8

Completeness

We recall the definition of a Cauchy sequence. Let (X, d) be a given metric space and let (xn ) be a sequence of points of X. Then (xn ) (xn ) is a Cauchy sequence 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 Cauchy sequence (xn ) in X converges to some point of X. A subset A of X is called complete if A as a metric subspace of (X, d) is complete, that is, if every Cauchy sequence (xn ) in A converges to a point in A. By the above example, not every metric space is complete; (0, 1) with the standard 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 and Propositions 8.1 Recall that if (Xi , di ), 1 ≤ i ≤ m, are metric spaces and X = X1 × . . . × Xn , then d(x, y) = max dj (xi , yi ) 1≤j≤n

where x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ X, defines a metric on X. The pair (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.

49

Proof. Let xn = (x1n , . . . , xm n ) and (xn ) be a Cauchy sequence in (X, d). Then for a given ε > 0 there exists k such that d(xn , xm ) < ε for all n, m ≥ k. Since dj (xjn , xjm ) ≤ 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, that xn → x in X. Since n X

max{|x1 − y1 | , . . . , |xn − yn |} ≤ ≤

√

i=1

|xi − yi |2

!1/2

n max{|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

P∞

k=1 |xk | < ∞. Then ℓ1 is a vector space if addition and multiplication by a number are defined as follows, (xn ) + (yn ) = (xn + yn ), α(xn ) = (αxn ).

If x = (xk ) ∈ ℓ1 , then kxk = k(xk )k :=

∞ X

|xk |

k=1 P∞ k=1

defines a norm on ℓ1 and d(x, y) = kx − yk = |xk − yk | defines a distance on ℓ1 . We claim that (ℓ1 , d) is a complete metric space. (n) Indeed, let (Xn ) ⊂ ℓ1 be a Cauchy sequence. Then with Xn (xk ) we have ∞ X (n) (n) (m) (m) xk − xk ≤ xl − xl = kXn − Xm k . l=1

(n)

Hence for every k ≥ 1, the sequence (xk ) is Cauchy in R and since R with (n) the standard metric is complete, the sequence (xk ) converges to some xk . Set X = (xk ). We suspect that X is the limit in ℓ1 of the sequence (Xn ). To see this we first show that X ∈ ℓ1 . Since (Xn ) is Cauchy in ℓ1 , there is K such that kXn − Xm k < 1 for all n, m ≥ K. In particular, N ∞ ∞ X X (n) X (n) (K) (K) xk ≤ xk − xk + xk

k=1

k=1

k=1

≤ kXn − XK k + kXK k < 1 + kXK k

50

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

k=1

|xk | ≤ 1 + kXK k

and taking limit as N → ∞ we get ∞ X

k=1

|xk | ≤ 1 + kXK k .

So, X ∈ ℓ1 . Next we show that kXn − Xk → 0 as n → ∞. Given ε > 0, there is K such that kXn − Xm k < 1 for all n, m ≥ K. Consequently, for every N ≥ 1 we have N X (n) (m) xk − xk ≤ kXn − Xm k < ε, k=1

for all n, m ≥ K.

With n > K and N fixed, we let m → ∞ to find that N X (n) xk − xk ≤ ε.

k=1

Since this is true for every N ,

kXn − Xk =

∞ X (n) xk − xk ≤ ε

k=1

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 same metric 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 the Cauchy condition in X, and since (X, d) is complete, there exists x ∈ X such that xn → 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 show that x ∈ Y . Since (xn ) converges in X, it satisfies the Cauchy condition in X and so, it also satisfies the Cauchy condition in Y . Since (Y, d) is complete, it converges to 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. 51

Let (X, d) and (Y, ρ) be metric spaces. A function f : X → Y is said to 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 bounded and by Cb (X, Y ) the space of bounded continuous function f : X → Y . If Y = 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 the same 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 show that 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 that D(fn , f ) ≤ ε

for n ≥ N .

It remains to show that f is bounded. Since fN is bounded, there are y ∈ Y and r > 0 such that fN (X) ⊂ Br (y). From (1) and the triangle inequality, we obtain ρ(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. Next we 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 is continuous. 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.

