Functional Analysis. L. Pedro Poitevin. June 6, 2007

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1 Functional Analysis L. Pedro Poitevin June 6, Metric saces Definition 1. Let X be a set. A function d : X X [0, ) is said to be a metric on X if for all x, y, z X, the following conditions are satisfied: 1. d(x, y) 0; 2. d(x, y) = 0 if and only if x = y; 3. d(x, y) = d(y, x); 4. d(x, z) d(x, y) + d(y, z). If X is a set and d is a metric on X, we say that (X, d) is a metric sace. Definition 2. A subsace (Y, ˆd) of a metric sace (X, d) is obtained by taking a subset Y X and restricting d to Y Y ; thus the metric ˆd on Y is the restriction ˆd = d Y Y. This restriction ˆd is called the metric induced on Y by d. Definition 3. Let (X, d) be a metric sace. Given a oint x o X and a real number r > 0, we define three tyes of sets: B(x 0, r) = {x X d(x, x 0 ) < r} (called the oen ball of radius r centered at x 0 ), B[x 0, r] = {x X d(x, x 0 ) r} (called the closed ball of radius r centered at x 0 ), and S(x 0, r) = {x X d(x, x 0 ) = r} (called the shere of radius r centered at x 0 ). Definition 4. Let (X, d) be a metric sace. A subset A of X is said to be oen if it contains a ball about each of its oints. A subset C of X is said to be closed if its comlement (in X) is oen. Definition 5. Let (X, d) be a metric sace, and let A be a subset of X. A oint x A is said to be an interior oint of A if there exists r > 0 such that B(x, r) A. The set of all interior oints of A is denoted by Int(A) and called the interior of A. Remark 6. Let (X, d) be a metric sace, and denote by τ the collection of all oen sets in X. Then τ has the following roerties: 1. The union of any collection of members of τ is a member of τ (i.e. τ is closed under arbitrary unions); 2. The intersection of finitely many members of τ is a member of τ (i.e. τ is closed under finite intersections).

2 Definition 7. If X is a set and τ is a collection of subsets of X closed under both arbitrary unions and finite intersections, then (X, τ) is said to be a toological sace and τ is called a toology on X. Remark 8. If (X, d) is a metric sace and τ denotes the collection of all oen sets in X (according to d), then (X, τ) is a toological sace. Definition 9. Let (X, d) and (Y, ˆd) be metric saces. A maing T : X Y is said to be continuous at a oint x 0 X if for every ɛ > 0 there exists a δ > 0 such that ˆd(T x, T x 0 ) < ɛ whenever d(x, x 0 ) < δ. The maing T is said to be continuous if it is continuous at every oint of X. Theorem 10. Let (X, d) and (Y, ˆd) be metric saces. A maing T : X Y is continuous if and only if the inverse image of any oen set in Y is an oen set in X. Definition 11. Let (X, d) be a metric sace, and let A be a subset of X. Then a oint x 0 X is said to be an accumulation oint of A if for every r > 0, B(x 0, r)\{x 0 } A. The union of A with the set of all its accumulation oints is called the closure of A and is denoted by Ā. Definition 12. Let (X, d) be a metric sace. A subset A of X is said to be dense in X if Ā = X. The metric sace (X, d) is said to be searable if X has a countable subset which is dense in X. Definition 13. Let (x n ) be a sequence in a metric sace (X, d), and let x X. We say that the sequence (x n ) converges to x if lim n d(x n, x) = 0. If (x n ) does not converge to any oint of X, then (x n ) is said to diverge. Definition 14. A sequence (x n ) in a metric sace (X, d) is a bounded sequence if there exist x 0 X and r > 0 such that {x n n N} B(x 0, r). Lemma 15. Let (X, d) be a metric sace. Then: 1. A convergent sequence in X is bounded and its limit is unique. 2. If x n x and y n y in X, then d(x n, y n ) d(x, y). Definition 16. Let (x n ) be a sequence in a metric sace (X, d). We say that (x n ) is Cauchy if for all ɛ > 0, there is N N such that for all m, n N, d(x m, x n ) ɛ. Definition 17. A metric sace (X, d) is said to be comlete if every Cauchy sequence in X converges in X. Theorem 18. Let (X, d) be a metric sace. If A is a nonemty subset of X, then: 1. A oint x X belongs to Ā if and only if there exists a sequence (x n) in A such that x n x. 2. The set A is closed if and only if whenever (x n ) is a convergent sequence in A converging to x, it must be the case that x A. Theorem 19. Let (X, d) be a comlete metric sace. Let Y be a subset of X. If we endow Y with the induced metric ˆd it inherits from X, then (Y, ˆd) is itself comlete as a metric sace if and only if the set Y is closed in X (with resect to the original metric d). Theorem 20. A maing T : X Y of a metric sace (X, d) into a metric sace (Y, ˆd) is continuous at a oint x 0 X if and only if x n x 0 imlies T x n T x 0. 2

