8 Completeness - Pennsylvania State University Completeness We recall the deﬁnition of a Cauchy...

8 Completeness

We recall the definition of a Cauchy sequence. Let (X, d) be a given metricspace and let (xn) be a sequence of points of X. Then (xn) (xn) is a Cauchysequence if for every ε > 0 there exists N ∈ N such that

d(xn, xm) < ε for all n,m ≥ N .

Properties of Cauchy sequences are summarized in the following propositions

Proposition 8.1. (i) If (xn) is a Cauchy sequence, then (xn) is bounded.

(ii) If (xn) is convergent, then (xn) is a Cauchy sequence.

(iii) If (xn) is Cauchy and it contains a convergent subsequence, then (xn)converges.

A Cauchy sequence need not converge. For example, consider the sequence(1/n) in the metric space ((0, 1), | · |). Clearly, the sequence is Cauchy in(0, 1) but does not converge to any point of the interval.

Definition 8.2. A metric space (X, d) is called complete if every Cauchysequence (xn) in X converges to some point of X. A subset A of X is calledcomplete if A as a metric subspace of (X, d) is complete, that is, if everyCauchy sequence (xn) in A converges to a point in A.

By the above example, not every metric space is complete; (0, 1) with thestandard metric is not complete.

Theorem 8.3. The space R with the standard metric is complete.

Theorem 8.3 is a consequence of the Bolzano-Weierstrass theorem andPropositions 8.1Recall that if (Xi, di), 1 ≤ i ≤ m, are metric spaces and X = X1 × . . .×Xn,then

d(x, y) = max1≤j≤n

dj(xi, yi)

where x = (x1, . . . , xn), y = (y1, . . . , yn) ∈ X, defines a metric on X. Thepair (X, d) is called the product of (Xi, di).

Theorem 8.4. If (Xi, di) are complete metric spaces for i = 1, . . . ,m, then

the product (X, d) is a complete metric space.

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Proof. Let xn = (x1n, . . . , xm

n ) and (xn) be a Cauchy sequence in (X, d). Thenfor a given ε > 0 there exists k such that d(xn, xm) < ε for all n,m ≥ k.Since

dj(xjn, xj

m) ≤ d(xn, xm) < ε,

it follows that {xjn} is Cauchy in (Xj , dj) for j = 1, . . . m. Since (Xj , dj) is

complete, for j = 1, . . . ,m there exists xj ∈ Xj such that xjn → xj . Then

setting x = (x1, . . . , xm), we see, in view of the above definition of d, thatxn → x in X. �

Since

max{|x1 − y1| , . . . , |xn − yn|} ≤(

n∑

i=1

|xi − yi|2)1/2

≤√

nmax{|x1 − y1| , . . . , |xn − yn|},

it follows from Theorem 8.3 and Proposition 8.4 that Rn with the Euclidean

metric is complete.

Example 8.5. Denote by ℓ1 the set of all real sequences (xk) satisfying∑

∞

k=1 |xk| <∞. Then ℓ1 is a vector space if addition and multiplication by a number are definedas follows,

(xn) + (yn) = (xn + yn), α(xn) = (αxn).

If x = (xk) ∈ ℓ1, then

‖x‖ = ‖(xk)‖ :=

∞∑

k=1

|xk|

defines a norm on ℓ1 and d(x, y) = ‖x − y‖ =∑

∞

k=1 |xk − yk| defines a distance onℓ1. We claim that (ℓ1, d) is a complete metric space.

Indeed, let (Xn) ⊂ ℓ1 be a Cauchy sequence. Then with Xn(x(n)k

) we have

∣

∣

∣x

(n)k

− x(m)k

∣

∣

∣≤

∞∑

l=1

∣

∣

∣x

(n)l

− x(m)l

∣

∣

∣= ‖Xn − Xm‖ .

Hence for every k ≥ 1, the sequence (x(n)k

) is Cauchy in R and since R with

the standard metric is complete, the sequence (x(n)k

) converges to some xk. SetX = (xk). We suspect that X is the limit in ℓ1 of the sequence (Xn). To seethis we first show that X ∈ ℓ1. Since (Xn) is Cauchy in ℓ1, there is K such that‖Xn − Xm‖ < 1 for all n, m ≥ K. In particular,

N∑

k=1

∣

∣

∣x

(n)k

∣

∣

∣≤

∞∑

k=1

∣

∣

∣x

(n)k

− x(K)k

∣

∣

∣+

∞∑

k=1

∣

∣

∣x

(K)k

∣

∣

∣

≤ ‖Xn − XK‖ + ‖XK‖ < 1 + ‖XK‖

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for every N ≥ 1. Fixing N and taking limit as n → ∞ we get

N∑

k=1

|xk| ≤ 1 + ‖XK‖

and taking limit as N → ∞ we get

∞∑

k=1

|xk| ≤ 1 + ‖XK‖ .