52

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 is continuous, 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 called a 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 complete 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 (f n (x)) converges to u. Proof. Claim: If a, b ∈ X, then d(a, b) ≤ Indeed,

1 [d(a, f (a)) + d(b, f (b))]. 1−c

(2)

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 point since if f (a) = a and f (b) = b, then the right side of (2) is equal to 0 implying that d(a, b) = 0. To prove existence of a fixed point, take any x ∈ X. Since f is a contraction with the constant c, the n-fold composition f n is a contraction with the constant cn . Using this and (2) with a = f n (x) and b = f n (b), and noticing that f (a) = f n+1 (x) = f n (f (x)), f (b) = f m (f (x)), we conclude 1 d(f n (x), f m (x)) = d(a, b) ≤ d(a, f (a)) + d(b), b)] 1−c 1 d(f n (x)f n (f (x))) + d(f m (x), f m (f (x))] = 1−c cn + cm ≤ d(x, f (x)) 1−c

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

53

8.0.2

Application of Banach Fixed Point Theorem

Here is an application of the Banach fixed point theorem to the local existence 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 ) ∈ U there 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 function y : [x0 − δ, x0 + δ] → R such that y(x) = y0 +

Z

x

f (t, y(t))dt.

x0

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 closed and 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 and αM < b. Denote by X the set of all continuous functions g : I → J. The set X with the metric ρ(g, h) = sup{|g(x) − h(x)|, x ∈ I} is a complete metric space. For g ∈ X, let Z x

f (t, g(t))dt.

(T g)(x) = y0 +

x0

Then T g : I → R is continuous since if x1 , x2 ∈ I and x2 > x1 , then Z x2 Z x2 |f (t, g(t))| dt ≤ M |x2 − x1 | . |(T g)(x2 ) − (T g)(x1 )| = f (t, g(t))dt ≤ x1

x1

For x0 ≤ x ≤ x0 + δ, Z |(T g)(x) − y0 | =

x

x0

Z f (t, g(t))dt ≤

x

x0

|f (t, g(t))|dt ≤ M |x − x0 | < M δ < b

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

54

x ∈ [x0 , x0 + δ], Z x |(T g)(x) − (T h)(x)| = [f (t, g(t)) − f (t, h(t))] dt x Z x0 |f (t, g(t)) − f (t, h(t))| dt ≤ x0

≤ α|x − x0 |d(g, h) < αδd(g, h).

Similarly, |(T g)(x) − (T h)(x)| ≤ α|x − x0 |d(g, h) < αδd(g, h) for x ∈ [x0 − δ, x0 ]. Since αδ < 1, T is a contraction and in view of Banach’s fixed point theorem there exists a unique continuous function y : I → J such that Z x f (t, y(t))dt. y(x) = (T y)(x) = y0 + x0

55

Completeness

We recall the definition of a Cauchy sequence. Let (X, d) be a given metric space and let (xn ) be a sequence of points of X. Then (xn ) (xn ) is a Cauchy sequence 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 Cauchy sequence (xn ) in X converges to some point of X. A subset A of X is called complete if A as a metric subspace of (X, d) is complete, that is, if every Cauchy sequence (xn ) in A converges to a point in A. By the above example, not every metric space is complete; (0, 1) with the standard 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 and Propositions 8.1 Recall that if (Xi , di ), 1 ≤ i ≤ m, are metric spaces and X = X1 × . . . × Xn , then d(x, y) = max dj (xi , yi ) 1≤j≤n

where x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ X, defines a metric on X. The pair (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.

49

Proof. Let xn = (x1n , . . . , xm n ) and (xn ) be a Cauchy sequence in (X, d). Then for a given ε > 0 there exists k such that d(xn , xm ) < ε for all n, m ≥ k. Since dj (xjn , xjm ) ≤ 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, that xn → x in X. Since n X

max{|x1 − y1 | , . . . , |xn − yn |} ≤ ≤

√

i=1

|xi − yi |2

!1/2

n max{|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

P∞

k=1 |xk | < ∞. Then ℓ1 is a vector space if addition and multiplication by a number are defined as follows, (xn ) + (yn ) = (xn + yn ), α(xn ) = (αxn ).

If x = (xk ) ∈ ℓ1 , then kxk = k(xk )k :=

∞ X

|xk |

k=1 P∞ k=1

defines a norm on ℓ1 and d(x, y) = kx − yk = |xk − yk | defines a distance on ℓ1 . We claim that (ℓ1 , d) is a complete metric space. (n) Indeed, let (Xn ) ⊂ ℓ1 be a Cauchy sequence. Then with Xn (xk ) we have ∞ X (n) (n) (m) (m) xk − xk ≤ xl − xl = kXn − Xm k . l=1

(n)

Hence for every k ≥ 1, the sequence (xk ) is Cauchy in R and since R with (n) the standard metric is complete, the sequence (xk ) converges to some xk . Set X = (xk ). We suspect that X is the limit in ℓ1 of the sequence (Xn ). To see this we first show that X ∈ ℓ1 . Since (Xn ) is Cauchy in ℓ1 , there is K such that kXn − Xm k < 1 for all n, m ≥ K. In particular, N ∞ ∞ X X (n) X (n) (K) (K) xk ≤ xk − xk + xk

k=1

k=1

k=1

≤ kXn − XK k + kXK k < 1 + kXK k

50

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

k=1

|xk | ≤ 1 + kXK k

and taking limit as N → ∞ we get ∞ X

k=1

|xk | ≤ 1 + kXK k .

