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9.1 Contractive Mappings

We begin this chapter with a few results on single-valued contractive mappings, which will be used in subsequent sections.

Let (X,ρ) be a complete metric space. Recall that an operator A:X→X is said to be nonexpansive if

$$\rho(Ax,Ay) \le\rho(x,y) \quad\mbox{for all } x,y \in X. $$

We denote by \(\mathfrak{A}\) the set of all nonexpansive operators A:X→X. We assume that X is bounded and set

$$d(X)=\sup\bigl\{ \rho(x,y): x,y \in X\bigr\} <\infty. $$

We equip the set \(\mathfrak{A}\) with the metric \(\rho_{\mathfrak{A}}\) defined by

$$ \rho_\mathfrak{A}(A,B):=\sup\bigl\{ \rho(Ax,Bx): x \in X\bigr\} ,\quad A,B \in \mathfrak{A}. $$
(9.1)

It is clear that the metric space \((\mathfrak{A},\rho_{\mathfrak{A}})\) is complete.

Denote by \(\mathcal{A}\) the set of all sequences \(\{A_{t}\}_{t=1}^{\infty}\), where \(A_{t} \in\mathfrak{A}\), t=1,2,… . A member of \(\mathcal{A}\) will occasionally be denoted by boldface A.

For the set \(\mathcal{A}\) we define a metric \(\rho_{\mathcal{A}}\) by

$$ \rho_\mathcal{A}\bigl(\{A_t\}_{t=1}^{\infty}, \{B_t\}_{t=1}^{\infty }\bigr)=\sup \bigl\{ \rho(A_tx,B_tx): t=1,2,\ldots\mbox{ and } x \in X\bigr\} . $$
(9.2)

Clearly, the metric space \((\mathcal{A},\rho_{\mathcal{A}})\) is also complete.

A sequence \(\{A_{t}\}_{t=1}^{\infty} \in\mathcal{A}\) is called contractive if there exists a decreasing function ϕ:[0,d(X)]→[0,1] such that

$$ \phi(t)<1 \quad\mbox{for all } t \in\bigl(0,d(X)\bigr] $$
(9.3)

and

$$ \rho(A_tx,A_ty) \le\phi\bigl(\rho(x,y) \bigr)\rho(x,y) \quad\mbox{for all } x,y \in X \mbox{ and all integers } t \ge1. $$
(9.4)

An operator \(A \in{ \mathfrak{A}} \) is called contractive if the sequence \(\{A_{t}\}_{t=1}^{\infty}\) with A t =A, t=1,2,… , is contractive.

It is known that the iterates of any contractive mapping converge to its unique fixed point (see Chap. 3). The following theorem, which was obtained in [144], extends this result to infinite products.

Theorem 9.1

Assume that the sequence \(\{A_{t}\}_{t=1}^{\infty}\) is contractive and that ε>0. Then there exists a natural number N such that for each integer T≥N, each mapping h:{1,…,T}→{1,2,…} and each x,y∈X,

$$ \rho(A_{h(T)}\cdot\cdots\cdot A_{h(1)}x, A_{h(T)}\cdot\cdots\cdot A_{h(1)}y) \le \varepsilon . $$
(9.5)

Proof

There exists a decreasing function ϕ:[0,d(X)]→[0,1] such that inequalities (9.3) and (9.4) hold. Choose a natural number N>4 such that

$$ d(X)\phi(\varepsilon )^N<\varepsilon . $$
(9.6)

Assume that T≥N is an integer, h:{1,…,T}→{1,2,…} and that x,y∈X are given. We intend to show that (9.5) holds. Assume it does not. Then

$$ \begin{aligned}[b] &\rho(x,y) > \varepsilon \quad\mbox{and}\quad \rho(A_{h(n)}\cdot \cdots\cdot A_{h(1)}x, A_{h(n)}\cdot\cdots\cdot A_{h(1)}y) > \varepsilon , \\ &\quad n=1,\ldots, N. \end{aligned} $$
(9.7)

It follows from (9.7) and (9.4) that

$$\rho(A_{h(1)}x,A_{h(1)}y) \le \phi\bigl(\rho(x,y)\bigr)\rho(x,y) \le\phi(\varepsilon ) \rho(x,y) $$

and that for all integers i=1,…,N−1,

$$ \begin{aligned} &\rho(A_{h(i+1)}A_{h(i)}\cdot\cdots\cdot A_{h(1)}x, A_{h(i+1)}A_{h(i)}\cdot\cdots\cdot A_{h(1)}y) \\ &\quad\le \phi(\varepsilon ) \rho(A_{h(i)}\cdot\cdots\cdot A_{h(1)}x, A_{h(i)}\cdot\cdots\cdot A_{h(1)}y). \end{aligned} $$

When combined with (9.6), this inequality implies that

$$\rho(A_{h(N)}\cdot\cdots\cdot A_{h(1)}x, A_{h(N)}\cdot \cdots\cdot A_{h(1)}y) \le\phi(\varepsilon )^N \rho(x,y) \le d(X) \phi(\varepsilon )^N<\varepsilon , $$

a contradiction. This completes the proof of Theorem 9.1. □

Corollary 9.2

Assume that the sequence \(\{A_{t}\}_{t=1}^{\infty}\) is contractive. Then

$$\rho(A_{h(T)}\cdot\cdots\cdot A_{h(1)}x, A_{h(T)}\cdot \cdots\cdot A_{h(1)}y) \to0 \quad\textit{as}\ T \to \infty, $$

uniformly in h:{1,2,…}→{1,2,…} and in x,y∈X.

We remark in passing that such results are called weak ergodic theorems in the population biology literature [43].

9.2 Star-Shaped Spaces

We say that a complete metric space (X,ρ) is star-shaped if it contains a point x ∗∈X with the following property:

For each x∈X, there exists a mapping

$$ t \to t x \oplus(1-t)x_* \in X,\quad t \in(0,1), $$
(9.8)

such that for each t∈(0,1) and each x,y∈X,

$$ \rho\bigl(tx \oplus(1-t)x_*, ty \oplus(1-t)x_*\bigr) \le t \rho(x,y) $$
(9.9)

and

$$ \rho\bigl(tx \oplus(1-t)x_*, x\bigr) \le (1-t)\rho(x,x_*). $$
(9.10)

For each \(A \in{ \mathfrak{A}}\) and each γ∈(0,1), define \(A_{\gamma} \in{ \mathfrak{A}}\) by

$$ A_{\gamma}x=(1-\gamma)Ax \oplus\gamma x_*,\quad x \in X. $$
(9.11)

For each \(\mathbf{A}=\{A_{t}\}_{t=1}^{\infty} \in\mathcal{A}\), let \(\mathbf{A}_{\gamma}=\{A_{\gamma t}\}_{t=1}^{\infty}\), where

$$ A_{\gamma t}x= (1-\gamma) A_t x \oplus\gamma x_*, \quad x \in X, t=1,2,\ldots. $$
(9.12)

Theorem 9.3

Assume that \(\mathcal{B}\) is a closed subset of \(\mathcal{A}\) such that for each \(\mathbf{A} \in\mathcal{B}\) and each γ∈(0,1), the sequence \(\mathbf{A}_{\gamma} \in\mathcal{B}\). Then there exists a set \(\mathcal{F}\) which is a countable intersection of open and everywhere dense subsets of \(\mathcal{B}\) (with the relative topology) such that each \(\mathbf{A} \in\mathcal{F}\) is contractive.

Proof

It follows from (9.10) that for each \(\mathbf{A}=\{A_{t}\}_{t=1}^{\infty} \in\mathcal{B}\), each γ∈(0,1) and each x∈X,

$$\rho(A_{\gamma t}x,A_tx) \le\gamma\rho(A_tx,x_*). $$

This implies that A γ →A in \(\mathcal{B}\) as γ→0+ and that the set \(\{\mathbf{A}_{\gamma}: \mathbf{A} \in\mathcal{B}, \gamma\in(0,1)\}\) is everywhere dense in \(\mathcal{B}\).

Let \(\mathbf{A}=\{A_{t}\}_{t=1}^{\infty} \in\mathcal{B} \) and γ∈(0,1) be given. Inequality (9.9) implies that

$$ \rho(A_{\gamma t}x,A_{\gamma t}y) \le(1-\gamma)\rho(x,y) $$
(9.13)

for all x,y∈X and all integers t≥1. For each integer i≥1, choose a positive number

$$ \delta(\mathbf{A},\gamma,i) <(4i)^{-1}d(X)\gamma $$
(9.14)

and define

$$ U(\mathbf{A},\gamma,i)=\bigl\{ \mathbf{B} \in\mathcal{B}: \rho_\mathcal{A} (\mathbf{A}_{\gamma},\mathbf{B})<\delta(\mathbf{A},\gamma,i) \bigr\} . $$
(9.15)

Let i≥1 be an integer. We claim that the following property holds:

P(1):

For each B∈U(A,γ,i), each x,y∈X satisfying ρ(x,y)≥i −1 d(X) and each integer t≥1, the inequality ρ(B t x,B t y)≤(1−γ/2)ρ(x,y) is valid.

Indeed, assume that B∈U(A,γ,i), the points x,y∈X satisfy

$$ \rho(x,y) \ge i^{-1}d(X), $$
(9.16)

and that t≥1 is an integer. It follows from the definition of U(A,γ,i) (see (9.15) and (9.14)), (9.13) and (9.16) that

$$ \begin{aligned} \rho(B_tx,B_ty) &\le \rho(A_{\gamma t}x,A_{\gamma t}y)+ 2\delta(\mathbf{A},\gamma,i) \\ &<2\delta(\mathbf{A},\gamma,i)+(1-\gamma)\rho(x,y) \le(1-\gamma)\rho(x,y) +(2i)^{-1}\gamma d(X) \\ &\le(1-\gamma)\rho(x,y)+2^{-1} \gamma\rho(x,y) \le(1-\gamma/2) \rho(x,y). \end{aligned} $$

Thus

$$ \rho(B_tx,B_ty) \le(1-\gamma/2) \rho(x,y). $$
(9.17)

Now define

$$ \mathcal{F}:=\bigcap_{i=1}^{\infty} \bigcup\bigl\{ U(\mathbf{A}, \gamma,i): \mathbf{A} \in\mathcal{B}, \gamma\in(0,1)\bigr\} . $$
(9.18)

It is clear that \(\mathcal{F}\) is a countable intersection of open and everywhere dense subsets of \(\mathcal{B}\) (equipped with the relative topology). We claim that any \(\mathbf{B} \in\mathcal{F}\) is contractive. To show this, assume that i is a natural number. There exist \(\mathbf{A} \in\mathcal{B}\) and γ∈(0,1) such that B∈U(A,γ,i). By property P(1), for each x,y∈X satisfying ρ(x,y)≥i −1 d(X) and each integer t≥1, inequality (9.17) holds. Since i is an arbitrary natural number we conclude that B is contractive. Theorem 9.3 is proved. □

Theorem 9.4

Assume that \({\mathfrak{B}}\) is a closed subset of \({\mathfrak{A}}\) such that for each \(A \in{ \mathfrak{B}}\) and each γ∈(0,1), the mapping \(A_{\gamma} \in{ \mathfrak{B}}\). Then there exists a set \(\mathcal{F}\) which is a countable intersection of open and everywhere dense subsets of \({\mathfrak{B}}\) (with the relative topology) such that each \(A \in\mathcal{F}\) is contractive.

Proof

For each \(A \in{ \mathfrak{B}}\) denote by Q(A) the sequence \(\mathbf{A}=\{A_{t}\}_{t=1}^{\infty}\) with A t =A, t=1,2,… . Set

$$\mathcal{B}=\bigl\{ Q(A): A \in{ \mathfrak{B}}\bigr\} . $$

It is easy to see that \(\mathcal{B}\) is a closed subset of \(\mathcal{A}\) and that for each \(\mathbf{A} \in\mathcal{B}\) and each γ∈(0,1), the sequence \(\mathbf{A}_{\gamma} \in\mathcal{B}\). Now Theorem 9.4 follows from Theorem 9.3 and the equality

$$\rho_{\mathfrak{A}}(A,B)=\rho_\mathcal{A}\bigl(Q(A),Q(B)\bigr). $$

 □

9.3 Convergence of Iterates of Set-Valued Mappings

Assume that (E,∥⋅∥) is a Banach space, K is a nonempty, bounded and closed subset of E, and there exists θ∈K such that for each point x∈K,

$$t x+ (1-t)\theta\in K, \quad t \in(0,1). $$

We consider the star-shaped complete metric space K with the metric ∥x−y∥, x,y∈K. Denote by S(K) the set of all nonempty closed subsets of K. For x∈K and A⊂K, set

$$\rho(x,A)=\inf\bigl\{ \Vert x-y\Vert : y \in A\bigr\} , $$

and for each A,B∈S(K), let

$$ H(A,B)=\max\Bigl\{ \sup_{x \in A}\rho(x,B), \sup _{y \in B}\rho(y,A)\Bigr\} . $$
(9.19)

We equip the set S(K) with the Hausdorff metric H(⋅,⋅). It is well known that the metric space (S(K),H) is complete. Clearly, {θ}∈S(K).

