforx\in E,\alpha \in \left[0,1\right].$$

If t and T are commuting, then Fix(t) ∩ Fix(T) ≠ ∅.

Shahzad and Markin [2] improved Theorem 1.1 by removing the assumption that the nonexpansive multivalued mapping T in that theorem has a convex-valued contractive approximation. They also noted that Theorem 1.1 needs the additional assumption that T(·) ∩ E ≠ ∅ for that result to be valid.

Theorem 1.2. [[2], Theorem 4.2] Let X be a complete CAT(0) space, and E a bounded closed and convex subset of X. Assume t : E → E and T : E → 2 ^X are nonexpansive mappings with T(x) a compact convex subset of X and T(x) ∩ E ≠ ∅ for each x ∈ E. If the mappings t and T commute, then Fix(t) ∩ Fix(T) ≠ ∅.

Dhompongsa et al. [3] extended Theorem 1.1 to uniform convex Banach spaces.

Theorem 1.3. [[3], Theorem 4.2] Let E be a nonempty bounded closed and convex subset of a uniform convex Banach space X. Assume t : E → E and T : E → 2 ^Eare nonexpansive mappings with T(x) a nonempty compact convex subset of E. If t and T are commuting, then Fix(t) ∩ Fix(T) ≠ ∅.

The result has been improved, generalized, and extended under various assumptions. See for examples, [[4], Theorem 3.3], [[5], Theorem 3.4], [[6], Theorem 3.9], [[7], Theorem 4.7], [[8], Theorem 5.3], [[9], Theorem 5.2], [[10], Theorem 3.5], [[11], Theorem 4.2], [[12], Theorem 3.8], [[13], Theorem 3.1].

Recall that a bounded closed and convex subset E of a Banach space X has the fixed point property for nonexpansive mappings (FPP) (respectively, for multivalued nonexpansive mappings (MFPP)) if every nonexpansive mapping of E into E has a fixed point (respectively, every nonexpansive mapping of E into 2 ^E with compact convex values has a fixed point). The following concepts and result were introduced and proved by Bruck [14, 15]. For a bounded closed and convex subset E of a Banach space X, a mapping t : E → X is said to satisfy the conditional fixed point property (CFP) if either t has no fixed points, or t has a fixed point in each nonempty bounded closed convex set that leaves t invariant. A set E is said to have the hereditary fixed point property for nonexpansive mappings (HFPP) if every nonempty bounded closed convex subset of E has the fixed point property for nonexpansive mappings; E is said to have the conditional fixed point property for nonexpansive mappings (CFPP) if every nonexpansive t : E → E satisfies (CFP).

Theorem 1.4. [[15], Theorem 1] Let E be a nonempty closed convex subset of a Banach space X. Suppose E is weakly compact or bounded and separable. Suppose E has both (FPP) and (CFPP). Then for any commuting family S of nonexpansive self-mappings of E, the set F(S) of common fixed points of S is a nonempty nonexpansive retract of E.

The object of this paper is to extend Theorems 1.3 and 1.4 for a commuting family S of nonexpansive mappings one of which is multivalued. As consequences,

• (i) Theorem 1.3 is extended to a bigger class of Banach spaces while a class of mappings is no longer finite;

• (ii) Theroem 1.4 is extended so that one of its members in S can be multivalued.

2 Preliminaries

Let E be a nonempty subset of a Banach space X. A mapping t : E → X is said to be nonexpansive if

$$\left|\right|tx-ty\left|\right|\le \left|\right|x-y\left|\right|,x,y\in E.$$

The set of fixed points of t will be denoted by Fix(t) := {x ∈ E : tx = x}. A subset C of E is said to be t-invariant if t(C) ⊂ C. As usual, B(x, ε) = {y ∈ X : ||x - y|| < ε} stands for an open ball. For a subset A and ε > 0, the ε-neighborhood of A is defined as

$$B_{\epsilon }\left(A\right)=\left\{y\in X:\left|\right|x-y\left|\right|<\epsilon ,\mathsf{\text{for some }}x\in A\right\}=\bigcup _{x\in A}B\left(x,\epsilon \right).$$

Note that if A is convex, then B _ε (A) is also convex. We write $Ā$ for the closure of A.

