32 I The Expanding Universe of Numbers
In example (ii), the completeness ofFn 2 is trivial, since any fundamental sequence
is ultimately constant.
In example (iii), the completeness ofC(I)with respect to the first definition of
distance follows from the completeness ofRand the fact that the limit of a uniformly
convergent sequence of continuous functions is again a continuous function.
However,C(I)is not complete with respect to either of the two alternative defini-
tions of distance. It is possible also for a sequence to converge with respect to the two
alternative definitions of distance, but not with respect to the first definition. Similarly,
a sequence may converge in the first alternative metric, but not even be a fundamental
sequence in the second.
The completions of the metric spaceC(I)with respect to the two alternative met-
rics may actually be identified with spaces offunctions. The completion for the first
alternative metric is the setL(I)of allLebesgue measurablefunctionsf:I →R
such that
∫b
a|f(x)|dx<∞,functions which take the same value at all points ofI, except for a set of measure zero,
being identified. The completionL^2 (I)for the second alternative metric is obtained
by replacing
∫b
a|f(x)|dxby∫b
a|f(x)|(^2) dxin this statement.
It may be shown that the metric spaces of examples (iv)–(vi) are all complete. In
example (vi), the strong triangle inequality implies that{an}is a fundamental sequence
if (and only if) d(an+ 1 ,an)→0asn→∞.
LetEbe an arbitrary metric space andf:E→EamapofEinto itself. A point
x ̄∈Eis said to be afixed pointoffiff(x ̄)= ̄x. A useful property of complete metric
spaces is the followingcontraction principle, which was first established in the present
generality by Banach (1922), but was previously known in more concrete situations.
Proposition 26Let E be a complete metric space and let f:E→EbeamapofE
into itself. If there exists a real numberθ, with 0 <θ< 1 , such that
d(f(x′),f(x′′))≤θd(x′,x′′) for all x′,x′′∈E,
then the map f has a unique fixed pointx ̄∈E.
Proof It is clear that there is at most one fixed point, since 0≤d(x′,x′′)≤θd(x′,x′′)
impliesx′=x′′. To prove that a fixed point exists we use themethod of successive
approximations.
Choose anyx 0 ∈Eand define the sequence{xn}recursively by
xn=f(xn− 1 )(n≥ 1 ).
For anyk≥1wehave
d(xk+ 1 ,xk)=d(f(xk),f(xk− 1 ))≤θd(xk,xk− 1 ).
Applying thisktimes, we obtain
d(xk+ 1 ,xk)≤θkd(x 1 ,x 0 ).