For each subset A∈S(K) and each t∈[0,1], define

$$ t A \oplus(1-t){\theta}:=\bigl\{ tx +(1-t)\theta: x \in A\bigr\} \in S(K). $$
(9.20)

It is easy to see that the complete metric space (S(K),H) is star-shaped.

Denote by \({\mathfrak{A}}\) the set of all nonexpansive operators T:S(K)→S(K). For the set \({\mathfrak{A}}\) we consider the metric \(\rho_{\mathfrak{A}}\) defined by

$$ \rho_{\mathfrak{A}}(T_1,T_2):=\sup\bigl\{ H\bigl(T_1(A),T_2(A)\bigr): A \in S(K)\bigr\} ,\quad T_1,T_2 \in{ \mathfrak{A}}. $$
(9.21)

Denote by \(\mathcal{M}\) the set of all mappings T:K→S(K) such that

$$ H\bigl(T(x),T(y)\bigr) \le\Vert x-y\Vert, \quad x,y \in K. $$
(9.22)

A mapping \(T \in\mathcal{M}\) is called contractive if there exists a decreasing function ϕ:[0,d(K)]→[0,1] such that

$$ \phi(t)<1 \quad\mbox{for all } t \in\bigl(0,d(K)\bigr] $$
(9.23)

and

$$ H\bigl(T(x),T(y)\bigr) \le\phi\bigl(\Vert x-y\Vert \bigr)\Vert x-y\Vert \quad\mbox{for all } x,y \in K. $$
(9.24)

Assume that \(T \in\mathcal{M}\). For each A∈S(K), denote by \(\tilde{T}(A)\) the closure of the set ⋃{T(x):x∈A} in the norm topology.

Proposition 9.5

Assume that \(T \in\mathcal{M}\). Then the mapping \(\tilde{T}\) belongs to \({\mathfrak{A}}\).

Proof

Let A,B∈S(K). We claim that

$$ H\bigl(\tilde{T}(A),\tilde{T}(B)\bigr) \le H(A,B). $$
(9.25)

Given ε>0, there exist \(x_{1} \in\tilde{T}(A)\) and \(x_{2} \in\tilde{T}(B)\) such that

$$ \max\bigl\{ \rho\bigl(x_1,\tilde{T}(B)\bigr),\rho \bigl(x_2,\tilde{T}(A)\bigr)\bigr\} +\varepsilon /2 > H\bigl(\tilde{T}(A), \tilde{T}(B)\bigr). $$
(9.26)

We may assume that

$$\rho\bigl(x_1,\tilde{T}(B)\bigr) \ge\rho\bigl(x_2,\tilde{T}(A)\bigr). $$

Therefore

$$ \rho\bigl(x_1,\tilde{T}(B)\bigr)+\varepsilon /2 > H\bigl( \tilde{T}(A),\tilde{T}(B)\bigr). $$
(9.27)

We may assume that x 1∈T(A). There exist points x 0∈A such that x 1∈T(x 0) and y 0∈B such that

$$\Vert x_0-y_0\Vert < \rho(x_0,B)+ \varepsilon /2 \le H(A,B) +\varepsilon /2. $$

Therefore inequality (9.22) implies that

$$\rho\bigl(x_1,\tilde{T}(B)\bigr) \le\rho\bigl(x_1,T(y_0) \bigr) \le H\bigl(T(x_0),T(y_0)\bigr) \le\Vert x_0-y_0\Vert <H(A,B)+\varepsilon /2. $$

Now (9.27) yields

$$H\bigl(\tilde{T}(A),\tilde{T}(B)\bigr)<H(A,B)+\varepsilon . $$

Since ε is an arbitrary positive number, we conclude that (9.25) holds. Proposition 9.5 is proved. □

Proposition 9.6

Assume that \(T \in\mathcal{M}\). Then the mapping \(\tilde{T}\) is contractive if and only if the mapping T is contractive.

Proof

It is clear that T is contractive if \(\tilde{T}\) is contractive. Assume now that the mapping T is contractive. Then there exists a decreasing function ϕ:[0,d(K)]→[0,1] such that (9.23) and (9.24) hold.

Let A,B∈S(K). We assert that

$$ H\bigl(\tilde{T}(A),\tilde{T}(B)\bigr) \le \max\bigl\{ 1/2,\phi \bigl(H(A,B)/4\bigr)\bigr\} H(A,B). $$
(9.28)

To see this, we may assume that H(A,B)>0 and that

$$ H\bigl(\tilde{T}(A),\tilde{T}(B)\bigr) > H(A,B)/2. $$
(9.29)

Let

$$ \varepsilon \in\bigl(0,H(A,B)/4\bigr). $$
(9.30)

By the definition of the Hausdorff metric, there exist \(x_{1} \in \tilde{T}(A)\) and \(x_{2} \in\tilde{T}(B)\) such that

$$ \max\bigl\{ \rho\bigl(x_1,\tilde{T}(B)\bigr),\rho \bigl(x_2,\tilde{T}(A)\bigr)\bigr\} +\varepsilon /2 > H\bigl(\tilde{T}(A), \tilde{T}(B)\bigr). $$
(9.31)

We may assume that

$$\rho\bigl(x_1,\tilde{T}(B)\bigr) \ge\rho\bigl(x_2,\tilde{T}(A)\bigr). $$

Therefore

$$ \rho\bigl(x_1,\tilde{T}(B)\bigr)+\varepsilon /2 > H\bigl( \tilde{T}(A),\tilde{T}(B)\bigr). $$
(9.32)

We may also assume that x 1∈T(A). There exist x 0∈A such that x 1∈T(x 0) and y 0∈B such that

$$ \Vert x_0-y_0\Vert < \rho(x_0,B)+\varepsilon /2 \le H(A,B) +\varepsilon /2. $$
(9.33)

Therefore (9.24) implies that

$$\begin{aligned} \rho\bigl(x_1,\tilde{T}(B)\bigr) & \le\rho\bigl(x_1,T(y_0)\bigr) \le H\bigl(T(x_0),T(y_0) \bigr) \le\phi\bigl(\Vert x_0-y_0\Vert \bigr)\Vert x_0-y_0\Vert \\ &\le\phi\bigl(\Vert x_0-y_0\Vert \bigr) \bigl(H(A,B)+\varepsilon /2 \bigr). \end{aligned}$$
(9.34)

Combining this with (9.32), we see that

$$ -\varepsilon /2 + H\bigl(\tilde{T}(A),\tilde{T}(B)\bigr)< \phi\bigl(\Vert x_0-y_0\Vert \bigr) \bigl(H(A,B)+\varepsilon /2\bigr). $$
(9.35)

It follows from (9.22), (9.32), (9.29) and (9.30) that

$$ \begin{aligned} \Vert x_0-y_0\Vert &\ge H \bigl(T(x_0),T(y_0)\bigr) \ge\rho\bigl(x_1,T(y_0) \bigr) \ge\rho\bigl(x_1,\tilde{T}(B)\bigr) \\ &\ge -\varepsilon /2 + H\bigl(\tilde{T}(A),\tilde{T}(B)\bigr) >-\varepsilon /2+H(A,B)/2 \ge H(A,B)/4. \end{aligned} $$

Thus

$$\Vert x_0-y_0\Vert \ge H(A,B)/4. $$

Combining this last inequality with (9.35), we can deduce that

$$-\varepsilon /2 + H\bigl(\tilde{T}(A),\tilde{T}(B)\bigr)< \phi\bigl(H(A,B)/4\bigr) \bigl(H(A,B)+\varepsilon /2\bigr). $$

Since ε is an arbitrary positive number, we conclude that

$$H\bigl(\tilde{T}(A),\tilde{T}(B)\bigr)\le \phi\bigl(H(A,B)/4\bigr) \bigl(H(A,B) \bigr). $$

This completes the proof of Proposition 9.6. □

We equip the set \(\mathcal{M}\) with the metric \(\rho_{\mathcal{M}}\) defined by

$$ \rho_\mathcal{M}(T_1,T_2):=\sup \bigl\{ H\bigl(T_1(x),T_2(x)\bigr): x \in K\bigr\} ,\quad T_1,T_2 \in\mathcal{M}. $$
(9.36)

It is not difficult to verify that the metric space \((\mathcal{M},\rho_{\mathcal{M}})\) is complete.

For each \(T \in\mathcal{M}\), set \(P(T)=\tilde{T}\). It is easy to see that for each \(T_{1},T_{2} \in\mathcal{M}\),

$$ \rho_{\mathfrak{A}}\bigl(P(T_1),P(T_2) \bigr)=\rho_\mathcal{M}(T_1,T_2). $$
(9.37)

Denote

$$ {\mathfrak{B}}=\bigl\{ P(T): T \in\mathcal{M}\bigr\} . $$
(9.38)

It is clear that the metric spaces \(({\mathfrak{B}}, \rho_{\mathfrak{A}})\) and \((\mathcal{M}, \rho_{\mathcal{M}})\) are isometric.

For each \(T \in{ \mathfrak{A}}\) and each γ>0, define

$$T_{\gamma}(A)=(1-\gamma)T(A) \oplus\gamma{\theta}. $$

It is easy to see that \(T_{\gamma} \in{ \mathfrak{A}}\) for each \(T \in {\mathfrak{A}} \) and each γ>0, and moreover, \(T_{\gamma} \in{ \mathfrak{B}}\) if \(T \in{ \mathfrak{B}}\). Now we can apply Theorem 9.4 and obtain the following result.

Theorem 9.7

There exists a set \(\mathcal{F}\) which is a countable intersection of open and everywhere dense subsets of \((\mathcal{M},\rho_{\mathcal{M}})\) such that each \(T \in\mathcal{F}\) is contractive.

Theorem 3.1 and Proposition 9.6 imply the following result.

Theorem 9.8

Assume that the operator \(T \in\mathcal{M}\) is contractive. Then there exists a unique set A T ∈S(K) such that \(\tilde{T}(A_{T})=A_{T}\) and \((\tilde{T})^{n}(B) \to A_{T}\) as n→∞, uniformly for all B∈S(K).

Let \(T \in\mathcal{M}\). A sequence \(\{x_{n}\}_{n=1}^{N} \subset K\) with N≥1 (respectively, \(\{x_{n}\}_{n=1}^{\infty} \subset K\)) is called a trajectory of T if x i+1∈T(x i ), i=1,…,N−1 (respectively, i=1,2,…).

Theorem 9.8 leads to the following results.

Theorem 9.9

Let the operator \(T \in\mathcal{M}\) be contractive and let the set A T ∈S(K) be as guaranteed by Theorem 9.8. Then for each ε>0, there exists a natural number n such that for each trajectory \(\{x_{i}\}_{i=1}^{n} \subset K\) of T, ρ(x n ,A T )<ε.

Theorem 9.10

Let the operator \(T \in\mathcal{M}\) be contractive and let the set A T ∈S(K) be as guaranteed by Theorem 9.8. Then for each ε>0, there exists a natural number n such that for each z∈K and each x∈A T , there exists a trajectory \(\{x_{i}\}_{i=1}^{n} \subset K\) of T such that x 1=z and ρ(x n ,x)<ε.

Corollary 9.11

Let the operator \(T \in\mathcal{M}\) be contractive and let the set A T ∈S(K) be as guaranteed by Theorem 9.8. Then for each x∈A T , there is a trajectory \(\{x_{i}\}_{i=1}^{\infty} \subset A_{T}\) such that x 1=x and lim inf i→∞∥x i −x∥=0.

Corollary 9.12

Let the operator \(T \in\mathcal{M}\) be contractive and let the set A T ∈S(K) be as guaranteed by Theorem 9.8. Assume that the set A T is separable. Then for each x∈A T , there is a trajectory \(\{x_{i}\}_{i=1}^{\infty} \subset A_{T}\) such that x 1=x and for each y∈A T , lim inf i→∞∥x i −y∥=0.

9.4 Existence of Fixed Points

We consider a complete metric space of nonexpansive set-valued mappings acting on a closed and convex subset of a Banach space with a nonempty interior, and show that a generic mapping in this space has a fixed point. We then prove analogous results for two complete metric spaces of set-valued mappings with convex graphs. These results were obtained in [145].