We shall denote by 2 ^E the family of all subsets of E, CB(E) the family of all nonempty closed bounded subsets of E and denote by KC(E) the family of all nonempty compact convex subsets of E. Let H(·,·) be the Hausdorff distance defined on CB(X), i.e.,

$$H\left(A,B\right):=max\left\{\underset{a\in A}{sup}dist\left(a,B\right),\underset{b\in B}{sup}dist\left(b,A\right)\right\},A,B\in CB\left(X\right),$$

where dist(a, B) := inf{||a - b|| : b ∈ B} is the distance from the point a to the subset B.

A multivalued mapping T : E → CB(X) is said to be nonexpansive if

$$H\left(Tx,Ty\right)\le \left|\right|x-y\left|\right|\mathsf{\text{for}}\mathsf{\text{all}}x,y\in E.$$

T is said to be upper semi-continuous if for each x₀ ∈ E, for each neighborhood U of Tx₀, there exists a neighborhood V of x₀ such that Tx ⊂ U for each x ∈ V. Clearly, every upper semi-continuous mapping T has a closed graph, i.e., for each sequence {x _n } ⊂ E converging to x₀ ∈ E, for each y _n ∈ Tx _n with y _n → y₀, one has y₀ ∈ Tx₀. Fix(T ) is the set of fixed points of T, i.e., Fix(T):= {x ∈ E : x ∈ Tx}. A subset C of E is said to be T-invariant if Tx ∩ C ≠ ∅ for all x ∈ C. For λ ∈ (0, 1), we say that a multivalued mapping T : E → CB(X) satisfies condition (C _λ ) if λdist(x, Tx) ≤ ||x - y|| implies H(Tx, Ty) ≤ ||x - y|| for x, y ∈ E. The following example shows that a mapping T satisfying condition (C _λ ) for some λ ∈ (0, 1) can be discontinuous:

Let λ ∈ (0, 1) and $a=\frac{2\left(\lambda +1\right)}{\lambda \left(\lambda +2\right)}$. Define a mapping $T:\left[0,\frac{2}{\lambda }\right]\to KC\left(\left[0,\frac{2}{\lambda }\right]\right)$ by

$$Tx=\left(\begin{array}{cc}\left\{\frac{x}{2}\right\}\hfill & \mathsf{\text{if}}x\ne \frac{2}{\lambda },\hfill \\ \left[\frac{1}{\lambda },a\right]\hfill & \mathsf{\text{if}}x=\frac{2}{\lambda }.\hfill \end{array}\right.$$

Clearly, $\frac{1}{\lambda }<a<\frac{2}{\lambda }$ and T is nonexpansive on $\left[0,\frac{2}{\lambda }\right)$. Thus, we only verify that, for $x\in \left[0,\frac{2}{\lambda }\right)$,

$$\lambda dist\left(x,Tx\right)\le \left|\right|x-\frac{2}{\lambda }\left|\right|\Rightarrow H\left(Tx,T\frac{2}{\lambda }\right)\le \left|\right|x-\frac{2}{\lambda }\left|\right|$$

(2.1)

and

$$\lambda dist\left(\frac{2}{\lambda },T\frac{2}{\lambda }\right)\le \left|\right|\frac{2}{\lambda }-x\left|\right|\Rightarrow H\left(T\frac{2}{\lambda },Tx\right)\le \left|\right|\frac{2}{\lambda }-x\left|\right|.$$

(2.2)

If $\lambda dist\left(x,Tx\right)\le \left|\right|x-\frac{2}{\lambda }\left|\right|$, then $x\le \frac{4}{\lambda \left(\lambda +2\right)}$ and

$$H\left(Tx,T\frac{2}{\lambda }\right)=a-\frac{x}{2}\le \frac{2}{\lambda }-x=\left|\right|x-\frac{2}{\lambda }\left|\right|.$$

