Nonlinear Control Sy.. - Free

2.10. CONTRACTION MAPPING 55
and then, by induction
d(xn+l,xn) < End(xl,xo)
(2.24)
Now suppose m > n > N. Then, by successive applications of the triangle inequality we
obtain
m
d(xn, xm) E d(xt, xi-1)
t=n+1
< (Cn + `n+l +... + m-1) d(xi, xo)
d(xn, xm) < n(1 + 1 + ... + Cm-n-1) d(xl, x0). (2.25)
The series (1 + x + x2 + ) converges to 1/(1 - x), for all x < 1. Therefore, noticing
that all the summands in equation (2.25) are positive, we have:
defining 1 £ d(xl, xo) = E, we have that
1+e+...+Cm-n-1 <
1
(1 - E)
d(xn, xm) < n d(xl, xo)
d(xn, xm) < e n, m > N.
In other words, {xn} is a Cauchy sequence. Since the metric space (X, d) is complete,
{xn} has a limit, limn,,,.(xn) = x, for some x E X. Now, since f is a contraction, f is
continuous. It follows that
f(x) f(xn) = rnn (xn+1) = x.
Moreover, we have seen that xn E S C X, Vn, and since S is closed it follows that x E S.
Thus, the existence of a fixed point is proved. To prove uniqueness suppose x and y are
two different fixed points. We have
f(x)=x, f(y)=y (2.26)
d(f(x),f(y)) = d(x, y) by (2.26) (2.27)
= d(x, y) < d(x, y) because f is a contraction (2.28)
where < 1. Since (2.28) can be satisfied if and only if d(x, y) = 0, we have x = y. This
completes the proof.