Let (X,∥⋅∥) be a Banach space and denote by S(X) the set of all nonempty, closed and convex subsets of X. For x∈X and A⊂X, set

$$\rho(x,A)=\inf\bigl\{ \Vert x-y\Vert : y \in A\bigr\} , $$

and for each A,B∈S(X), let

$$ H(A,B)=\max\Bigl\{ \sup_{x \in A}\rho(x,B), \sup _{y \in B}\rho(y,A)\Bigr\} . $$
(9.39)

The interior of a subset A⊂X will be denoted by \(\operatorname{int}(A)\). For each x∈X and each r>0, set B(x,r)={y∈X:∥y−x∥≤r}. For the set S(X) we consider the uniformity determined by the following base:

$$ \mathcal{G}(n)=\bigl\{ (A,B) \in S(X) \times S(X): H(A,B) \le n^{-1}\bigr\} , $$
(9.40)

n=1,2,… . It is well known that the space S(X) with this uniformity is metrizable and complete. We endow the set S(X) with the topology induced by this uniformity.

Assume now that K is a nonempty, closed and convex subset of X and denote by S(K) the set of all A∈S(X) such that A⊂K. It is clear that S(K) is a closed subset of S(X). We equip the topological subspace S(K)⊂S(X) with its relative topology.

Denote by \(\mathcal{M}_{ne}\) the set of all mappings T:K→S(K) such that T(x) is bounded for all x∈K and

$$ H\bigl(T(x),T(y)\bigr) \le\Vert x-y\Vert, \quad x,y \in K. $$
(9.41)

In other words, the set \(\mathcal{M}_{ne}\) consists of those nonexpansive set-valued self-mappings of K which have nonempty, bounded, closed and convex point images.

Fix θ∈K. For the set \(\mathcal{M}_{ne}\) we consider the uniformity determined by the following base:

$$ \begin{aligned}[b] \mathcal{E}(n)={}&\bigl\{ (T_1,T_2) \in\mathcal{M}_{ne} \times\mathcal{M}_{ne}: H \bigl(T_1(x),T_2(x)\bigr) \le n^{-1} \\ &\mbox{for all } x \in K \mbox{ satisfying } \Vert x-\theta\Vert \le n\bigr\} , \quad n=1,2,\ldots. \end{aligned} $$
(9.42)

It is not difficult to verify that the space \(\mathcal{M}_{ne}\) with this uniformity is metrizable and complete.

The following result is well known [45, 102]; see also [116].

Theorem 9.13

Assume that T:K→S(K), γ∈(0,1), and

$$H\bigl(T(x),T(y)\bigr) \le\gamma\Vert x-y\Vert, \quad x,y \in K. $$

Then there exists x T ∈K such that x T ∈T(x T ).

The existence of fixed points for set-valued mappings which are merely nonexpansive is more delicate and was studied by several authors. See, for example, [67, 94, 119] and the references therein.

We prove the following result which shows that if \(\operatorname{int}(K)\) is nonempty, then a generic nonexpansive mapping does have a fixed point.

Theorem 9.14

Assume that \(\operatorname{int}(K) \neq \emptyset\). Then there exists an open and everywhere dense set \(\mathcal{F} \subset \mathcal{M}_{ne}\) with the following property: for each \(\widehat{S} \in\mathcal{F}\), there exist \(\bar{x} \in K\) and a neighborhood \(\mathcal{U}\) of \(\widehat{S}\) in \(\mathcal{M}_{ne}\) such that \(\bar{x} \in S(\bar{x})\) for each \(S \in\mathcal{U}\).

For our second result we assume, in addition, that the closed and convex subset K⊂X is bounded. Denote by \(\mathcal{M}_{a}\) the set of all mappings T:K→S(K) such that

$$ \alpha T(x_1)+(1-\alpha)Tx_2 \subset T \bigl(\alpha x_1+(1-\alpha)x_2\bigr) $$
(9.43)

for each x 1,x 2∈K and all α∈(0,1). In other words, the set \(\mathcal{M}_{a}\) consists of all set-valued self-mappings of K with convex graphs. Note that convex-valued mappings and, in particular, mappings with convex graphs, as well as spaces of convex sets, find application in several areas of mathematics. See, for example, [54, 90, 92, 166, 168, 169, 177] and the references mentioned there. We denote by \(\mathcal{M}_{ac}\) the set of all those continuous mappings T:K→S(K) which belong to \(\mathcal{M}_{a}\).

For the set \(\mathcal{M}_{a}\) we consider the uniformity determined by the following base:

$$ \begin{aligned}[b] \mathcal{E}_a(n)={}&\bigl\{ (T_1,T_2) \in\mathcal{M}_a \times\mathcal{M}_a: H \bigl(T_1(x),T_2(x)\bigr)\le n^{-1} \\ &\mbox{for all } x \in K\bigr\} , \quad n=1,2,\ldots. \end{aligned} $$
(9.44)

It is easy to see that the space \(\mathcal{M}_{a}\) with this uniformity is metrizable and complete. It is clear that \(\mathcal{M}_{ac}\) is a closed subset of \(\mathcal{M}_{a}\). We endow the topological subspace \(\mathcal{M}_{ac} \subset\mathcal{M}_{a}\) with its relative topology and prove the following result [145].

Theorem 9.15

Assume that K is bounded and \(\operatorname{int}(K) \neq \emptyset\). Then there exists an open and everywhere dense subset \(\mathcal{F}_{a}\) of \(\mathcal{M}_{a}\) with the following property: for each \(\widehat{S} \in\mathcal{F}_{a}\), there exist \(\bar{x} \in K\) and a neighborhood \(\mathcal{U}\) of \(\widehat{S}\) in \(\mathcal{M}_{a}\) such that \(\bar{x} \in S(\bar{x})\) for each \(S \in\mathcal{U}\).

Moreover, \(\mathcal{F} _{a}\) contains an open and everywhere dense subset \(\mathcal{F}_{ac}\) of \(\mathcal{M}_{ac}\).

Usually a generic result is obtained when it is shown that the set of “good” points in a complete metric space contains a dense G δ subset. Note that our results are stronger because in each one of them we construct an open and everywhere dense subset of “good” points.

In both Theorems 9.14 and 9.15 we assume that the interior of K is nonempty. The following proposition, which will be proved in the next section, shows that this situation is typical.

Proposition 9.16

The set of all elements of S(X) (respectively, S b (X)) with a nonempty interior contains an open and everywhere dense subset of S(X) (respectively, S b (X)).

9.5 An Auxiliary Result and the Proof of Proposition 9.16

We need the following auxiliary result (see Proposition 5.1 of [179] for the finite dimensional case). If (Y,∥⋅∥) is a normed linear space, x∈Y and r>0, then we denote by B(x,r) the closed ball of radius r centered at x.

Lemma 9.17

Let (Y,∥⋅∥) be a normed linear space and let r>0 be given. Assume that C is a closed and convex subset of Y such that for all y∈B(0,r),

$$ \inf_{x \in C}\Vert y-x\Vert \le r. $$
(9.45)

Then 0∈C.

Proof

If 0∉C, then by the separation theorem there exists a bounded linear functional l∈Y ∗ such that ∥l∥=1 and

$$p=\inf\bigl\{ l(x): x \in C\bigr\} >0. $$

There is y 0∈B(0,r) such that l(−y 0)>r−p/2. By (9.45), there is x 0∈C such that ∥y 0−x 0∥<r+p/2. Now we have

$$ \begin{aligned} p &\le l(x_0)=l(y_0)+l(x_0-y_0) < -r+p/2+\Vert x_0-y_0\Vert \\ &< -r +p/2+ r +p/2=p. \end{aligned} $$

Since we have reached a contradiction, we conclude that the origin does belong to C. □

Proof of Proposition 9.16

Let A∈S(X) and ε>0 be given. Denote by \(\tilde{A}\) the closure of the set A+{y∈X:∥y∥≤ε}. Clearly, \(\tilde{A} \in S(X)\) (if A∈S b (X), then \(\tilde{A} \in S_{b}(X)\)) and \(H(A,\tilde{A}) \le \varepsilon \). To complete the proof, it is sufficient to show that each B∈S(X) for which \(H(\tilde{A},B) \le \varepsilon /2 \) has a nonempty interior.

To this end, let B∈S(X) and \(H(B,\tilde{A}) \le \varepsilon /2\). We claim that each point of A belongs to the interior of B. To see this, let x∈A and y∈B(x,ε/2).

Then B−y is a closed and convex subset of X, \(B(0,\varepsilon /2 ) \subset\tilde{A}-y\) and \(H(B-y,\tilde{A}-y) \le \varepsilon /2\). By Lemma 9.17, 0∈B−y and y∈B. Thus B(x,ε/2)⊂B. This completes the proof of Proposition 9.16. □

9.6 Proof of Theorem 9.14

Fix \(x_{*} \in \operatorname{int}(K)\). There exists r ∗∈(0,1) such that

$$ B(x_*,r_*) \subset K. $$
(9.46)

Let \(T \in\mathcal{M}_{ne}\) and γ∈(0,1) be given. Define T γ :K→S(K) by

$$ T_{\gamma}(x)=(1-\gamma)Tx+\gamma x_*,\quad x \in K. $$
(9.47)

It is clear that \(T_{\gamma} \in\mathcal{M}_{ne}\) and H(T γ (x),T γ (y))≤γ∥x−y∥ for all x,y∈K. By Theorem 9.13, there exists a point x T,γ ∈K such that

$$ T_{\gamma}(x_{T,\gamma})=x_{T,\gamma}. $$
(9.48)

Consider the set

$$T_{\gamma}(K)=\bigcup\bigl\{ T_{\gamma}(y): y \in K\bigr\} \subset\bigl\{ (1-\gamma)y+ \gamma x_*: y \in K\bigr\} . $$

It follows from this inclusion and (9.46) that for each z∈T γ (K),

$$ B(z,\gamma r_*) \subset K. $$
(9.49)

For each x∈K, denote by \(\tilde{T}_{\gamma}(x)\) the closure of T γ (x)+B(0,γr ∗) in the norm topology. By (9.49), \(\tilde{T}_{\gamma}(x) \in S(K)\) for all x∈K. It is easy to see that \(\tilde{T}_{\gamma} \in\mathcal{M}_{ne}\). By (9.48),

$$ B(x_{T,\gamma},\gamma r_*) \subset\tilde{T}_{\gamma}(x_{T,\gamma}). $$
(9.50)

Since the point images of the nonexpansive mapping T are bounded, the image under T of any bounded subset of K is also bounded. Therefore \(\tilde{T}_{\gamma} \to T\) as γ→0+.

Let \(T \in\mathcal{M}_{ne}\) and γ∈(0,1). There exists an open neighborhood U(T,γ) of \(\tilde{T}_{\gamma}\) in \(\mathcal{M}_{ne}\) such that for each S∈U(T,γ),

$$ H\bigl(\tilde{T}_{\gamma}(x_{T,\gamma}),S(x_{T,\gamma}) \bigr) \le\gamma r_*. $$
(9.51)

Define

$$\mathcal{F}:=\bigcup\bigl\{ U(T,\gamma): T \in\mathcal{M}_{ne}, \gamma\in (0,1)\bigr\} . $$

It is clear that \(\mathcal{F}\) is an open and everywhere dense subset of \(\mathcal{M}_{ne}\).

Assume that \(\widehat{S} \in\mathcal{F}\). There exist a mapping \(T \in \mathcal{M}_{ne}\) and a number γ∈(0,1) such that \(\widehat{S} \in U( T,\gamma)\). Let S∈U(T,γ). Then (9.51) and (9.50) hold. Consider now the sets \(\tilde{T}_{\gamma}(x_{T,\gamma})-x_{T,\gamma}\) and S(x T,γ )−x T,γ . By (9.51),

$$ H\bigl(\tilde{T}_{\gamma}(x_{T,\gamma})-x_{T,\gamma}, S(x_{T,\gamma})-x_{T,\gamma}\bigr) \le\gamma r_*. $$
(9.52)

By (9.50),

$$ B(0,\gamma r_*) \subset \tilde{T}_{\gamma}(x_{T,\gamma})-x_{T,\gamma}. $$
(9.53)

It follows from (9.52), (9.53) and Lemma 9.17 that 0∈S(x T,γ )−x T,γ . In other words, x T,γ ∈S(x T,γ ) and Theorem 9.14 is proved.

