Introduction to Banach Spaces

Transcription

1 CHAPTER 8 Introduction to Banach Saces 1. Uniform and Absolute Convergence As a rearation we begin by reviewing some familiar roerties of Cauchy sequences and uniform limits in the setting of metric saces. Definition 1.1. A metric sace is a air (X, ρ), where X is a set and ρ is a real-valued function on X X which satisfies that, for any x, y, z X, (a) ρ(x, y) 0 and ρ(x, y) = 0 if and only if x = y, (b) ρ(x, y) = ρ(y, x), (c) ρ(x, z) ρ(x, y) + ρ(y, z). (Triangle inequality) The function ρ is called the metric on X. Any metric sace has a natural toology induced from its metric. A subset U of X is said to be oen if for any x U there exists some r > 0 such that B r (x) U. Here B r (x) = {y X : ρ(x, y) < r} is the oen ball of radius r centered at x. It is an easy exercise to show that oen balls are indeed oen and the collection of oen sets is indeed a toology, called the metric toology. On the contrary, there are toological saces whose toology can be defined by some metric. In this case we say the toology is metrizable. Definition 1.2. A sequence {x n } in a metric sace (X, ρ) is said to be a Cauchy Sequence if ε > 0, N N such that ρ(x n, x m ) < ε whenever n, m N. The metric sace (X, ρ) is said to be comlete if every Cauchy sequence is convergent. Definition 1.3. Let (X, ρ) be a metric sace. For any nonemty set A X, the diameter of the set A is defined by diam(a) = su{ρ(x, y) : x, y A}. The set A is said to be bounded if its diameter is finite. Otherwise, we say it is unbounded. Let S be a nonemty set. We say a function f : S X is bounded if its image f(s) is a bounded set. Equivalently, it is bounded if for any x X, there exists M > 0 such that ρ(f(s), x) M for any s S. We say f is unbounded if it is not bounded. Definition 1.4. Given a sequence {f n } of functions from S to X. ointwise to the function f : S X if s S, ε > 0, N s N such that ρ(f n (s), f(s)) < ε, n N. We say {f n } converges In this case, the function f is called the ointwise limit. We say {f n } converges uniformly to a function f : S X if the above N s is indeendent of s; that is, ε > 0, N N such that ρ(f n (s), f(s)) < ε, n N, s S. 115

2 INTRODUCTION TO BANACH SPACES The function f is called the uniform limit of {f n }. When S is countably infinite, the function f above is a sequence in X, and {f n } is a sequence of sequences in X; or in other words, {f n (m)} is a double sequence in X. The next two theorems highlight some imortant features of Cauchy sequences and uniform convergence. Theorem 1.1. (Cauchy Sequences) Consider sequences in a metric sace (X, ρ). (a) Any convergent sequence is a Cauchy sequence. (b) Any Cauchy sequence is bounded. (c) If a subsequence of a Cauchy sequence converges, then the Cauchy sequence converges to the same limit. Proof. (a) Suose {x n } converges to x. Given ε > 0, there is an N N such that ρ(x n, x) < ε/2 for any n N. The sequence {x n } is Cauchy because ρ(x n, x m ) < ρ(x n, x) + ρ(x, x m ) < ε for any n, m N. (b) Let {x n } be a Cauchy sequence. Choose N N such that ρ(x n, x m ) < 1 for all n, m N. Then for any x X, ρ(x n, x) ρ(x n, x N ) + ρ(x N, x) < max{ρ(x 1, x N ), ρ(x 2, x N ),, ρ(x N 1, x N ), 1} + ρ(x N, x), where the last equation is a finite bound indeendent of n. (c) Let {x n } be a Cauchy sequence with a subsequence {x n } converging to x. Given ε > 0, choose K, N N such that ρ(x n, x) < ε 2 for any K, ρ(x n, x m ) < ε for any n, m N. 2 Taing n such that K and n N, then This shows that {x n } converges to x. ρ(x n, x) ρ(x n, x n ) + ρ(x n, x) < ε for any n N. Theorem 1.2. (Uniform Convergence) Given a sequence of functions {f n } from a nonemty set S to a metric sace (X, ρ). Suose {f n } converges uniformly to a function f : S X. (a) If each f n is bounded, then so is f. (b) Assume S is a toological sace, E S. If each f n is continuous on E, then so is f. Proof. (a) Choose N N such that ρ(f(s), f n (s)) < 1 for any n N and s S. Given x X, choose M > 0 such that ρ(f N (s), x) < M for any s S. Then f is bounded since ρ(f(s), x) ρ(f(s), f N (s)) + ρ(f N (s), x) 1 + M for any t S. (b) Given ε > 0, e E. Choose N N such that ρ(f(s), f n (s)) < ε/3 for any n N and s S. For this articular N, f N is continuous at e, and so there is a neighborhood U of e such that ρ(f N (e), f N (u)) < ε/3 whenever u U. Then ρ(f(e), f(u)) ρ(f(e), f N (e)) + ρ(f N (e), f N (u)) + ρ(f N (u), f(u)) < ε u U. Therefore f is continuous at e, and is continuous on E since e E is arbitrary.