3 1.1 Comact metric saces Definition 21. Let (x n ) n N be a sequence in a set S. If k : N N is an increasing ma (i.e. k(m) < k(n) whenever m < n), then the sequence (x kn ) n N is said to be a subsequence of (x n ). Definition 22. A metric sace X is said to be comact if every sequence in X has a convergent subsequence. 2 Introduction to Banach saces In these notes, we will restrict our attention to vector saces over R. Many of the results resented have natural extensions to the case of saces over the comlex field, but in the interest of roviding as gentle an introduction to functional analysis as ossible, we have decided to omit the latter. Definition 23. A nonnegative function on a vector sace X is called a norm on X if for every x, y X, and every λ R, 1. x 0; 2. x = 0 if and only if x = 0; 3. λx = λ x ; 4. x + y x + y. If X is a vector sace and is a norm on X, we say that (X, ) is a normed sace. A Banach sace is a normed sace that is comlete in the canonical metric defined by ρ(x, y) = x y for x, y X. Remark 24. All the toological notions in Banach saces will refer to the canonical metric defined by the norm, unless otherwise stated. Let (X, ) be a normed sace. B X = {x X x 1} denotes the closed unit ball of X; S X = {x X x = 1} denotes the unit shere of X; If M X, then san(m) denotes the linear san of M, i.e. the intersection of all linear subsaces of X containing M; equivalently, san(m) is the smallest linear subsace of X containing M; san(m) stands for the closed linear hull of M; If M X, the convex hull of M will be denoted by conv(m), and the closed convex hull of M will be denoted by conv(m). For subsets A, B X and a scalar α, we also write A + B = {a + b a A, b B} and αa = {αa a A}. By a subsace we always mean a linear subsace. Let Y be a subsace of (X, ). By (Y, ) we denote Y endowed with the restriction of to Y. 3

4 Exercise 25. Let Y be a subsace of a Banach sace (X, ). Then (Y, ) is a Banach sace if and only if Y is closed in X. Hint. Any convergent sequence in Y is a Cauchy sequence in X. Examle 26. C[0, 1] denotes the vector sace of all scalar-valued continuous functions on [0, 1], endowed with the norm f = su{ f(t) t [0, 1]} = max{ f(t) t [0, 1]}. Exercise 27. Prove that C[0, 1] is a Banach sace. Remark 28. The revious exercise was solved in class. Definition 29. The sace l n denotes the n-dimensional vector sace of all n-tules of scalars, endowed with the suremum norm defined for x = (x 1,..., x n ) l n by x = max{ x i i = 1,..., n}. Remark 30. An argument similar to that used for C[0, 1] ensures that l n is indeed a Banach sace. Lemma 31. Let, q > 1 be such that = 1. ( and q are called conjugate exonents q in this situation.) Then for any a, b 0 we have that ab a + bq q. Remark 32. If, q are conjugate exonents, then 1 = +q, q = +q, and ( 1)(q 1) = q 1. In articular, u = t 1 imlies t = u q 1. Later on, we will need to also note that ( 1)q =. Remark 33. The revious lemma was roved in class. Theorem 34 (Hölder inequality). Let, q > 1 be such that q a k, b k R, for k = 1,..., n, we have ( n n ) 1 ( ) 1 q n a k b k a k b k. k=1 k=1 k=1 = 1. Then for all Remark 35. In case = q = 2, the above inequality is known as the Cauchy-Schwarz inequality. Remark 36. Hölder inequality was roved in class. Theorem 37 (Minkowski inequality). Let [1, ). Then for all a k, b k R, k = 1,..., n, we have ( n k=1 a k + b k ) 1 ( n k=1 Remark 38. Minkowski inequality was roved in class. ) 1 ( n ) 1 a k + b k. k=1 4