9.7 Proof of Theorem 9.15

Lemma 9.18

Let \(T \in\mathcal{M}_{a}\) and ε>0 be given. Then there exist points z 1∈K and z 2∈T(z 1) such that ∥z 1−z 2∥≤ε.

Proof

Consider any sequence \(\{y_{i}\}_{i=1}^{\infty} \subset K\) such that y i+1∈T(y i ), i=0,1,… . Choose a natural number n such that

$$n \varepsilon >2\sup\bigl\{ \Vert x\Vert : x \in K\bigr\} . $$

Set \(z_{1}=n^{-1}\sum_{i=0}^{n-1}y_{i}\) and \(z_{2}=n^{-1}\sum_{i=1}^{n} y_{i}\). It is clear that z 2∈T(z 1). By the choice of n,

$$\Vert z_1-z_2\Vert \le n^{-1}\Vert y_n-y_0\Vert \le2n^{-1} \sup\bigl\{ \Vert x\Vert : x \in K\bigr\} <\varepsilon , $$

as asserted. Lemma 9.18 is proved. □

Fix \(x_{*} \in \operatorname{int}(K)\). There exists r ∗∈(0,1) such that

$$ B(x_*,r_*) \subset K. $$
(9.54)

Let \(T \in\mathcal{M}_{a}\) and γ∈(0,1) be given. Define T γ :K→S(K) by

$$ T_{\gamma}(x)=(1-\gamma)Tx+\gamma x_*,\quad x \in K. $$
(9.55)

It is obvious that \(T_{\gamma} \in\mathcal{M}_{a}\) and \(T_{\gamma} \in\mathcal{M}_{ac}\) if \(T \in\mathcal{M}_{ac}\).

Consider now the set

$$T_{\gamma}(K)=\bigcup\bigl\{ T_{\gamma}(y): y \in K\bigr\} \subset\bigl\{ (1-\gamma)y+ \gamma x_*: y \in K\bigr\} . $$

It follows from this inclusion and (9.54) that for each z∈T γ (K),

$$ B(z,\gamma r_*) \subset K. $$
(9.56)

For each x∈K denote by \(\tilde{T}_{\gamma}(x)\) the closure of T γ (x)+B(0,γr ∗) in the norm topology. Clearly, \(\tilde{T}_{\gamma} \in\mathcal{M}_{a}\) and \(\tilde{T}_{\gamma} \in \mathcal{M}_{ac}\) if \(T \in\mathcal{M}_{ac}\). By Lemma 9.18, there exist x T,γ ∈K and \(\bar{x}_{T,\gamma} \in T_{\gamma}(x_{T,\gamma})\) such that

$$\Vert \bar{x}_{T,\gamma}-x_{T,\gamma}\Vert \le2^{-1}\gamma r_*. $$

It follows from this inequality and the definition of \(\tilde{T}_{\gamma} (x_{T,\gamma})\) that

$$ B\bigl(x_{T,\gamma},2^{-1}\gamma r_*\bigr) \subset \tilde{T}_{\gamma}(x_{T,\gamma}). $$
(9.57)

There exists an open neighborhood U(T,γ) of \(\tilde{T}_{\gamma}\) in \(\mathcal{M}_{a}\) such that for each S∈U(T,γ),

$$ H\bigl(\tilde{T}_{\gamma}(x_{T,\gamma}),S(x_{T,\gamma}) \bigr) \le 2^{-1}\gamma r_*. $$
(9.58)

Note that \(\tilde{T}_{\gamma} \to T\) as γ→0+.

Define

$$\mathcal{F}_a:=\bigcup\bigl\{ U(T,\gamma): T \in\mathcal{M}_a, \gamma\in (0,1)\bigr\} $$

and

$$\mathcal{F}_{ac}:=\Bigl[\bigcup\bigl\{ U(T,\gamma): T \in\mathcal{M}_{ac}, \gamma\in(0,1)\bigr\} \Bigr] \cap\mathcal{M}_{ac}. $$

It is clear that \(\mathcal{F}_{a}\) is an open and everywhere dense subset of \(\mathcal{M}_{a}\), and \(\mathcal{F}_{ac}\) is an open and everywhere dense subset of \(\mathcal{M}_{ac}\).

Assume that \(\widehat{S} \in\mathcal{F}_{a}\). There exist \(T \in \mathcal{M}_{a}\) and γ∈(0,1) such that \(\widehat{S} \in U( T,\gamma)\). Let S∈U(T,γ). Then (9.58) and (9.57) hold. Consider the sets \(\tilde{T}_{\gamma}(x_{T,\gamma})-x_{T,\gamma}\) and S(x T,γ )−x T,γ . By (9.58),

$$ H\bigl(\tilde{T}_{\gamma}(x_{T,\gamma})-x_{T,\gamma}, S(x_{T,\gamma})-x_{T,\gamma}\bigr) \le2^{-1}\gamma r_*. $$
(9.59)

By (9.57),

$$ B\bigl(0,2^{-1}\gamma r_*\bigr) \subset \tilde{T}_{\gamma}(x_{T,\gamma})-x_{T,\gamma}. $$
(9.60)

It follows from (9.59), (9.60) and Lemma 9.17 that 0∈S(x T,γ )−x T,γ and x T,γ ∈S(x T,γ ). This completes the proof of Theorem 9.15.

9.8 An Extension of Theorem 9.15

Consider the complete uniform space S(X) defined in the previous section. Assume that K is a nonempty, closed and convex (not necessarily bounded) subset of X. Denote by \({\mathfrak{M}}_{a}\) the set of all mappings T:K→S(X) such that

$$ \alpha Tx_1+ (1-\alpha)Tx_2 \subset T \bigl(\alpha x_1+(1-\alpha)x_2\bigr) $$
(9.61)

for all x 1,x 2∈K and each α∈(0,1). As we have already mentioned, such mappings find application in many areas. We denote by \({\mathfrak{M}}_{ac}\) the set of all continuous mappings T:K→S(X) which belong to \({\mathfrak{M}}_{a}\).

For the set \({\mathfrak{M}}_{a}\) we consider two uniformities, strong and weak, and the strong and weak topologies generated by them. (The weak uniformity is weaker than the strong one.) The strong uniformity is determined by the following base:

$$ \begin{aligned}[b] \mathcal{E}_s(n)={}&\bigl\{ (T_1,T_2) \in{ \mathfrak{M}}_a \times{ \mathfrak{M}}_a: H \bigl(T_1(x),T_2(x)\bigr)\le n^{-1} \\ &\mbox{for all } x \in K\bigr\} ,\quad n=1,2,\ldots. \end{aligned} $$
(9.62)

It is not difficult to see that the space \({\mathfrak{M}}_{a}\) with this uniformity is metrizable and complete, and that \({\mathfrak{M}}_{ac}\) is a closed subset of \({\mathfrak{M}}_{a}\).

Fix θ∈K. For the set \({\mathfrak{M}}_{a}\) we also consider the weak uniformity determined by the following base:

$$ \begin{aligned}[b] \mathcal{E}_w(n)={}&\bigl\{ (T_1,T_2) \in{ \mathfrak{M}}_a \times{ \mathfrak{M}}_a: H\bigl(T_1(x),T_2(x)\bigr) \le n^{-1} \\ &\mbox{for all } x \in K \mbox{ satisfying } \Vert x-\theta\Vert \le n\bigr\} , \quad n=1,2,\ldots. \end{aligned} $$
(9.63)

It is not difficult to verify that the space \({\mathfrak{M}}_{a}\) with this weaker uniformity is also metrizable and complete, and that \({\mathfrak{M}}_{ac}\) is, once again, a closed subset of \({\mathfrak{M}}_{a}\).

Denote by \({\mathfrak{M}}_{a}^{*}\) the set of all \(T \in{ \mathfrak{M}}_{a}\) such that there exists a bounded sequence \(\{x_{i}\}_{i=0}^{\infty} \subset K\) with x i+1∈T(x i ), i=0,1,… . Set \({\mathfrak{M}}_{ac}^{*} ={\mathfrak{M}}_{a}^{*} \cap{ \mathfrak{M}}_{ac}\). Denote by \(\bar {\mathfrak{M}}_{a}^{*s}\) the closure of \({\mathfrak{M}}_{a}^{*}\) in the space \({\mathfrak{M}}_{a}\) with the strong topology, by \(\bar{\mathfrak{M}}_{a}^{*w}\) the closure of \({\mathfrak{M}}_{a}^{*}\) in the space \({\mathfrak{M}}_{a}\) with the weak topology, by \(\bar{\mathfrak{M}}_{ac}^{*s}\) the closure of \({\mathfrak{M}}_{ac}^{*}\) in the space \({\mathfrak{M}}_{a}\) with the strong topology and by \(\bar{\mathfrak{M}}_{ac}^{*w}\) the closure of \({\mathfrak{M}}_{ac}^{*}\) in the space \({\mathfrak{M}}_{a}\) with the weak topology. We equip the topological subspaces \(\bar{\mathfrak{M}}_{a}^{*s}\), \(\bar {\mathfrak{M}}_{a}^{*w}\), \(\bar{\mathfrak{M}}_{ac}^{*s}\), \(\bar{\mathfrak{M}}_{ac}^{*w} \subset{ \mathfrak{M}}_{a}\) with both the weak and strong relative topologies.

In this section we prove the following result [145].

Theorem 9.19

There exists an open everywhere dense (in the weak topology) subset \(\mathcal{F}_{a}^{w}\) of \(\bar{\mathfrak{M}}_{a}^{*w}\) with the following property: for each \(A \in\mathcal{F}_{a}^{w}\), there exist z ∗∈K and a neighborhood \(\mathcal{W}\) of A in \({\mathfrak{M}}_{a}\) with the weak topology such that z ∗∈S(z ∗) for each \(S \in\mathcal{W}\). Moreover, there exists an open (in the weak topology) and everywhere dense (in the strong topology) subset \(\mathcal{F}_{a}^{s}\) of \(\bar{\mathfrak{M}}_{a}^{*s}\), an open (in the weak topology) and everywhere dense (in the strong topology) subset \(\mathcal{F}_{ac}^{s}\) of \(\bar{\mathfrak{M}}_{ac}^{*s}\), and an open everywhere dense (in the weak topology) subset \(\mathcal{F}_{ac}^{w}\) of \(\bar{\mathfrak{M}}_{ac}^{*w}\) such that \(\mathcal{F}^{s}_{ac} \subset \mathcal{F}_{a}^{s} \subset\mathcal{F}_{a}^{w}\) and \(\mathcal{F}_{ac}^{s} \subset \mathcal{F}_{ac}^{w} \subset\mathcal{F}_{a}^{w}\).

In the proof of Theorem 9.19 we will use the following auxiliary result (cf. Lemma 9.18).

Lemma 9.20

Let \(T \in{ \mathfrak{M}}^{*}_{a}\) and ε>0 be given. Then there exist z 1∈K and z 2∈T(z 1) such that ∥z 1−z 2∥≤ε.

Proof of Theorem 9.19

Let \(T \in{ \mathfrak{M}}_{a}\) and γ∈(0,1) be given. For each x∈K, denote by T γ (x) the closure of Tx+B(0,γ) in the norm topology. Clearly, \(T_{\gamma} \in{ \mathfrak{M}}_{a}\) and \(T_{\gamma} \in{ \mathfrak{M}}_{ac}\) if \(T \in{ \mathfrak{M}}_{ac}\). It is easy to see that for each \(T \in {\mathfrak{M}}_{a}\), T γ →T as γ→0+ in the strong topology.

