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Unformatted text preview: Fixed Points 97 Theorem 7.12. A normed vector space X is complete if and only if every obso
lutely summable series in X is summoble. That is, X is complete if and only if
2:; x,, converges in X wheneVer 23:1 lixnll < co. pnoor. First suppOSe that X is complete, and let (x,,) be a sequence in X for
"I which 2:; [lxnll < 00. If we write s", = “:1 x" for the sequence of partial
sums, then, for m > n, the triangle inequality yields In 5 Z "on.
k =n+l \ r—\
L.“ “5m "" gull : Since the partial sums of 23:1 i.r,, form a convergent (and hence Cauchy) se
quence, we have that 2:1an “ka —> 0 as m, n —> 00. Thus, (3,.) is also a Cauchy
sequence and, as such, converges in X. Next suppose that absolutely summabie series in X are summable, and let (x,,)
be a Cauchy sequence in X. As always, it is enough to ﬁnd a subsequence of
(x,,) that converges. To this end, choose a subsequence (rm) for which "rm, —
xnkll < 24‘ for all It. (How?) Then, in particular, 2:, ﬁrm. d x”, ] converges.
Consequently, the series Zﬁﬁxm, —x,,k) converges in X. As we remarked earlier,
this means that the sequence rum+l = x,,l + Ziljlttnm  x“) converges in X. [II There is never too much of a good thing: Note that Theorem 7.12 gives us yet another
characterization of completeness in IR. The familiar fact that every absolutely summable
series of real numbers is summable is actually equivalent to the least upper bound axiom. EXERCISES 31. If 2:11;}, is a convergent series in a normed vector space X, show that
00 < 00 llzn=1 x11 _ Zn=1 “xi? 32. Use Theorem 7.12 to prove that 31 is complete. 33. Let 5 denote the vector space of all ﬁnitely nonzero real sequences; that is,
x = (35") E s if x“ = 0 for all but ﬁnitely many it. Show that s is not complete under the sup norm llxllm = sup,l Ix" i. 34. Prove that a normed vector space X is complete if and only if every sequence
(x,,) in X satisfying liar“ — xn+1 < 2’”, for all n, converges to a point of X. 35. Prove that a normed vector space X is complete if and only if its closed unit
ball B := {x E X : x 5 1} is complete. Fixed Points “'9 Completeness is a useful property to have around if you are interested in solving equa
tions. How so? Well, think about the sorts of tricks that we use in R. How, for example, 98 Completeness would you compute rx/i. “by hand”? You would most likely start by finding an approxi—
mate solution to the equation x2 = 2 and then look for ways to improve your estimate. Most numerical techniques give, in fact, a sequence of “better and better” approxi
mate solutions, where “better and better” typic ally means that the error in approximation
gets smaller. The completeness of R affords us such luxuries; we can effectively pro
claim the existence of solutions without necessarily ﬁnding thernt Once we have a
Cauchy Sequence of approximate solutions, completeness will ﬁnish the job. The same holds true in any complete space. We can effectively solve certain “ab
stract” equations by simply displaying a Cauchy sequence of approximate solutions.
One such technique, called the method of successive approximations. is used in the
standard proof of existence for solutions to differential equations and is generally cred
ited to Picard in 1890. (But the technique itself goes back at least to Liouville, who
ﬁrst published it in 1838, and it may have even been known to Cauchy.) We will see an
example of this method shortly. The modern metric space version of the method of successive approximations was
explicitly stated in Banach’s thesis in 1922. In this setting it is most often referred to as
Banach ’s contraction mapping principle. Amap f : M e M on a metric space (M, a!)
is called a contraction (or, better still, a strict contraction) if there is some constant or
withO g at < 1 such that d(f(.r}, f(y)) 3 ard(x, y) is satisﬁed for all r, y e M. That is,
a contraction shrinks the distance between pairs of points by a factor strictly less than
1. Please note that any contraction is automatically continuous (since it is Lipschitz). Banach‘s approach seeks to solve an “abstract” equation of the form f (x) = it (this
is more general than it might appear). That is, we look for a ﬁxed point for f . If f is
a contraction deﬁned on a complete metric space, we can even prescribe a sequence of approximate solutions: Theorem 7.13. Let (M, a’ ) be a complete metric space, and let f : M —> M be a
(strict) contraction. Thea, f has a unique ﬁxed point. Moreover; given any point
x0 6 M, the sequence of functional iterates (f " (x0)) always converges to theﬁxea’ point for f. [The notation f " means the composition of f with itself n times: f o f o   o f. For
example, f 10:} = f ( f [20), f 30:) = f ( f 200), and, so on. The sequence of functional
iterates (f "(10) is called the orbit of .1: under f .] PROOF. Let x0 be any point in M, and consider the sequence (f "been. If (f "(xOD converges, we are done. Indeed, if x = limnam f “(3%), then, since
f is continuous, we have f (x) = lim,Hm f { f "(x0)) = lim,,_,m f "+'(_.r0) =
lirn,,_,90 f "(1:0) = it. And this x is unique, for if y is also a ﬁxed point for f, then d(.r, y) = d(f(x), f(y)) '5 ord(x. y), which forcés a’(x, y) = D.