Hence (2.1) holds. If $\lambda dist\left(\frac{2}{\lambda },T\frac{2}{\lambda }\right)\le \left|\right|\frac{2}{\lambda }-x\left|\right|$, then $x\le \frac{4}{\lambda \left(\lambda +2\right)}$ and

$$H\left(T\frac{2}{\lambda },Tx\right)=a-\frac{x}{2}\le \frac{2}{\lambda }-x=\left|\right|\frac{2}{\lambda }-x\left|\right|.$$

Thus (2.2) holds. Therefore, T satisfies condition (C _λ ). Clearly, T is upper semi-continuous but not continuous (and hence T is not nonexpansive).

For a multivalued mapping T : E → CB(X), a sequence {x _n } in E of a Banach space X for which lim_n→∞dist(x _n , Tx _n ) = 0 is called an approximate fixed point sequence (afps for short) for T.

Let (M, d) be a metric space. A geodesic path joining x ∈ X to y ∈ X is a map c from a closed interval [0, r] ⊂ ℝ to X such that c(0) = x, c(r) = y and d(c(t), c(s)) = |t - s| for all s, t ∈ [0, r]. The mapping c is an isometry and d(x, y) = r. The image of c is called a geodesic segment joining x and y which when unique is denoted by seg[x, y]. A metric space (M, d) is said to be of hyperbolic type if it is a metric space that contains a family L of metric segments (isometric images of real line bounded segments) such that (a) each two points x, y in M are endpoints of exactly one member seg[x, y] of L, and (b) if p, x, y ∈ M and m ∈ seg[x, y] satisfies d(x, m) = αd(x, y) for α ∈ [0, 1], then d(p, m) ≤ (1 - α)d(p, x) + αd(p, y). M is said to be metrically convex if for any two points x, y ∈ M with x ≠ y there exists z ∈ M, x ≠ z ≠ y, such that d(x, y) = d(x, z) + d(z, y). Obviously, every metric space of hyperbolic type is always metrically convex. The converse is true provided that the space is complete: If (M, d) is a complete metric space and metrically convex, then (M, d) is of hyperbolic type (cf. [[16], Page 24]). Clearly, every nonexpansive retract is of hyperbolic type.

Proposition 2.1. [[17], Proposition 2] Suppose (M, d) is of hyperbolic type, let {α _n } ⊂ [0, 1), if {x _n } and {y _n } are sequences in M which satisfy for all i, n,

(i) x_n+1∈ seg[x _n , y _n ] with d(x _n , x_n+1) = α _n d(x _n , y _n ),

(ii) d(y_n+1, y _n ) ≤ d(x_n+1, x _n ),

(iii) d(y_i+n, x _i ) ≤ d < ∞,

(iv) α _n ≤ b < 1, and

(v)${\sum }_{s=0}^{\infty }{\alpha }_s=+\infty$.

Then lim_n→∞d(y _n , x _n ) = 0.

Let E be a nonempty bounded closed subset of a Banach space X and {x _n } a bounded sequence in X. For x ∈ X, define the asymptotic radius of {x _n } at x as the number

$$r\left(x,\left\{x_n\right\}\right)=\underset{n\to \infty }{limsup}\left|\right|x_n-x\left|\right|.$$

Let

$$r\left(E,\left\{x_n\right\}\right)=inf\left\{r\left(x,\left\{x_n\right\}\right):x\in E\right\}$$

and

$$A\left(E,\left\{x_n\right\}\right)=\left\{x\in E:r\left(x,\left\{x_n\right\}\right)=r\left(E,\left\{x_n\right\}\right)\right\}.$$

The number r(E, {x _n }) and the set A(E, {x _n }) are, respectively, called the asymptotic radius and asymptotic center of {x _n } relative to E. The sequence {x _n } is called regular relative to E if r(E, {x _n }) = r(E, {x_n′}) for each subsequence {x_n′} of {x _n }. It is well known that: every bounded sequence contains a subsequence that is regular relative to a given set (see [[16], Lemma 15.2] or [[18], Theorem 1]). Further, {x _n } is called asymptotically uniform relative to E if A(E, {x _n }) = A(E, {x_n′}) for each subsequence {x_n′} of {x _n }. It was noted in [16] that if E is nonempty and weakly compact, then A(E, {x _n }) is nonempty and weakly compact, and if E is convex, then A(E, {x _n }) is convex.

