The Real And Complex Number Systems

3. Here is another proof on (b).
Proof: Given any point x S, and thus consider f n x S. Then there is a
convergent subsequence f nk x, say its limit p, sinceS is compact. Consider
dfp, p d f limf nk x , limf nk x
k k
and
Note that
d limff nk x, limf nk x by continuity of f at p
k k
limdf nk1 x, f nk x 1
k
df nk1 x, f nk x ... df 2 f nk1 x, ff nk1 x. 2
limdf 2 f nk1 x, ff nk1 x
k
d limf 2 f nk1 x, limff nk1 x
k k
d f 2 limf nk1 x , f limf nk1 x by continuity of f 2 and f at p
k k
df 2 p, fp. 3
So, by (1)-(3), we know that
fp, fp df 2 p, fp p fp
by hypothesis
dfx, fy dx, y
where x y. Hence, f has a unique fixed point p by (a) in Exercise.
Note. 1.Ifx n x, andy n y, then dx n , y n dx, y. Thatis,
lim n
dx n , y n d lim n
x n , lim n
y n .
Proof: Consider
dx n , y n dx n , x dx, y dy, y n and
dx, y dx, x n dx n , y n dy n , y,
then
|dx n , y n dx, y| dx, x n dy, y n 0.
So,wehaveproveit.
2. The reader should compare the method with Exercise 4.72.
4.72 Assume that f satisfies the condition in Exercise 4.71. If x S, letp 0 x,
p n1 fp n ,andc n dp n , p n1 for n 0.
(a) Prove that c n is a decreasing sequence, and let c limc n .
Proof: Consider
c n1 c n dp n1 , p n2 dp n , p n1
dfp n , fp n1 dp n , p n1
dp n , p n1 dp n , p n1
0,
so c n is a decreasing sequence. And c n has a lower bound 0, by Completeness of R,
we know that c n is a convergent sequence, say c limc n .