3 1. UNIFORM AND ABSOLUTE CONVERGENCE 117 Definition 1.5. A vector sace V over field F is called a normed vector sace (or normed sace) if there is a real-valued function on V, called the norm, such that for any x, y V and any α F, (a) x 0 and x = 0 if and only if x = 0. (b) αx = α x. (c) x + y x + y. (Triangle inequality) A norm of V defines a metric ρ on V via ρ(x, y) = x y. All concets from metric and toological saces are alicable to normed saces. There are multile ways of choosing norms once a norm is selected. A trivial one is to multily the original norm by a ositive constant. Concets lie neighborhood, convergence, and comleteness are indeendent of the choice of these two norms, and so we shall consider them equivalent norms. A more recise characterization of equivalent norms is as follows. Definition 1.6. Let V be a vector sace with two norms,. We say these two norms are equivalent if there exists some constant c > 0 such that 1 c x x c x for any x V. Examle 1.1. The Euclidean sace F n, F = R or C, with the standard norm defined by x = ( x x n 2 ) 1 2 is a normed sace. Now consider two other norms of F n defined by x = max{ x 1,, x n }, called the su norm; x 1 = x x n, called the 1-norm. Verifications for axioms of norms are comletely straightforward. In the case of su norm, balls in R n are actually cubes in R n with faces arallel to coordinate axes. In the case of 1-norm, balls in R n are cubes in R n with vertices on coordinate axes. These norms are equivalent since x x x 1 n x. Definition 1.7. A comlete normed vector sace is called a Banach sace. Examle 1.2. Consider the Euclidean sace F n, F = R or C, with the standard norm. The normed sace (R n, ) is comlete since every Cauchy sequence is bounded and every bounded sequence has a convergent subsequence with limit in R n (the Bolzano-Weierstrass theorem). The saces (R n, 1 ) and (R n, ) are also Banach saces since these norms are equivalent. Examle 1.3. Given a nonemty set X and a normed sace (Y, ) over field F. The sace of functions from X to Y form a vector sace over F, where addition and scalar multilication are defined in a trivial manner: Given two functions f, g, and two scalars α, β F, define αf + βg by (αf + βg)(x) = αf(s) + βg(s), x X. Let b(x, Y ) be the subsace consisting of bounded functions from X to Y. Define a real-valued function on b(x, Y ) by f = su f(x). x X