A Banach space X is said to satisfy the Kirk-Massa condition if the asymptotic center of each bounded sequence of X in each bounded closed and convex subset is nonempty and compact. A more general space than spaces satisfying the Kirk-Massa condition is a space having property (D). Property (D) introduced in [19] is defined as follows:

Definition 2.2. [[19], Definition 3.1] A Banach space X is said to have property (D) if there exists λ ∈ [0, 1) such that for any nonempty weakly compact convex subset E of X, any sequence {x _n } ⊂ E that is regular and asymptotically uniform relative to E, and any sequence {y _n } ⊂ A(E, {x _n }) that is regular and asymptotically uniform relative to E we have

$$r\left(E,\left\{y_n\right\}\right)\le \lambda r\left(E,\left\{x_n\right\}\right).$$

Theorem 2.3. [[19], Theorem 3.6] Let E be a nonempty weakly compact convex subset of a Banach space X that has property (D). Assume that T : E → KC(E) is a nonexpansive mapping. Then, T has a fixed point.

A direct consequence of Theorem 2.3 is that every weakly compact convex subset of a space having property (D) has both (MFPP) for multivalued nonexpansive mappings and (CFPP). The class of spaces having property (D) contains several well-known ones including k-uniformly rotund, nearly uniformly convex, uniformly convex in every direction spaces, and spaces satisfying Opial condition (see [3, 19–23] and references therein).

The following useful result is due to Bruck:

Theorem 2.4. [[14], Theorem 1] Let E be a nonempty closed convex subset of a Banach space X. Suppose E is locally weakly compact and F is a nonempty subset of E. Let N(F) = {f|f} : E → E is nonexpansive and fx = x for all x ∈ F}. Suppose that for each z in E, there exists h in N(F) such that h(z) ∈ F. Then, F is a nonexpansive retract of E.

3 Main results

We first state three main results:

Theorem 3.1. Let E be a weakly compact convex subset of a Banach space X. Suppose E has (MFPP) and (CFPP). Let S be any commuting family of nonexpansive self-mappings of E. If T : E → KC(E) is a multivalued nonexpansive mapping that commutes with every member of S. Then, F(S) ∩ Fix(T) ≠ ∅.

Theorem 3.2. Let X be a Banach space satisfying the Kirk-Massa condition and let E be a weakly compact convex subset of X. Let S be any commuting family of nonexpansive self-mappings of E. Suppose T : E → KC(E) is a multivalued mapping satisfying condition (C _λ ) for some λ ∈ (0, 1) that commutes with every member of S. If T is upper semi-continuous, then F(S) ∩ Fix(T) ≠ ∅.

Theorem 3.3. Let E be a weakly compact convex subset of a Banach space X. Suppose E has (MFPP) and (CFPP). Let S be any commuting family of nonexpansive self-mappings of E. If T : E → KC(E) is a multivalued nonexpansive mapping that commutes with every member of S. Suppose in addition that T satisfies.

(i) there exists a nonexpansive mapping s : E → E such that sx ∈ Tx for each x ∈ E,

(ii) Fix(T) = {x ∈ E : Tx = {x}} ≠ ∅.

Then, F(S) ∩ Fix(T) is a nonempty nonexpansive retract of E.

Remark 3.4.

• (i) As corollaries, the results in Theorems 3.1 and 3.3 are valid for spaces X having property (D).

• (ii) Theorem 3.3 can be viewed as a generalization of Theorem 1.4 of Bruck for weakly compact convex domains.

