Introduction to Functional Analysis

Transcription

1 Introduction to Functional Analysis Carnegie Mellon University, , Spring 2014 Acknowledgements These notes are based on the lecture course given by Irene Fonseca but may differ from the exact lecture notes. The lecturer is not responsible for errors in these notes: any comments and corrections should be sent to me at Contents 1 Banach spaces 2 Homeworks 7 Last updated: Tuesday 21st January

2 1 Banach spaces Definition 1.1. A vector space (or linear space) over a field K is a nonempty set X, whose elements are called vectors, together with two operations + : X X X, : K X X called addition and scalar multiplication, respectively, satisfying the following properties: (a) (X, +) is an abelian group; (b) s(tx) = (st)x for all s, t K and x X; (c) 1x = x for all x X; (d) s(x + y) = sx + sy for all s K and x, y X; (e) (s + t)x = sx + tx for all s, t K and x X. In this course we ll take K = R or C. Some examples of vector spaces over R include: ˆ R n for n N; ˆ R[x] = {a 0 + a 1 x + + a n x n : n 0, a i R}; ˆ C(X) = {continuous functions f : X R} for X some topological space. Definition 1.2. Let X be a vector space and let E X. (i) E is convex if, for all x, y E and 0 t 1 we have (1 t)x + ty E; (ii) E is a subspace of X if, for all x, y E and s, t K we have sx + ty E. Note that if E X is a subspace then E is convex, though the reverse need not necessarily be true. Proposition 1.3. Let X be a vector space and let E = {E α : α I} be a family of convex sets, where I is any indexing set. Then ˆ α I E α is convex; ˆ If E is totally ordered by (i.e. if α, β I then either E α E β or E β E α ) then α I E α is convex. 2

3 Proof. (i). Let x, y α I E α. For any α I, we have x, y E α, and hence for any 0 t 1, so by convexity (1 t)x + ty E α. Since this holds for all α, (1 t)x + ty α I E α. (ii). Let x, y α I E α. Then there exist α, β I such that x E α and y E β. Without loss of generality, E α E β, and hence x E β. Since E β is convex, (1 t)x + ty E β for all 0 t 1. But then (1 t)x + ty α I E α. Definition 1.4. Let t 1,..., t n K and x 1,..., x n X, and let x = n i=1 t ix i. (i) x is called a linear combination of the x i ; (ii) If each t i 0 and n i=1 t i = 1 then x is called a convex combination of the x i. Proposition 1.5. Let X be a vector space. A subset E X be a subspace (resp. convex subset) if and only if E is closed under taking linear combinations (resp. convex combinations) of its elements. Proof. Exercise. Definition 1.6. Let X be a vector space and E X. The convex hull (or convex envelope) of E, denoted co(e), is the intersection of all convex subsets of X containing E, i.e. co(e) = {E X : E is convex, E E } Definition 1.7. Let X be a vector space and E X. The linear span of E, denoted span(e), is the intersection of all subspaces of X containing E, i.e. span(e) = {E X : E is a subspace ofx, E E } Note that the convex hull and linear span of a subset are both well-defined, since X contains E and is both convex and a subspace of X. Note that the convex hull is well-defined since X is a convex subset of X containing E. Proposition 1.8. Let X be a vector space and E X be a nonempty subset. Then { n } n co(e) = t i x i : n N, x i E, t i 0, t i = 1 i=1 { n } span(e) = t i x i : n N, x i E, t i K i=1 i=1 Definition 1.9. Let X be a vector space over K. A norm on X is a function : X R such that 3

4 (i) x 0 for all x X; (ii) x = 0 if and only if x = 0; (iii) αx = α x for all α K and x X; (iv) x + y x + y for all x, y X. We call the pair (X, ) a normed space. For notation s sake, we will sometimes just write X instead of (X, ), and we may write X for to distinguish it from a norm on another space. Exercise If (i)(ii)(iii) in Definition 1.9 hold then (iv) holds if and only if (iv ) holds, where (iv ) is the assertion that x y x y for all x, y X. Definition Let X be a set. A function d : X X R is a metric on X if (i) d(x, x) = 0 for all x X; (ii) d(x, y) = d(y, x) for all x, y X; (iii) d(x, z) d(x, y) + d(y, z) for all x, y, z X. We call the pair (X, d) a metric space. Every normed space X induces a metric ρ X : X X X by setting We call ρ X the metric induced by. ρ X (x, y) = x y Although every norm gives rise to a metric, not every metric gives rise to a norm. example, X = R is a vector space over R. There is a metric on X defined by { 1 if x y d(x, y) = 0 if x = y For but this is not induced by a norm. If it were then we d have which is nonsense. 1 = d(2, 0) = 2 0 = 2(1 0) = = 2d(1, 0) = 2 Definition Let X be a vector space. X is a Banach space if it is complete with respect to its norm, i.e. given any sequence {x n } n N X such that then there exists x X such that x n x. ε > 0 n 0 N m, n n 0 x n x m < ε 4

