Functional Analysis, Math 7320 Lecture Notes from August taken by Yaofeng Su

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1 Functional Analysis, Math 7320 Lecture Notes from August taken by Yaofeng Su 1 Essentials of Topology 1.1 Continuity Next we recall a stronger notion of continuity: Definition. Let (X, d X ), (Y, d Y ) be metric spaces, A map f : X Y is called uniformly continuous, if ɛ > 0, δ > 0, for all p, q X, d X (p, q) < δ, we have d Y (f(p), f(q)) < ɛ Theorem. Let X, Y be normed spaces, and d X, d Y are the metric induced by the norm of X, Y, If A : X Y is linear continuous map, then A is uniformly continuous. Proof. Two steps to verify uniform continuity: (1) Since A is continuous at 0, we have ɛ > 0 δ > 0, s.th. p X, d X (p, 0) < δ, d Y (A(p), 0) < ɛ. (2) We use this to prove uniform continuity: Let ɛ > 0 be given and choose δ > 0 as in (1), then p, q X, d X (p, q) < δ, we have z = p q satisfies d X (z, 0) < δ, d Y (A(z), 0) < ɛ. Now using the linearity of A gives d Y (A(p), A(q)) = A(p q) Y = d Y (A(p q), 0) < ɛ A is uniformly continuous. 1

2 Next we make precise in which way vector space and norm structures are compatible: Lemma. Let X, Y be norm space, then (x, y) X Y = x X + y Y is a norm. Proof. We verify three properties of norm: (1) Verify positive definiteness: (x, y) X Y = x X + y Y = 0, implies by x X 0 and y Y 0 and the positive definiteness of the norms on X and Y x = 0, y = 0. (x, y) = 0. (2) Verify scaling property from the homogeneity of the norms on X and Y : λ(x, y) X Y = λx X + λy Y = λ ( x X + y Y ) = λ (x, y) X Y. λ(x, y) X Y = λ (x, y) X Y. (3) Verify triangle inequality from that of the norms on X and Y : (x, y)+(p, q) X Y = x+p X + y+q Y x X + p X + y Y + q Y = (x, y) X Y + (p, q) X Y. (x, y) + (p, q) X Y (x, y) X Y + (p, q) X Y. this completes the proof Theorem. Let (X,. ) be a norm space, then f : X X X, (x, y) x+y is uniformly continuous. Scalar product: : K X X : (λ, x) λx is continuous but not uniformly continuous. Proof. We prove this three statements: (1) Since For ɛ > 0, let δ = ɛ > 0, then x, p X, y, q Y, with (x, y) (p, q) X Y = x p X + y q Y < δ, we have f is uniformly continuous. (x + y) (p + q) X Y x p X + y q Y < δ = ɛ. 2

3 (2) Since it is in metric space, we only need to verify the sequence convergence: let λ n, λ 0 K, x n, x 0 X, with x n x 0, λ n λ 0. by convergence of x n, we have for ɛ = 1, n 0 > 0 s.th. n > n 0, x n x 0 1. n > n 0, x n x n x 0 + x x 0. Hence, we can set K = max{ x n, x 0 + 1, n n 0 }, then by convergence of λ n, x n, we have ɛ, n 1 N s.th. n > n 1, λ n λ 0 ɛ/2k. and ɛ, n 2 > n 1, n > n 2, x n x 0 ɛ/2( λ 0 + 1). ɛ, n 2, n > n 2, λ n x n λ 0 x 0 (λ n λ 0 ) x n + λ 0 (x n x 0 ) ɛ/2 + ɛ/2 = ɛ. this completes the proof. (3) Scalar product is not uniform. Assume this were the case, then for a given ɛ > 0, there would be δ > 0 such that any pair of points at distance at most δ would be mapped to a pair at a distance at most ɛ. Take x 0 a non zero element, and let ɛ = 2 x 0. Then take λ 1,n = n, λ 2,n = n + 1/n, x 1,n = nx 0, x 2,n = nx 0, then (λ 1,n, x 1,n ) (λ 2,n, x 2,n )) = 1/n 0, which becomes smaller than any δ > 0 but for any n λ 1,n x 1,n λ 2,n x 2,n = x 0 and hence λ 1,n x 1,n λ 2,n x 2,n = x 0 > ɛ. By contradiction, the scalar product is not uniformly continuous. 1.2 Completeness Next we talk about completeness: Definition. Let (X, τ) be topological space, (x n ) n N X, we say x n x in X, if for each U U x (neighourhood of x), n 0 N s.t. n n 0, x n U Remark. For metric spaces, this implies the usual form of convergence: ɛ, n 0 N, n n 0, d(x n, x) < ɛ. 3