Let \(T \in{ \mathfrak{M}}_{a}^{*}\) and γ∈(0,1). By Lemma 9.20, there exists x T,γ ∈K such that

$$ B\bigl(x_{T,\gamma},2^{-1}\gamma\bigr) \subset T_{\gamma}(x_{T,\gamma}). $$
(9.64)

There also exists an open neighborhood U(T,γ) of T γ in \({\mathfrak{M}}_{a}\) with the weak topology such that for each S∈U(T,γ),

$$ H\bigl(T_{\gamma}(x_{T,\gamma}),S(x_{T,\gamma}) \bigr) \le 2^{-1}\gamma. $$
(9.65)

Define

$$\begin{gathered} \mathcal{F}_a^s:=\Bigl[\bigcup\bigl\{ U(T,\gamma): T \in{ \mathfrak{M}}_a^*, \gamma\in(0,1)\bigr\} \Bigr] \cap\bar{\mathfrak{M}}_a^{*s}, \\ \mathcal{F}_a^w:=\Bigl[\bigcup\bigl\{ U(T,\gamma): T \in{ \mathfrak{M}}_a^*, \gamma\in(0,1)\bigr\} \Bigr] \cap\bar{\mathfrak{M}}_a^{*w}, \\ \mathcal{F}_{ac}^s:=\Bigl[\bigcup\bigl\{ U(T,\gamma): T \in{ \mathfrak{M}}_{ac}^*, \gamma\in(0,1)\bigr\} \Bigr] \cap\bar{\mathfrak{M}}_{ac}^{*s} \end{gathered}$$

and

$$\mathcal{F}_{ac}^w:=\Bigl[\bigcup\bigl\{ U(T,\gamma): T \in{ \mathfrak{M}}_{ac}^*, \gamma\in(0,1)\bigr\} \Bigr] \cap\bar{\mathfrak{M}}_{ac}^{*w}. $$

Clearly, \(\mathcal{F}^{s}_{ac} \subset\mathcal{F}_{a}^{s} \subset \mathcal{F}_{a}^{w}\) and \(\mathcal{F}_{ac}^{s} \subset\mathcal{F}_{ac}^{w} \subset \mathcal{F}_{a}^{w}\). It is easy to see that \(\mathcal{F}_{a}^{s}\) is an open (in the weak topology) and everywhere dense (in the strong topology) subset of \(\bar{\mathfrak{M}}_{a}^{s*}\), \(\mathcal{F}_{a}^{w}\) is an open everywhere dense (in the weak topology) subset of \(\bar{\mathfrak{M}}_{a}^{w*}\), \(\mathcal{F}_{ac}^{s}\) is an open (in the weak topology) and everywhere dense (in the strong topology) subset of \(\bar{\mathfrak{M}}_{ac}^{*s}\), and \(\mathcal{F}^{w}_{ac}\) is an open everywhere dense (in the weak topology) subset of \(\bar{\mathfrak{M}}_{ac}^{*w}\).

Assume that \(A\in\mathcal{F}_{a}^{w}\). Then there exist \(T \in{ \mathfrak{M}}_{a}^{*}\) and γ∈(0,1) such that A∈U(T,γ). By (9.64),

$$ B\bigl(0,2^{-1}\gamma\bigr) \subset T_{\gamma}(x_{T,\gamma})-x_{T,\gamma}. $$
(9.66)

Let S∈U(T,γ). By (9.65),

$$ H\bigl(T_{\gamma}(x_{T,\gamma})-x_{T,\gamma}, S(x_{T,\gamma})-x_{T,\gamma}\bigr) \le2^{-1}\gamma. $$
(9.67)

It follows from (9.66), (9.67) and Lemma 9.17 that 0∈S(x T,γ )−x T,γ and x T,γ ∈S(x T,γ ). This completes the proof of Theorem 9.19. □

9.9 Generic Existence of Fixed Points

Let (X,d) be a complete metric space. For x∈X and a nonempty subset A of X, set d(x,A)=inf a∈A d(x,a).

In the space X, an open ball and a closed ball of center a and radius r>0 are denoted by S X (a,r) and S X [a,r], respectively.

Set

$$\mathcal{B}(X)=\{A \subset X: A \mbox{ is nonempty closed and bounded}\}. $$

The space \(\mathcal{B}(X)\) is equipped with the Hausdorff metric

$$h(A,B)=\max\Bigl\{ \sup_{a \in A}d(a,B), \sup_{b \in B}d(b,A) \Bigr\} ,\quad A,B \in\mathcal{B}(X). $$

Note that h(⋅,⋅) is, in fact, defined for all pairs of nonempty subsets of X (not necessarily bounded and closed).

A map \(F:X \to\mathcal{B}(X)\) is said to be nonexpansive (respectively, strictly contractive with a constant L F ∈[0,1)) if it satisfies

$$\begin{aligned} &h\bigl(F(x),F(y)\bigr) \le d(x,y) \qquad\bigl(\mbox{resp. } h\bigl(F(x),F(y) \bigr) \le L_Fd(x,y)\bigr) \\ &\quad\mbox{for all } x,y \in X. \end{aligned} $$

The set \(\operatorname{fix}(F)=\{x \in X: x \in F(x)\} \) is called the fixed point set of F.

We say that most (or typical) elements of X have a given property P if the set \(\tilde{X}\) of all x∈X having P is residual in X, i.e., \(X \setminus\tilde{X}\) is of the first Baire category in X.

Let E be a real Banach space with norm ∥⋅∥. Set

$$\mathcal{X}(E)=\{ A \subset E: A \mbox{ is nonempty and compact}\} $$

and

$$\mathcal{E} (E)=\{A \subset E: A \mbox{ is nonempty, compact and convex}\}. $$

The spaces \(\mathcal{X} (E)\) and \(\mathcal{E}(E)\) are equipped with the Hausdorff metric h under which each one of them is complete.

For any star-shaped set A⊂E, \(\operatorname{st}(A)\) denotes the set of all a∈A such that ta+(1−t)x∈A for every x∈A and t∈[0,1].

In this section we prove that most compact-valued nonexpansive map from a closed bounded star-shaped subset of a Banach space E into itself have fixed points. This result was obtained in [53].

Let E be a real Banach space. For a nonempty, closed, bounded and star-shaped set D⊂E, define

$$\mathcal{X}_D=\bigl\{ A \in\mathcal{X}(E): A \subset D\bigr\} . $$

Under the Hausdorff metric h the space \(\mathcal{X}_{D}\) is complete. Set

$$\begin{gathered} \mathcal{M}=\{F:D \to\mathcal{X}_D: F \mbox{ is nonexpansive}\}, \\ \mathcal{N}=\{G : D \to\mathcal{X}_D: G \mbox{ is strictly contractive}\}. \end{gathered}$$

The space \(\mathcal{M}\) is equipped with the metric of uniform convergence

$$ \rho(F_1,F_2)=\sup_{x \in D}h \bigl(F_1(x),F_2(x)\bigr),\quad F_1,F_2 \in\mathcal{M} $$
(9.68)

under which it is complete.

Given \(F:D \to\mathcal{X}_{D}\) and \(A \in\mathcal{X}_{D}\), set

$$ \varPhi _F(A)=\bigcup_{x \in A}F(x). $$
(9.69)

Lemma 9.21

Let \(F:D \to\mathcal{X}_{D}\) satisfy

$$h\bigl(F(x),F(y)\bigr)\le L_F\Vert x-y\Vert \quad (L_f \ge0) \ \textit{for all}\ x,y \in D. $$

Then (9.69) defines a map \(\varPhi _{F}:\mathcal{X}_{D} \to\mathcal{X}_{D}\) satisfying

$$ h\bigl(\phi_F(A),\phi_F(B)\bigr)\le L_Fh(A,B) \quad\textit{for all}\ A,B \in \mathcal{X}_D. $$
(9.70)

Proof

It is evident that \(\varPhi _{F}(A) \in\mathcal{X}_{D}\) for each \(A \in\mathcal{X}_{D}\). To prove (9.70), let \(A,B \in \mathcal{X}_{D}\). Let u∈ϕ F (A). Then u∈F(x) for some x∈A. Since B is compact, there is a point y∈B such that ∥x−y∥=d(x,B). We have

$$ \begin{aligned} d\bigl(u,\varPhi _F(B)\bigr)&\le d\bigl(u,F(y)\bigr) \le d\bigl(u,F(x)\bigr)+h\bigl(F(x),F(y)\bigr) \\ &\le L_F\Vert x-y\Vert \le L_Fd(x,B). \end{aligned} $$

Thus

$$ d\bigl(u,\varPhi _F(B)\bigr)\le L_Fh(A,B) \quad\mbox{for each } u \in \varPhi _F(A) $$
(9.71)

and similarly,

$$ d\bigl(u,\varPhi _F(A)\bigr) \le L_Fh(A,B) \quad\mbox{for each } u \in \varPhi _F(B). $$
(9.72)

Combining (9.71) and (9.72), we get (9.70), as asserted. □

Lemma 9.22

Let \(F,G \in\mathcal{M}\) be such that ρ(F,G)<δ, where δ>0. Then

$$ h\bigl(\varPhi _F(A),\varPhi _G(A)\bigr)<\delta \quad\textit{for each}\ A \in\mathcal{X}_D. $$
(9.73)

Proof

Let \(A \in\mathcal{X}_{D}\) and ε>0 be given. Since A is compact, and F and G are uniformly continuous, there exist a finite set \(\{a_{i}\}_{i=1}^{N} \subset A\) and σ>0 such that, setting A i =A∩S D [a i ,σ], one has

$$h\bigl(F(x),F(a_i)\bigr)\le \varepsilon ,\qquad h\bigl(G(x),G(a_i) \bigr) \le \varepsilon \quad\mbox{for every } x \in A_i, i=1,2,\ldots,N. $$

Hence

$$h\bigl(\varPhi _F(A_i),F(a_i)\bigr) \le \varepsilon ,\qquad h\bigl(\varPhi _G(A_i),G(a_i) \bigr)\le \varepsilon ,\quad i=1,\ldots,N. $$

Therefore

$$ \begin{aligned} h\bigl(\varPhi _F(A),\varPhi _G(A)\bigr)={}&h \Biggl(\bigcup_{i=1}^N\varPhi _F(A_i), \bigcup_{i=1}^N\phi_G(A_i) \Biggr)\le\max_{1 \le i \le N}h\bigl(\varPhi _F(A_i), \varPhi _G(A_i)\bigr) \\ \le{}&\max_{1 \le i \le N}\bigl[h\bigl(\varPhi _F(A_i),F(a_i) \bigr)+h\bigl(F(a_i),G(a_i)\bigr) \\ &{}+h\bigl(G(a_i), \varPhi _G(A_i)\bigr)\bigr] \\ \le{}& 2\varepsilon +\rho(F,G), \end{aligned} $$

which implies h(Φ F (A),Φ G (A))≤ρ(F,G)<δ. Since \(A \in\mathcal{X}_{D}\) is arbitrary, inequality (9.73) indeed holds as claimed. □

Lemma 9.23

The set \(\mathcal{N}\) is dense in \(\mathcal{M}\).

Proof

Let \(F \in\mathcal{M}\). For a natural number n, define \(G_{n}:D \to\mathcal{X}_{D}\) by

$$G_n(x)=n^{-1}a+\bigl(1-n^{-1}\bigr)F(x),\quad x \in D, $$

where \(a \in\operatorname{st}(D)\). Since \(G_{n} \in\mathcal{N}\) and ρ(G n ,F)→0 as n→∞, the result follows. □

Lemma 9.24

Let \(G \in\mathcal{N}\) and let ε>0 be given. Then there exists 0<δ G (ε)<ε such that

$$ \begin{aligned}[b] &\textit{if}\ F \in S_\mathcal{M}\bigl(G, \delta_{G}(\varepsilon )\bigr),\ \textit{then}\ h\bigl(\varPhi _F^n(A), \varPhi _G^n(A)\bigr)<\varepsilon \ \textit{for every}\ A \in \mathcal{X}_D \\ &\textit{and all natural numbers}\ n. \end{aligned} $$
(9.74)

Proof

Let \(G \in\mathcal{N}\) be strictly contractive with constant 0≤L G <1 and let ε>0 be given. By Lemma 9.21, \(\varPhi _{G}:\mathcal{X}_{D} \to\mathcal{X}_{D}\) is strictly contractive with the same constant L G . We claim that (9.74) holds with δ G (ε)=δ, where 0<δ<(1−L G )ε.

Let \(F \in S_{\mathcal{M}}(G,\delta)\). By Lemma 9.22, (9.73) is satisfied. Let \(A \in\mathcal{X}_{D}\) be arbitrary. By (9.73),

$$h\bigl(\varPhi _F\bigl(\varPhi _F(A)\bigr), \varPhi _G\bigl(\varPhi _F(A)\bigr)\bigr) \le\delta $$

and thus

$$ \begin{aligned} h\bigl(\varPhi ^2_F(A), \varPhi ^2_G(A)\bigr) \le{}&h\bigl(\varPhi _F\bigl( \varPhi _F(A)\bigr),\varPhi _G\bigl(\varPhi _F(A)\bigr) \bigr) \\ &{}+h\bigl(\varPhi _G\bigl(\varPhi _F(A)\bigr),\varPhi _G\bigl(\varPhi _G(A)\bigr)\bigr) \\ <{}&\delta+L_Gh\bigl(\varPhi _F(A),\varPhi _G(A) \bigr)\le\delta(1+L_g). \end{aligned} $$

Using induction, we obtain, for any natural number n,

$$h\bigl(\varPhi _F^n(A),\varPhi _G^n(A) \bigr) \le\delta\bigl(1+L_G+\cdots+L_G^{n-1} \bigr). $$

Thus

$$ \begin{aligned} &h\bigl(\varPhi _F^n(A), \varPhi _G^n(A)\bigr) \le\delta(1-L_G)^{-1} \\ &\quad\mbox{for every } A \in\mathcal{X}_D \mbox{ and any natural number } n. \end{aligned} $$

Since δ<(1−L G )ε, (9.74) holds, as claimed. □

Put

$$\mathcal{M}_0=\bigl\{ F \in\mathcal{M}: \operatorname{fix}(F) \mbox{ is compact nonempty}\bigr\} . $$

Theorem 9.25

The set \(\mathcal{M}_{0}\) is residual in \(\mathcal{M}\).

