The Real And Complex Number Systems

Proof: If p and p are fixed points of f where p p , then by hypothesis, we have
dp, p dfp, fp dp, p
which is absurb. So, f has at most one fixed point.
Let f : 0, 1/2 0, 1/2 by fx x 2 . Then we have
|fx fy| |x 2 y 2 | |x y||x y| |x y|.
However, f has no fixed point since if it had, say its fixed point p, then
p 2 p p 1 0, 1/2 or p 0 0, 1/2.
(b) If S is compact, prove that f has exactly one fixed point. Hint. Show that
gx dx, fx attains its minimum on S.
Proof: Let g dx, fx, and thus show that g is continuous on a compact set S as
follows. Since
dx, fx dx, y dy, fy dfy, fx
dx, y dy, fy dx, y
2dx, y dy, fy
dx, fx dy, fy 2dx, y *
and change the roles of x, andy, wehave
dy, fy dx, fx 2dx, y **
Hence, by (*) and (**), we have
|gx gy| |dx, fx dy, fy| 2dx, y for all x, y S. ***
Given 0, there exists a /2 such that as dx, y , x, y S, wehave
|gx gy| 2dx, y by (***).
So, we have proved that g is uniformly continuous on S.
So, consider min xS gx gp, p S. We show that gp 0 dp, fp. Suppose
NOT, i.e., fp p. Consider
df 2 p, fp dfp, p gp
which contradicts to gp is the absolute maximum. Hence, gp 0 p fp. Thatis,
f has a unique fixed point in S by (a).
(c)Give an example with S compact in which f is not a contraction.
Solution: Let S 0, 1/2, andf x 2 : S S. Then we have
|x 2 y 2 | |x y||x y| |x y|.
So, this f is not contraction.
Remark: 1. In (b), the Choice of g is natural, since we want to get a fixed point. That
is, fx x. Hence, we consider the function g dx, fx.
2. Here is a exercise that makes us know more about Remark 1. Let f : 0, 1 0, 1
be a continuous function, show that there is a point p such that fp p.
Proof: Consider gx fx x, then g is a continuous function defined on 0, 1.
Assume that there is no point p such that gp 0, that is, no such p so that fp p. So,
by Intermediate Value Theorem, we know that gx 0 for all x 0, 1, orgx 0
for all x 0, 1. Without loss of generality, suppose that gx 0 for all x 0, 1 which
is absurb since g1 f1 1 0. Hence, we know that there is a point p such that
fp p.