Definition 3.5. Let E be a nonempty bounded closed and convex subset of a Banach space X. Let t : E → E be a single valued mapping, T : E → KC(E) a multivalued mapping. Then, t and T are said to be commuting mappings if tTx ⊂ Ttx for all x ∈ E.

If in Theorem 2.4, we put F = Fix(t) where t : E → E is nonexpansive, then it was noted in [[15], Remark 1] that a retraction c ∈ N(F) can be chosen so that cW ⊂ W for all t-invariant closed and convex subsets W of E. With the same proof, we can show that the same result is valid for F = F(S). In this case, we define N(F(S)) = {f | f : E → E is nonexpansive, Fix(f) ⊃ F(S), f(W) ⊂ W whenever W is a closed convex S-invariant subset of E}. Here, by an "S-invariant"subset, we mean a subset that is left invariant under every member of S.

Lemma 3.6. Let E be a nonempty weakly compact convex subset of a Banach space X and let S be any commuting family of nonexpansive self-mappings of E. Suppose that E has (FPP) and (CFPP). Then, F(S) is a nonempty nonexpansive retract of E, and a retraction c can be chosen so that every S-invariant closed and convex subset of E is also c-invariant.

Proof. Note by Theorem 1.4 that F(S) is nonempty. According to Theorem 2.4, it suffices to show that for each z in E, there exists h in N(F(S)) such that h(z) ∈ F(S).

Let z ∈ E and K = {f(z)|f ∈ N(F(S))} ⊂ E. Since K is weakly compact convex and invariant under every member in S, we obtain by Theorem 1.4 that F(S)∩K ≠ ∅, i.e., there exists h in N(F(S)) such that h(z) ∈ F(S). Theorem 2.4 then implies that F(S) is a nonexpansive retract of E, where a retraction is chosen from N(F(S)).   □

Proof of Theorem 3.1 Let c be a nonexpansive retraction of E onto F(S) obtained in Lemma 3.6. Set Ux := Tcx for x ∈ E. Clearly,

$$H\left(Ux,Uy\right)=H\left(Tcx,Tcy\right)\le \left|\right|cx-cy\left|\right|\le \left|\right|x-y\parallel \mathsf{\text{for}}x,y\in E.$$

Thus, U is nonexpansive, and since E has (MFPP), there exists p ∈ Up = Tcp. Since Tcp is S-invariant, by the property of c, Tcp is also c-invariant, i.e., cp ∈ Tcp. Therefore, F(S) ∩ Fix(T) ≠ ∅. □

The following proposition is needed for a proof of Theorem 3.2.

Proposition 3.7. Let A be a compact convex subset of a Banach space X and let a nonempty subset F of A be a nonexpansive retract of A. Suppose a mapping U : A → KC(A) is upper semi-continuous and satisfies:

(i) c(Ux) ⊂ Ux for all x ∈ F where c is a nonexpansive retraction of A onto F, and

(ii) F is U -invariant.

Then, U has a fixed point in F.

Proof. Let ε > 0. Since F is compact, there exists a finite ε-dense subset {z₁, z₂, ..., z _n } of F , i.e., $F\subset {\bigcup }_{i=1}^{n}B\left(z_i,\frac{\epsilon }{2}\right)$. Put $K:=\overline{co}\left(z_1,z_2,\dots ,z_n\right)$ and define $Vx={\overline{B}}_{\epsilon }\left(Ucx\right)\cap K$ for x ∈ K. Clearly, V : K → KC(K). For x ∈ K, cx ∈ F thus by (ii) there exists y ∈ Ucx ∩ F. Then, choose z _i for some i such that $\left|\right|z_i-y\left|\right|\le \frac{\epsilon }{2}$. Therefore, $z_i\in {\bar{B}}_{\epsilon }\left(Ucx\right)\cap K$, i.e., V x is nonempty for x ∈ K. We now show that V is upper semi-continuous. Let {x _n } be a sequence in K converging to some x ∈ K and y _n ∈ V x _n with y _n → y. Choose a _n ∈ Ucx _n such that ||y _n - a _n || ≤ ε. As A is compact, we may assume that a _n → a for some a ∈ A. By upper semi-continuity of U, a ∈ Ucx. Consider

$$\left|\right|y-a\left|\right|\le \left|\right|y-y_n\left|\right|+\left|\right|y_n-a_n\left|\right|+\left|\right|a_n-a\left|\right|.$$