5 Definition 39. Let [1, ). The sace l n denotes the n-dimensional vector sace R n, endowed with the norm defined for x = (x 1,..., x n ) l n by x = ( n x i ) 1. Remark 40. Note that is indeed a norm on l n. Definition 41. Let [1, ). The sace l denotes the vector sace of all scalar-valued sequences x = (x n ) n=1 satisfying x n <, endowed with the norm x = ( x i ) 1. Exercise 42. Prove that (l, ) is a normed vector sace. Definition 43. Let x = (x n ) be a sequence of reals. su(x) = {n x n 0}. We define the suort of x by Definition 44. The sace l denotes the vector sace of all bounded scalar-valued sequences with the norm defined for x = (x n ) l by x = su{ x n n N}. The sace c 00 denotes the subsace of l consisting of all x = (x n ) such that su(x) is finite. The sace c denotes the subsace of l consisting of all x = (x n ) such that lim n x n exists and is finite. The sace c 0 denotes the subsace of l consisting of all x = (x n ) such that lim n x n = 0. Remark 45. Note that c 0 is the closure of c 00 in l. Note also that if x = (x n ) belongs to c 0 or c 00, then x = max{ x n n N}. Theorem 46. For [1, ], the sace l is a Banach sace. Proof. The case = is similar to that of C[0, 1], and we omit it here. Consider [1, ). Let (x (k) ) k=1 a Cauchy sequence in l, where x (k) = (x (k) i ). Given ɛ > 0, find k 0 such that for every k, l k 0, ( x (k) i x (l) i ) 1 ɛ. ( ) In articular, x (k) i x (l) i ɛ for every k, l k 0 and i N, whence the sequence (x (k) i ) k=1 converges to some x i for every i N. Put x = (x i ). We will show that x l. Because the sequence (x (k) i ) k=1 is convergent, ( ) 1 we know there exists a constant C > 0 such that x(k) i C for every k. Therefore ( n ) 1 x(k) i C for every n, k N. By letting k, we get that ( n x i ) 1 C for every n N. Therefore ( x i ) 1 C and x l. 5

6 We will now show that x (k) x in l. Given ɛ > 0, we let l in ( ) and get ) 1 x(k) i x i ɛ for every n N and every k k 0. We let n to obtain ( n ( ) 1 x (k) i x i for every k k 0. Therefore x (k) x in l. = x (k) x ɛ Remark 47. It is routine to show that c and c 0 are Banach saces, by showing them to be closed subsaces of l. Exercise 48. Prove that c 00 is not comlete. Definition 49. A ma T : (M, d) (N, d ) between metric saces is called Lischitz if there is C > 0 such that d (T (x), T (y)) Cd(x, y) for all x, y M. Remark 50. In the case of normed saces M and N, this inequality becomes T (x) T (y) N C x y M. Proosition 51. Let (X, X ) and (Y, Y ) be normed saces, and let T be a linear maing from X to Y. The following statements are equivalent: 1. T is continuous on X. 2. T is continuous at the origin. 3. There is C > 0 such that T (x) Y C x X for every x X. 4. T is Lischitz. 5. T (B X ) is a bounded set in Y. Proof. (1) (2) follows from the linearity of T. (3) (4) follows similarly since T (x) T (y) Y = T (x y) Y C x y X. If (2) is true, given ɛ > 0, there is δ > 0 such that T (x) Y ɛ whenever x X δ. If x X \ {0}, then δ x x X = δ, and thus ( ) x Y T δ ɛ. x X Thus T (x) Y ɛ x δ X for every x X, which shows (3). On the other hand, (3) clearly imlies (2) with δ = ɛ. C Assuming (3), we obtain that T (x) Y C whenever x, y B X, so T (B X ) is bounded. On the other hand, if T (B X ) is bounded and, say T (x) Y C for every x B X, then for every x X \ {0}, we have ( ) x Y T C x X so T (x) Y C x X. This shows the equivalence of (3) and (4). 6