So, X ∈ ℓ1 . Next we show that kXn − Xk → 0 as n → ∞. Given ε > 0, there is K such that kXn − Xm k < 1 for all n, m ≥ K. Consequently, for every N ≥ 1 we have N X (n) (m) xk − xk ≤ kXn − Xm k < ε, k=1

for all n, m ≥ K.

With n > K and N fixed, we let m → ∞ to find that N X (n) xk − xk ≤ ε.

k=1

Since this is true for every N ,

kXn − Xk =

∞ X (n) xk − xk ≤ ε

k=1

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 same metric 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 the Cauchy condition in X, and since (X, d) is complete, there exists x ∈ X such that xn → 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 show that x ∈ Y . Since (xn ) converges in X, it satisfies the Cauchy condition in X and so, it also satisfies the Cauchy condition in Y . Since (Y, d) is complete, it converges to 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. 51

Let (X, d) and (Y, ρ) be metric spaces. A function f : X → Y is said to 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 bounded and by Cb (X, Y ) the space of bounded continuous function f : X → Y . If Y = 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 the same 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 show that 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 that D(fn , f ) ≤ ε

for n ≥ N .

It remains to show that f is bounded. Since fN is bounded, there are y ∈ Y and r > 0 such that fN (X) ⊂ Br (y). From (1) and the triangle inequality, we obtain ρ(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. Next we 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 is continuous. 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.

52

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 is continuous, 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 called a 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 complete 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 (f n (x)) converges to u. Proof. Claim: If a, b ∈ X, then d(a, b) ≤ Indeed,

1 [d(a, f (a)) + d(b, f (b))]. 1−c

(2)

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 point since if f (a) = a and f (b) = b, then the right side of (2) is equal to 0 implying that d(a, b) = 0. To prove existence of a fixed point, take any x ∈ X. Since f is a contraction with the constant c, the n-fold composition f n is a contraction with the constant cn . Using this and (2) with a = f n (x) and b = f n (b), and noticing that f (a) = f n+1 (x) = f n (f (x)), f (b) = f m (f (x)), we conclude 1 d(f n (x), f m (x)) = d(a, b) ≤ d(a, f (a)) + d(b), b)] 1−c 1 d(f n (x)f n (f (x))) + d(f m (x), f m (f (x))] = 1−c cn + cm ≤ d(x, f (x)) 1−c

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

53

8.0.2

Application of Banach Fixed Point Theorem

Here is an application of the Banach fixed point theorem to the local existence 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 ) ∈ U there 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 function y : [x0 − δ, x0 + δ] → R such that y(x) = y0 +

Z

x

f (t, y(t))dt.

x0

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 closed and 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 and αM < b. Denote by X the set of all continuous functions g : I → J. The set X with the metric ρ(g, h) = sup{|g(x) − h(x)|, x ∈ I} is a complete metric space. For g ∈ X, let Z x

f (t, g(t))dt.

(T g)(x) = y0 +

x0

Then T g : I → R is continuous since if x1 , x2 ∈ I and x2 > x1 , then Z x2 Z x2 |f (t, g(t))| dt ≤ M |x2 − x1 | . |(T g)(x2 ) − (T g)(x1 )| = f (t, g(t))dt ≤ x1

x1

For x0 ≤ x ≤ x0 + δ, Z |(T g)(x) − y0 | =

x

x0

Z f (t, g(t))dt ≤

x

x0

|f (t, g(t))|dt ≤ M |x − x0 | < M δ < b

The same inequality holds for x0 − δ ≤ x ≤ x0 , and so T g ∈ 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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x ∈ [x0 , x0 + δ], Z x |(T g)(x) − (T h)(x)| = [f (t, g(t)) − f (t, h(t))] dt x Z x0 |f (t, g(t)) − f (t, h(t))| dt ≤ x0

≤ α|x − x0 |d(g, h) < αδd(g, h).

Similarly, |(T g)(x) − (T h)(x)| ≤ α|x − x0 |d(g, h) < αδd(g, h) for x ∈ [x0 − δ, x0 ]. Since αδ < 1, T is a contraction and in view of Banach’s fixed point theorem there exists a unique continuous function y : I → J such that Z x f (t, y(t))dt. y(x) = (T y)(x) = y0 + x0

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