5 Some examples of Banach spaces include: (a) R n is a Banach space when endowed with the Euclidean norm, which is defined by (x 1,..., x n ) = ( n (b) The set C([0, 1]) of continuous functions [0, 1] R is naturally a vector space over R by defining addition and scalar multiplication pointwise. Then C([0, 1]) is a Banach space when endowed with the l norm, which is defined by i=1 x 2 i ) 1 2 f = sup{ f(t) : 0 t 1} Note that this norm is well-defined since [0, 1] is compact, so any function f : [0, 1] R is bounded. Proposition Let (X, ) be a Banach space and Y X be a subspace. Then Y is a Banach space under the restriction of to Y if and only if Y is a closed subset of X with respect to the induced metric ρ. Proof. ( ). Suppose Y is Banach and let y Y. Then there exists a sequence {y n } n N Y such that y n y. Since {y n } n N converges it is Cauchy, so since Y is Banach, there exists y Y such that y n y. But y = y since 0 y y y y n + y y n = 0 as n so y Y. Hence Y = Y and Y is closed. ( ). Suppose Y = Y and let {y n } n N be a Cauchy sequence in Y. Since X is a Banach space, there exists y X such that y n y. But then y Y, and since Y = Y, this implies that y Y. So Y is Banach. Note that in the proof of Proposition 1.13, the ( ) direction did not use the fact that X is Banach. Thus any closed subspace of a normed space is a Banach space. Remark We will see that in a finite-dimensional normed space, every subspace is closed, hence Banach. This is not so for the infinite-dimensional case, as illustrated by the following example. Example Let X{{x n } n N R : x n converges in R}, and define {x n } n N = sup{ x n : n N} It is an easy exercise to see that (X, ) is a Banach space. 5

6 Define Y = {{x n } n N R : x n = 0 for all but finitely many n} X Again it is easy to see that Y is a subspace of X. However Y is not a closed subspace. Consider the sequence {y m } m N Y defined by { 1 yn m = n if n m 0 otherwise for example y 7 = {1, 12, 13, 14, 15, 16, 17 }, 0, 0,... Then {y m } m N converges in X to the sequence y = {y n } n N defined by y n = 1 n. So y Y. However y Y since y has no nonzero terms. Thus Y is not closed, and by Proposition 1.13 Y is not Banach. We will see later that there is in fact no norm on Y that would make it a Banach space. 6

7 INTRODUCTION TO FUNCTIONAL ANALYSIS Assignment # 1, Due Thursday, January 30, Let X be a normed space of dimension 2. Prove that the unit sphere B(0, 1); = {x X : x = 1} is path connected (hence connected). 2. Let X be a normed space, and let A L(X) (we abbreviate L(X) := L(X; X)). Prove that the following linear subspaces are closed in L(X): (i) {B L(X) : AB = 0}, i.e., the set of all the right divisors for A. (ii) {B L(X) : AB = BA}, i.e., the centralizer of A. (iii) Let G X and let Y be a normed space. Prove that M := {U L(X; Y ) : G keru} is a closed linear subspace of L(X; Y ). 3. Let X be a normed space. Characterize all the linear subspaces L X which contain a ball. 4. Let X and Y be normed spaces, and let T : X Y and T n : X Y, n N, be linear operators. (i) Prove that the sets and A := {x X : T n x T x} n B := {x X : {T n x} n N is not a Cauchy sequence} are either empty or dense in X. (ii) Let A X be such that X \ A is a linear subspace. Prove that A is either dense or empty. 5. Let (X 1, d 1 ) and (X 2, d 2 ) be two metric spaces and let f : X 1 X 2 be a continuous surjective function such that for all x, y X 1. d 1 (x, y) d 2 (f(x), f(y)) (i) If X 1 is complete then X 2 is complete. Prove or give a counterexample. (ii) If X 2 is complete then X 1 is complete. Prove or give a counterexample. 1