Proof

For \(G \in\mathcal{N}\) and any natural number k, let \(S_{\mathcal{M}}(G,\delta_{G}(1/k))\), where δ G (1/k)<1/k exists according to Lemma 9.24. Define

$$\mathcal{M}^*:=\bigcap_{k=1}^{\infty}\bigcup _{G \in\mathcal{N}}S_\mathcal{M}\bigl(G,\delta_G(1/k) \bigr). $$

Clearly, \(\mathcal{M}^{*}\) is residual in \(\mathcal{M}\), since \(\mathcal{M}^{*}\) is the countable intersection of sets which are open and, by Lemma 9.23, dense in \(\mathcal{M}\). The theorem is an immediate consequence of the following assertion.

Claim.

$$\mathcal{M}^* \subset\mathcal{M}_0. $$

Let \(F \in\mathcal{M}^{*}\) be given. By the definition of \(\mathcal{M}^{*}\), there exists a sequence \(\{G_{k}\}_{k=1}^{\infty} \subset\mathcal{N}\) such that

$$ F \in S_\mathcal{M}\bigl(G_k, \delta_{G_k}(1/k)\bigr) \quad\mbox{for every natural number }k. $$
(9.75)

Thus by Lemma 9.24, for each natural number k,

$$ \begin{aligned}[b] &h\bigl(\varPhi _F^n(A),\varPhi _{G_k}^n(A) \bigr)<1/k \\ &\quad\mbox{for every } A \in \mathcal{X}_D \mbox{ and every natural number } n. \end{aligned} $$
(9.76)

According to Lemma 9.21, \(\varPhi _{G_{k}}:\mathcal{X}_{D} \to\mathcal{X}_{D}\) is strictly contractive, for \(G_{k} \in\mathcal{N}\), and hence for each natural number k, there exists \(Z_{k}\in\mathcal{X}_{D}\) such that

$$Z_k=\varPhi _{G_k}(Z_k). $$
(j):

\(\{Z_{k}\}_{k=1}^{\infty} \subset\mathcal{X}_{D}\) is a Cauchy sequence.

To see this, let ε>0 be given. Let k,k′>4/ε be arbitrary natural numbers and let \(A \in\mathcal{X}_{D}\). Since \(\varPhi _{G_{k}}^{n}(A) \to Z_{k}\) and \(\varPhi _{G_{k'}}^{n}(A) \to Z_{k'}\) as n→∞, there exists a natural number m such that

$$ \begin{aligned}[b] &h\bigl(\varPhi ^n_{G_k}(A),Z_k\bigr)< \varepsilon /4,\qquad h\bigl(\varPhi _{G_{k'}}^n(A),Z_{k'} \bigr)<\varepsilon \\ &\quad\mbox{for every integer }n \ge m. \end{aligned} $$
(9.77)

In view of (9.77) and (9.76), one has

$$ \begin{aligned} h(Z_k,Z_{k'}) \le{}& h \bigl(Z_k,\varPhi _{G_k}^m(A)\bigr)+h\bigl( \varPhi _{G_k}^m(A),\varPhi _F^m(A)\bigr) \\ &{}+h\bigl(\varPhi _F^m(A), \varPhi _{G_{k'}}^m(A) \bigr)+h\bigl(\varPhi ^m_{G_{k'}}(A),Z_{k'}\bigr) \\ <{}&\varepsilon /4+1/k+\bigl(k'\bigr)^{-1} +\varepsilon /4< \varepsilon , \end{aligned} $$

for 1/k+1/k′<ε/2. As k,k′>4/ε are arbitrary, (j) is proved.

Since \(\{Z_{k}\}_{k=1}^{\infty} \subset\mathcal{X}_{D}\) is a Cauchy sequence and \(\mathcal{X}_{D}\) is a complete metric space, there exists \(Z \in\mathcal{X}_{D}\) such that Z k →Z as k→∞

(jj):

For each \(A \in\mathcal{X}_{D}\), the sequence \(\{\varPhi _{F}^{n}(A)\}\) converges to Z as n→∞. Moreover, Z=Φ F (Z) is the unique fixed point of Φ F .

Let \(A \in\mathcal{X}_{D}\). Given ε>0, fix a natural number k>3/ε large enough so that h(Z k ,Z)<ε/3. Hence by (9.76), for every natural number n, one has

$$ \begin{aligned} h\bigl(\varPhi _F^n(A),Z\bigr) &\le h\bigl( \varPhi _F^n(A),\varPhi _{G_k}^n(A)\bigr)+ h \bigl(\varPhi _{G_k}^n(A),Z_k\bigr)+ h(Z_k,Z) \\ &<1/k +h\bigl(\varPhi _{G_k}^n(A),Z_k\bigr) + \varepsilon /3. \end{aligned} $$

Since \(h(\varPhi _{G_{k}}^{n}(A),Z_{k})\) tends to zero as n→∞, there is a natural number n 0 such that \(h(\varPhi _{G_{k}}^{n}(A),Z_{k})<\varepsilon /3\) for all n≥n 0. Moreover, 1/k<ε/3, and thus

$$h\bigl(\varPhi _F^n(A),Z\bigr)<\varepsilon \quad\mbox{for every } n \ge n_0. $$

This shows that \(\varPhi _{F}^{n}(A) \to Z\) as n→∞. The second statement of (jj) is obvious.

(jjj):

The fixed point set \(\operatorname{fix}(F)\) is a nonempty compact subset of D.

First we show that the set \(\operatorname{fix}(F)\) is nonempty. As \(G_{k} \in\mathcal{N}\), by Nadler’s theorem [102], for each natural number k, there is a point a k ∈D such that

$$ a_k \in G_k(a_k), \quad k=1,2, \ldots. $$
(9.78)

For each natural number k,

$$ a_k \in \varPhi _{G_k}^n(a_k) \quad\mbox{for every natural number } n. $$
(9.79)

This is obvious if n=1 because \(\varPhi _{G_{k}}(a_{k})=G_{k}(a_{k})\). Assuming that (9.79) is valid for n, then for n+1 one has \(a_{k} \in \varPhi _{G_{k}}(a_{k} )\subset \varPhi _{G_{k}}(\varPhi _{G_{k}}^{n}(a_{k}))=\varPhi _{G_{k}}^{n+1}(a_{k})\) and thus (9.79) holds for every natural number n. Since \(\varPhi _{G_{k}}^{n} (a_{k}) \to Z_{k}\) as n→∞, it follows that a k ∈Z k . On the other hand, Z k →Z implies d(a k ,Z)→0 as k→∞. Since Z is compact, there is a subsequence \(\{a_{k_{n}}\}_{n=1}^{\infty}\) which converges to some a∈D.

We have a∈F(a). In fact, (9.75) implies that

$$\begin{aligned} &h\bigl(F(x),G_{k_n}(x)\bigr)<\delta_{G_{k_n}}(1/k_n) \\ &\quad\mbox{for every } x \in D \mbox{ and any natural number } n. \end{aligned}$$
(9.80)

In view of (9.78) and (9.80), one has

$$ \begin{aligned} d\bigl(a,F(a)\bigr) \le{}& \Vert a-a_{k_n}\Vert +d \bigl(a_{k_n},G_{k_n}(a_{k_n})\bigr) \\ &{}+h\bigl(G_{k_n}(a_{k_n}),F(a_{k_n})\bigr)+h \bigl(F(a_{k_n}),F(a)\bigr) \\ <{}&\Vert a-a_{k_n}\Vert +\delta_{G_{k_n}}(1/k_n)+h \bigl(F(a_{k_n}),F(a)\bigr)\le 2\Vert a-a_{k_n}\Vert +1/k_n \end{aligned} $$

because \(\delta_{G_{k_{n}}}(1/k_{n})<1/k_{n}\). As n→∞, the right-hand side tends to zero and thus d(a,F(a))=0, i.e. a∈F(a). Hence \(\operatorname{fix}(F)\neq\emptyset\), as claimed. It remains to show that \(\operatorname{fix} (F)\) is compact. To see this, let \(x \in \operatorname{fix} (F)\). Then \(x \in \varPhi _{F}^{n}(x)\) for every natural number n. Since by (jj), \(\varPhi _{F}^{n}(x) \to Z\) as n→∞, it follows that z∈Z. Thus \(\operatorname{fix} (F) \subset Z\), which implies that \(\operatorname{fix}(F)\) is compact for so is Z and \(\operatorname{fix}(F)\) is closed. Hence (jjj) holds. Therefore \(F \in\mathcal{M}_{0}\). This completes the proof of the claim and of Theorem 9.25 itself. □

For a nonempty, closed, bounded and star-shaped set D⊂E, let

$$ \mathcal{E}_D=\bigl\{ A \in\mathcal{E}(E): A \subset D \bigr\} . $$
(9.81)

When endowed with the Hausdorff metric h, the space \(\mathcal{E}_{d}\) is complete. Define

$$ \mathcal{U}=\{F: D \to\mathcal{E}_D: F \mbox{ is nonexpansive}\}. $$
(9.82)

The set \(\mathcal{U}\) is endowed with the metric ρ of uniform convergence (9.68) under which it is complete. Set

$$ \mathcal{U}_0=\bigl\{ F \in\mathcal{U}: \operatorname{fix}(F) \mbox{ is nonempty and compact}\bigr\} . $$
(9.83)

Using the same argument as in the proof of Theorem 9.25, one can also prove the following result.

Theorem 9.26

The set \(\mathcal{U}_{0}\) is residual in \(\mathcal{U}\).

9.10 Topological Structure of the Fixed Point Set

In this section, which is based on [53], we study the topological structure of the fixed point set for a typical compact-and convex-valued nonexpansive map from a closed, convex and bounded subset of a Banach space into itself.

Let E be a real Banach space. In this section D denotes a closed, convex and bounded subset of E with a nonempty interior \(\operatorname{int}(D)\). Set

$$S=\bigl\{ x \in E: \Vert x\Vert \le1\bigr\} . $$

Let \(\mathcal{E}_{D}\), \(\mathcal{U}\) and \(\mathcal{U}_{0}\) be given by (9.81), (9.82) and (9.83) with D as above.

Define

$$ \mathcal{U}_1:=\bigl\{ F \in\mathcal{U}: \mbox{ there is } \alpha_F \mbox{ such that } F(x)+\sigma_F S \subset D, x \in D\bigr\} . $$
(9.84)

Lemma 9.27

The set \(\mathcal{U}_{1}\) is open and dense in \(\mathcal{U}\).

Proof

First we show that \(\mathcal{U}_{1}\) is open in \(\mathcal{U}\). Let \(F \in\mathcal{U}_{1}\) and let σ F >0 be the corresponding number in (9.84). For 0<ε<σ F /2 we have \(S_{\mathcal{U}}(F,\varepsilon )\subset \mathcal{U}_{1}\). In fact, each \(G \in S_{\mathcal{U}}(F,\varepsilon )\) satisfies G(x)⊂F(z)+εS for each x∈D. Thus, taking σ G =σ F /2, one has

$$G(x)+\sigma_GS\subset F(x)+(\varepsilon +\sigma_G)S\subset F(x)+\sigma_FS \subset D \quad\mbox{for each } x \in D. $$

Hence \(S_{\mathcal{U}}(F,\varepsilon ) \subset\mathcal{U}_{1}\). Thus \(\mathcal{U}_{1}\) is indeed open in \(\mathcal{U}\).