So, X ∈ ℓ1. Next we show that ‖Xn − X‖ → 0 as n → ∞. Given ε > 0, there isK such that ‖Xn − Xm‖ < 1 for all n, m ≥ K. Consequently, for every N ≥ 1 wehave

N∑

k=1

∣

∣

∣x

(n)k

− x(m)k

∣

∣

∣≤ ‖Xn − Xm‖ < ε, for all n, m ≥ K.

With n > K and N fixed, we let m → ∞ to find that

N∑

k=1

∣

∣

∣x(n)k

− xk

∣

∣

∣ ≤ ε.

Since this is true for every N ,

‖Xn − X‖ =∞∑

k=1

∣

∣

∣x

(n)k

− xk

∣

∣

∣≤ ε

for n > K. Hence d(Xn, X) → 0 and since X ∈ ℓ1, the space ℓ1 is complete.

A subspace of a complete metric space may not be complete. For example,R with the standard metric is complete but (0, 1) equipped with the samemetric is not complete.

Proposition 8.6. If (X, d) is a complete metric space and Y is a closed

subspace of X, then (Y, d) is complete.

Proof. Let (xn) be a Cauchy sequence of points in Y . Then (xn) also satisfies theCauchy condition in X , and since (X, d) is complete, there exists x ∈ X such thatxn → x. But Y is also closed, so x ∈ Y showing that Y is complete. �

Proposition 8.7. If (X, d) is a metric space, Y ⊂ X and (Y, d) is complete,

then Y is closed.

Proof. Let (xn) be a sequence of points in Y such that xn → x. We have to showthat x ∈ Y . Since (xn) converges in X , it satisfies the Cauchy condition in X andso, it also satisfies the Cauchy condition in Y . Since (Y, d) is complete, it convergesto some point in Y , say to y ∈ Y . Since any sequence can have at most one limit,x = y. So x ∈ Y and Y is closed. �

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Let (X, d) and (Y, ρ) be metric spaces. A function f : X → Y is saidto be bounded if the image f(X) is contained in a bounded subset of Y .Denote by B(X,Y ) the set of all functions f : X → Y which are boundedand by Cb(X,Y ) the space of bounded continuous function f : X → Y . IfY = R, we simply write B(X) and Cb(X) instead of B(X,Y ) and Cb(X, R),respectively. We have Cb(X,Y ) ⊂ B(X,Y ). For f, g ∈ B(X,Y ), we set

D(f, g) := sup{ρ(f(x), g(x)) | x ∈ X},

where ρ denotes the metric on Y . The metric on Cb(X,Y ) is defined in thesame way.

Theorem 8.8. Suppose that (Y, ρ) is a complete metric space. Then the

spaces B(X,Y ) and (Cb(X,Y ),D) are complete.

Proof. The verification that D is a metric is left as an exercise. It suffices to showthat B(X, Y ) is a complete and that Cb(X, Y ) is a closed subset of B(X, Y ). Let{fn} be a Cauchy sequence in B(X, Y ). Then (fn(x)) is a Cauchy sequence in (Y, ρ)for every x ∈ X . Since, by assumption, (Y, ρ) is complete, the sequence (fn(x))converges in Y . Define f(x) = limn→∞ fn(x). Given ε > 0, there is N such that

ρ(fn(x), fm(x)) < ε for n, m ≥ N.

Fix n ≥ N and let m tend to ∞. Since the function z → ρ(fn(x), z) is continuous,it follows that

ρ(fn(x), f(x)) ≤ ε for n ≥ N (1)

which implies thatD(fn, f) ≤ ε for n ≥ N.

It remains to show that f is bounded. Since fN is bounded, there are y ∈ Y andr > 0 such that fN (X) ⊂ Br(y). From (1) and the triangle inequality, we obtainρ(f(x), y) ≤ ρ(f(x), fN (x)) + ρ(fN(x), y) < ε + r showing that f(X) ⊂ Br+ε(y)and that f is bounded. Hence the space (B(X, Y ), D) is complete as claimed. Nextwe shall show that Cb(X, Y ) is a closed subset of B(X, Y ). Take any sequence(fn) ⊂ Cb(X, Y ) such that D(fn, f) → 0 as n → ∞. We have to show that f iscontinuous. Fix x0 ∈ X and let ε > 0. Then there is N ∈ N such that

D(fn, f) < ε/3 for all n ≥ N .

In particular, in view of definition of D,

ρ(fN (x), f(x)) < ε/3 for all n ≥ N .

Since fN is continuous at x0, there is δ > 0 such that

ρ(fN (x), fN (x0)) < ε/3 for all x satisfying d(x0, x) < δ.