So our goal is clear: We need to show that (f”(xo)) is a Cauchy sequence. But: at (f"+‘tx5). f"(xo)) s cm' [Ft—:37, f”‘*(xo))
E 052d (f'HlYo}. f"_2(Y0)) E it"dlflxcls xo} = Ct!" Fixed Points 99 And so form _>_ n the triangle inequality yields I“ m a! (f"*+‘(xo). f"(xo)) : 2d (kaOfo), ﬂit—0)) g c 20:". (7.1) k=lr .E=n But since 0 5 a < 1, we have 221:“ or" —> O as m, n —> 00. (Why?) Thus, (f"(_r0})
is Cauchy. CI Note that the proof of Theorem 7.13 even gives us a rough estimate for the error in
approximation. If we pick an initial “guess” x0 for the ﬁxed point x, then, by letting
in —> oo in equation (7.1}, we get a I! dﬁf”(xo).. x) s were). x0) Zak = woo), x0) k=n 1—“ Example 7.14 Suppose that f : [(1. b] —> [(1, b I is continuous on in, b ], differentiable on (a. b),
and has If ’(x) 5 or < 1 for all a < x < b. Then it follows from the mean value
theorem that f(x) — f(y) s alx — yl for all x, y e (a, b] and, hence, that f has
a unique ﬁxed point. See Figures 7.1 and 7.2. Thecase0<f’<1. [ HEX y=flxl Figure
7 .1 Figure
7 .2 F ixcd Points 101 to at least six places. Roughly speaking, each iteration increases the accuracy by
one decimal place. Not bad.
M
E X E R C I S E 46. Extend the result in Example 7.15 as follows: Suppose that F : [n, b] —> IR
is continuous on [(2, b1, differentiable in (a, b}, and satisﬁes [7(a) < O, F (b) > 0,
and 0 < K1 5 F ’(x) 5 Kg. Show that there is a unique solution to the equation
FCx) = 0. [Hintz Consider the equation f (x) = x, where f (x) = x '— AFUC) for some suitably chosen L]
M Under suitable conditions on f , the same technique can be applied to the problem
of existence and uniqueness of the solution to the initial value problem: 31’: fowl. ya?) = J'n For example, if f is continuous in some rectangle containing (0, yo) in its interior, and
if f is Lipschitz in its second variable, I f (.r, y)  f (x, zjl 5 K ly — zl, for some constant
K, then a unique solution exists — at least in some small neighborhood of x = 0. This
fact was ﬁrst observed by Lipschitz himseif (hence the name Lipschitz condition), but
Lipschitz did not have metric spaces at his disposal. Most modern proofs use some
form of Banach‘s contraction mapping principle (often in the form of the method of successive approximations}.
We will not give the full details of the proof here, but we will at least show how Banach’s theorem enters the picture. For this we will want to rephrase the problem as
a ﬁxed—point problem on some complete metric space. First notice that by integrating both sides of the differential equation we get ‘
YOC) : yo +f f0, MGM? (I Z 0).
0 That is, we need a ﬁxed point for the map go 1—) F (go), where (Freon (x) = yo + f ftt. new.
0 For simplicity, let's assume that f is deﬁned and continuous on all of R2 (and still
Lipschitz in its second variable). Then the integral on the ti ghthand side of this formula
is well deﬁned for any continuous function go. Letis consider F as a map on C [0, 5],
where 8 > 0 will be specified shortly. Next we’ll check that F is a Lipschitz map on C[0, 6]. For any 0 5 x 5 a, note that fo' In. sound? — for re, we» dz I(F(qo)) (x) — (Farm on =
s [0' lfU. tom} — ftt. we)» dz 5 K f halt) — an): J:
G
5 Kx max lth — an): 05: 5.x E K3 “<0 —l1’so 100 Completeness EXERCISES 36. The function f (x) = x2 has two obvious ﬁxed points: p0 = 0 and p1 = 1.
Show that there is a G < 8 < 1 such that f(x) ~— poi < lx — pgl whenever
Ix — pol < 6,.t # po.Conc1udethatf"(x) —> p0 whenever litpol < 3,.r # p0.