4 INTRODUCTION TO BANACH SPACES It is clearly a norm on b(x, Y ), also called the su norm. Convergence with resect to the su norm is clearly the same as uniform convergence. If (Y, ) is a Banach sace, then any Cauchy sequence {f n } in b(x, Y ) converges ointwise to some function f : X Y, since {f n (x)} is a Cauchy sequence in Y for any fixed x X. In fact, the convergence f n f is uniform. To see this, let ε > 0 be arbitrary. Choose N N such that f n f m < ε/2 whenever n, m N. For any x X, there exists some m x N such that f mx (x) f(x) < ε/2. Then for any n N, f n (x) f(x) f n (x) f mx (x) + f mx (x) f(x) < ε. This roves that the convergence f n f is uniform. By Theorem 1.2(a), f b(x, Y ), and so the sace b(x, Y ) with the su norm is a Banach sace. Examle 1.4. Let (X, T ) be a toological sace and let (Y, ) be a Banach sace over F = R or C. Denote by C(X, Y ) the sace of continuous functions from X to Y. Let C b (X, Y ) = C(X, Y ) b(x, Y ), the sace of bounded continuous functions from X to Y. Given any sequence {f n } in C b (X, Y ) which converges uniformly to f b(x, Y ). By Theorem 1.2(b), f C b (X, Y ). This shows that C b (X, Y ) is a closed subsace of b(x, Y ), and is therefore a Banach sace (see Exercise 1.1). Definition 1.8. A series =1 a in a normed sace X is said to be convergent (or summable) if its artial sum n =1 a converges to some s X as n. We say =1 a is absolutely convergent (or absolutely summable) if =1 a <. In the following we rove some useful criteria for comleteness and uniform convergence of series. Theorem 1.3. A normed sace X is comlete if and only if every absolutely convergent series is convergent. Proof. Suose X is comlete, =1 a is absolutely convergent. We need to show the convergence of s n = n =1 a. Given ε > 0, choose N N such that =N a < ε, then s n s m < ε whenever n, m N. Thus {s n } n=1 is a Cauchy sequence, and so it converges. Conversely, suose every absolutely convergent series in X converges. Let {s n } n=1 be a Cauchy sequence in X. By Theorem 1.1 it suffices to show that {s n } n=1 has a convergent subsequence {s n } =1. Now choose n such that n < n +1, s n s n+1 < 1 2 for any N. Then the series s n1 + ( ) =1 sn+1 s n converges absolutely, so that it converges to some s X. This imlies that s n = s n1 + 1 ( ) j=1 snj+1 s nj converges to s as, comleting the roof. Corollary 1.4. (Weierstrass M-test) Let b(x, Y ) be the sace bounded functions from a nonemty set X to a Banach sace (Y, ). Given a sequence of functions {f n } n=1 in b(x, Y ). If f n M n for any n N and n=1 M n converges, then n=1 f n converges uniformly. Proof. The assumtion says that n=1 f n is absolutely convergent. By Theorem 1.3 (and Examle 1.3), the series converges in b(x, Y ), imlying that the convergence is uniform.

5 Exercises. 2. THE l SPACE Show that a subset of a comlete metric sace is comlete if and only if it is closed Let Σ 2 = {0, 1} N, the sace of infinite sequences of {0, 1}; that is, Given λ > 1, a, b Σ 2, let Σ 2 = {(a 1, a 2,...) : a = 0 or 1 for each }. ρ λ (a, b) = Show that (Σ 2, ρ λ ) is a comlete metric sace. =1 a b λ Consider the sace of real sequences s. Let a b ρ(a, b) = 2 (1 + a b ). Show that (s, ρ) is a comlete metric sace. = Consider the sace BV [a, b] of functions on [a, b] with bounded variations. For any f BV [a, b], let f = f(a) + V b a (f). Show that (BV [a, b], ) is a Banach sace. Is it searable? 2. The l Sace Definition 2.1. Let F = R or C. Given 0 < <, define l = {a = (a 1, a 2, ) : a F for any, a < }, a = l = {a = (a 1, a 2, ) : a F for any, su a < }, a = su a ) 1, The sace l consists of bounded sequences in F. Addition and multilication of sequences are defined comonentwise: (a 1, a 2, ) + (b 1, b 2, ) = (a 1 + b 1, a 2 + b 2, ) (a 1, a 2, ) (b 1, b 2, ) = (a 1 b 1, a 2 b 2, ). Clearly l with any 0 < is a vector sace, since αa = α a for any α F and a, b l a + b (2 max{ a, b }) 2 a. ( a + b ),. a, b l su a + b su a + su b. When 1, the function is a norm on l. This follows from Theorem 2.1 below. Theorem 2.1. Given 1, q with q = 1. Let a, b be sequences of comlex numbers. (a) (Young s inequality) If u, v 0, 1 <, q <, then uv u + vq q.