7 Definition 52. Let X and Y be normed saces, and let T be a linear ma from X into Y. T is called a bounded linear oerator if T (B X ) is bounded in Y. We define the oerator norm T = su{ T (x) Y x B X }. The sace B(X, Y ) denotes the vector sace of all bounded linear oerators from X into Y, endowed with the oerator norm. In the case X = Y we ut B(X) = B(X, X). Remark 53. Note that T is the smallest number C > 0 that satisfies condition (3) in the revious roosition. Remark 54. If T : X Y is a bounded linear oerator, then the kernel Ker(T ) = {x X T (x) = 0} is a closed subsace of X. Proosition 55. Let X and Y be normed linear saces. If Y is a Banach sace, then B(X, Y ) is also a Banach sace. Remark 56. Oerators in B(X, R) are called continuous linear functionals. We have that f = su{ f(x) x B X } for f B(X, R). Definition 57. An oerator T B(X, Y ) is called a linear isomorhism (or just an isomorhism) if T is a bijection and T 1 B(X, Y ). Remark 58. Note that the inverse ma of a linear bijection is always a linear ma. Definition 59. Normed saces X and Y are said to be linearly isomorhic (or simly isomorhic) if there is a linear isomorhism T of X onto Y. Remark 60. It is easy to see that an isomorhism T carries Cauchy sequences onto Cauchy sequences. Therefore, if X and Y are isomorhic normed saces and X is a Banach sace, then Y must be a Banach sace as well. Definition 61. An oerator T B(X, Y ) is called a (linear) isomorhism from X into Y if T is an isomorhism from X onto a subset T (X) of Y. Remark 62. Clearly, T (X) is a subsace of Y, and since it is a comlete sace, it must be closed in Y. Definition 63. An oerator T B(X, Y ) is called a (linear) isometry if T (x) Y = x X for every x X. Saces X and Y are called (linearly) isometric if there exists a linear isometry T of X onto Y. Definition 64. Let X be a vector sace equied with two norms. Denote X 1 = (X, 1 ) and X 2 = (X, 2 ). The norms 1 and 2 are said to be equivalent if the identity maing I X : X X defined by I X (x) = x is an isomorhism between the saces X 1 and X 2, that is to say, if there exist constants c, C > 0 such that for every x X. c x 2 x 1 C x 2 Definition 65. Let X, Y be normed saces. We introduce the normed sace X Y called a direct (toological) sum of X and Y, that consists of all ordered airs (x, y) X Y, endowed with the coordinatewise vector sace oerations and with the norm (x, y) = x X + y Y. 7

8 Remark 66. X and Y are isometric to subsaces {(x, 0) x X} and {(0, y) y Y } of X Y. Let Y be a closed subsace of a normed sace X. For x X, we consider the coset x relative to Y defined by x = {z X x z Y } = {x + y y Y }. The sace X/Y = { x x X} of all cosets, together with addition and scalar multilication defined by x + ŷ = x + y and λ x = λx, is clearly a vector sace. Remark 67. It is easy to check that x = inf{ y y x} makes X/Y into a normed sace. Proof. For any z x we have that x = inf{ z y y Y } := dist(z, Y ). Therefore x = 0 if and only if x Y, because Y is closed. If Y is a subsace of X, then dist(αx, Y ) = α dist(x, Y ). Therefore λ x = λ x. The triangle inequality follows since if x 1, x 2 are in X and y 1, y 2 are in Y, then x 1 + x 2 (y 1 + y 2 ) x 1 y 1 + x 2 y 2. Therefore dist(x 1 + x 2, Y ) dist(x 1, Y ) + dist(x 2, Y ). Definition 68. Let Y be a closed subsace of a normed sace X. The sace X/Y endowed with the canonical norm x = inf{ x x x} is called the quotient sace of X with resect to Y. Proosition 69. Let Y be a closed subsace of a Banach sace X. Banach sace. Then X/Y is a Proof. Assume that ( x n ) is a Cauchy sequence in X/Y. Choose a subsequence ( x nk ) such that x nk x nk+1 < 2 k. This choice is ossible because for arbitrary x nk x nk, the distance of x nk to x nk +1 is less than 2 k. It is standard to check that (x nk ) is Cauchy in X and therefore x nk x for some x X (since X is comlete). Since x nk x nk+1 x nk x nk+1 for every k, we get that x nk x. Consequently, x n x in X/Y. Remark 70. It is easy to check that (X Y )/Y is isomorhic to X and (X Y )/X is isomorhic to Y. (Note: Identify Y with the closed subsace {(0, y) y Y } of X Y.) Proosition 71. Let X be a vector sace. If X is finite-dimensional, then any two norms on X are equivalent. In articular, all finite-dimensional normed saces are Banach saces, and every normed sace of dimension n is isomorhic to l2 n. Remark 72. The revious roosition was roved in class. An imortant fact that was established in the rocess of roving it is that the unit ball of any finite-dimensional normed sace is comact. Remark 73. Consequently, if X is a Banach sace and Y is a finite-dimensional subsace of X, then Y is closed in X. Lemma 74 (Riesz). Let X be a normed sace. If Y is a roer closed subsace of X, then for every ɛ > 0 there is x S X such that dist(x, Y ) 1 ɛ. 8

