# Ngô Quốc Anh

## December 30, 2008

### Nonexpansive Maps and a fixed point theorem due to Browder, Gohde and Kirk

Filed under: Các Bài Tập Nhỏ, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 23:41

Let $(X,d)$ be a metric space with $C \subset X$. Recall that a mapping $F: C \to X$ is non-expansive if $F$ satisfies

$d(F(x),F(y) \leqslant d(x,y)$

for all $x,y \in X$.

Theorem. Let $C$ be a nonempty, closed, convex subset of a normed linear space $E$ with $F: C \to C$ nonexpansive and $F(C)$ a subset of a compact set of $C$. Then $F$ has a fixed point.

Proof. Let $x_0 \in C$. For $n\geqslant 2$ define

$\displaystyle F_n = \left( 1-\frac{1}{n}\right)F +\frac{1}{n} x_0$

Since $C$ is convex and $x_0 \in C$, we see that $F_n: C \to C$ and it is clear that $F_n: C \to C$ is a contradiction. Therefore each $F_n$ has a unique fixed point $x_n \in C$. That is

$\displaystyle x_n = F_n(x_n)=\left( 1-\frac{1}{n}\right)F(x_n) +\frac{1}{n} x_0$.

In addition, since $F(C)$ lies in a compact subset of $C$, there exist a subsequence $S$ of integers and a $u \in C$ with $F(x_n) \to u$ as $n \to \infty$ in $S$.

Thus $x_n \to u$ as $n \to \infty$ in $S$. By continuity$F(x_n) \to F(u)$, as $n \to \infty$ in $S$ which claims that $u=F(u)$.

The main theorem of this topic is a result proved independently by Browder, Gohde and Kirk in 1965. We state it as follows.

Theorem. Let $C$ be a nonempty, closed, bounded, convex set in a (real) Hilbert space $H$. Then each nonexpansive map $F : C \to C$ has at least one fixed point.

Remark. Notice that uniqueness need not hold as the example, $F(x) = x, x\in C = [0, 1]$, shows.

Remark. In fact the above Theorem, it is enough to assume that $H$ is a uniformly convex Banach space.

In the proof of the above Theorem, we will need the following two technical results.

Claim 1. Let $H$ be a Hilbert space with $u, v \in H$, and let $r, R$ be constants with $0 \leqslant r \leqslant R$. If there exists an $x \in H$ with

$\displaystyle \|u-x \| \leq R, \| v -x\| \leq R, \quad \|\frac{u+v}{2} - x\| \geq r$

then

$\displaystyle \|u-v\| \leqslant 2\sqrt{R^2-r^2}$.

Proof. This comes from the so-called parallelogram law, that is,

$\displaystyle\|u-v\|^2 = 2\|u-x\|^2 + 2\|v -x\|^2 - \| (u-x)+(v-x)\|^2$.

Claim 2. Let $H$ be a Hilbert space, $C\subset H$ a bounded set and $F : C\to C$ a nonexpansive map. Suppose $x \in C, y \in C$ and $a = \frac{x + y}{2}\in C$. Let $\delta (C)$ denote the diameter of $C$ and let $\varepsilon \leqslant \delta (C)$ with $\|x-F(x)\| \leqslant \varepsilon$ and $\|y - F(y)\|\leqslant \varepsilon$. Then

$\displaystyle\|a-F(a)\| \leq 2\sqrt{\varepsilon} \sqrt{2\delta(C)}$.

Proof of main theorem. Assume that $0\in C$. (A modified argument from the one given below holds for any $x_0 \in C$, therefore for simplicity we let $x_0 = 0$.) Also assume that $F(0) \ne 0$ (otherwise we are finished). For each $n \geqslant 2$, notice that

$\displaystyle F_n = \left( 1-\frac{1}{n}\right)F : C \to C$

is a contradiction. Now the first theorem guarantees that there exists a unique $x_n \in C$ with

$\displaystyle x_n= F_n(x_n)=\left( 1- \frac{1}{n}\right) F(x_n)$.

Thus,

$\displaystyle \|x_n - F(x_n) \| = \frac{1}{n} \|F(x_n)\| \leq \frac{1}{n} \delta(C)$.

For each $n \geqslant 2$, let

$\displaystyle {Q_n} = \left\{ {x \in C:x - F(x) \leqslant \frac{1}{n}\delta (C)} \right\}$.

Now

$\displaystyle Q_2 \supseteq Q_3 \supseteq ... \supseteq Q_n \dots$

is a decreasing sequence of nonempty closed sets. Let

$\displaystyle {d_n} = \inf \left\{ {\left\| x \right\|:x \in {Q_n}} \right\}$

and since the $Q_n$s are decreasing we have

$\displaystyle d_2 \leq d_3 \leq... \leq d_n \leq...$

with $d_i \leqslant \delta (C)$ for each $i \geqslant 2$. Consequently, ${d_n} \uparrow d$ with $d \leqslant \delta (C)$.

Next let

$\displaystyle A_n =Q_{8n^2} \cap \overline{B(0, d+1/n)}$,

where

$\displaystyle B\left( {0,d + \frac{1}{n}} \right) = \left\{ {x \in H:\left\| x \right\| < d + \frac{1}{n}} \right\}$.

Now $A_n$ is a decreasing sequence of closed, nonemplty sets. We now show that

$\displaystyle\lim_{n\to\infty} \delta(A_n)=0$.

To see this, let . Then

$\displaystyle \|u-0\| \leq d+\frac{1}{n}, \quad \|v-0\| \leq d+\frac{1}{n}$.

Also since we have

$\displaystyle \|u -F(u) \| \leq \frac{1}{8n^2} \delta(C), \quad \| v-F(v)\| \leq \frac{1}{8n^2} \delta(C)$.

Thus

$\displaystyle\left\| {\frac{{u + v}}{2} - F\left( {\frac{{u + v}}{2}} \right)} \right\| \leqslant 2\sqrt {2\delta (C)} \sqrt {\frac{1}{{8{n^2}}}\delta (C)} = \frac{1}{n}\delta (C)$.

This implies

$\displaystyle \frac{u+v}{2} \in Q_n$

and

$\displaystyle \left\| \frac{u+v}{2} - 0\right\| \geq d_n$.

It is easy to check that $r : H \to \overline{B}_r$ is non-expansive. As a result

$r \circ F: \overline{B}_r \to \overline{B}_r$

is a non-expansive map. This guarantees that there exists $x \in \overline{B}_r$ with $r(F(x))=x$. If $F(x) \in \overline{B}_r$, then

$x=r(F(x))=F(x)$

and $F$ has a fixed point. If $F(x)$ does not belong to $\overline{B}_r$ then

$\displaystyle x=r(F(x)) = r \frac{F(x)}{\| F(x) \|} = \lambda F(x)$

with

$\displaystyle\lambda = \frac{r}{\|F(x)\|} <1$.

Source: Ravi P. Agarwal, Maria Meehan, Donal O’Regan, Fixed point theory and applications, Cambridge University Press, 2001.

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