Next we show that \(\mathcal{U}_{1}\) is dense in \(\mathcal{U}\). Let \(F \in \mathcal{U}\), 0<ε<1 and let \(a \in \operatorname{int}(D)\). Then S E [a,θ]⊂D for some θ>0. Fix λ such that 0<λ<ε/(2M), where M=sup x∈D ∥x∥+1, and define \(G:D \to\mathcal{E}_{D}\) by

$$G(x)=\lambda a+(1-\lambda)F(x), \quad x \in D. $$

Clearly, \(G \in\mathcal{U}\), and ρ(G,F)<ε since for each x∈D,

$$ \begin{aligned} h\bigl(\lambda a+(1-\lambda)F(x),F(x)\bigr)&=h\bigl(\lambda a+(1-\lambda)F(x),\lambda F(x)+(1-\lambda)F(x)\bigr) \\ &\le\lambda h\bigl(a,F(x)\bigr)<2\lambda M<\varepsilon . \end{aligned} $$

Furthermore, taking 0<σ G <λθ, for each x∈D, one has

$$G(x)+\sigma_G S\subset \lambda\bigl(a+\sigma_G \lambda^{-1}S\bigr)+(1-\lambda)F(x)\subset\lambda S_E[a, \theta]+(1-\lambda)D\subset D $$

and thus G∈U 1. Since ρ(G,F)<ε it follows that \(\mathcal{U}_{1}\) is dense in \(\mathcal{U}\), as asserted. □

Set

$$\mathcal{B}_D=\{A \subset D: A \mbox{ is nonempty, closed and convex}\}. $$

The following result is a special case of a theorem due to Ricceri [165].

Lemma 9.28

Let \(F:D \to\mathcal{B}_{D}\) be strictly contractive. Then the fixed point set \(\operatorname{fix}(F)\) of F is a nonempty absolute retract.

We call the subset of a metric space an R δ -set if it is the intersection of a descending sequence of absolute retracts.

Theorem 9.29

The fixed point set \(\operatorname{fix}(F)\) of most \(F \in\mathcal{U}\) is a nonempty and compact R δ -set.

Proof

Let \(\mathcal{U}_{0}\) and \(\mathcal{U}_{1}\) be defined by (9.83) and (9.84), respectively. By Theorem 9.26 and Lemma 9.27, the set \(\mathcal{U}^{*}=\mathcal{U}_{0}\cap U_{1}\) is residual in \(\mathcal{U}\). Our theorem is therefore an immediate consequence of the following assertion.

Claim. For each \(F \in\mathcal{U}^{*}\), the set \(\operatorname{fix}(F)\) is a nonempty and compact R δ -set.

Let \(F \in\mathcal{U}^{*}\). Since \(F \in\mathcal{U}_{1}\), there exists σ F >0 such that

$$ F(x)+\sigma_FS \subset D \quad\mbox{for each } x \in D. $$
(9.85)

Let \(a \in \operatorname{int}(D)\). Then S E [a,θ]⊂D for some θ>0. For a natural number n, define \(G_{n}:D \to\mathcal{E}_{D}\) by

$$G_n(x)=\bigl(2^{-n}\bigr)a+\bigl(1-2^{-n} \bigr)F(x),\quad x \in D. $$

In addition, for a natural number n, set

$$ Q_n(x)=G_n(x)+(1/n)S,\quad x \in D. $$
(9.86)

Let n 0 be a natural number such that n≥n 0 implies (1/n)/(1−2−n)<σ F . For n≥n 0 and x∈D, one has

$$ \begin{aligned} Q_n(x)&=\bigl(2^{-n}\bigr)a+ \bigl(1-2^{-n}\bigr)\bigl[F(x)+(1/n) \bigl(1-(1/2)^n \bigr)^{-1}S\bigr] \\ &\subset2^{-n}a+\bigl(1-2^{-n}\bigr) \bigl(F(x)+ \sigma_FS\bigr)\subset 2^{-n}a+\bigl(1-2^{-n} \bigr)D \end{aligned} $$

by (9.84).

Therefore Q n (x)⊂D and thus Q n (x)∈B D for each x∈D. It follows that for each natural number n≥n 0, (9.86) defines a map \(Q_{n}:D \to\mathcal{B}_{D}\) which is a strict contraction and, moreover, ρ(Q n ,F)→0 as n→∞. Fix n 1≥n 0 so that

$$ n(n+1)2^{-n}M<1 \quad\mbox{for every } n \ge n_1, \mbox{ where } M=\sup_{x \in D}\Vert x\Vert. $$
(9.87)

We claim that for each integer n≥n 1, one has

$$ F(x) \subset Q_{n+1}(x) \subset Q_n(x) \quad \mbox{for every } x \in D. $$
(9.88)

To see this, let n≥n 1 and x∈D be arbitrary. Then

$$ \begin{aligned} &Q_{n+1}(x)+(1/n)S \\ &\quad=\bigl(1-2^{-n-1} \bigr)F(x)+2^{-n-1}a+(n+1)^{-1}S+(1/n)S \\ &\quad=\bigl(1-2^{-n}\bigr)F(x)+\bigl(2^{-n}-2^{-n-1} \bigr)F(x)+2^{-n}a-\bigl(2^{-n}+2^{n+1} \bigr)a \\ &\qquad{}+(n+1)^{-1}S+ n^{-1}S \end{aligned} $$

and thus

$$ Q_{n+1}(x) +(1/n)S\subset Q_n(x)+2^{-n} 2^{-1}\bigl(F(x)-a\bigr)+(n+1)^{-1}S. $$
(9.89)

Now,

$$\bigl(F(x)-a\bigr)/2 \subset(D-a)/2 \subset MS $$

and hence by (9.87),

$$ 2^{-n}\bigl(F(x)-a\bigr) \subset\bigl(n(n+1) \bigr)^{-1}S. $$
(9.90)

Combining (9.90) with (9.89), we obtain

$$Q_{n+1}(x)+(1/n)S\subset Q_n(x)+(1/n)S. $$

It now follows from Radström’s cancellation law [113] that

$$Q_{n+1}(x) \subset Q_n(x). $$

It remains to be shown that F(x)⊂Q n+1 x. Clearly,

$$ \begin{aligned} F(x)+(n+1)^{-1}S={}&\bigl(1-2^{-n-1} \bigr)F(x)+2^{-n-1}F(x)+2^{-n-1}a \\ &{}-2^{-n-1}a+(n+1)^{-1}S \\ ={}&Q_{n+1}(x)+2^{-n}\bigl(F(x)-a\bigr)/2 \subset Q_{n+1}(x)+(n+1)^{-1}S, \end{aligned} $$

since by (9.90),

$$\bigl(1/2^n\bigr) \bigl(F(x)-a\bigr)/2\subset\bigl(n(n+1) \bigr)^{-1}S \subset(n+1)^{-1}S. $$

Therefore by Radström’s cancellation law F(x)⊂Q n+1(x) and thus (9.89) is valid.

For each integer n≥n 1, \(Q_{n}:D \to\mathcal{B}_{D}\) is a strict contraction and by Lemma 9.27 its fixed point set \(\operatorname{fix}(Q_{n})\) is a nonempty absolute retract. On the other hand, the set \(\operatorname{fix}(F)\) is nonempty and compact because \(F \in\mathcal{U}^{*} \subset\mathcal{U}_{0}\). By (9.88),

$$\operatorname{fix}(F) \subset\operatorname{fix}(Q_{n+1}) \subset \operatorname{fix}(Q_n) \quad\mbox{for every } n \ge n_1, $$

which implies that

$$\operatorname{fix}(F) \subset\bigcap_{n \ge n_1} \operatorname{fix}(Q_n). $$

On the other hand, let \(x \in\operatorname{fix}(Q_{n})\) for every n≥n 1. Then x∈F(x) because

$$d\bigl(x,F(x)\bigr) \le d\bigl(x,Q_n(x)\bigr)+h\bigl(Q_n(x),F(x) \bigr)\le\rho(Q_n,F) $$

and ρ(Q n ,F)→∞ as n→∞. Hence

$$\operatorname{fix}(F)=\bigcap_{n\ge n_1} \operatorname{fix}(Q_n) $$

and thus \(\operatorname{fix}(F)\) is a nonempty and compact R δ -set. Therefore our claim is valid and this completes the proof of Theorem 9.29. □

9.11 Approximation of Fixed Points

In this section, which is based on [53], we consider iterative schemes for approximating fixed points of closed-valued strict contractions in metric spaces.

Throughout this and the next section of this chapter, (X,ρ) is a complete metric space and T:X→2X∖{∅} is a strict contraction such that T(x) is a closed set for each x∈X. Thus T satisfies

$$ h\bigl(T(x),T(y)\bigr) \le c\rho(x,y) \quad\mbox{for all } x,y \in X, $$
(9.91)

where 0≤c<1.

For each x∈X and each nonempty set A⊂X, let

$$\rho(x,A)=\inf\bigl\{ \rho(x,y): y \in A\bigr\} . $$

Theorem 9.30

Let T:X→2X∖{∅} be a strict contraction such that T(x) is a closed set for each x∈X and T satisfies (9.91) with 0≤c<1. Assume that x 0∈X, \(\{\varepsilon _{i}\}_{i=0}^{\infty}\subset(0,\infty)\), \(\sum_{i=0}^{\infty} \varepsilon _{i}<\infty\), and that for each integer i≥0,

$$ x_{i+1}\in T(x_i),\qquad \rho(x_i,x_{i+1})\le \rho\bigl(x_i,T(x_i) \bigr)+\varepsilon _i. $$
(9.92)

Then \(\{x_{i}\}_{i=0}^{\infty}\) converges to a fixed point of T.

Proof

First, we claim that \(\{x_{i}\}_{i=0}^{\infty}\) is a Cauchy sequence. Indeed, let i≥0 be an integer. Then by (9.92) and (9.91),

$$\rho(x_{i+1},x_{i+2}) \le \rho\bigl(x_{i+1},T(x_{i+1}) \bigr)+\varepsilon _{i+1}\le h\bigl(T(x_i),T(x_{i+1}) \bigr)+\varepsilon _{i+1} $$

and

$$ d(x_{i+1},x_{i+2}) \le c \rho(x_i,x_{i+1})+ \varepsilon _{i+1}. $$
(9.93)

By (9.93),

$$\rho(x_1,x_2) \le c\rho(x_0,x_1)+ \varepsilon _1 $$

and

$$ \rho(x_2,x_3) \le c \rho(x_1,x_2)+ \varepsilon _2\le c^2\rho(x_0,x_1)+c \varepsilon _1+\varepsilon _2. $$
(9.94)

Now we use induction to show that for each integer n≥1,

$$ \rho(x_n,x_{n+1}) \le c^n \rho(x_0,x_1)+\sum_{i=0}^{n-1}c^i \varepsilon _{n-i}. $$
(9.95)

In view of (9.94), inequality (9.95) is valid for n=1,2.

Assume that k≥1 is an integer and that (9.95) holds for n=k. When combined with (9.93), this implies that

$$ \begin{aligned} \rho(x_{k+1},x_{k+2}) &\le c \rho(x_k,x_{k+1})+\varepsilon _{k+1}\le c^{k+1} \rho(x_0,x_1)+\sum_{i=0}^{k-1}c^{i+1} \varepsilon _{k-i}+\varepsilon _{k+1} \\ &=c^{k+1}\rho(x_0,x_1)+\sum _{i=0}^kc^i\varepsilon _{k+1-i}. \end{aligned} $$

Thus (9.95) holds with n=k+1 and therefore (9.95) holds for all integers n≥1. By (9.95),

$$ \begin{aligned} \sum_{n=1}^{\infty} \rho(x_n,x_{n+1}) &\le \sum_{n=1}^{\infty} \Biggl(c^n\rho(x_0,x_1)+\sum _{i=1}^nc^{n-i}\varepsilon _i \Biggr) \\ &\le \rho(x_0,x_1)\sum_{n=1}^{\infty}c^n+ \sum_{i=1}^{\infty}\Biggl(\sum _{j=0}^{\infty}c^j\Biggr) \varepsilon _i \\ &\le\Biggl(\sum_{n=0}^{\infty}c^n \Biggr)\Biggl[\rho(x_0,x_1)+\sum _{n=1}^{\infty }\varepsilon _n\Biggr]<\infty. \end{aligned} $$

Thus \(\{x_{n}\}_{n=0}^{\infty}\) is indeed a Cauchy sequence and there exists

$$ x_*=\lim_{n \to\infty}x_n. $$
(9.96)

We claim that x ∗∈T(x ∗). Let ε>0 be given. By (9.96), there is an integer n 0≥1 such that for each integer n≥n 0,

$$ \rho(x_n,x_*) \le \varepsilon /8. $$
(9.97)

Let n≥n 0 be an integer. By (9.91),

$$ h\bigl(T(x_n),T(x_*)\bigr) \le c \rho(x_n,x_*) \le c \varepsilon /8. $$
(9.98)

By (9.92),

$$x_{n+1} \in T(x_n). $$

When combined with (9.98), this implies that

$$\rho\bigl(x_{n+1},T(x_*)\bigr) \le c\varepsilon /8. $$

Hence there is

$$ y \in T(x_*) $$
(9.99)

such that ρ(x n+1,y)≤εc/4. Together with (9.97) and (9.99), this implies that

$$\rho\bigl(x_*,T(x_*)\bigr) \le\rho(x_*,y) \le\rho(x_*,x_{n+1})+\rho (x_{n+1},y) \le \varepsilon /8+\varepsilon /4. $$

Since ε is an arbitrary positive number, we conclude that

$$x_* \in T(x_*), $$

as claimed. Theorem 9.30 is proved. □

Theorem 9.31

Let T:X→2X∖{∅} be a strict contraction such that T(x) is a closed set for all x∈X and T satisfies (9.91) with 0≤c<1. Let ε>0 be given. Then there exists δ>0 such that if x∈X and ρ(x,T(x))<δ, then there is \(\bar{x} \in X\) such that \(\bar{x} \in T(\bar{x})\) and \(\rho(x,\bar{x})\le \varepsilon \).