9 Proof. Choose an arbitrary element ẑ X/Y satisfying 1 > ẑ > 1 ɛ. Pick any z ẑ with z 1, and set x = z z. Then dist(x, Y ) = dist(z, Y ) z = ẑ z ẑ > 1 ɛ. Theorem 75. Let X be a normed sace. X is finite-dimensional if and only if the unit ball B X of X is comact. Proof. If X is finite-dimensional, then B X is comact, as we saw in the roof of roosition 71. If X is infinite-dimensional, we find by induction (using Riesz lemma), an infinite sequence x n S X such that dist(x n, san(x 1,..., x n 1 ) > 1/2. Then dist(x m, x n ) > 1/2 for all m n, and therefore (x n ) cannot have a convergent subsequence. Hence B X is not comact. Proosition The saces c and c 0 are searable. 3. The sace l is not searable. 1. If [1, ), then the sace l is searable. Proof. 1. Consider in l the family F formed by all finitely suorted vectors with rational coefficients. Then F is countable. (It is a countable union of countable sets.) We will show that F is dense in l. Given x l and ɛ > 0, let us choose n 0 > 0 such that x i ɛ 2 i=n 0 and then find rational numbers r 1,..., r n0 1 such that x i r i ɛ 2n 0 for i = 1,..., n 0 1. Then s = (r 1, r 2,..., r n0 1, 0,...) is in F and s x = n 0 1 Therefore F is dense in l. x i r 2 + i=n 0 x i n 0 1 ɛ 2n 0 + ɛ 2 < ɛ. 2. The roof of the searability of c 0 is similar to that of the searability of l. The case of the sace c needs only one adjustment namely, we define F as the family of vectors with rational coefficients such that the vectors are eventually constant. 3. Assume that l is searable, and let D be a dense countable set in l. The cardinal number of the family F of all subsets of N is the cardinality of the continuum; in articular, it is uncountable. For every F F, let χ F denote the characteristic function of F in N. If F 1, F 2 F and F 1 F 2, then χ F1 χ F2 χ F1 (n) χ F2 (n) = 1 for some n (F 1 \ F 2 ) (F 2 \ F 1 ). For each F F, let d F D be chosen such that χ F d F < 1. If F 4 1 F 2, then d F1 d F2 > 1. Indeed, if we had 4 d F1 d F2 1, then 4 χ F1 χ F2 χ F1 d F1 + d F1 d F2 + d F2 χ F2 3 4 < 1, 9