8 (iii) Analyze (i) and (ii) if, in addition, X 1 and X 2 are normed spaces, f is linear, and d 1 and d 2 are the distances associated to the norms in X 1 and X 2, respectively. 6. Let f : C([0, 1]) R be given by f(g) := g(1) for all g C([0, 1]). (i) Prove that f is continuous if on C([0, 1]) we consider the L norm, i.e., g := max x [0,1] g(x) for g C([0, 1]). (ii) Prove that f is not continuous if on C([0, 1]) we consider the L p norm, ( 1 1/p 1 p <, i.e., g p := dx) 0 g(x) p for g C([0, 1]). 7. Let X be a complex normed space and let T : X X be a linear and continuous operator.. (i) Prove that if there is a linear and continuous operator S : X X such that T = T ST then the range of T is a closed set. (ii) Deduce that if there is a polynomial P C[z], P (z) = N n=0 a nz n, N N, a n, z C, such that P (T ) = 0 then: 1. If a 0 0 then the range of T is closed. 2. If N n=m a nt n = 0 for some m > 0 such that a m 0 then ranget m ranget. 2

9 INTRODUCTION TO FUNCTIONAL ANALYSIS Assignment # 2, Due Thursday, February 6, 2014 In this homework you will be using the following results Banach-Steinhaus Theorem: Let X be a Banach space a Y a normed space. Let {T α } α I L(X; Y ) (I is any set of indexes) be such that sup T α x Y < + for all x X. α I Then sup T α x L(X;Y ) < +. α I Open Mapping Theorem: Let X and Y be Banach spaces and let T L(X; Y ) be surjective. Then T is an open map, i.e., if U X is open then so is T (U). Closed Graph Theorem: Let X and Y be Banach spaces and assume that T : X Y is linear and Graph(T ) := {(x, T (x)) : x X} is a closed subset of X Y (a normed space with (x, y) X Y := x X + y Y ). Then T is continuous. and this corollary of the Hahn-Banach Theorem Proposition: Let X be a normed space, and let x X. Then x X = max{l(x) : L X, L X 1}. 1. (i) Prove that l is not separable. { N } (ii) Let X := i=1 a iχ Ai : N N, a i K, A i N are pairwise disjoint, where for A N, χ A = {a n } n N is defined by { 1 if n A, a n := 0 if n / A. Prove that X is dense in l. (iii) Let x = {x n } n N be a sequence of real numbers that belongs to l p, for all p p 0, for some p 0 1. Prove that lim x p = x. p 1

10 2. Let X and Y be normed spaces and let T L(X; Y ) be such that T (X) is closed. Prove that there exists a constant C > 0 such that for every y T (X) there exists x X such that y = T (X) and x X C y Y. 3. Let X, Y and Z be normed spaces, and let B : X Y Z be a bilinear operator. (i) Assume that X or Y are Banach spaces. Prove that if B is separately continuous then it is continuous. (ii) Assume that X and Y are Banach spaces. If B n : X Y Z, n N, are bilinear continuous operators and B n (x, y) B(x, y) as n and for all (x, y) X Y, prove that B is also continuous. 4. Let ξ = {ξ n } n N l and assume that ξ n < ξ for all n N. Define L : l 1 R by L(x) := ξ n x n for all x = {n n } n N l 1. n=1 (i) Prove that L is a linear continuous functional and L ξ. (ii) Prove that there is no element x in the closed unit ball of l 1 such that L(x) = ξ. 5. (i) Prove that for all L (l ) there is a unique finitely additive measure µ : P(N) K with bounded variation such that where χ A = {a n } n N and Prove also that µ(a) = L(χ A ) for all A N, (1) a n := { 1 if n A, 0 if n / A. ( µ (N) it the total variation of the measure µ ). µ (N) L (2) (i) Conversely, show that for every finitely additive measure µ : P(N) K with bounded variation there exists a unique element L (l ) such that (1) and (2) hold. Hint: Use Problem 1-(ii). 2

11 6. Let 1 < p < + and let X be a Banach space. Denote be w p (X) the set of all sequences {x n } n N X with the property that {L(x n )} n N l p for all L X. For {x n } n N w p (X) define {x n } n N weak := sup L X, L 1 ( n=1 L(x n ) p ) 1 p. Prove that (w p (X), weak ) is a Banach space (in particular, show that if {x n } n N w p (X) then {x n } n N weak < + ). 7. (i) Let Y be a Banach space, Z a normed space, and B n : Y Z, n N, linear and continuous operators such that for every sequence {y n } n N with y n 0 it follows that B n (y n ) 0. Prove that sup B n < +. n N (ii) Let ϕ n : [0, 1] R, n N, be continuous functions. Prove that sup sup n N x [0,1] ϕ n (x) < + if and only if for every sequence {f n } n N of continuous functions f n : [0, 1] R coverging uniformly on [0, 1] it follows that ϕ n f n 0 uniformly on [0, 1]. 3