Proof

Choose a positive number δ such that

$$ 4\delta(1-c)^{-1}<\varepsilon . $$
(9.100)

Consider

$$ x \in X \quad\mbox{such that}\quad \rho\bigl(x,T(x)\bigr)<\delta. $$
(9.101)

Set

$$ x_0=x. $$
(9.102)

By (9.101), there is

$$ x_1 \in T(x_0) $$
(9.103)

such that

$$ \rho(x_0,x_1)<\delta. $$
(9.104)

For each integer n≥1, choose

$$ x_{n+1} \in T(x_n) $$
(9.105)

such that

$$ \rho(x_{n+1},x_n)\le\rho \bigl(x_n,T(x_n)\bigr) (1+c)/(2c). $$
(9.106)

By (9.91), (9.103), (9.105) and (9.106), for each integer n≥1,

$$\rho(x_n,x_{n+1})\le(1+c)h\bigl(T(x_{n-1}),T(x_n) \bigr)/(2c)\le \bigl((1+c)/2\bigr)\rho(x_n,x_{n-1}). $$

When combined with (9.104), this implies that for each integer n≥1,

$$ \rho(x_n,x_{n+1}) \le\bigl[(1+c)/2 \bigr]^n\rho(x_0,x_1) \le \bigl[(1+c)/2 \bigr]^n\delta. $$
(9.107)

Therefore

$$\sum_{n=0}^{\infty}\rho(x_n,x_{n+1})< \infty, $$

\(\{x_{n}\}_{n=0}^{\infty}\) is a Cauchy sequence and there exists \(\bar{x} \in X\) such that

$$ \bar{x}=\lim_{n \to\infty}x_n. $$
(9.108)

Since x n+1∈T(x n ) for all integers n≥0, (9.108) implies that

$$\bar{x} \in T(\bar{x}). $$

By (9.100), (9.107) and (9.108),

$$ \begin{aligned} \rho(x_0,\bar{x})& =\lim_{n \to\infty} \rho(x_0,x_n)\le \sum_{n=0}^{\infty} \rho(x_i,x_{i+1}) \\ &\le\sum_{i=0}^{\infty}\bigl[(1+c)/2 \bigr]^i\delta=2\delta/(1-c)<\varepsilon /2. \end{aligned} $$

This completes the proof of Theorem 9.31. □

The conclusions of the following two theorems hold uniformly for all those relevant sequences \(\{x_{i}\}_{i=0}^{\infty}\) the initial point of which lies in a closed ball of center θ∈X and radius M>0.

Theorem 9.32

Let T:X→2X∖{∅} be a strict contraction such that T(x) is a closed set for all x∈X and T satisfies (9.91) with 0≤c<1. Fix θ∈X. Let ε>0 and M>0 be given. Then there exist δ∈(0,ε) and an integer n 0≥1 with the following property:

for each sequence \(\{x_{i}\}_{i=0}^{\infty} \subset X\) such that ρ(x 0,θ)≤M and such that for each integer n≥0,

$$x_{n+1} \in T(x_n) \quad\textit{and}\quad \rho(x_{n+1},x_n)\le\delta+\rho\bigl(x_n,T(x_{n}) \bigr), $$

we have

$$\rho(x_{n+1},x_n) < \varepsilon \quad\textit{for all integers}\ n \ge n_0. $$

Proof

Choose δ∈(0,1) such that

$$ \delta(1-c)^{-1}<\varepsilon /2 $$
(9.109)

and a natural number n 0 such that

$$ c^{n_0}\bigl(2M+1+\rho\bigl(\theta, a(\theta)\bigr)\bigr)< \varepsilon /2. $$
(9.110)

Let x 0∈X,

$$ \{x_n\}_{n=0}^{\infty} \subset X, \qquad\rho(x_0,\theta) \le M, $$
(9.111)

and assume that for each integer n≥0,

$$ x_{n+1} \in T(x_n), \qquad \rho(x_{n+1},x_n) \le \rho\bigl(x_n,T(x_n) \bigr)+\delta. $$
(9.112)

We now estimate ρ(x 0,T(x 0)). By (9.91) and (9.111),

$$ \begin{aligned}[b] \rho\bigl(x_0,T(x_0) \bigr)&\le \rho(x_0,\theta)+\rho\bigl(\theta,T(\theta)\bigr)+h \bigl(T(\theta),T(x_0)\bigr) \\ &\le \rho(x_0,\theta)+\rho\bigl(\theta,T(\theta)\bigr)+\rho( \theta,x_0)\le 2M+\rho\bigl(\theta,T(\theta)\bigr). \end{aligned} $$
(9.113)

By (9.112) and (9.113),

$$ \rho(x_0,x_1) \le\rho \bigl(x_0,T(x_0)\bigr)+\delta\le 2M+1+\rho\bigl( \theta,T(\theta)\bigr). $$
(9.114)

By (9.112) and (9.91), for each integer n≥0,

$$ \begin{aligned}[b] \rho(x_{n+2},x_{n+1}) &\le \rho\bigl(x_{n+1},T(x_{n+1})\bigr)+\delta \\ &\le h\bigl(T(x_n),T(x_{n+1})\bigr)+\delta\le c \rho(x_n,x_{n+1})+\delta. \end{aligned} $$
(9.115)

Next, we show by induction that for each integer n≥1,

$$ \rho(x_{n+1},x_n) \le\delta\sum _{i=0}^{n-1}c^i+c^n \rho(x_0,x_1). $$
(9.116)

By (9.115), inequality (9.116) holds for n=1. Assume that k≥1 is an integer and that (9.116) holds with n=k. Then by (9.115),

$$\rho(x_{k+2},x_{k+1}) \le c\rho(x_k,x_{k+1})+ \delta\le\delta \sum_{i=0}^kc^i+c^{k+1} \rho(x_0,x_1). $$

Thus (9.116) holds with n=k+1 and therefore it holds for all integers n≥1. By (9.116) and (9.114), for all natural numbers n,

$$ \rho(x_{n+1},x_n) \le\delta (1-c)^{-1}+c^n\bigl(2M+1+\rho\bigl(\theta,T(\theta)\bigr)\bigr). $$
(9.117)

Finally, by (9.117), (9.109) and (9.110), we obtain, for all integers n≥n 0,

$$\rho(x_n,x_{n+1})\le\delta(1-c)^{-1}+c^{n_0}\bigl(2M+1+ \rho\bigl(\theta,T (\theta)\bigr)\bigr) <\varepsilon . $$

Theorem 9.32 is proved. □

Theorems 9.30 and 9.31 imply the following additional result.

Theorem 9.33

Let T:X→2X∖{∅} be a strict contraction such that T(x) is a closed set for all x∈X and T satisfies (9.91) with 0≤c<1. Let positive numbers ε and M be given. Then there exist δ>0 and an integer n 0≥1 such that if a sequence \(\{x_{i}\}_{i=0}^{\infty}\subset X\) satisfies

$$\rho(x_0,\theta)\le M,\qquad x_{n+1}\in T (x_n) \quad\textit{and}\quad \rho(x_n,x_{n+1})\le \rho\bigl(x_n,T(x_n)\bigr)+\delta $$

for all integers n≥0, then for each integer n≥n 0, there is a point y∈X such that y∈T(y) and ρ(y,x n )<ε.

The following example shows that Theorem 9.33 cannot be improved in the sense that the fixed point y, the existence of which is guaranteed by the theorem, is not, in general, the same for all integers n≥n 0.

Example 9.34

Let X=[0,1], ρ(x,y)=|x−y| and T(x)=[0,1] for all x∈[0,1]. Let δ>0 be given. Choose a natural number k such that 1/k<δ. Put

$$\begin{gathered} x_0=0,\qquad x_i=i/k,\quad i=0,\ldots, k, \\ x_{i+k}=1-i/k,\quad i=0,\ldots,k, \end{gathered}$$

and for all integers p≥0 and any i∈{0,…,2k}, put

$$x_{2pk+i}=x_i. $$

Then \(\{x_{i}\}_{i=0}^{\infty} \subset X\) and for any integer i≥0, we have

$$x_{i+1}\in T(x_i)\quad \mbox{and}\quad |x_i-x_{i+1}| \le k^{-1}<\delta. $$

On the other hand, for all x∈X and any integer p≥0,

$$\max\bigl\{ |x-x_i|: i=2kp,\ldots, 2pk+2k\bigr\} \ge1/2. $$

9.12 Approximating Fixed Points in Caristi’s Theorem

We begin this section by recalling the following two versions of Caristi’s fixed point theorem [36].

Theorem 9.35

([82], Theorem 3.9)

Suppose that (X,ρ) is a complete metric space and T:X→X is a continuous mapping which satisfies for some ϕ:X→[0,∞),

$$\rho(x,Tx) \le\phi(x)-\phi(Tx),\quad x \in X. $$

Then \(\{T^{n}x\}_{n=1}^{\infty} \) converges to a fixed point of T for each x∈X.

Theorem 9.36

([82], Theorem 4.1)

Suppose that (X,ρ) is a complete metric space, ϕ:X→R 1 is a lower semicontinuous function which is bounded from below, and T:X→X satisfies

$$\rho(x,Tx) \le\phi(x)-\phi(Tx),\quad x \in X. $$

Then T has a fixed point.

We now present and prove a set-valued analog of Caristi’s theorem with computational errors.

Theorem 9.37

Assume that (X,ρ) is a complete metric space, T:X→2X∖{∅}, \(\operatorname{graph}(T):=\{(x,y) \in X \times X: y \in T(x)\}\) is closed, ϕ:X→R 1∪{∞} is bounded from below, and that for each x∈X,

$$ \inf\bigl\{ \phi(y)+\rho(x,y): y \in T(x)\bigr\} \le\phi(x). $$
(9.118)

Let \(\{\varepsilon _{n}\}_{n=0}^{\infty} \subset(0,\infty)\), \(\sum_{n=0}^{\infty} \varepsilon _{n}<\infty\), and let x 0∈X satisfy ϕ(x 0)<∞. Assume that for each integer n≥0,

$$ x_{n+1}\in T(x_n) $$
(9.119)

and

$$ \phi(x_{n+1})+\rho(x_n,x_{n+1})\le \inf\bigl\{ \phi(y)+\rho(x,y): y \in T(x_n)\bigr\} + \varepsilon _n. $$
(9.120)

Then \(\{x_{n}\}_{n=0}^{\infty}\) converges to a fixed point of T.

Proof

Clearly, ϕ(x n )<∞ for all integers n≥0. By (9.120), for each integer n≥0,

$$ \begin{aligned}[b] \rho(x_n,x_{n+1})&\le- \phi(x_{n+1})+\varepsilon _n+\inf\bigl\{ \phi(y)+\rho(x,y): y \in T(x_n)\bigr\} \\ &\le-\phi(x_{n+1})+\phi(x_n)+\varepsilon _n. \end{aligned} $$
(9.121)

By (9.121), for each integer m≥1,

$$ \begin{aligned} \sum_{i=0}^{m} \rho(x_i,x_{i+1}) &\le \phi(x_0)- \phi(x_m)+\sum_{i=0}^{\infty} \varepsilon _i \\ &\le\phi(x_0)-\inf(\phi)+\sum_{i=0}^{\infty} \varepsilon _i<\infty. \end{aligned} $$

Thus \(\{x_{i}\}_{i=0}^{\infty} \) is a Cauchy sequence and there exists \(\bar{x}=\lim_{i \to\infty} x_{i}\). Since the graph of T is closed, it follows that

$$(\bar{x},\bar{x})=\lim_{i \to\infty} (x_i,x_{i+1}) \in\operatorname{graph}(T). $$

This completes the proof of Theorem 9.37. □