4 For metric space, we will have a weaker notion of convergence called Cauchy property Definition. Let (X, d) be metric space, a sequence (x n ) n N is called Cauchy sequence if: ɛ, n 0 N, n, m n 0, d(x n, x m ) ɛ Remark. A convergence sequence is Cauchy by triangle inequality: if ɛ, n 0, n > n 0, d(x n, x) < ɛ/2. then ɛ, n 0, m, n > n 0, d(x n, x m ) d(x n, x) + d(x m, x) < ɛ/2 + ɛ/2 = ɛ. but converse is not true for me spacex like open interval equipped with usual metric Definition. (X, d) be metric space is called complete if each Cauchy sequence converges in X. A complete normed space is called Banach Space Remark. A subset of a complete metric space is completeness iff it is closed(sees next lemma for proof) Lemma. Let (X, d) be complete metric space and Y X, then (Y, d) is complete iff Y is closed in X Proof. We prove it by two steps: (1) if Y is a closed in X, cauchy sequence (x n ) n N Y. Since X is complete, Since Y is closed, the limit Y is complete. (x n ) n N X. x X, x n x. x Y. (2) Assume Y is complete and take any x Y. Then there is a sequence (x n ) n N in Y such that x n x X Since this sequence converges in X, then then ɛ, n 0 N, n n 0, d(x n, x) ɛ/2. ɛ, n 0 N, n, m n 0, d(x n, x m ) d(x m, x) + d(x n, x) ɛ. this sequence is al cauchy in Y. So it is Cauchy in Y and by the completeness of Y, it converges in Y to the limit x Y. Thus the limit x Y, Y is closed. 4

5 Example. We give me concrete examples: (a) K = Cor R are complete metric space. (b) Q R is not complete. (c) (K n,. p ), 1 p is Banach space. (d) l p, 1 p with norm x p = ( x p ) 1/p is a Banach Space as well as c 0 l (e) Bounded function space B(X, K) is a Banach Space equipped with.. (f) if X is a norm space, Y is Banach Space, then B(X, Y ) is Banach Space with the operator norm. Proof. Now we prove all of them: (a) For number field K like C, R, since C = R 2, it deduced to the case of R, but I do not how to prove R is complete because R is originally defined as the equivalence class of cauchy sequence in Q, that is, R is defined as the completeness of Q. (b) We prove that Q is not closed in R. Every irrational number can be expressed as binary form: x = a i /2 i, a i = 0/1. consider the finite term: x n = n a i /2 i, a i = 0/1. we note that x n x and x n Q, which complete the proof. (c) This is special case of (d), we turn to (d) first. (How is (c) embedded in (d)?) You need to use that it is a closed subset to relegate it to (d). (d) For p <, first we prove it is a norm space. (1) if x p = ( x n,) p 1/p = 0, then n, x n = 0, x = 0. (2) λx p = ( λx n,) p 1/p = λ x p. (3) by Minkowskii inequality, x + y p = ( x n, + y n, p ) 1/p ( x n, p ) 1/p + ( y n, p ) 1/p = x p + y p. 5

6 this complete the proof of norm properties. We turn to completeness: let (x n ) n N l p is a cauchy sequence, that is: for each, we have ɛ > 0, n 0 N, n, m n 0, ( x n, x m, p ) 1/p < ɛ/2. ɛ > 0, n 0 N, n, m n 0, ( x n, x m, p ) 1/p < ɛ/2. that is, for each, (x n, ) n N is cauchy sequence. For fixed n n 0, we apply Fatou lemma to when m, ( x n, y p ) 1/p = ( y, x n, y. ( x n, x m, p ) 1/p < ɛ. lim m x n, x m, ) p 1/p lim m ( x n, x m, p ) 1/p ɛ/2. that is ɛ > 0, n 0 N, n n 0, ( x n, y p ) 1/p ɛ/2 < ɛ. Moreover, by Minkowski s inequality: ( y p ) 1/p ( x n0, y p ) 1/p + ( x n0, p ) 1/p < ɛ + ( x n0, p ) 1/p <. y l p. Thus, we have shown in l p, x n y. This complete the proof of completeness for case p <. A similar proof can be applied to the case when p = : For p =, first we prove it is a norm space. (1) if x = sup n x n = 0, then n, x n = 0, x = 0. (2) λx = sup n λx n = λ x. 6

7 (3) triangle inequality: x + y = sup n x n + y n sup n this complete the proof of norm properties. We turn to completeness: let (x n ) n N l is a cauchy sequence, that is: for each, we have x n + sup y n = x + y n ɛ > 0, n 0 N, n, m n 0, sup x n, x m, < ɛ/2. ɛ > 0, n 0 N, n, m n 0, x n, x m, < ɛ/2. that is, for each, (x n, ) n N is cauchy sequence. For fixed n n 0, we let m, that is y, x n, y. ɛ, n 0, n > n 0,, x n, y = lim m x n, x m, ɛ/2 < ɛ. Moreover, by triangular inequality: y l. sup ɛ > 0, n 0 N, n n 0, sup x n, y ɛ/2 < ɛ. y sup Thus, we have shown in l, x n0, y + sup x n y. This complete the proof of completeness for case p =. l is al complete. c 0 is closed with respect to the norm in l : Assume a sequence in c 0 and a limit in l that is x n x. x n0, < ɛ + sup x n0, <. ɛ, n 0, n > n 0,, x n, x < ɛ/2. Fix n n 0 then letting, we have by lim x n, = 0 that lim x = lim x n, x ɛ/2 < ɛ. 7

8 x c 0, that is c 0 is a closed space in l. Consequently, c 0 is al Banach Space. Returning to (c), consider K n l p i.e. K n = {x l p : x = 0, n + 1}. assume in l p, we have x n K n x, that is: ɛ > 0, n 0 N, n, m n 0, ( x n, x p ) 1/p < ɛ/2. then we have: ɛ > 0, n 0 N, n, m n 0, ( x p ) 1/p = ( x n, x p ) 1/p < ɛ/2. n+1 n+1 that means: x = 0, n + 1. x K n i.e. K n is closed subspace in l p, K n is complete. 8