By letting n → ∞, we obtain ||y - a|| ≤ ε, i.e., y ∈ V x and the proof that V is upper semi-continuous is complete. By Kakutani fixed point theorem, there exists p _ε ∈ V p _ε , that is, ||p _ε - b _ε || ≤ ε for some b _ε ∈ Ucp _ε .

By the assumption on U, we see that cb _ε ∈ Ucp _ε and ||cp _ε - cb _ε || ≤ || p _ε - b _ε || ≤ ε. Taking $\epsilon =\frac{1}{n}$ and write q _n for $c{p}_{\frac{1}{n}}$ and b _n for $c{b}_{\frac{1}{n}}$, we obtain a sequence {q _n } ⊂ F and b _n ∈ Uq _n ∩F with ||q _n - b _n || → 0. By the compactness of F, we assume that q _n → q and b _n → b. It is seen that q = b ∈ Uq.   □

Proof of Theorem 3.2 As observed earlier, E has both (FPP) and (CFPP), thus we start with a nonexpansive retraction c of E onto F(S) obtained by Lemma 3.6. For each x ∈ F(S), Tx is invariant under every member of S and Tx is convex, thus Tx is c-invariant. Clearly, c is a nonexpansive retraction of Tx onto Tx ∩ F(S) that is nonempty by Theorem 1.4.

Next, we show that there exists an afps for T in F(S). Let x₀ ∈ F (S). Since Tx₀ ∩ F (S) ≠ ∅, we can choose y₀ ∈ Tx₀ ∩ F (S). Since F (S) is of hyperbolic type, there exists x₁ ∈ F (S) such that

$$\left|\right|x_0-x_1\left|\right|=\lambda \left|\right|x_0-y_0\left|\right|\mathsf{\text{and}}\left|\right|x_1-y_0\left|\right|=\left(1-\lambda \right)\left|\right|x_0-y_0\left|\right|.$$

Choose y′₁ ∈ Tx₁ for which ||y _o - y′₁|| = dist(y₀, Tx₁). Set y₁ = cy′₁. Clearly, ||y₀ - y₁|| = ||cy₀ - cy′₁|| ≤ ||y₀ - y′₁||. Therefore, we can choose y₁ ∈ Tx₁ ∩ F (S) so that ||y₀ - y₁|| = dist(y₀, Tx₁). In this way, we will find a sequence {x _n } ⊂ F (S) satisfying

$$\left|\right|x_n-x_{n+1}\left|\right|=\lambda \left|\right|x_n-y_n\left|\right|\mathsf{\text{and}}\left|\right|x_{n+1}-y_n\left|\right|=\left(1-\lambda \right)\left|\right|x_n-y_n\left|\right|,$$

where y _n ∈ Tx _n ∩ F (S) and ||y _n - y_n+1|| = dist(y _n , Tx_n+1).

Since λdist(x _n , Tx _n ) ≤ λ||x _n - y _n || = ||x _n - x_n+1||,

$$\left|\right|y_n-y_{n+1}\left|\right|\le H\left(T{x}_n,T{x}_{n+1}\right)\le \left|\right|x_n-x_{n+1}\left|\right|.$$

From Proposition 2.1, lim_n→∞||y _n - x _n || = 0 and {x _n } is an afps for T in F(S). Assume that {x _n } is regular relative to E. By the Kirk-Massa condition, A := A(E, {x _n }) is assumed to be nonempty compact and convex. Define Ux = Tx ∩ A for x ∈ A. We are going to show that Ux is nonempty for each x ∈ A. First, let r := r(E, {x _n }). If r = 0 and if x ∈ A, then x _n → x and y _n → x. Using upper semi-continuity of T , we see that x ∈ Tx, i.e., F(S) ∩ Fix(T) ≠ ∅.

