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.

1. Chào bạn, mình có bài tập này không biết bạn có muốn giúp mình ko?
Giả sử X là không gian metrix đủ và f:X–>X là ánh xạ liên tục thỏa mãn điều kiện:
d(x,f(x))R+.
CMR dãy {f^n(x)} hội tụ đến điểm bất động của f với mỗi x thuộc X

Comment by Violet — October 23, 2009 @ 10:52

• $d(x,f(x))R+$ là cái gì bạn?

Comment by Ngô Quốc Anh — October 23, 2009 @ 11:26

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