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a contracting function in a complete metric space has one fixed point (proof)

created by ariels

(idea) by ariels (1.9 d) (print)   ?   (I like it!) 1 C! Sat Sep 02 2000 at 6:40:29

Let X be a complete metric space, and let f: X->X be a contracting function (we shall be using the definitions and symbols of my writeup in that node, so you might want to read it). Our aim is to show that f has a unique fixed point. The proof gives a taste of constructive methods in complete metric spaces; non-constructive methods also abound, but have a vastly different character. In particular, we shall give a method which converges to the fixed point of f; this method is in fact used, e.g. to draw fractals or solve differential equations.

Uniqueness is easy: suppose x,y are 2 fixed points of f. Then

d(x,y) = d(f(x),f(y)) <= c d(x,y)
with c<1, so d(x,y)=0 and x=y -- f has at most one fixed point.

For existence, first note that any contracting function must be continuous. Pick any x0 in X, and define xn=f(xn-1). If we can prove that this sequence converges, by applying f and using its continuity we see that

f(lim xn) = lim f(xn) = lim xn+1 = lim xn,
so the limit is a (the) fixed point of f

Write d=d(x0,x1). Then utilising the definition of a contracting function n times, d(xn+1,xn) <= d cn. But this means that if m>n then d(xm,xn) < d cn / (1-c), so our sequence is a Cauchy sequence. Thus it has a limit, which we saw is the fixed point of f.

Note that the above is a procedure for converging to the fixed point! Furthermore, we see that the rate of convergence is respectable -- a constant number of digits for each iteration.


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contracting function Editor Log: September 1, 2000 complete Wikipedia Mathematics
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