This means that p0 is an attracting ﬁxed point for f; every orbit that starts out near
0 converges to 0. In contrast, ﬁnd a 8 > 0 such that if 1x — pgl < 8,x ,=’ p1, then
 f (x) — pll > [x — pll. This means that p: is a repelling ﬁxed point for f; orbits that start out near 1 are pushed away from 1. In fact, given any x 7—5 1, we have
f"(x) 71+ 1. . 37. Suppose that f : (a, b) —> (a. b) has a ﬁxed point p in (ct, b} and that f is
differentiable at p. If I f ’(p) < l, prove that p is an attracting ﬁxed point for f. If
i f ’(p) > 1, prove that p is a repelling ﬁxed point for f. 38.
(a) Let f (.r) = arctanx. Show‘ that f ’(0) = l and that 0 is an attracting ﬁxed point for f.
(b) Let g(:c} = x3 + 1:. Show that g’(0) = l and that O is a repelling ﬁxed point for g.
(c) Let Mar) 2 x2 + 1/4. Show that It ’(1/2) = 1 and that 1/2 is a ﬁxed point for II that is neither attracting nor repelling. 39. The cubic x3 — x — 1 has a unique real root 3:0 with 1 < x0 < 2. Find it! [Hinti
Iterating the function f (x) = x3 — I won't work.f Why?} Example 7 .15 We’ll Show how Theorem 7.13 can be used to ﬁnd an estimate for, say, 3/3. That
is, we’ll solve the equation F(.r) = x3 — 5 = 0. Non,’ it is clear that I < «3/5 < 2, so
let’s consider F as a map on the interval [1, 2 1. And since the equation F (x) = 0
isn’t quite appropriate, let’s consider the equivalent equation f (x) = x, Where
f (x) = x —— lFLr) for some suitably chosen .1 e R. The claim is that it’s possible
to ﬁnd A > 0 such that (i) f : [1.2] —> [1, 2], and (ii) 1f’(:t)l E a < 1 for
1 < .r < 2. In fact, abitof experimentation will convince you that anyO «c it < 1/6 will do. Let’s try A = Us. Table 7.1 displays a few iterations of the scheme
x,.+1 = f Ex”) = xH — (x?I — 5)/8, starting with x0 = 1.5. The last value is accurate Table 7.1
x" f (In)
L5 L703125
L703125 L7106070513
L7106070518 L7099147354 {7099147854
L7099818467
L7099753773
L7099760016 L7099818467
L7099753773
L7099760016
L7099759414 102 Completeness It follows that IIFUD) — FWNICO 5 K6 "go — pllm. 'Ihus, F is a contraction on C[0, 5]
provided that 6 is chosen to satisfy K6 < I and, in this case, F has a unique ﬁxed point inC[0,6]. Example 7.16 Consider the initial value problem y ’ = 2x(l + y), y(0) = 0. By integrating both
sides of the differential equation, we see that we need a function go satisfying
rp(x) = f; 2t(l +qa(t)) (it = (F (p))(x). The method of successive approximations
amounts to taking an initial “guess” at the solution, say on E 0, and iterating F.
Thus, elm : f; 2m + (Dd: = .9. Next, gages) = f; 2:(1+:2)d: = x2 + We
Another iteration would yield gogtr) = x3 + x4/2 + x5/6. And so on. Finally,
induction yields 00 x2!“ .2
gm) = Z? = 6" —~ 1.
15:1 ' This solution is valid on all of R (and agrees, naturally, with the solution obtained
by separation of variables). EXERCISES 41. Let M be complete and let f : M —> M be continuous. If f k is a strict con
traction for some integer k > 1, show that f has a unique ﬁxed point. 42. Deﬁne r : C[0, 1] —> C[O, 1]by (T(f))(:t) = fgfomr. Show thatT is
not a strict contraction while T3 is. What is the fixed point of T?
43. Show that each of the hypotheses of the contraction mapping principle is nec
essary by ﬁnding examples of a space M and a map f : M —> M having no ﬁxed
point where: (a) M is incomplete (but f is still a strict contraction). (h) f satisﬁes only d( f (x), f(y)) < d(x, y) for all .1: 7i y (buts/1 is still complete}. Completions Completeness is a central theme in this book; it will return frequently. It may comfort
you to know that every metric space can he “completed.” In effect, this means that by
tacking on a few “missing” limit points We can make an incomplete space complete.
While the approach that we will take may not suggest anything so simple as adding a
few points here and there, it is nevertheless the picture to bear in mind. In time, all will be made clear!
First, a deﬁnition. A metric space (115', d) is called a completion for (M, a‘ ) if (i) (M, a?) is complete, and
(ii) (M, d) is isometric to a dense subset of (112?, d). ...
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