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 continuityF(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_ns 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.

2 Comments »

  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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