Therefore, we assume for the rest of the proof that r > 0. Let x ∈ A. If for some subsequence $\left\{x_{n_k}\right\}$ of {x _n }, $\lambda dist\left(x_{n_k},T{x}_{n_k}\right)\ge \left|\right|x_{n_k}-x\left|\right|$ for each k, we have

$$0=\underset{n\to \infty }{limsup}\lambda dist\left(x_{n_k},T{x}_{n_k}\right)\ge \underset{n\to \infty }{limsup}\left|\right|x_{n_k}-x\left|\right|\ge r$$

since {x _n } is regular relative to E and this is a contradiction. Therefore,

$$\lambda dist\left(x_n,T{x}_n\right)\le \left|\right|x_n-x\left|\right|\mathsf{\text{for sufficiently large }}n.$$

(3.1)

Now, we show that Ux is nonempty. Choose y _n ∈ Tx _n so that ||x _n - y _n || = dist(x _n , Tx _n ) and choose z _n ∈ Tx such that ||y _n - z _n || = dist(y _n , Tx). As Tx is compact, we may assume that {z _n } converges to z ∈ Tx. Using (3.1) and the fact that T satisfies condition (C _λ ), we have

$$\begin{array}{lll}\hfill \left|\right|x_n-z\left|\right|& \le \left|\right|x_n-y_n\left|\right|+\left|\right|y_n-z_n\left|\right|+\left|\right|z_n-z\left|\right|& \hfill \text{(1)}\\ =\left|\right|x_n-y_n\left|\right|+dist\left(y_n,Tx\right)+\left|\right|z_n-z\left|\right|& \hfill \text{(2)}\\ \le \left|\right|x_n-y_n\left|\right|+H\left(T{x}_n,Tx\right)+\left|\right|z_n-z\left|\right|& \hfill \text{(3)}\\ \le \left|\right|x_n-y_n\left|\right|+\left|\right|x_n-x\left|\right|+\left|\right|z_n-z\left|\right|\mathsf{\text{for sufficiently large }}n.& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$$

Taking lim sup_n→∞in the above inequalities to obtain

$$\underset{n\to \infty }{limsup}\left|\right|x_n-z\left|\right|\le \underset{n\to \infty }{limsup}\left|\right|x_n-x\left|\right|=r$$

that implies that z ∈ Ux proving that Ux is nonempty as claimed.

Now, we show that U is upper semi-continuous. Let {z _k } be a sequence in A converging to some z ∈ A and y _k ∈ Uz _k with y _k → y. Consider the following estimates:

$$\underset{n\to \infty }{limsup}\left|\right|x_n-y\left|\right|\le \underset{n\to \infty }{limsup}\left|\right|x_n-y_k\left|\right|+\underset{n\to \infty }{limsup}\left|\right|y_k-y\left|\right|=r\left(E,\left\{x_n\right\}\right)+\underset{n\to \infty }{limsup}\left|\right|y_k-y\left|\right|\mathsf{\text{for each }}k.$$

Letting k → ∞, it follows that

$$\underset{n\to \infty }{limsup}\left|\right|x_n-y\left|\right|\le r\left(E,\left\{x_n\right\}\right).$$

Hence y ∈ A. From upper semi-continuity of T, y ∈ Tz. Therefore, y ∈ Uz and thus U is upper semi-continuous. Put F := F(S) ∩ A. Since A is c-invariant, it is clear that F is a nonexpansive retract of A by the retraction c. Now, if x ∈ F, then Ux is S-invariant which implies Ux is c-invariant. Therefore, condition (i) in Proposition 3.7 is justified. To verify condition (ii), we let x ∈ F. Take y ∈ Ux. It is obvious that cy ∈ Ux ∩ F(S), so U satisfies condition (ii) of Proposition 3.7. Upon applying Proposition 3.7 we obtain a fixed point in F of U and thus of T and we are done.   □