10 a contradiction. Therefore, the ma F d F is one-to-one and mas an uncountable set into a countable set, which is a contradiction. Thus l is not searable. Proosition 77. The sace C[0, 1] is searable. Proof. Let P be the collection of olynomials on [0, 1] with rational coefficients. Such a collection forms an algebra in C[0, 1], searates the oints of [0, 1] and contains a constant function. By the Stone-Weierstrass theorem, the closure of P in C[0, 1] is C[0, 1]. Proosition 78. B(l 2 ) is not searable. Remark 79. Proved in class. 3 Hilbert saces Definition 80. An inner roduct (or a dot roduct) on a vector sace X is a scalar-valued function, on X X such that 1. for every y X, the function x x, y is linear; 2. x, y = y, x for every x, y X, where the bar denotes comlex conjugation; 3. x, x 0 for every x X; 4. x, x = 0 if and only if x = 0. Note that 0, y = y, 0 = 0 for all y X. Remark 81. Since we will be working only over the real numbers, the second condition above becomes x, y = y, x for all x, y X. All the statements that aear from now on in this section have true analogues in the more general setting in which the scalar field is C, but we will only state results for the case in which the scalar field is R, and this will simlify the roofs a little bit. Theorem 82 (Cauchy-Schwarz inequality). Let, be an inner roduct on a vector sace X. 1. For x, y X, we have that x, y x, x y, y. 2. The function x = x, x is a norm on X. Remark 83. One consequence of the above theorem is that, is a continuous maing from (X, ) (X, ) into the scalar field. In articular, for a fixed vector y X, x x, y is a continuous linear functional on X. Definition 84. A Banach sace H is called a Hilbert sace if there is an inner roduct, on H such that x = x, x for every x H. Exercise 85. Let (H,,, ) be a Hilbert sace. Prove that for every x, y H, we have x + y 2 + x y 2 = 2 x y 2. 10

11 This equation is referred to as the arallelogram equality. Prove moreover the following so-called olarization identity x, y = 1 4 ( x + y 2 x y 2). Exercise 86. Let (X, ) be a Banach sace over R for which the arallelogram equality is satisfied. Prove that if one defines, by the olarization identity, then,, turns out to be an inner roduct on X. Moreover, in this situation x 2 = x, x, and so (X,,, ) is a Hilbert sace. Remark 87. The arallelogram equality gives that l n 2, l 2, and L 2 are Hilbert saces. Definition 88. Let H be a Hilbert sace. Let x, y H. We say that x is orthogonal to y, denoted x y, if x, y = 0. Let M H. We say that x is orthogonal to M, denoted x M, if x is orthogonal to every vector y from M. Definition 89. Let F be a subsace of a Hilbert sace H. The set F = {h H h F } is called the orthogonal comlement of F in H. Proosition 90. If F is a subset of H, then F is a closed subsace of H. Remark 91. Let H be a Hilbert sace. Note that if F is a subsace of H, F F = {0}. This means that every element z F + F has a unique exression in the form z = x + y with x F and y F. We can also see that the orthogonality gives z 2 = x+y, x+y = x 2 + y 2. It follows that T : F F H defined by T (x, y) = x+y is an isomorhism of F F onto F + F H. Theorem 92 (Riesz). Let F be a subsace of a Hilbert sace. If F is closed, then F +F = H. Remark 93. Note that for any closed subsace F of a Hilbert sace H, we have that ( F ) = F Definition 94. Let H be a Hilbert sace and let S H. S is called an orthonormal set if s 1, s 2 = 0 whenever s 1 s 2 S and {s, s} = 1 for every s S. A maximal orthonormal set (in the sense of inclusion) in H is called an orthonormal basis of H. Theorem 95. Every Hilbert sace has an orthonormal basis. Theorem 96. Every searable infinite-dimensional Hilbert sace H has an orthonormal basis (e i ). Moreover, if (e i ) is an orthonormal basis of H, then for every x H, x = x, e i e i. The numbers x, e i are called Fourier coefficients, and x = x, e i e i is called the Fourier exansion of x or the Fourier series for x. Proosition 97. Let (e i ) be an orthonormal set in a Hilbert sace H, and let x H. 1. x, e i 2 x 2 (The Bessel inequality); 11

12 2. If (e i ) is an orthonormal basis of H, then x 2 = x, e i 2 (The Parseval equality); 3. If the Parseval equality holds for every x H, then (e i ) is an orthonormal basis of H. 4. If san ((e i ) ) = H, then (e i ) is an orthonormal basis of H. Theorem 98 (Riesz, Fischer). Every searable infinite-dimensional Hilbert sace H is linearly isometric to l 2. 12