6 INTRODUCTION TO BANACH SPACES (b) (Hölder s Inequality for l ) If a l, b l q, then ab l 1 and (c) (Minowsi s Inequality for l ) ab 1 a b q. a + b a + b. 1 Proof. It would be convenient to write q = (q = if = 1, q = 1 if = ). The curve y = x 1 can be alternatively written x = y q 1. Part (a) follows easily by observing U = u 0 x 1 dx = u, V = v 0 y q 1 dy = vq q, U + V uv. For art (b), the cases = 1, q = and =, q = 1 are obvious. Consider 1 <, q <. The cases a = 0 or b q = 0 are also obvious, so we assume 0 < a, b q <. Let A = a a, B = b b q. By Young s inequality, A B ( A + B q ) = A + B q q = 1 q q + 1 q = 1, ab 1 = a b = a b q A B a b q. The cases = 1 and = for (c) are obvious. For 1 < <, (c) follows easily from (b): a + b = a + b a + b 1 a + a + b 1 b ) 1 a + b = a + b 1 ( a + b ). ) 1 ( ) 1 a + a + b b ) 1 Theorem 2.2. For any 1, (l, ) is a Banach sace. Proof. Comleteness of l is a secial case of Examle 1.3. Consider 1 <. Let {a (n) } i=1 be a Cauchy sequence in l. For each, {a (n) } n=1 is a Cauchy sequence of real numbers since 1 a (n) j j = a (n). Then there is a sequence a = (a 1, a 2, ) such that, for each, j=1 a (n) a R as n.

7 Given ε > 0, M N, there exists some N N such that 2. THE l SPACE 121 ( M =1 ) 1 ( Let m, then let M, we find =1 ) 1 < ε n > m N. ( a (n) a = =1 ) 1 a ε for any n N. Thus a (n) a 0 as n, and so a a a (n) + a (n) <, a l. This verifies comleteness of l. Theorem 2.3. The sace l is searable if 1 <, and the sace l is not searable. Proof. The sace l is not searable because it has an uncountable subset s = {a = (a 1, a 2,...) l : a n = 0 or 1 n} and a b = 1 for any a b s. Consider 1 <. Let D be the set of finite sequences with rational coordinates. Clearly D is countable. Given a l and any ε > 0, we can choose N N such that =N+1 a < ε /2. Now choose b 1,, b N Q such that N =1 a b < ε /2. Let b = (b 1,, b N, 0, 0, ) D. Then N a b = a b = a b + a < ε. =1 =1 =N+1 Thus a b < ε. This shows that D is dense since ε > 0 is arbitrary. Exercises Consider the l sace with 0 < < 1. Verify that ρ (a, b) = =1 a b is a metric on l. Prove that (l, ρ ) is a comlete searable metric sace Prove the following generalization of Young s inequality: Given 1 < 1,, n < with = 1. If u 1,, u n 0, then n =1 1 u 1 u n u un n. n Use it to formulate a generalization of Hölder s inequality for l Consider sequences of real numbers. Show that the sace c 0 of sequences converging to zero with su norm is a Banach sace, and for any 1 < q, a l, Are these norms on l equivalent? l l q c 0, a a q a Exlain why the set s in the roof of Theorem 2.3 is uncountable.