Proof of Theorem 3.3 By (i) and (ii), it is seen that Fix(T) = Fix(s). Note by Theorem 3.1 that F(S) ∩ Fix(s) is nonempty. Let c be a retraction from E onto F(S) obtained by Lemma 3.6. Here, c belongs to the set N(F(S)) = {f | f : E → E is nonexpansive, Fix(f) ⊃ F(S), f(W) ∩ W whenever W is a closed convex S-invariant subset of E}. Put F = F(S) ∩ Fix(s) and let N(F) = {f | f : E → E is nonexpansive, Fix(f) ⊃ F}. Let z ∈ E and consider the weakly compact and convex set K := {f(z)|f ∈ N(F)}. It is left to show that h(z) ∈ F for some h ∈ N(F). Since K is S-invariant, K is therefore c-invariant. It is evident that K is s-invariant. Thus sc : K → K. Therefore, sc has a fixed point, say x, in K, i.e., sc(x) = x. By (i), sc(x) ∈ Tcx. Since Tcx is c-invariant, we have cx ∈ Tcx. That is cx ∈ Fix(T) = Fix(s). Hence scx = x = cx, i.e., cx ∈ F(S) ∩ Fix(s), and the proof is complete.   □

When S consists of only the identity mapping of E, we immediately have the following corollary:

Corollary 3.8. Let E be a weakly compact convex subset of a Banach space X. Suppose E has (MFPP). If T : E → KC(E) is a multivalued nonexpansive mapping satisfying.

(i) there exists a nonexpansive mapping s : E → E such that sx ∈ Tx for each x ∈ E,

(ii) Fix(T) = {x ∈ E : Tx = {x}} ≠ ∅.

Then Fix(T) is a nonempty nonexpansive retract of E.

Of course, when T is single valued, condition (i) is satisfied. Even a very simple example shows that condition (ii) in Corollary 3.8 may not be dropped.

Example 3.9. Let X be the Hilbert space ℝ²with the usual norm, and let f : [0, 1] → [0, 1] be a continuous function that is strictly concave,$f\left(0\right)=\frac{1}{2}$and f(1) = 1. Moreover let f′(x) ≤ 1 for x ∈ [0, 1]. Let T : [0, 1]² → KC([0, 1]²) be defined by T(x, y) = [0, x] × [f(x), 1]. We show that T is nonexpansive, but Fix(T)≠ {x : Tx = {x}} and Fix(T) is not metrically convex. If (x₁, y₁), (x₂, y₂) ∈ [0, 1]², then

$$H\left(T\left(x_1,y_1\right),T\left(x_2,y_2\right)\right)=|x_1-x_2|\le \left|\right|\left(x_1,y_1\right)-\left(x_2,y_2\right)\left|\right|.$$

Hence T is nonexpansive. However,$a=\left(0,\frac{1}{2}\right)$is a fixed point but Ta ≠ {a}. Finally, Fix(T) is not metrically convex since, putting b = (1, 1), we see that b ∈ Tb, but$\frac{a+b}{2}=\left(\frac{1}{2},\frac{3}{4}\right)\notin T\frac{a+b}{2}$since f is strictly concave.

In [[14], Lemma 6] it was stated that: Let E be a nonempty weakly compact convex subset of a Banach space X. Suppose E has (HFPP). Suppose F is a nonempty nonexpansive retract of E and t : E → E is a nonexpansive mapping which leaves F invariant. Then Fix(t) ∩ F is a nonempty nonexpansive retract of E.

Here, we have a multivalued version (with a similar proof) of this result.

Corollary 3.10. Let E and T be as in Corollary 3.8. Suppose F is a nonexpansive retract of E by a retraction c. If Tx is c-invariant for each x ∈ F, then Fix(T) ∩ F is a nonempty nonexpansive retract of E.