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Hence, if d(x0, x) < δ, then

ρ(f(x), f(x0)) ≤ ρ(f(x), fN (x)) + ρ(fN (x), fN (x0)) + ρ(f(x)0, fN (x0))

≤ 2D(fN , f) + ρ(fN (x), fN (x0)) < 2(ε/2) + ε/3 = ε.

This means that f is continuous at x0 and since x0 was an arbitrary point, f iscontinuous, i.e., f ∈ Cb(X, Y ) as claimed. �

8.0.1 Banach Fixed Point Theorem

Definition 8.9. Let (X, d) be a metric space. A map f : X → X is calleda contraction if there is constant c ∈ (0, 1) such that

d(f(x), f(y)) ≤ cd(x, y)

for all x, y ∈ X.

Theorem 8.10 (Banach Fixed Point Theorem). Let (X, d) be a com-

plete metric space and f : X → X a contraction. Then there exists exactly

one point u ∈ X such that f(u) = u. Moreover, for every x ∈ X, the

sequence (fn(x)) converges to u.

Proof. Claim: If a, b ∈ X , then

d(a, b) ≤ 1

1 − c[d(a, f(a)) + d(b, f(b))]. (2)

Indeed,

d(a, b) ≤ d(a, f(a)) + d(f(a), f(b)) + d(f(b), b)

≤ d(a, f(a)) + cd(a, b)) + d(f(b), b)

from which we conclude (2) Using (2), the map f can have at most one fixed pointsince if f(a) = a and f(b) = b, then the right side of (2) is equal to 0 implyingthat d(a, b) = 0. To prove existence of a fixed point, take any x ∈ X . Since f isa contraction with the constant c, the n-fold composition fn is a contraction withthe constant cn. Using this and (2) with a = fn(x) and b = fn(b), and noticingthat f(a) = fn+1(x) = fn(f(x)), f(b) = fm(f(x)), we conclude

d(fn(x), fm(x)) = d(a, b) ≤ 1

1 − c

[

d(a, f(a)) + d(b), b)]

=1

1 − c

[

d(fn(x)fn(f(x))) + d(fm(x), fm(f(x))]

≤ cn + cm

1 − cd(x, f(x))

This implies that the sequence (fn(x)) is Cauchy In view of the completeness of X ,there is u such that fn(x) → u. In view of the continuity of f , f(fn(x)) convergesto f(u). On the other hand f(fn(x)) = fn+1(x) converges to u and so, f(u) = uas required. �

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8.0.2 Application of Banach Fixed Point Theorem

Here is an application of the Banach fixed point theorem to the local exis-tence of solutions of ordinary differential equations.

Theorem 8.11 (Picard’s Theorem). Let U be an open subset of R2 and

let f : U → R be a continuous function which satisfies the Lipschitz condition

with respect to the second variable, that is,

|f(x, y1) − f(x, y2)| ≤ α|y1 − y2|

for all (x, y1), (x, y2) ∈ U , and some α > 0. Then for a given (x0, y0) ∈ Uthere is δ > 0 so that the differential equation

y′(x) = f(x, y(x))

has a unique solution y : [x0 − δ, x0 + δ] → R such that y(x0) = y0.

Proof. Note that it is enough to show that there are δ > 0 and a unique functiony : [x0 − δ, x0 + δ] → R such that

y(x) = y0 +

∫

x

x0

f(t, y(t))dt.

Fix (x0, y0) ∈ U . Then we find δ > 0 and b > 0 such that if I = [x0 − δ, x0 + δ]and J = [y0 − b, y0 + b], then I × J ⊂ U . Since f is continuous and I × J is closedand bounded, f is bounded on I × J . That is, |f(x, y)| ≤ M for some M and all(x, y) ∈ U . Replacing δ by a smaller number we may assume that αδ < 1 andαM < b. Denote by X the set of all continuous functions g : I → J . The set Xwith the metric ρ(g, h) = sup{|g(x)− h(x)|, x ∈ I} is a complete metric space. Forg ∈ X , let

(Tg)(x) = y0 +

∫

x

x0

f(t, g(t))dt.

Then Tg : I → R is continuous since if x1, x2 ∈ I and x2 > x1, then

|(Tg)(x2) − (Tg)(x1)| =

∣

∣

∣

∣

∫

x2

x1

f(t, g(t))dt

∣

∣

∣

∣

≤∫

x2

x1

|f(t, g(t))| dt ≤ M |x2 − x1| .

For x0 ≤ x ≤ x0 + δ,

|(Tg)(x) − y0| =

∣

∣

∣

∣

∫ x

x0

f(t, g(t))dt

∣

∣

∣

∣

≤∫ x

x0

|f(t, g(t))|dt ≤ M |x − x0| < Mδ < b

The same inequality holds for x0 − δ ≤ x ≤ x0, and so Tg ∈ X for any g ∈ X .Since f is Lipschitz with respect to the second variable, we obtain for g, h ∈ X and

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