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					Course 07130023/MAAB3202

Functional Analysis
Lecture 6

By Yisheng Huang

yishengh@suda.edu.cn

School of Mathematical Sciences Soochow University

Lecture 6
Complete Metric Spaces
Contents
1. Concept of completeness. 2. Examples of complete metric spaces. 3. Properties of complete metric spaces. 4. A Characterization of completeness.

Concept of Completeness
« A metric space (X, d) is said to be complete if every Cauchy sequence (i.e., d(xn , xm ) → 0 as n, m → ∞) in X converges to an element of X. A set A ⊂ X is complete in (X, d) if every Cauchy sequence of A converges to an element of A.

Examples of Complete and Incomplete Metric Spaces
m The spaces R, C, Rn and C[a, b] are complete. These follow from the Cauchy convergent criterion in the course of Mathematical Analysis. m The space
∞

is complete.

Proof : In fact, suppose that {xn } is a Cauchy sequence of ∞ , where xn = (n) (n) {ξk } such that |ξk | Kn for k ∈ N, Kn is a constant for each n ∈ N. Now for each ε > 0 there exists N ∈ N such that d(xn , xm ) = sup |ξk − ξk | < ε
k (n) (m)

for all n, m > N,

and then

|ξk − ξk | < ε

(n)

(m)

for all k ∈ N, for all n, m > N.
(1) (2) (n)

(1)

Hence for each k ∈ N the sequence of numbers {ξk , ξk , · · · , ξk , · · · } is a Cauchy sequence of R or C, so that by the completeness of R or C it converges to some ξk as n goes to infinity. Let x = {ξk }. We will show that x ∈ ∞ and xn → x as n → ∞. Indeed, letting m → ∞ in (1), we get 18

|ξk − ξk | Therefore for all k ∈ N |ξk | |ξk
(N +1)

(n)

ε

for all k ∈ N, n > N.

(2)

− ξk | + |ξk

(N +1)

|

ε + KN +1 ,

which implies that x = {ξk } ∈
(n) k

∞.

It follows from (2) that ε for all n > N,

sup |ξk − ξk | i.e., d(xn , x) ε for all n > N . So
∞

is complete.

m The interval (0, 1), as a subspace of R, is incomplete. m The set Q, as a subspace of R, is incomplete. m The set of all continuous functions on [a, b] with the metric
b

d1 (x, y) = is incomplete.

|x(t) − y(t)| dt,
a

Properties of Complete Metric Spaces
« Theorem Suppose that (X, d) is a metric space, and that A ⊂ X.

(1) If A is complete, then A is closed. (2) If X is complete and A is closed, then A is complete. Proof (1) Suppose that x ∈ A, then there exists a sequence {xn } in A such that xn → x as n → ∞. So {xn } is a Cauchy sequence. Since A is complete, we must have x ∈ A. Hence A ⊂ A, so that A = A and then A is closed. (2) Suppose that {xn } is a Cauchy sequence in A. Then it is a Cauchy sequence in X. Since X is complete, it follows that xn → x ∈ X as n → ∞. Noting that x is a limit point of A (i.e., x ∈ A), and so x must belong to A since A is closed. It follows that {xn } converges in A. « An Application The spaces c, c0 , as closed subspaces of
∞,

are complete.

A Characterization of Completeness: Nested Closed Balls

« Concept A nest of closed balls is a sequence {Sn := S(xn , rn )} of closed balls such that S1 ⊃ S2 ⊃ · · · and the diameter d(Sn ) → 0 (i.e., radius rn → 0) as n → ∞. 19

« Theorem Suppose that (X, d) is a metric space. Then (X, d) is complete if and only if every nest of closed balls has a nonempty intersection. Proof : Let X be complete and let {Sn } be a nest. Denote xn be the center of Sn , then {xn } is a Cauchy sequence in X, and hence xn → x as n → ∞ for some x ∈ X since X is complete. Therefore x is a limit point of Sn , so that x ∈ Sn since Sn is closed. It follows that x ∈ ∩ {Sn : n ∈ N} and so ∩ {Sn : n ∈ N} contains at least one point. Actually ∩{Sn : n ∈ N} is a single point set. Conversely, let every nest have nonempty intersection. Take any Cauchy sequence {xn }. Choose n1 such that d(xn , xn1 ) < 1/2, n > n1 ; n2 > n1 such that d(xn , xn2 ) < 1/4, n > n2 , · · · . Determine S1 = S(xn1 , 1) = {x ∈ X : d(x, xn1 ) 1}, S2 = S(xn2 , 1/2), · · · . Then we obtain n1 < n2 < n3 < · · · and S1 ⊃ S2 ⊃ S3 ⊃ · · · with d(Sn ) → 0 as n → ∞. Hence ∩{Sn : n ∈ N} = {x} for some x ∈ X. So {xn } has a convergent subsequence. Exercise 1.11 implies the whole {xn } converges in X, i.e., X is complete.

Exercises 1.24-1.25, 1.27-1.29 for minimum requirements. Exercises 1.26, 1.32 for option.

Solutions of some Exercises by Students:
1.1 b = 0 ⇒ a = kb for some k ∈ C. Denote k := x + iy with x, y ∈ R, then |a + b| = |a| + |b| ⇔ x = x2 + y 2 ⇔ k = x 0. 1.6 The “supremum” part: Since
∞ n→∞

lim d(xn , 0) = lim

n→∞

k=1

1 n n = lim = 1, k1+n n→∞ 1 + n 2
1−ε ε ,

it follows that for each ε : 0 < ε < 1 we may choose n0 ∈ N such that n0 > so that d(xn0 , 0) > 1 − ε, i.e., sup d(x, y) = 1. 1.9 (⇐) part can be deduced trivially.

(⇒) Suppose that ρ(x, y) = |f (x) − f (y)| is a metric on R, then ρ(x, y) = 0 ⇐⇒ x = y, it follows that x = y =⇒ f (x) = f (y). We claim that if x < y < z then f (x) < f (y) < f (z) or f (x) > f (y) > f (z). Indeed, if not, we have e.g f (x) > f (y) and f (z) > f (y). Assume that f (z) > f (x), thus f (z) > f (x) > f (y). Consider the continuous function f on the closed interval [y, z], by the intermediate value theorem, there exists t ∈ (y, z) such that f (t) = f (x) and then t = x, but this is impossible since t > y > x. For other cases, e.g. f (x) < f (y), f (z) < f (y) when x > y > z, we can prove them similarly. 20

1.23 Suppose A is a subspace of X. Let D = {xn : n ∈ N} is a countable dense subset of X (if D is finite, let xs = xs+1 = . . . ). We construct a countable dense subset of A in the following way: for every m ∈ Z+ , we choose a point xm,i in B(xi , 1/m) ∩ A (if B(xi , 1/m) ∩ A = ∅ , ignore it) and let Em = {xm,i : i ∈ N}. Clearly Em is countable and so is the set E = ∪m∈Z+ Em . We claim that E is the dense subset of A. In fact, for every ε > 0 and x ∈ A considering M = [2/ε] + 2. It follows that x must belong to some ball of the set {B(xi , 1/M ) : i ∈ N}. Let x ∈ B(xi , 1/M ). Therefore there exists a point y ∈ E such that y ∈ B(xi , 1/M ). It follows the construction of E that d(x, y) < 1/M + 1/M < ε. Hence A is separable.

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